CS计算机代考程序代写 ER AI Euclid's Elements of Geometry

Euclid’s Elements of Geometry

EUCLID’S ELEMENTS OF GEOMETRY

The Greek text of J.L. Heiberg (1883–1885)

from Euclidis Elementa, edidit et Latine interpretatus est I.L. Heiberg, in aedibus

B.G. Teubneri, 1883–1885

edited, and provided with a modern English translation, by

Richard Fitzpatrick

First edition – 2007
Revised and corrected – 2008

ISBN 978-0-6151-7984-1

Contents

Introduction 4

Book 1 5

Book 2 49

Book 3 69

Book 4 109

Book 5 129

Book 6 155

Book 7 193

Book 8 227

Book 9 253

Book 10 281

Book 11 423

Book 12 471

Book 13 505

Greek-English Lexicon 539

Introduction

Euclid’s Elements is by far the most famous mathematical work of classical antiquity, and also has the distinction
of being the world’s oldest continuously used mathematical textbook. Little is known about the author, beyond

the fact that he lived in Alexandria around 300 BCE. The main subjects of the work are geometry, proportion, and

number theory.

Most of the theorems appearing in the Elements were not discovered by Euclid himself, but were the work of

earlier Greek mathematicians such as Pythagoras (and his school), Hippocrates of Chios, Theaetetus of Athens, and

Eudoxus of Cnidos. However, Euclid is generally credited with arranging these theorems in a logical manner, so as to
demonstrate (admittedly, not always with the rigour demanded by modern mathematics) that they necessarily follow

from five simple axioms. Euclid is also credited with devising a number of particularly ingenious proofs of previously
discovered theorems: e.g., Theorem 48 in Book 1.

The geometrical constructions employed in the Elements are restricted to those which can be achieved using a

straight-rule and a compass. Furthermore, empirical proofs by means of measurement are strictly forbidden: i.e.,
any comparison of two magnitudes is restricted to saying that the magnitudes are either equal, or that one is greater

than the other.

The Elements consists of thirteen books. Book 1 outlines the fundamental propositions of plane geometry, includ-
ing the three cases in which triangles are congruent, various theorems involving parallel lines, the theorem regarding

the sum of the angles in a triangle, and the Pythagorean theorem. Book 2 is commonly said to deal with “geometric

algebra”, since most of the theorems contained within it have simple algebraic interpretations. Book 3 investigates
circles and their properties, and includes theorems on tangents and inscribed angles. Book 4 is concerned with reg-

ular polygons inscribed in, and circumscribed around, circles. Book 5 develops the arithmetic theory of proportion.
Book 6 applies the theory of proportion to plane geometry, and contains theorems on similar figures. Book 7 deals

with elementary number theory: e.g., prime numbers, greatest common denominators, etc. Book 8 is concerned with

geometric series. Book 9 contains various applications of results in the previous two books, and includes theorems
on the infinitude of prime numbers, as well as the sum of a geometric series. Book 10 attempts to classify incommen-

surable (i.e., irrational) magnitudes using the so-called “method of exhaustion”, an ancient precursor to integration.

Book 11 deals with the fundamental propositions of three-dimensional geometry. Book 12 calculates the relative
volumes of cones, pyramids, cylinders, and spheres using the method of exhaustion. Finally, Book 13 investigates the

five so-called Platonic solids.

This edition of Euclid’s Elements presents the definitive Greek text—i.e., that edited by J.L. Heiberg (1883–

1885)—accompanied by a modern English translation, as well as a Greek-English lexicon. Neither the spurious

books 14 and 15, nor the extensive scholia which have been added to the Elements over the centuries, are included.
The aim of the translation is to make the mathematical argument as clear and unambiguous as possible, whilst still

adhering closely to the meaning of the original Greek. Text within square parenthesis (in both Greek and English)

indicates material identified by Heiberg as being later interpolations to the original text (some particularly obvious or
unhelpful interpolations have been omitted altogether). Text within round parenthesis (in English) indicates material

which is implied, but not actually present, in the Greek text.

My thanks to Mariusz Wodzicki (Berkeley) for typesetting advice, and to Sam Watson & Jonathan Fenno (U.

Mississippi), and Gregory Wong (UCSD) for pointing out a number of errors in Book 1.

ELEMENTS BOOK 1

Fundamentals of Plane Geometry Involving
Straight-Lines

STOIQEIWN aþ. ELEMENTS BOOK 1VOroi. Definitions
αʹ. Σημεῖόν ἐστιν, οὗ μέρος οὐθέν. 1. A point is that of which there is no part.
βʹ. Γραμμὴ δὲ μῆκος ἀπλατές. 2. And a line is a length without breadth.
γʹ. Γραμμῆς δὲ πέρατα σημεῖα. 3. And the extremities of a line are points.
δʹ. Εὐθεῖα γραμμή ἐστιν, ἥτις ἐξ ἴσου τοῖς ἐφ᾿ ἑαυτῆς 4. A straight-line is (any) one which lies evenly with

σημείοις κεῖται. points on itself.
εʹ. ᾿Επιφάνεια δέ ἐστιν, ὃ μῆκος καὶ πλάτος μόνον ἔχει. 5. And a surface is that which has length and breadth
ϛʹ. ᾿Επιφανείας δὲ πέρατα γραμμαί. only.
ζʹ. ᾿Επίπεδος ἐπιφάνειά ἐστιν, ἥτις ἐξ ἴσου ταῖς ἐφ᾿ 6. And the extremities of a surface are lines.

ἑαυτῆς εὐθείαις κεῖται. 7. A plane surface is (any) one which lies evenly with
ηʹ. ᾿Επίπεδος δὲ γωνία ἐστὶν ἡ ἐν ἐπιπέδῳ δύο γραμμῶν the straight-lines on itself.

ἁπτομένων ἀλλήλων καὶ μὴ ἐπ᾿ εὐθείας κειμένων πρὸς 8. And a plane angle is the inclination of the lines to
ἀλλήλας τῶν γραμμῶν κλίσις. one another, when two lines in a plane meet one another,
θʹ. ῞Οταν δὲ αἱ περιέχουσαι τὴν γωνίαν γραμμαὶ εὐθεῖαι and are not lying in a straight-line.

ὦσιν, εὐθύγραμμος καλεῖται ἡ γωνία. 9. And when the lines containing the angle are
ιʹ. ῞Οταν δὲ εὐθεῖα ἐπ᾿ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς straight then the angle is called rectilinear.

γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν 10. And when a straight-line stood upon (another)
ἐστι, καὶ ἡ ἐφεστηκυῖα εὐθεῖα κάθετος καλεῖται, ἐφ᾿ ἣν straight-line makes adjacent angles (which are) equal to
ἐφέστηκεν. one another, each of the equal angles is a right-angle, and
ιαʹ. Ἀμβλεῖα γωνία ἐστὶν ἡ μείζων ὀρθῆς. the former straight-line is called a perpendicular to that
ιβʹ. ᾿Οξεῖα δὲ ἡ ἐλάσσων ὀρθῆς. upon which it stands.
ιγʹ. ῞Ορος ἐστίν, ὅ τινός ἐστι πέρας. 11. An obtuse angle is one greater than a right-angle.
ιδʹ. Σχῆμά ἐστι τὸ ὑπό τινος ἤ τινων ὅρων περιεχόμενον. 12. And an acute angle (is) one less than a right-angle.
ιεʹ. Κύκλος ἐστὶ σχῆμα ἐπίπεδον ὑπὸ μιᾶς γραμμῆς 13. A boundary is that which is the extremity of some-

περιεχόμενον [ἣ καλεῖται περιφέρεια], πρὸς ἣν ἀφ᾿ ἑνὸς thing.
σημείου τῶν ἐντὸς τοῦ σχήματος κειμένων πᾶσαι αἱ 14. A figure is that which is contained by some bound-
προσπίπτουσαι εὐθεῖαι [πρὸς τὴν τοῦ κύκλου περιφέρειαν] ary or boundaries.
ἴσαι ἀλλήλαις εἰσίν. 15. A circle is a plane figure contained by a single line
ιϛʹ. Κέντρον δὲ τοῦ κύκλου τὸ σημεῖον καλεῖται. [which is called a circumference], (such that) all of the
ιζʹ. Διάμετρος δὲ τοῦ κύκλου ἐστὶν εὐθεῖά τις διὰ τοῦ straight-lines radiating towards [the circumference] from

κέντρου ἠγμένη καὶ περατουμένη ἐφ᾿ ἑκάτερα τὰ μέρη one point amongst those lying inside the figure are equal
ὑπὸ τῆς τοῦ κύκλου περιφερείας, ἥτις καὶ δίχα τέμνει τὸν to one another.
κύκλον. 16. And the point is called the center of the circle.
ιηʹ. ῾Ημικύκλιον δέ ἐστι τὸ περιεχόμενον σχῆμα ὑπό τε 17. And a diameter of the circle is any straight-line,

τῆς διαμέτρου καὶ τῆς ἀπολαμβανομένης ὑπ᾿ αὐτῆς περι- being drawn through the center, and terminated in each
φερείας. κέντρον δὲ τοῦ ἡμικυκλίου τὸ αὐτό, ὃ καὶ τοῦ direction by the circumference of the circle. (And) any

κύκλου ἐστίν. such (straight-line) also cuts the circle in half.†

ιθʹ. Σχήματα εὐθύγραμμά ἐστι τὰ ὑπὸ εὐθειῶν πε- 18. And a semi-circle is the figure contained by the
ριεχόμενα, τρίπλευρα μὲν τὰ ὑπὸ τριῶν, τετράπλευρα δὲ τὰ diameter and the circumference cuts off by it. And the
ὑπὸ τεσσάρων, πολύπλευρα δὲ τὰ ὑπὸ πλειόνων ἢ τεσσάρων center of the semi-circle is the same (point) as (the center
εὐθειῶν περιεχόμενα. of) the circle.
κʹ. Τῶν δὲ τριπλεύρων σχημάτων ἰσόπλευρον μὲν 19. Rectilinear figures are those (figures) contained

τρίγωνόν ἐστι τὸ τὰς τρεῖς ἴσας ἔχον πλευράς, ἰσοσκελὲς by straight-lines: trilateral figures being those contained
δὲ τὸ τὰς δύο μόνας ἴσας ἔχον πλευράς, σκαληνὸν δὲ τὸ by three straight-lines, quadrilateral by four, and multi-
τὰς τρεῖς ἀνίσους ἔχον πλευράς. lateral by more than four.
καʹ ῎Ετι δὲ τῶν τριπλεύρων σχημάτων ὀρθογώνιον μὲν 20. And of the trilateral figures: an equilateral trian-

τρίγωνόν ἐστι τὸ ἔχον ὀρθὴν γωνίαν, ἀμβλυγώνιον δὲ τὸ gle is that having three equal sides, an isosceles (triangle)
ἔχον ἀμβλεῖαν γωνίαν, ὀξυγώνιον δὲ τὸ τὰς τρεῖς ὀξείας that having only two equal sides, and a scalene (triangle)
ἔχον γωνίας. that having three unequal sides.

STOIQEIWN aþ. ELEMENTS BOOK 1
κβʹ. Τὼν δὲ τετραπλεύρων σχημάτων τετράγωνον μέν 21. And further of the trilateral figures: a right-angled

ἐστιν, ὃ ἰσόπλευρόν τέ ἐστι καὶ ὀρθογώνιον, ἑτερόμηκες triangle is that having a right-angle, an obtuse-angled
δέ, ὃ ὀρθογώνιον μέν, οὐκ ἰσόπλευρον δέ, ῥόμβος δέ, ὃ (triangle) that having an obtuse angle, and an acute-
ἰσόπλευρον μέν, οὐκ ὀρθογώνιον δέ, ῥομβοειδὲς δὲ τὸ τὰς angled (triangle) that having three acute angles.
ἀπεναντίον πλευράς τε καὶ γωνίας ἴσας ἀλλήλαις ἔχον, ὃ 22. And of the quadrilateral figures: a square is that
οὔτε ἰσόπλευρόν ἐστιν οὔτε ὀρθογώνιον· τὰ δὲ παρὰ ταῦτα which is right-angled and equilateral, a rectangle that
τετράπλευρα τραπέζια καλείσθω. which is right-angled but not equilateral, a rhombus that
κγʹ. Παράλληλοί εἰσιν εὐθεῖαι, αἵτινες ἐν τῷ αὐτῷ which is equilateral but not right-angled, and a rhomboid

ἐπιπέδῳ οὖσαι καὶ ἐκβαλλόμεναι εἰς ἄπειρον ἐφ᾿ ἑκάτερα that having opposite sides and angles equal to one an-
τὰ μέρη ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις. other which is neither right-angled nor equilateral. And

let quadrilateral figures besides these be called trapezia.
23. Parallel lines are straight-lines which, being in the

same plane, and being produced to infinity in each direc-
tion, meet with one another in neither (of these direc-

tions).

† This should really be counted as a postulate, rather than as part of a definition.AÊt mata. Postulates
αʹ. ᾿Ηιτήσθω ἀπὸ παντὸς σημείου ἐπὶ πᾶν σημεῖον 1. Let it have been postulated† to draw a straight-line

εὐθεῖαν γραμμὴν ἀγαγεῖν. from any point to any point.
βʹ. Καὶ πεπερασμένην εὐθεῖαν κατὰ τὸ συνεχὲς ἐπ᾿ 2. And to produce a finite straight-line continuously

εὐθείας ἐκβαλεῖν. in a straight-line.
γʹ. Καὶ παντὶ κέντρῳ καὶ διαστήματι κύκλον γράφεσθαι. 3. And to draw a circle with any center and radius.
δʹ. Καὶ πάσας τὰς ὀρθὰς γωνίας ἴσας ἀλλήλαις εἶναι. 4. And that all right-angles are equal to one another.
εʹ. Καὶ ἐὰν εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὰς ἐντὸς 5. And that if a straight-line falling across two (other)

καὶ ἐπὶ τὰ αὐτὰ μέρη γωνίας δύο ὀρθῶν ἐλάσσονας ποιῇ, straight-lines makes internal angles on the same side
ἐκβαλλομένας τὰς δύο εὐθείας ἐπ᾿ ἄπειρον συμπίπτειν, ἐφ᾿ (of itself whose sum is) less than two right-angles, then
ἃ μέρη εἰσὶν αἱ τῶν δύο ὀρθῶν ἐλάσσονες. the two (other) straight-lines, being produced to infinity,

meet on that side (of the original straight-line) that the
(sum of the internal angles) is less than two right-angles

(and do not meet on the other side).‡

† The Greek present perfect tense indicates a past action with present significance. Hence, the 3rd-person present perfect imperative >Hit sjw
could be translated as “let it be postulated”, in the sense “let it stand as postulated”, but not “let the postulate be now brought forward”. The

literal translation “let it have been postulated” sounds awkward in English, but more accurately captures the meaning of the Greek.
‡ This postulate effectively specifies that we are dealing with the geometry of flat, rather than curved, space.KoinaÈ ênnoiai. Common Notions
αʹ. Τὰ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα. 1. Things equal to the same thing are also equal to
βʹ. Καὶ ἐὰν ἴσοις ἴσα προστεθῇ, τὰ ὅλα ἐστὶν ἴσα. one another.
γʹ. Καὶ ἐὰν ἀπὸ ἴσων ἴσα ἀφαιρεθῇ, τὰ καταλειπόμενά 2. And if equal things are added to equal things then

ἐστιν ἴσα. the wholes are equal.
δʹ. Καὶ τὰ ἐφαρμόζοντα ἐπ᾿ ἀλλήλα ἴσα ἀλλήλοις ἐστίν. 3. And if equal things are subtracted from equal things

εʹ. Καὶ τὸ ὅλον τοῦ μέρους μεῖζόν [ἐστιν]. then the remainders are equal.†

4. And things coinciding with one another are equal
to one another.

5. And the whole [is] greater than the part.

† As an obvious extension of C.N.s 2 & 3—if equal things are added or subtracted from the two sides of an inequality then the inequality remains

STOIQEIWN aþ. ELEMENTS BOOK 1
an inequality of the same type. aþ. Proposition 1
᾿Επὶ τῆς δοθείσης εὐθείας πεπερασμένης τρίγωνον To construct an equilateral triangle on a given finite

ἰσόπλευρον συστήσασθαι. straight-line.

∆ Α

Β Ε BA ED

῎Εστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ. Let AB be the given finite straight-line.
Δεῖ δὴ ἐπὶ τῆς ΑΒ εὐθείας τρίγωνον ἰσόπλευρον So it is required to construct an equilateral triangle on

συστήσασθαι. the straight-line AB.
Κέντρῳ μὲν τῷ Α διαστήματι δὲ τῷ ΑΒ κύκλος Let the circle BCD with center A and radius AB have

γεγράφθω ὁ ΒΓΔ, καὶ πάλιν κέντρῳ μὲν τῷ Β διαστήματι δὲ been drawn [Post. 3], and again let the circle ACE with
τῷ ΒΑ κύκλος γεγράφθω ὁ ΑΓΕ, καὶ ἀπὸ τοῦ Γ σημείου, center B and radius BA have been drawn [Post. 3]. And
καθ᾿ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι, ἐπί τὰ Α, Β σημεῖα let the straight-lines CA and CB have been joined from

ἐπεζεύχθωσαν εὐθεῖαι αἱ ΓΑ, ΓΒ. the point C, where the circles cut one another,† to the
Καὶ ἐπεὶ τὸ Α σημεῖον κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου, points A and B (respectively) [Post. 1].

ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ· πάλιν, ἐπεὶ τὸ Β σημεῖον κέντρον And since the point A is the center of the circle CDB,
ἐστὶ τοῦ ΓΑΕ κύκλου, ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ. ἐδείχθη δὲ AC is equal to AB [Def. 1.15]. Again, since the point
καὶ ἡ ΓΑ τῇ ΑΒ ἴση· ἑκατέρα ἄρα τῶν ΓΑ, ΓΒ τῇ ΑΒ ἐστιν B is the center of the circle CAE, BC is equal to BA
ἴση. τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα· καὶ ἡ ΓΑ ἄρα [Def. 1.15]. But CA was also shown (to be) equal to AB.
τῇ ΓΒ ἐστιν ἴση· αἱ τρεῖς ἄρα αἱ ΓΑ, ΑΒ, ΒΓ ἴσαι ἀλλήλαις Thus, CA and CB are each equal to AB. But things equal
εἰσίν. to the same thing are also equal to one another [C.N. 1].
᾿Ισόπλευρον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον. καὶ συνέσταται Thus, CA is also equal to CB. Thus, the three (straight-

ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης τῆς ΑΒ. ὅπερ ἔδει lines) CA, AB, and BC are equal to one another.
ποιῆσαι. Thus, the triangle ABC is equilateral, and has been

constructed on the given finite straight-line AB. (Which

is) the very thing it was required to do.

† The assumption that the circles do indeed cut one another should be counted as an additional postulate. There is also an implicit assumption

that two straight-lines cannot share a common segment.bþ. Proposition 2†
Πρὸς τῷ δοθέντι σημείῳ τῇ δοθείσῃ εὐθείᾳ ἴσην εὐθεῖαν To place a straight-line equal to a given straight-line

θέσθαι. at a given point (as an extremity).
῎Εστω τὸ μὲν δοθὲν σημεῖον τὸ Α, ἡ δὲ δοθεῖσα εὐθεῖα Let A be the given point, and BC the given straight-

ἡ ΒΓ· δεῖ δὴ πρὸς τῷ Α σημείῳ τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ line. So it is required to place a straight-line at point A
ἴσην εὐθεῖαν θέσθαι. equal to the given straight-line BC.
᾿Επεζεύχθω γὰρ ἀπὸ τοῦ Α σημείου ἐπί τὸ Β σημεῖον For let the straight-line AB have been joined from

εὐθεῖα ἡ ΑΒ, καὶ συνεστάτω ἐπ᾿ αὐτῆς τρίγωνον ἰσόπλευρον point A to point B [Post. 1], and let the equilateral trian-
τὸ ΔΑΒ, καὶ ἐκβεβλήσθωσαν ἐπ᾿ εὐθείας ταῖς ΔΑ, ΔΒ gle DAB have been been constructed upon it [Prop. 1.1].

STOIQEIWN aþ. ELEMENTS BOOK 1
εὐθεῖαι αἱ ΑΕ, ΒΖ, καὶ κέντρῳ μὲν τῷ Β διαστήματι δὲ τῷ And let the straight-lines AE and BF have been pro-
ΒΓ κύκλος γεγράφθω ὁ ΓΗΘ, καὶ πάλιν κέντρῳ τῷ Δ καὶ duced in a straight-line with DA and DB (respectively)
διαστήματι τῷ ΔΗ κύκλος γεγράφθω ὁ ΗΚΛ. [Post. 2]. And let the circle CGH with center B and ra-

dius BC have been drawn [Post. 3], and again let the cir-

cle GKL with center D and radius DG have been drawn
[Post. 3].

∆

᾿Επεὶ οὖν τὸ Β σημεῖον κέντρον ἐστὶ τοῦ ΓΗΘ, ἴση ἐστὶν Therefore, since the point B is the center of (the cir-
ἡ ΒΓ τῇ ΒΗ. πάλιν, ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ τοῦ cle) CGH , BC is equal to BG [Def. 1.15]. Again, since
ΗΚΛ κύκλου, ἴση ἐστὶν ἡ ΔΛ τῇ ΔΗ, ὧν ἡ ΔΑ τῇ ΔΒ ἴση the point D is the center of the circle GKL, DL is equal
ἐστίν. λοιπὴ ἄρα ἡ ΑΛ λοιπῇ τῇ ΒΗ ἐστιν ἴση. ἐδείχθη δὲ to DG [Def. 1.15]. And within these, DA is equal to DB.
καὶ ἡ ΒΓ τῇ ΒΗ ἴση· ἑκατέρα ἄρα τῶν ΑΛ, ΒΓ τῇ ΒΗ ἐστιν Thus, the remainder AL is equal to the remainder BG
ἴση. τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα· καὶ ἡ ΑΛ [C.N. 3]. But BC was also shown (to be) equal to BG.
ἄρα τῇ ΒΓ ἐστιν ἴση. Thus, AL and BC are each equal to BG. But things equal
Πρὸς ἄρα τῷ δοθέντι σημείῳ τῷ Α τῇ δοθείσῃ εὐθείᾳ to the same thing are also equal to one another [C.N. 1].

τῇ ΒΓ ἴση εὐθεῖα κεῖται ἡ ΑΛ· ὅπερ ἔδει ποιῆσαι. Thus, AL is also equal to BC.
Thus, the straight-line AL, equal to the given straight-

line BC, has been placed at the given point A. (Which
is) the very thing it was required to do.

† This proposition admits of a number of different cases, depending on the relative positions of the point A and the line BC. In such situations,

Euclid invariably only considers one particular case—usually, the most difficult—and leaves the remaining cases as exercises for the reader.gþ. Proposition 3
Δύο δοθεισῶν εὐθειῶν ἀνίσων ἀπὸ τῆς μείζονος τῇ For two given unequal straight-lines, to cut off from

ἐλάσσονι ἴσην εὐθεῖαν ἀφελεῖν. the greater a straight-line equal to the lesser.
῎Εστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι ἄνισοι αἱ ΑΒ, Γ, ὧν Let AB and C be the two given unequal straight-lines,

μείζων ἔστω ἡ ΑΒ· δεῖ δὴ ἀπὸ τῆς μείζονος τῆς ΑΒ τῇ of which let the greater be AB. So it is required to cut off
ἐλάσσονι τῇ Γ ἴσην εὐθεῖαν ἀφελεῖν. a straight-line equal to the lesser C from the greater AB.
Κείσθω πρὸς τῷ Α σημείῳ τῇ Γ εὐθείᾳ ἴση ἡ ΑΔ· καὶ Let the line AD, equal to the straight-line C, have

κέντρῳ μὲν τῷ Α διαστήματι δὲ τῷ ΑΔ κύκλος γεγράφθω been placed at point A [Prop. 1.2]. And let the circle
ὁ ΔΕΖ. DEF have been drawn with center A and radius AD
Καὶ ἐπεὶ τὸ Α σημεῖον κέντρον ἐστὶ τοῦ ΔΕΖ κύκλου, [Post. 3].

STOIQEIWN aþ. ELEMENTS BOOK 1
ἴση ἐστὶν ἡ ΑΕ τῇ ΑΔ· ἀλλὰ καὶ ἡ Γ τῇ ΑΔ ἐστιν ἴση. And since point A is the center of circle DEF , AE
ἑκατέρα ἄρα τῶν ΑΕ, Γ τῇ ΑΔ ἐστιν ἴση· ὥστε καὶ ἡ ΑΕ is equal to AD [Def. 1.15]. But, C is also equal to AD.
τῇ Γ ἐστιν ἴση. Thus, AE and C are each equal to AD. So AE is also

equal to C [C.N. 1].

∆

Ε
Β

Δύο ἄρα δοθεισῶν εὐθειῶν ἀνίσων τῶν ΑΒ, Γ ἀπὸ τῆς Thus, for two given unequal straight-lines, AB and C,
μείζονος τῆς ΑΒ τῇ ἐλάσσονι τῇ Γ ἴση ἀφῄρηται ἡ ΑΕ· ὅπερ the (straight-line) AE, equal to the lesser C, has been cut
ἔδει ποιῆσαι. off from the greater AB. (Which is) the very thing it was

required to do.dþ. Proposition 4
᾿Εὰν δύο τρίγωνα τὰς δύο πλευρὰς [ταῖς] δυσὶ πλευραῖς If two triangles have two sides equal to two sides, re-

ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην spectively, and have the angle(s) enclosed by the equal
ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ τὴν straight-lines equal, then they will also have the base
βάσιν τῂ βάσει ἴσην ἕξει, καὶ τὸ τρίγωνον τῷ τριγώνῳ ἴσον equal to the base, and the triangle will be equal to the tri-
ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται angle, and the remaining angles subtended by the equal
ἑκατέρα ἑκατέρᾳ, ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν. sides will be equal to the corresponding remaining an-

gles.

∆

Γ Ε Ζ FB

C E

῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς Let ABC and DEF be two triangles having the two
τὰς ΑΒ, ΑΓ ταῖς δυσὶ πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα sides AB and AC equal to the two sides DE and DF , re-
ἑκατέραν ἑκατέρᾳ τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ spectively. (That is) AB to DE, and AC to DF . And (let)
καὶ γωνίαν τὴν ὑπὸ ΒΑΓ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴσην. λέγω, the angle BAC (be) equal to the angle EDF . I say that
ὅτι καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν, καὶ τὸ ΑΒΓ the base BC is also equal to the base EF , and triangle
τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ABC will be equal to triangle DEF , and the remaining
ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾿ ἃς angles subtended by the equal sides will be equal to the
αἱ ἴσαι πλευραὶ ὑποτείνουσιν, ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ, corresponding remaining angles. (That is) ABC to DEF ,
ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ. and ACB to DFE.

᾿Εφαρμοζομένου γὰρ τοῦ ΑΒΓ τριγώνου ἐπὶ τὸ ΔΕΖ For if triangle ABC is applied to triangle DEF ,† the
τρίγωνον καὶ τιθεμένου τοῦ μὲν Α σημείου ἐπὶ τὸ Δ σημεῖον point A being placed on the point D, and the straight-line

STOIQEIWN aþ. ELEMENTS BOOK 1
τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΔΕ, ἐφαρμόσει καὶ τὸ Β σημεῖον AB on DE, then the point B will also coincide with E,
ἐπὶ τὸ Ε διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΔΕ· ἐφαρμοσάσης δὴ on account of AB being equal to DE. So (because of)
τῆς ΑΒ ἐπὶ τὴν ΔΕ ἐφαρμόσει καὶ ἡ ΑΓ εὐθεῖα ἐπὶ τὴν ΔΖ AB coinciding with DE, the straight-line AC will also
διὰ τὸ ἴσην εἶναι τὴν ὑπὸ ΒΑΓ γωνίαν τῇ ὑπὸ ΕΔΖ· ὥστε καὶ coincide with DF , on account of the angle BAC being
τὸ Γ σημεῖον ἐπὶ τὸ Ζ σημεῖον ἐφαρμόσει διὰ τὸ ἴσην πάλιν equal to EDF . So the point C will also coincide with the
εἶναι τὴν ΑΓ τῇ ΔΖ. ἀλλὰ μὴν καὶ τὸ Β ἐπὶ τὸ Ε ἐφηρμόκει· point F , again on account of AC being equal to DF . But,
ὥστε βάσις ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ἐφαρμόσει. εἰ γὰρ τοῦ point B certainly also coincided with point E, so that the
μὲν Β ἐπὶ τὸ Ε ἐφαρμόσαντος τοῦ δὲ Γ ἐπὶ τὸ Ζ ἡ ΒΓ βάσις base BC will coincide with the base EF . For if B coin-
ἐπὶ τὴν ΕΖ οὐκ ἐφαρμόσει, δύο εὐθεῖαι χωρίον περιέξουσιν· cides with E, and C with F , and the base BC does not
ὅπερ ἐστὶν ἀδύνατον. ἐφαρμόσει ἄρα ἡ ΒΓ βάσις ἐπὶ τὴν coincide with EF , then two straight-lines will encompass

ΕΖ καὶ ἴση αὐτῇ ἔσται· ὥστε καὶ ὅλον τὸ ΑΒΓ τρίγωνον an area. The very thing is impossible [Post. 1].‡ Thus,
ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον ἐφαρμόσει καὶ ἴσον αὐτῷ ἔσται, the base BC will coincide with EF , and will be equal to
καὶ αἱ λοιπαὶ γωνίαι ἐπὶ τὰς λοιπὰς γωνίας ἐφαρμόσουσι καὶ it [C.N. 4]. So the whole triangle ABC will coincide with
ἴσαι αὐταῖς ἔσονται, ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ἡ δὲ ὑπὸ the whole triangle DEF , and will be equal to it [C.N. 4].
ΑΓΒ τῇ ὑπὸ ΔΖΕ. And the remaining angles will coincide with the remain-
᾿Εὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς [ταῖς] δύο ing angles, and will be equal to them [C.N. 4]. (That is)

πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν γωνίαν τῇ ABC to DEF , and ACB to DFE [C.N. 4].
γωνίᾳ ἴσην ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, Thus, if two triangles have two sides equal to two
καὶ τὴν βάσιν τῂ βάσει ἴσην ἕξει, καὶ τὸ τρίγωνον τῷ sides, respectively, and have the angle(s) enclosed by the
τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς equal straight-line equal, then they will also have the base
γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ equal to the base, and the triangle will be equal to the tri-
ὑποτείνουσιν· ὅπερ ἔδει δεῖξαι. angle, and the remaining angles subtended by the equal

sides will be equal to the corresponding remaining an-

gles. (Which is) the very thing it was required to show.

† The application of one figure to another should be counted as an additional postulate.

‡ Since Post. 1 implicitly assumes that the straight-line joining two given points is unique.eþ. Proposition 5
Τῶν ἰσοσκελῶν τριγώνων αἱ τρὸς τῇ βάσει γωνίαι ἴσαι For isosceles triangles, the angles at the base are equal

ἀλλήλαις εἰσίν, καὶ προσεκβληθεισῶν τῶν ἴσων εὐθειῶν αἱ to one another, and if the equal sides are produced then
ὑπὸ τὴν βάσιν γωνίαι ἴσαι ἀλλήλαις ἔσονται. the angles under the base will be equal to one another.

∆

῎Εστω τρίγωνον ἰσοσκελὲς τὸ ΑΒΓ ἴσην ἔχον τὴν Let ABC be an isosceles triangle having the side AB
ΑΒ πλευρὰν τῇ ΑΓ πλευρᾷ, καὶ προσεκβεβλήσθωσαν ἐπ᾿ equal to the side AC, and let the straight-lines BD and
εὐθείας ταῖς ΑΒ, ΑΓ εὐθεῖαι αἱ ΒΔ, ΓΕ· λέγω, ὅτι ἡ μὲν CE have been produced in a straight-line with AB and
ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΑΓΒ ἴση ἐστίν, ἡ δὲ ὑπὸ ΓΒΔ τῇ AC (respectively) [Post. 2]. I say that the angle ABC is
ὑπὸ ΒΓΕ. equal to ACB, and (angle) CBD to BCE.
Εἰλήφθω γὰρ ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ, καὶ For let the point F have been taken at random on BD,

ἀφῃρήσθω ἀπὸ τῆς μείζονος τῆς ΑΕ τῇ ἐλάσσονι τῇ ΑΖ and let AG have been cut off from the greater AE, equal

STOIQEIWN aþ. ELEMENTS BOOK 1
ἴση ἡ ΑΗ, καὶ ἐπεζεύχθωσαν αἱ ΖΓ, ΗΒ εὐθεῖαι. to the lesser AF [Prop. 1.3]. Also, let the straight-lines
᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΖ τῇ ΑΗ ἡ δὲ ΑΒ τῇ ΑΓ, FC and GB have been joined [Post. 1].

δύο δὴ αἱ ΖΑ, ΑΓ δυσὶ ταῖς ΗΑ, ΑΒ ἴσαι εἰσὶν ἑκατέρα In fact, since AF is equal to AG, and AB to AC,
ἑκατέρᾳ· καὶ γωνίαν κοινὴν περιέχουσι τὴν ὑπὸ ΖΑΗ· βάσις the two (straight-lines) FA, AC are equal to the two
ἄρα ἡ ΖΓ βάσει τῇ ΗΒ ἴση ἐστίν, καὶ τὸ ΑΖΓ τρίγωνον τῷ (straight-lines) GA, AB, respectively. They also encom-
ΑΗΒ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς pass a common angle, FAG. Thus, the base FC is equal
γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ to the base GB, and the triangle AFC will be equal to the
ὑποτείνουσιν, ἡ μὲν ὑπὸ ΑΓΖ τῇ ὑπὸ ΑΒΗ, ἡ δὲ ὑπὸ ΑΖΓ triangle AGB, and the remaining angles subtendend by
τῇ ὑπὸ ΑΗΒ. καὶ ἐπεὶ ὅλη ἡ ΑΖ ὅλῃ τῇ ΑΗ ἐστιν ἴση, ὧν the equal sides will be equal to the corresponding remain-
ἡ ΑΒ τῇ ΑΓ ἐστιν ἴση, λοιπὴ ἄρα ἡ ΒΖ λοιπῇ τῇ ΓΗ ἐστιν ing angles [Prop. 1.4]. (That is) ACF to ABG, and AFC
ἴση. ἐδείχθη δὲ καὶ ἡ ΖΓ τῇ ΗΒ ἴση· δύο δὴ αἱ ΒΖ, ΖΓ δυσὶ to AGB. And since the whole of AF is equal to the whole
ταῖς ΓΗ, ΗΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ of AG, within which AB is equal to AC, the remainder
ΒΖΓ γωνίᾳ τῃ ὑπὸ ΓΗΒ ἴση, καὶ βάσις αὐτῶν κοινὴ ἡ ΒΓ· BF is thus equal to the remainder CG [C.N. 3]. But FC
καὶ τὸ ΒΖΓ ἄρα τρίγωνον τῷ ΓΗΒ τριγώνῳ ἴσον ἔσται, καὶ was also shown (to be) equal to GB. So the two (straight-
αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα lines) BF , FC are equal to the two (straight-lines) CG,
ἑκατέρᾳ, ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν· ἴση ἄρα ἐστὶν GB, respectively, and the angle BFC (is) equal to the
ἡ μὲν ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἡ δὲ ὑπὸ ΒΓΖ τῇ ὑπὸ ΓΒΗ. angle CGB, and the base BC is common to them. Thus,
ἐπεὶ οὖν ὅλη ἡ ὑπὸ ΑΒΗ γωνία ὅλῃ τῇ ὑπὸ ΑΓΖ γωνίᾳ the triangle BFC will be equal to the triangle CGB, and
ἐδείχθη ἴση, ὧν ἡ ὑπὸ ΓΒΗ τῇ ὑπὸ ΒΓΖ ἴση, λοιπὴ ἄρα ἡ the remaining angles subtended by the equal sides will be
ὑπὸ ΑΒΓ λοιπῇ τῇ ὑπὸ ΑΓΒ ἐστιν ἴση· καί εἰσι πρὸς τῇ equal to the corresponding remaining angles [Prop. 1.4].
βάσει τοῦ ΑΒΓ τριγώνου. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΖΒΓ τῇ Thus, FBC is equal to GCB, and BCF to CBG. There-
ὑπὸ ΗΓΒ ἴση· καί εἰσιν ὑπὸ τὴν βάσιν. fore, since the whole angle ABG was shown (to be) equal
Τῶν ἄρα ἰσοσκελῶν τριγώνων αἱ τρὸς τῇ βάσει γωνίαι to the whole angle ACF , within which CBG is equal to

ἴσαι ἀλλήλαις εἰσίν, καὶ προσεκβληθεισῶν τῶν ἴσων εὐθειῶν BCF , the remainder ABC is thus equal to the remainder
αἱ ὑπὸ τὴν βάσιν γωνίαι ἴσαι ἀλλήλαις ἔσονται· ὅπερ ἔδει ACB [C.N. 3]. And they are at the base of triangle ABC.
δεῖξαι. And FBC was also shown (to be) equal to GCB. And

they are under the base.
Thus, for isosceles triangles, the angles at the base are

equal to one another, and if the equal sides are produced
then the angles under the base will be equal to one an-

other. (Which is) the very thing it was required to show.�þ. Proposition 6
᾿Εὰν τριγώνου αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν, καὶ If a triangle has two angles equal to one another then

αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι πλευραὶ ἴσαι ἀλλήλαις the sides subtending the equal angles will also be equal
ἔσονται. to one another.

Β Γ

∆ D

῎Εστω τρίγωνον τὸ ΑΒΓ ἴσην ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν Let ABC be a triangle having the angle ABC equal
τῇ ὑπὸ ΑΓΒ γωνίᾳ· λέγω, ὅτι καὶ πλευρὰ ἡ ΑΒ πλευρᾷ τῇ to the angle ACB. I say that side AB is also equal to side
ΑΓ ἐστιν ἴση. AC.

STOIQEIWN aþ. ELEMENTS BOOK 1
Εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ, ἡ ἑτέρα αὐτῶν μείζων For if AB is unequal to AC then one of them is

ἐστίν. ἔστω μείζων ἡ ΑΒ, καὶ ἀφῃρήσθω ἀπὸ τῆς μείζονος greater. Let AB be greater. And let DB, equal to
τῆς ΑΒ τῇ ἐλάττονι τῇ ΑΓ ἴση ἡ ΔΒ, καὶ ἐπεζεύχθω ἡ ΔΓ. the lesser AC, have been cut off from the greater AB
᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΒ τῇ ΑΓ κοινὴ δὲ ἡ ΒΓ, δύο δὴ [Prop. 1.3]. And let DC have been joined [Post. 1].

αἱ ΔΒ, ΒΓ δύο ταῖς ΑΓ, ΓΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ Therefore, since DB is equal to AC, and BC (is) com-
γωνία ἡ ὑπὸ ΔΒΓ γωνίᾳ τῇ ὑπὸ ΑΓΒ ἐστιν ἴση· βάσις ἄρα ἡ mon, the two sides DB, BC are equal to the two sides
ΔΓ βάσει τῇ ΑΒ ἴση ἐστίν, καὶ τὸ ΔΒΓ τρίγωνον τῷ ΑΓΒ AC, CB, respectively, and the angle DBC is equal to the
τριγώνῳ ἴσον ἔσται, τὸ ἔλασσον τῷ μείζονι· ὅπερ ἄτοπον· angle ACB. Thus, the base DC is equal to the base AB,
οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ· ἴση ἄρα. and the triangle DBC will be equal to the triangle ACB
᾿Εὰν ἄρα τριγώνου αἱ δὑο γωνίαι ἴσαι ἀλλήλαις ὦσιν, καὶ [Prop. 1.4], the lesser to the greater. The very notion (is)

αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι πλευραὶ ἴσαι ἀλλήλαις absurd [C.N. 5]. Thus, AB is not unequal to AC. Thus,

ἔσονται· ὅπερ ἔδει δεῖξαι. (it is) equal.†

Thus, if a triangle has two angles equal to one another
then the sides subtending the equal angles will also be

equal to one another. (Which is) the very thing it was

required to show.

† Here, use is made of the previously unmentioned common notion that if two quantities are not unequal then they must be equal. Later on, use

is made of the closely related common notion that if two quantities are not greater than or less than one another, respectively, then they must be

equal to one another. zþ. Proposition 7
᾿Επὶ τῆς αὐτῆς εὐθείας δύο ταῖς αὐταῖς εὐθείαις ἄλλαι On the same straight-line, two other straight-lines

δύο εὐθεῖαι ἴσαι ἑκατέρα ἑκατέρᾳ οὐ συσταθήσονται πρὸς equal, respectively, to two (given) straight-lines (which
ἄλλῳ καὶ ἄλλῳ σημείῳ ἐπὶ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα meet) cannot be constructed (meeting) at a different
ἔχουσαι ταῖς ἐξ ἀρχῆς εὐθείαις. point on the same side (of the straight-line), but having

the same ends as the given straight-lines.

ΒΑ

∆

Εἰ γὰρ δυνατόν, ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ δύο ταῖς For, if possible, let the two straight-lines AC, CB,
αὐταῖς εὐθείαις ταῖς ΑΓ, ΓΒ ἄλλαι δύο εὐθεῖαι αἱ ΑΔ, ΔΒ equal to two other straight-lines AD, DB, respectively,
ἴσαι ἑκατέρα ἑκατέρᾳ συνεστάτωσαν πρὸς ἄλλῳ καὶ ἄλλῳ have been constructed on the same straight-line AB,
σημείῳ τῷ τε Γ καὶ Δ ἐπὶ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα meeting at different points, C and D, on the same side
ἔχουσαι, ὥστε ἴσην εἶναι τὴν μὲν ΓΑ τῇ ΔΑ τὸ αὐτὸ πέρας (of AB), and having the same ends (on AB). So CA is
ἔχουσαν αὐτῇ τὸ Α, τὴν δὲ ΓΒ τῇ ΔΒ τὸ αὐτὸ πέρας ἔχου- equal to DA, having the same end A as it, and CB is
σαν αὐτῇ τὸ Β, καὶ ἐπεζεύχθω ἡ ΓΔ. equal to DB, having the same end B as it. And let CD
᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΑΔ, ἴση ἐστὶ καὶ γωνία ἡ have been joined [Post. 1].

ὑπὸ ΑΓΔ τῇ ὑπὸ ΑΔΓ· μείζων ἄρα ἡ ὑπὸ ΑΔΓ τῆς ὑπὸ Therefore, since AC is equal to AD, the angle ACD
ΔΓΒ· πολλῷ ἄρα ἡ ὑπὸ ΓΔΒ μείζων ἐστί τῆς ὑπὸ ΔΓΒ. is also equal to angle ADC [Prop. 1.5]. Thus, ADC (is)
πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΔΒ, ἴση ἐστὶ καὶ γωνία ἡ greater than DCB [C.N. 5]. Thus, CDB is much greater
ὑπὸ ΓΔΒ γωνίᾳ τῇ ὑπὸ ΔΓΒ. ἐδείχθη δὲ αὐτῆς καὶ πολλῷ than DCB [C.N. 5]. Again, since CB is equal to DB, the
μείζων· ὅπερ ἐστὶν ἀδύνατον. angle CDB is also equal to angle DCB [Prop. 1.5]. But
Οὐκ ἄρα ἐπὶ τῆς αὐτῆς εὐθείας δύο ταῖς αὐταῖς εὐθείαις it was shown that the former (angle) is also much greater

STOIQEIWN aþ. ELEMENTS BOOK 1
ἄλλαι δύο εὐθεῖαι ἴσαι ἑκατέρα ἑκατέρᾳ συσταθήσονται πρὸς (than the latter). The very thing is impossible.
ἄλλῳ καὶ ἄλλῳ σημείῳ ἐπὶ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα Thus, on the same straight-line, two other straight-
ἔχουσαι ταῖς ἐξ ἀρχῆς εὐθείαις· ὅπερ ἔδει δεῖξαι. lines equal, respectively, to two (given) straight-lines

(which meet) cannot be constructed (meeting) at a dif-

ferent point on the same side (of the straight-line), but
having the same ends as the given straight-lines. (Which

is) the very thing it was required to show.hþ. Proposition 8
᾿Εὰν δύο τρίγωνα τὰς δύο πλευρὰς [ταῖς] δύο πλευραῖς If two triangles have two sides equal to two sides, re-

ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει spectively, and also have the base equal to the base, then
ἴσην, καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἕξει τὴν ὑπὸ τῶν ἴσων they will also have equal the angles encompassed by the
εὐθειῶν περιεχομένην. equal straight-lines.

∆
Η

D
G

B
E

F
C

῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς Let ABC and DEF be two triangles having the two
τὰς ΑΒ, ΑΓ ταῖς δύο πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα sides AB and AC equal to the two sides DE and DF ,
ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ· respectively. (That is) AB to DE, and AC to DF . Let
ἐχέτω δὲ καὶ βάσιν τὴν ΒΓ βάσει τῇ ΕΖ ἴσην· λέγω, ὅτι καὶ them also have the base BC equal to the base EF . I say
γωνία ἡ ὑπὸ ΒΑΓ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση. that the angle BAC is also equal to the angle EDF .
᾿Εφαρμοζομένου γὰρ τοῦ ΑΒΓ τριγώνου ἐπὶ τὸ ΔΕΖ For if triangle ABC is applied to triangle DEF , the

τρίγωνον καὶ τιθεμένου τοῦ μὲν Β σημείου ἐπὶ τὸ Ε σημεῖον point B being placed on point E, and the straight-line
τῆς δὲ ΒΓ εὐθείας ἐπὶ τὴν ΕΖ ἐφαρμόσει καὶ τὸ Γ σημεῖον BC on EF , then point C will also coincide with F , on
ἐπὶ τὸ Ζ διὰ τὸ ἴσην εἶναι τὴν ΒΓ τῇ ΕΖ· ἐφαρμοσάσης δὴ account of BC being equal to EF . So (because of) BC
τῆς ΒΓ ἐπὶ τὴν ΕΖ ἐφαρμόσουσι καὶ αἱ ΒΑ, ΓΑ ἐπὶ τὰς ΕΔ, coinciding with EF , (the sides) BA and CA will also co-
ΔΖ. εἰ γὰρ βάσις μὲν ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ἐφαρμόσει, αἱ incide with ED and DF (respectively). For if base BC
δὲ ΒΑ, ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ, ΔΖ οὐκ ἐφαρμόσουσιν coincides with base EF , but the sides AB and AC do not
ἀλλὰ παραλλάξουσιν ὡς αἱ ΕΗ, ΗΖ, συσταθήσονται ἐπὶ τῆς coincide with ED and DF (respectively), but miss like
αὐτῆς εὐθείας δύο ταῖς αὐταῖς εὐθείαις ἄλλαι δύο εὐθεῖαι EG and GF (in the above figure), then we will have con-
ἴσαι ἑκατέρα ἑκατέρᾳ πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ ἐπὶ τὰ structed upon the same straight-line, two other straight-
αὐτὰ μέρη τὰ αὐτὰ πέρατα ἔχουσαι. οὐ συνίστανται δέ· lines equal, respectively, to two (given) straight-lines,
οὐκ ἄρα ἐφαρμοζομένης τῆς ΒΓ βάσεως ἐπὶ τὴν ΕΖ βάσιν and (meeting) at a different point on the same side (of
οὐκ ἐφαρμόσουσι καὶ αἱ ΒΑ, ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ, ΔΖ. the straight-line), but having the same ends. But (such
ἐφαρμόσουσιν ἄρα· ὥστε καὶ γωνία ἡ ὑπὸ ΒΑΓ ἐπὶ γωνίαν straight-lines) cannot be constructed [Prop. 1.7]. Thus,
τὴν ὑπὸ ΕΔΖ ἐφαρμόσει καὶ ἴση αὐτῇ ἔσται. the base BC being applied to the base EF , the sides BA
᾿Εὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς [ταῖς] δύο and AC cannot not coincide with ED and DF (respec-

πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν βάσιν τῇ βάσει tively). Thus, they will coincide. So the angle BAC will
ἴσην ἔχῃ, καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἕξει τὴν ὑπὸ τῶν also coincide with angle EDF , and will be equal to it
ἴσων εὐθειῶν περιεχομένην· ὅπερ ἔδει δεῖξαι. [C.N. 4].

Thus, if two triangles have two sides equal to two
side, respectively, and have the base equal to the base,

STOIQEIWN aþ. ELEMENTS BOOK 1
then they will also have equal the angles encompassed

by the equal straight-lines. (Which is) the very thing it

was required to show.jþ. Proposition 9
Τὴν δοθεῖσαν γωνίαν εὐθύγραμμον δίχα τεμεῖν. To cut a given rectilinear angle in half.

Β Γ

∆

Ζ F

B C

῎Εστω ἡ δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ ΒΑΓ. δεῖ Let BAC be the given rectilinear angle. So it is re-
δὴ αὐτὴν δίχα τεμεῖν. quired to cut it in half.
Εἰλήφθω ἐπὶ τῆς ΑΒ τυχὸν σημεῖον τὸ Δ, καὶ ἀφῃρήσθω Let the point D have been taken at random on AB,

ἀπὸ τῆς ΑΓ τῇ ΑΔ ἴση ἡ ΑΕ, καὶ ἐπεζεύχθω ἡ ΔΕ, καὶ and let AE, equal to AD, have been cut off from AC
συνεστάτω ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον τὸ ΔΕΖ, καὶ [Prop. 1.3], and let DE have been joined. And let the
ἐπεζεύχθω ἡ ΑΖ· λέγω, ὅτι ἡ ὑπὸ ΒΑΓ γωνία δίχα τέτμηται equilateral triangle DEF have been constructed upon
ὑπὸ τῆς ΑΖ εὐθείας. DE [Prop. 1.1], and let AF have been joined. I say that
᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΕ, κοινὴ δὲ ἡ ΑΖ, δύο δὴ the angle BAC has been cut in half by the straight-line

αἱ ΔΑ, ΑΖ δυσὶ ταῖς ΕΑ, ΑΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ. AF .
καὶ βάσις ἡ ΔΖ βάσει τῇ ΕΖ ἴση ἐστίν· γωνία ἄρα ἡ ὑπὸ For since AD is equal to AE, and AF is common,
ΔΑΖ γωνίᾳ τῇ ὑπὸ ΕΑΖ ἴση ἐστίν. the two (straight-lines) DA, AF are equal to the two
῾Η ἄρα δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ ΒΑΓ δίχα (straight-lines) EA, AF , respectively. And the base DF

τέτμηται ὑπὸ τῆς ΑΖ εὐθείας· ὅπερ ἔδει ποιῆσαι. is equal to the base EF . Thus, angle DAF is equal to
angle EAF [Prop. 1.8].

Thus, the given rectilinear angle BAC has been cut in

half by the straight-line AF . (Which is) the very thing it
was required to do.iþ. Proposition 10

Τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην δίχα τεμεῖν. To cut a given finite straight-line in half.
῎Εστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ· δεῖ δὴ τὴν Let AB be the given finite straight-line. So it is re-

ΑΒ εὐθεῖαν πεπερασμένην δίχα τεμεῖν. quired to cut the finite straight-line AB in half.
Συνεστάτω ἐπ᾿ αὐτῆς τρίγωνον ἰσόπλευρον τὸ ΑΒΓ, καὶ Let the equilateral triangle ABC have been con-

τετμήσθω ἡ ὑπὸ ΑΓΒ γωνία δίχα τῇ ΓΔ εὐθείᾳ· λέγω, ὅτι structed upon (AB) [Prop. 1.1], and let the angle ACB
ἡ ΑΒ εὐθεῖα δίχα τέτμηται κατὰ τὸ Δ σημεῖον. have been cut in half by the straight-line CD [Prop. 1.9].
᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, κοινὴ δὲ ἡ ΓΔ, δύο δὴ I say that the straight-line AB has been cut in half at

αἱ ΑΓ, ΓΔ δύο ταῖς ΒΓ, ΓΔ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ point D.
γωνία ἡ ὑπὸ ΑΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση ἐστίν· βάσις ἄρα For since AC is equal to CB, and CD (is) common,

STOIQEIWN aþ. ELEMENTS BOOK 1
ἡ ΑΔ βάσει τῇ ΒΔ ἴση ἐστίν. the two (straight-lines) AC, CD are equal to the two

(straight-lines) BC, CD, respectively. And the angle

ACD is equal to the angle BCD. Thus, the base AD
is equal to the base BD [Prop. 1.4].

Α
∆

BA
D

῾Η ἄρα δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ δίχα τέτμηται Thus, the given finite straight-line AB has been cut
κατὰ τὸ Δ· ὅπερ ἔδει ποιῆσαι. in half at (point) D. (Which is) the very thing it was

required to do.iaþ. Proposition 11
Τῇ δοθείσῃ εὐθείᾳ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου To draw a straight-line at right-angles to a given

πρὸς ὀρθὰς γωνίας εὐθεῖαν γραμμὴν ἀγαγεῖν. straight-line from a given point on it.

Α Β

∆ Γ Ε

C E

῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ τὸ δὲ δοθὲν σημεῖον Let AB be the given straight-line, and C the given
ἐπ᾿ αὐτῆς τὸ Γ· δεῖ δὴ ἀπὸ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ point on it. So it is required to draw a straight-line from
πρὸς ὀρθὰς γωνίας εὐθεῖαν γραμμὴν ἀγαγεῖν. the point C at right-angles to the straight-line AB.
Εἰλήφθω ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ, καὶ κείσθω Let the point D be have been taken at random on AC,

τῇ ΓΔ ἴση ἡ ΓΕ, καὶ συνεστάτω ἐπὶ τῆς ΔΕ τρίγωνον and let CE be made equal to CD [Prop. 1.3], and let the
ἰσόπλευρον τὸ ΖΔΕ, καὶ ἐπεζεύχθω ἡ ΖΓ· λέγω, ὅτι τῇ equilateral triangle FDE have been constructed on DE
δοθείσῃ εὐθείᾳ τῇ ΑΒ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου [Prop. 1.1], and let FC have been joined. I say that the
τοῦ Γ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ. straight-line FC has been drawn at right-angles to the
᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ, κοινὴ δὲ ἡ ΓΖ, δύο given straight-line AB from the given point C on it.

δὴ αἱ ΔΓ, ΓΖ δυσὶ ταῖς ΕΓ, ΓΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· For since DC is equal to CE, and CF is common,
καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ ἴση ἐστίν· γωνία ἄρα ἡ ὑπὸ the two (straight-lines) DC, CF are equal to the two
ΔΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ ἴση ἐστίν· καί εἰσιν ἐφεξῆς. ὅταν (straight-lines), EC, CF , respectively. And the base DF
δὲ εὐθεῖα ἐπ᾿ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας is equal to the base FE. Thus, the angle DCF is equal
ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν· ὀρθὴ to the angle ECF [Prop. 1.8], and they are adjacent.
ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΔΓΖ, ΖΓΕ. But when a straight-line stood on a(nother) straight-line

STOIQEIWN aþ. ELEMENTS BOOK 1
Τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ ἀπὸ τοῦ πρὸς αὐτῇ makes the adjacent angles equal to one another, each of

δοθέντος σημείου τοῦ Γ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ the equal angles is a right-angle [Def. 1.10]. Thus, each
ἦκται ἡ ΓΖ· ὅπερ ἔδει ποιῆσαι. of the (angles) DCF and FCE is a right-angle.

Thus, the straight-line CF has been drawn at right-

angles to the given straight-line AB from the given point
C on it. (Which is) the very thing it was required to do.ibþ. Proposition 12

᾿Επὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον ἀπὸ τοῦ δοθέντος To draw a straight-line perpendicular to a given infi-
σημείου, ὃ μή ἐστιν ἐπ᾿ αὐτῆς, κάθετον εὐθεῖαν γραμμὴν nite straight-line from a given point which is not on it.
ἀγαγεῖν.

∆

Α Β

Η Ε

G H

῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἄπειρος ἡ ΑΒ τὸ δὲ δοθὲν Let AB be the given infinite straight-line and C the
σημεῖον, ὃ μή ἐστιν ἐπ᾿ αὐτῆς, τὸ Γ· δεῖ δὴ ἐπὶ τὴν δοθεῖσαν given point, which is not on (AB). So it is required to
εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ, draw a straight-line perpendicular to the given infinite
ὃ μή ἐστιν ἐπ᾿ αὐτῆς, κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν. straight-line AB from the given point C, which is not on
Εἰλήφθω γὰρ ἐπὶ τὰ ἕτερα μέρη τῆς ΑΒ εὐθείας τυχὸν (AB).

σημεῖον τὸ Δ, καὶ κέντρῳ μὲν τῷ Γ διαστήματι δὲ τῷ ΓΔ For let point D have been taken at random on the
κύκλος γεγράφθω ὁ ΕΖΗ, καὶ τετμήσθω ἡ ΕΗ εὐθεῖα δίχα other side (to C) of the straight-line AB, and let the
κατὰ τὸ Θ, καὶ ἐπεζεύχθωσαν αἱ ΓΗ, ΓΘ, ΓΕ εὐθεῖαι· circle EFG have been drawn with center C and radius
λέγω, ὅτι ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ CD [Post. 3], and let the straight-line EG have been cut
τοῦ δοθέντος σημείου τοῦ Γ, ὃ μή ἐστιν ἐπ᾿ αὐτῆς, κάθετος in half at (point) H [Prop. 1.10], and let the straight-
ἦκται ἡ ΓΘ. lines CG, CH , and CE have been joined. I say that the
᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΗΘ τῇ ΘΕ, κοινὴ δὲ ἡ ΘΓ, δύο (straight-line) CH has been drawn perpendicular to the

δὴ αἱ ΗΘ, ΘΓ δύο ταῖς ΕΘ, ΘΓ ἴσαι εἱσὶν ἑκατέρα ἑκατέρᾳ· given infinite straight-line AB from the given point C,
καὶ βάσις ἡ ΓΗ βάσει τῇ ΓΕ ἐστιν ἴση· γωνία ἄρα ἡ ὑπὸ which is not on (AB).
ΓΘΗ γωνίᾳ τῇ ὑπὸ ΕΘΓ ἐστιν ἴση. καί εἰσιν ἐφεξῆς. ὅταν For since GH is equal to HE, and HC (is) common,
δὲ εὐθεῖα ἐπ᾿ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας the two (straight-lines) GH , HC are equal to the two
ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν, καὶ (straight-lines) EH , HC, respectively, and the base CG
ἡ ἐφεστηκυῖα εὐθεῖα κάθετος καλεῖται ἐφ᾿ ἣν ἐφέστηκεν. is equal to the base CE. Thus, the angle CHG is equal
᾿Επὶ τὴν δοθεῖσαν ἄρα εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ τοῦ to the angle EHC [Prop. 1.8], and they are adjacent.

δοθέντος σημείου τοῦ Γ, ὃ μή ἐστιν ἐπ᾿ αὐτῆς, κάθετος But when a straight-line stood on a(nother) straight-line
ἦκται ἡ ΓΘ· ὅπερ ἔδει ποιῆσαι. makes the adjacent angles equal to one another, each of

the equal angles is a right-angle, and the former straight-
line is called a perpendicular to that upon which it stands

[Def. 1.10].
Thus, the (straight-line) CH has been drawn perpen-

dicular to the given infinite straight-line AB from the

STOIQEIWN aþ. ELEMENTS BOOK 1
given point C, which is not on (AB). (Which is) the very

thing it was required to do.igþ. Proposition 13
᾿Εὰν εὐθεῖα ἐπ᾿ εὐθεῖαν σταθεῖσα γωνίας ποιῇ, ἤτοι δύο If a straight-line stood on a(nother) straight-line

ὀρθὰς ἢ δυσὶν ὀρθαῖς ἴσας ποιήσει. makes angles, it will certainly either make two right-
angles, or (angles whose sum is) equal to two right-

angles.

Ε
Α

∆ Β C

A
E

D B

Εὐθεῖα γάρ τις ἡ ΑΒ ἐπ᾿ εὐθεῖαν τὴν ΓΔ σταθεῖσα For let some straight-line AB stood on the straight-
γωνίας ποιείτω τὰς ὑπὸ ΓΒΑ, ΑΒΔ· λὲγω, ὅτι αἱ ὑπὸ ΓΒΑ, line CD make the angles CBA and ABD. I say that
ΑΒΔ γωνίαι ἤτοι δύο ὀρθαί εἰσιν ἢ δυσὶν ὀρθαῖς ἴσαι. the angles CBA and ABD are certainly either two right-
Εἰ μὲν οὖν ἴση ἐστὶν ἡ ὑπὸ ΓΒΑ τῇ ὑπὸ ΑΒΔ, δύο ὀρθαί angles, or (have a sum) equal to two right-angles.

εἰσιν. εἰ δὲ οὔ, ἤχθω ἀπὸ τοῦ Β σημείου τῇ ΓΔ [εὐθείᾳ] πρὸς In fact, if CBA is equal to ABD then they are two
ὀρθὰς ἡ ΒΕ· αἱ ἄρα ὑπὸ ΓΒΕ, ΕΒΔ δύο ὀρθαί εἰσιν· καὶ right-angles [Def. 1.10]. But, if not, let BE have been
ἐπεὶ ἡ ὑπὸ ΓΒΕ δυσὶ ταῖς ὑπὸ ΓΒΑ, ΑΒΕ ἴση ἐστίν, κοινὴ drawn from the point B at right-angles to [the straight-
προσκείσθω ἡ ὑπὸ ΕΒΔ· αἱ ἄρα ὑπὸ ΓΒΕ, ΕΒΔ τρισὶ ταῖς line] CD [Prop. 1.11]. Thus, CBE and EBD are two
ὑπὸ ΓΒΑ, ΑΒΕ, ΕΒΔ ἴσαι εἰσίν. πάλιν, ἐπεὶ ἡ ὑπὸ ΔΒΑ right-angles. And since CBE is equal to the two (an-
δυσὶ ταῖς ὑπὸ ΔΒΕ, ΕΒΑ ἴση ἐστίν, κοινὴ προσκείσθω ἡ gles) CBA and ABE, let EBD have been added to both.
ὑπὸ ΑΒΓ· αἱ ἄρα ὑπὸ ΔΒΑ, ΑΒΓ τρισὶ ταῖς ὑπὸ ΔΒΕ, ΕΒΑ, Thus, the (sum of the angles) CBE and EBD is equal to
ΑΒΓ ἴσαι εἰσίν. ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΒΕ, ΕΒΔ τρισὶ the (sum of the) three (angles) CBA, ABE, and EBD
ταῖς αὐταῖς ἴσαι· τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα· [C.N. 2]. Again, since DBA is equal to the two (an-
καὶ αἱ ὑπὸ ΓΒΕ, ΕΒΔ ἄρα ταῖς ὑπὸ ΔΒΑ, ΑΒΓ ἴσαι εἰσίν· gles) DBE and EBA, let ABC have been added to both.
ἀλλὰ αἱ ὑπὸ ΓΒΕ, ΕΒΔ δύο ὀρθαί εἰσιν· καὶ αἱ ὑπὸ ΔΒΑ, Thus, the (sum of the angles) DBA and ABC is equal to
ΑΒΓ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν. the (sum of the) three (angles) DBE, EBA, and ABC
᾿Εὰν ἄρα εὐθεῖα ἐπ᾿ εὐθεῖαν σταθεῖσα γωνίας ποιῇ, ἤτοι [C.N. 2]. But (the sum of) CBE and EBD was also

δύο ὀρθὰς ἢ δυσὶν ὀρθαῖς ἴσας ποιήσει· ὅπερ ἔδει δεῖξαι. shown (to be) equal to the (sum of the) same three (an-
gles). And things equal to the same thing are also equal
to one another [C.N. 1]. Therefore, (the sum of) CBE

and EBD is also equal to (the sum of) DBA and ABC.

But, (the sum of) CBE and EBD is two right-angles.
Thus, (the sum of) ABD and ABC is also equal to two

right-angles.

Thus, if a straight-line stood on a(nother) straight-
line makes angles, it will certainly either make two right-

angles, or (angles whose sum is) equal to two right-
angles. (Which is) the very thing it was required to show.

STOIQEIWN aþ. ELEMENTS BOOK 1
idþ. Proposition 14

᾿Εὰν πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ δύο If two straight-lines, not lying on the same side, make
εὐθεῖαι μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας adjacent angles (whose sum is) equal to two right-angles
δυσὶν ὀρθαῖς ἴσας ποιῶσιν, ἐπ᾿ εὐθείας ἔσονται ἀλλήλαις αἱ with some straight-line, at a point on it, then the two
εὐθεῖαι. straight-lines will be straight-on (with respect) to one an-

other.

Γ ∆

BC D

Πρὸς γάρ τινι εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ σημείῳ For let two straight-lines BC and BD, not lying on the
τῷ Β δύο εὐθεῖαι αἱ ΒΓ, ΒΔ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι same side, make adjacent angles ABC and ABD (whose
τὰς ἐφεξῆς γωνίας τὰς ὑπὸ ΑΒΓ, ΑΒΔ δύο ὀρθαῖς ἴσας sum is) equal to two right-angles with some straight-line
ποιείτωσαν· λέγω, ὅτι ἐπ᾿ εὐθείας ἐστὶ τῇ ΓΒ ἡ ΒΔ. AB, at the point B on it. I say that BD is straight-on with
Εἰ γὰρ μή ἐστι τῇ ΒΓ ἐπ᾿ εὐθείας ἡ ΒΔ, ἔστω τῇ ΓΒ respect to CB.

ἐπ᾿ εὐθείας ἡ ΒΕ. For if BD is not straight-on to BC then let BE be
᾿Επεὶ οὖν εὐθεῖα ἡ ΑΒ ἐπ᾿ εὐθεῖαν τὴν ΓΒΕ ἐφέστηκεν, straight-on to CB.

αἱ ἄρα ὑπὸ ΑΒΓ, ΑΒΕ γωνίαι δύο ὀρθαῖς ἴσαι εἰσίν· εἰσὶ δὲ Therefore, since the straight-line AB stands on the
καὶ αἱ ὑπὸ ΑΒΓ, ΑΒΔ δύο ὀρθαῖς ἴσαι· αἱ ἄρα ὑπὸ ΓΒΑ, straight-line CBE, the (sum of the) angles ABC and
ΑΒΕ ταῖς ὑπὸ ΓΒΑ, ΑΒΔ ἴσαι εἰσίν. κοινὴ ἀφῃρήσθω ἡ ABE is thus equal to two right-angles [Prop. 1.13]. But
ὑπὸ ΓΒΑ· λοιπὴ ἄρα ἡ ὑπὸ ΑΒΕ λοιπῇ τῇ ὑπὸ ΑΒΔ ἐστιν (the sum of) ABC and ABD is also equal to two right-
ἴση, ἡ ἐλάσσων τῇ μείζονι· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα angles. Thus, (the sum of angles) CBA and ABE is equal
ἐπ᾿ εὐθείας ἐστὶν ἡ ΒΕ τῇ ΓΒ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ to (the sum of angles) CBA and ABD [C.N. 1]. Let (an-
ἄλλη τις πλὴν τῆς ΒΔ· ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΓΒ τῇ ΒΔ. gle) CBA have been subtracted from both. Thus, the re-
᾿Εὰν ἄρα πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ mainder ABE is equal to the remainder ABD [C.N. 3],

δύο εὐθεῖαι μὴ ἐπὶ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας the lesser to the greater. The very thing is impossible.
δυσὶν ὀρθαῖς ἴσας ποιῶσιν, ἐπ᾿ εὐθείας ἔσονται ἀλλήλαις αἱ Thus, BE is not straight-on with respect to CB. Simi-
εὐθεῖαι· ὅπερ ἔδει δεῖξαι. larly, we can show that neither (is) any other (straight-

line) than BD. Thus, CB is straight-on with respect to
BD.

Thus, if two straight-lines, not lying on the same side,

make adjacent angles (whose sum is) equal to two right-
angles with some straight-line, at a point on it, then the

two straight-lines will be straight-on (with respect) to
one another. (Which is) the very thing it was required

to show.ieþ. Proposition 15
᾿Εὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὰς κατὰ κορυφὴν If two straight-lines cut one another then they make

γωνίας ἴσας ἀλλήλαις ποιοῦσιν. the vertically opposite angles equal to one another.

STOIQEIWN aþ. ELEMENTS BOOK 1
Δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ τεμνέτωσαν ἀλλήλας κατὰ For let the two straight-lines AB and CD cut one an-

τὸ Ε σημεῖον· λέγω, ὅτι ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΕΓ γωνία τῇ other at the point E. I say that angle AEC is equal to
ὑπὸ ΔΕΒ, ἡ δὲ ὑπὸ ΓΕΒ τῇ ὑπὸ ΑΕΔ. (angle) DEB, and (angle) CEB to (angle) AED.

∆

Γ D

᾿Επεὶ γὰρ εὐθεῖα ἡ ΑΕ ἐπ᾿ εὐθεῖαν τὴν ΓΔ ἐφέστηκε For since the straight-line AE stands on the straight-
γωνίας ποιοῦσα τὰς ὑπὸ ΓΕΑ, ΑΕΔ, αἱ ἄρα ὑπὸ ΓΕΑ, ΑΕΔ line CD, making the angles CEA and AED, the (sum
γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. πάλιν, ἐπεὶ εὐθεῖα ἡ ΔΕ ἐπ᾿ of the) angles CEA and AED is thus equal to two right-
εὐθεῖαν τὴν ΑΒ ἐφέστηκε γωνίας ποιοῦσα τὰς ὑπὸ ΑΕΔ, angles [Prop. 1.13]. Again, since the straight-line DE
ΔΕΒ, αἱ ἄρα ὑπὸ ΑΕΔ, ΔΕΒ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. stands on the straight-line AB, making the angles AED
ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΕΑ, ΑΕΔ δυσὶν ὀρθαῖς ἴσαι· αἱ and DEB, the (sum of the) angles AED and DEB is
ἄρα ὑπὸ ΓΕΑ, ΑΕΔ ταῖς ὑπὸ ΑΕΔ, ΔΕΒ ἴσαι εἰσίν. κοινὴ thus equal to two right-angles [Prop. 1.13]. But (the sum
ἀφῃρήσθω ἡ ὑπὸ ΑΕΔ· λοιπὴ ἄρα ἡ ὑπὸ ΓΕΑ λοιπῇ τῇ ὑπὸ of) CEA and AED was also shown (to be) equal to two
ΒΕΔ ἴση ἐστίν· ὁμοίως δὴ δειχθήσεται, ὅτι καὶ αἱ ὑπὸ ΓΕΒ, right-angles. Thus, (the sum of) CEA and AED is equal
ΔΕΑ ἴσαι εἰσίν. to (the sum of) AED and DEB [C.N. 1]. Let AED have
᾿Εὰν ἄρα δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὰς κατὰ κο- been subtracted from both. Thus, the remainder CEA is

ρυφὴν γωνίας ἴσας ἀλλήλαις ποιοῦσιν· ὅπερ ἔδει δεῖξαι. equal to the remainder BED [C.N. 3]. Similarly, it can
be shown that CEB and DEA are also equal.

Thus, if two straight-lines cut one another then they
make the vertically opposite angles equal to one another.

(Which is) the very thing it was required to show.i�þ. Proposition 16
Παντὸς τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης For any triangle, when one of the sides is produced,

ἡ ἐκτὸς γωνία ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν the external angle is greater than each of the internal and
μείζων ἐστίν. opposite angles.
῎Εστω τρίγωνον τὸ ΑΒΓ, καὶ προσεκβεβλήσθω αὐτοῦ Let ABC be a triangle, and let one of its sides BC

μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ· λὲγω, ὅτι ἡ ἐκτὸς γωνία ἡ ὑπὸ have been produced to D. I say that the external angle
ΑΓΔ μείζων ἐστὶν ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον τῶν ACD is greater than each of the internal and opposite
ὑπὸ ΓΒΑ, ΒΑΓ γωνιῶν. angles, CBA and BAC.
Τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε, καὶ ἐπιζευχθεῖσα ἡ ΒΕ Let the (straight-line) AC have been cut in half at

ἐκβεβλήσθω ἐπ᾿ εὐθείας ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΒΕ ἴση ἡ (point) E [Prop. 1.10]. And BE being joined, let it have

ΕΖ, καὶ ἐπεζεύχθω ἡ ΖΓ, καὶ διήχθω ἡ ΑΓ ἐπὶ τὸ Η. been produced in a straight-line to (point) F .† And let
᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΕ τῇ ΕΓ, ἡ δὲ ΒΕ τῇ ΕΖ, δύο EF be made equal to BE [Prop. 1.3], and let FC have

δὴ αἱ ΑΕ, ΕΒ δυσὶ ταῖς ΓΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· been joined, and let AC have been drawn through to
καὶ γωνία ἡ ὑπὸ ΑΕΒ γωνίᾳ τῇ ὑπὸ ΖΕΓ ἴση ἐστίν· κατὰ (point) G.
κορυφὴν γάρ· βάσις ἄρα ἡ ΑΒ βάσει τῇ ΖΓ ἴση ἐστίν, καὶ τὸ Therefore, since AE is equal to EC, and BE to EF ,
ΑΒΕ τρίγωνον τῷ ΖΕΓ τριγώνῳ ἐστὶν ἴσον, καὶ αἱ λοιπαὶ the two (straight-lines) AE, EB are equal to the two

STOIQEIWN aþ. ELEMENTS BOOK 1
γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, ὑφ᾿ (straight-lines) CE, EF , respectively. Also, angle AEB
ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν· ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΑΕ is equal to angle FEC, for (they are) vertically opposite
τῇ ὑπὸ ΕΓΖ. μείζων δέ ἐστιν ἡ ὑπὸ ΕΓΔ τῆς ὑπὸ ΕΓΖ· [Prop. 1.15]. Thus, the base AB is equal to the base FC,
μείζων ἄρα ἡ ὑπὸ ΑΓΔ τῆς ὑπὸ ΒΑΕ. ῾Ομοίως δὴ τῆς ΒΓ and the triangle ABE is equal to the triangle FEC, and
τετμημένης δίχα δειχθήσεται καὶ ἡ ὑπὸ ΒΓΗ, τουτέστιν ἡ the remaining angles subtended by the equal sides are
ὑπὸ ΑΓΔ, μείζων καὶ τῆς ὑπὸ ΑΒΓ. equal to the corresponding remaining angles [Prop. 1.4].

Thus, BAE is equal to ECF . But ECD is greater than

ECF . Thus, ACD is greater than BAE. Similarly, by
having cut BC in half, it can be shown (that) BCG—that

is to say, ACD—(is) also greater than ABC.

Β ∆
Γ

Α Ζ

Παντὸς ἄρα τριγώνου μιᾶς τῶν πλευρῶν προσεκ- Thus, for any triangle, when one of the sides is pro-
βληθείσης ἡ ἐκτὸς γωνία ἑκατέρας τῶν ἐντὸς καὶ ἀπε- duced, the external angle is greater than each of the in-
ναντίον γωνιῶν μείζων ἐστίν· ὅπερ ἔδει δεῖξαι. ternal and opposite angles. (Which is) the very thing it

was required to show.

† The implicit assumption that the point F lies in the interior of the angle ABC should be counted as an additional postulate.izþ. Proposition 17
Παντὸvς τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάσσονές For any triangle, (the sum of) two angles taken to-

εἰσι πάντῇ μεταλαμβανόμεναι. gether in any (possible way) is less than two right-angles.

Γ ∆

Β B

C D
῎Εστω τρίγωνον τὸ ΑΒΓ· λέγω, ὅτι τοῦ ΑΒΓ τριγώνου Let ABC be a triangle. I say that (the sum of) two

αἱ δύο γωνίαι δύο ὀρθῶν ἐλάττονές εἰσι πάντῃ μεταλαμ- angles of triangle ABC taken together in any (possible
βανόμεναι. way) is less than two right-angles.

STOIQEIWN aþ. ELEMENTS BOOK 1
᾿Εκβεβλήσθω γὰρ ἡ ΒΓ ἐπὶ τὸ Δ. For let BC have been produced to D.
Καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ ἐκτός ἐστι γωνία ἡ ὑπὸ And since the angle ACD is external to triangle ABC,

ΑΓΔ, μείζων ἐστὶ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ. it is greater than the internal and opposite angle ABC
κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ· αἱ ἄρα ὑπὸ ΑΓΔ, ΑΓΒ τῶν [Prop. 1.16]. Let ACB have been added to both. Thus,
ὑπὸ ΑΒΓ, ΒΓΑ μείζονές εἰσιν. ἀλλ᾿ αἱ ὑπὸ ΑΓΔ, ΑΓΒ the (sum of the angles) ACD and ACB is greater than
δύο ὀρθαῖς ἴσαι εἰσίν· αἱ ἄρα ὑπὸ ΑΒΓ, ΒΓΑ δύο ὀρθῶν the (sum of the angles) ABC and BCA. But, (the sum of)
ἐλάσσονές εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ὑπὸ ΒΑΓ, ACD and ACB is equal to two right-angles [Prop. 1.13].
ΑΓΒ δύο ὀρθῶν ἐλάσσονές εἰσι καὶ ἔτι αἱ ὑπὸ ΓΑΒ, ΑΒΓ. Thus, (the sum of) ABC and BCA is less than two right-
Παντὸvς ἄρα τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάσς- angles. Similarly, we can show that (the sum of) BAC

ονές εἰσι πάντῇ μεταλαμβανόμεναι· ὅπερ ἔδει δεῖξαι. and ACB is also less than two right-angles, and further
(that the sum of) CAB and ABC (is less than two right-
angles).

Thus, for any triangle, (the sum of) two angles taken
together in any (possible way) is less than two right-

angles. (Which is) the very thing it was required to show.ihþ. Proposition 18
Παντὸς τριγώνου ἡ μείζων πλευρὰ τὴν μείζονα γωνίαν In any triangle, the greater side subtends the greater

ὑποτείνει. angle.

Β
Γ

∆

B
C

῎Εστω γὰρ τρίγωνον τὸ ΑΒΓ μείζονα ἔχον τὴν ΑΓ For let ABC be a triangle having side AC greater than
πλευρὰν τῆς ΑΒ· λέγω, ὅτι καὶ γωνία ἡ ὑπὸ ΑΒΓ μείζων AB. I say that angle ABC is also greater than BCA.
ἐστὶ τῆς ὑπὸ ΒΓΑ· For since AC is greater than AB, let AD be made
᾿Επεὶ γὰρ μείζων ἐστὶν ἡ ΑΓ τῆς ΑΒ, κείσθω τῇ ΑΒ ἴση equal to AB [Prop. 1.3], and let BD have been joined.

ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΒΔ. And since angle ADB is external to triangle BCD, it
Καὶ ἐπεὶ τριγώνου τοῦ ΒΓΔ ἐκτός ἐστι γωνία ἡ ὑπὸ is greater than the internal and opposite (angle) DCB

ΑΔΒ, μείζων ἐστὶ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΔΓΒ· [Prop. 1.16]. But ADB (is) equal to ABD, since side
ἴση δὲ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ, ἐπεὶ καὶ πλευρὰ ἡ ΑΒ τῇ AB is also equal to side AD [Prop. 1.5]. Thus, ABD is
ΑΔ ἐστιν ἴση· μείζων ἄρα καὶ ἡ ὑπὸ ΑΒΔ τῆς ὑπὸ ΑΓΒ· also greater than ACB. Thus, ABC is much greater than
πολλῷ ἄρα ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ τῆς ὑπὸ ΑΓΒ. ACB.
Παντὸς ἄρα τριγώνου ἡ μείζων πλευρὰ τὴν μείζονα Thus, in any triangle, the greater side subtends the

γωνίαν ὑποτείνει· ὅπερ ἔδει δεῖξαι. greater angle. (Which is) the very thing it was required
to show.ijþ. Proposition 19

Παντὸς τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων In any triangle, the greater angle is subtended by the
πλευρὰ ὑποτείνει. greater side.
῎Εστω τρίγωνον τὸ ΑΒΓ μείζονα ἔχον τὴν ὑπὸ ΑΒΓ Let ABC be a triangle having the angle ABC greater

γωνίαν τῆς ὑπὸ ΒΓΑ· λέγω, ὅτι καὶ πλευρὰ ἡ ΑΓ πλευρᾶς than BCA. I say that side AC is also greater than side
τῆς ΑΒ μείζων ἐστίν. AB.

STOIQEIWN aþ. ELEMENTS BOOK 1
Εἰ γὰρ μή, ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ἢ ἐλάσσων· ἴση For if not, AC is certainly either equal to, or less than,

μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ· ἴση γὰρ ἂν ἦν καὶ γωνία ἡ AB. In fact, AC is not equal to AB. For then angle ABC
ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ· οὐκ ἔστι δέ· οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ would also have been equal to ACB [Prop. 1.5]. But it
τῇ ΑΒ. οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ· ἐλάσσων is not. Thus, AC is not equal to AB. Neither, indeed, is
γὰρ ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῆς ὑπὸ ΑΓΒ· οὐκ ἔστι AC less than AB. For then angle ABC would also have
δέ· οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ. ἐδείχθη δέ, ὅτι been less than ACB [Prop. 1.18]. But it is not. Thus, AC
οὐδὲ ἴση ἐστίν. μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ. is not less than AB. But it was shown that (AC) is not

equal (to AB) either. Thus, AC is greater than AB.

Γ C

Παντὸς ἄρα τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων Thus, in any triangle, the greater angle is subtended
πλευρὰ ὑποτείνει· ὅπερ ἔδει δεῖξαι. by the greater side. (Which is) the very thing it was re-

quired to show.kþ. Proposition 20
Παντὸς τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές In any triangle, (the sum of) two sides taken to-

εἰσι πάντῃ μεταλαμβανόμεναι. gether in any (possible way) is greater than the remaining
(side).

Β Γ

∆

῎Εστω γὰρ τρίγωνον τὸ ΑΒΓ· λέγω, ὅτι τοῦ ΑΒΓ For let ABC be a triangle. I say that in triangle ABC
τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ (the sum of) two sides taken together in any (possible
μεταλαμβανόμεναι, αἱ μὲν ΒΑ, ΑΓ τῆς ΒΓ, αἱ δὲ ΑΒ, ΒΓ way) is greater than the remaining (side). (So), (the sum
τῆς ΑΓ, αἱ δὲ ΒΓ, ΓΑ τῆς ΑΒ. of) BA and AC (is greater) than BC, (the sum of) AB

STOIQEIWN aþ. ELEMENTS BOOK 1
Διήχθω γὰρ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον, καὶ κείσθω τῇ ΓΑ and BC than AC, and (the sum of) BC and CA than

ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΓ. AB.
᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ, ἴση ἐστὶ καὶ γωνία For let BA have been drawn through to point D, and

ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ· μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ let AD be made equal to CA [Prop. 1.3], and let DC
ΑΔΓ· καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ μείζονα ἔχον τὴν ὑπὸ have been joined.
ΒΓΔ γωνίαν τῆς ὑπὸ ΒΔΓ, ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ Therefore, since DA is equal to AC, the angle ADC
μείζων πλευρὰ ὑποτείνει, ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων. ἴση is also equal to ACD [Prop. 1.5]. Thus, BCD is greater
δὲ ἡ ΔΑ τῇ ΑΓ· μείζονες ἄρα αἱ ΒΑ, ΑΓ τῆς ΒΓ· ὁμοίως than ADC. And since DCB is a triangle having the angle
δὴ δείξομεν, ὅτι καὶ αἱ μὲν ΑΒ, ΒΓ τῆς ΓΑ μείζονές εἰσιν, BCD greater than BDC, and the greater angle subtends
αἱ δὲ ΒΓ, ΓΑ τῆς ΑΒ. the greater side [Prop. 1.19], DB is thus greater than
Παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς BC. But DA is equal to AC. Thus, (the sum of) BA and

μείζονές εἰσι πάντῃ μεταλαμβανόμεναι· ὅπερ ἔδει δεῖξαι. AC is greater than BC. Similarly, we can show that (the
sum of) AB and BC is also greater than CA, and (the
sum of) BC and CA than AB.

Thus, in any triangle, (the sum of) two sides taken to-

gether in any (possible way) is greater than the remaining
(side). (Which is) the very thing it was required to show.kaþ. Proposition 21

᾿Εὰν τριγώνου ἐπὶ μιᾶς τῶν πλευρῶν ἀπὸ τῶν περάτων If two internal straight-lines are constructed on one
δύο εὐθεῖαι ἐντὸς συσταθῶσιν, αἱ συσταθεῖσαι τῶν λοιπῶν of the sides of a triangle, from its ends, the constructed
τοῦ τριγώνου δύο πλευρῶν ἐλάττονες μὲν ἔσονται, μείζονα (straight-lines) will be less than the two remaining sides
δὲ γωνίαν περιέξουσιν. of the triangle, but will encompass a greater angle.

ΓΒ

∆

Τριγώνου γὰρ τοῦ ΑΒΓ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΒΓ For let the two internal straight-lines BD and DC
ἀπὸ τῶν περάτων τῶν Β, Γ δύο εὐθεῖαι ἐντὸς συνεστάτωσαν have been constructed on one of the sides BC of the tri-
αἱ ΒΔ, ΔΓ· λέγω, ὅτι αἱ ΒΔ, ΔΓ τῶν λοιπῶν τοῦ τριγώνου angle ABC, from its ends B and C (respectively). I say
δύο πλευρῶν τῶν ΒΑ, ΑΓ ἐλάσσονες μέν εἰσιν, μείζονα δὲ that BD and DC are less than the (sum of the) two re-
γωνίαν περιέχουσι τὴν ὑπὸ ΒΔΓ τῆς ὑπὸ ΒΑΓ. maining sides of the triangle BA and AC, but encompass
Διήχθω γὰρ ἡ ΒΔ ἐπὶ τὸ Ε. καὶ ἐπεὶ παντὸς τριγώνου an angle BDC greater than BAC.

αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσιν, τοῦ ΑΒΕ ἄρα For let BD have been drawn through to E. And since
τριγώνου αἱ δύο πλευραὶ αἱ ΑΒ, ΑΕ τῆς ΒΕ μείζονές in any triangle (the sum of any) two sides is greater than
εἰσιν· κοινὴ προσκείσθω ἡ ΕΓ· αἱ ἄρα ΒΑ, ΑΓ τῶν ΒΕ, the remaining (side) [Prop. 1.20], in triangle ABE the
ΕΓ μείζονές εἰσιν. πάλιν, ἐπεὶ τοῦ ΓΕΔ τριγώνου αἱ δύο (sum of the) two sides AB and AE is thus greater than
πλευραὶ αἱ ΓΕ, ΕΔ τῆς ΓΔ μείζονές εἰσιν, κοινὴ προσκείσθω BE. Let EC have been added to both. Thus, (the sum
ἡ ΔΒ· αἱ ΓΕ, ΕΒ ἄρα τῶν ΓΔ, ΔΒ μείζονές εἰσιν. ἀλλὰ of) BA and AC is greater than (the sum of) BE and EC.
τῶν ΒΕ, ΕΓ μείζονες ἐδείχθησαν αἱ ΒΑ, ΑΓ· πολλῷ ἄρα αἱ Again, since in triangle CED the (sum of the) two sides
ΒΑ, ΑΓ τῶν ΒΔ, ΔΓ μείζονές εἰσιν. CE and ED is greater than CD, let DB have been added
Πάλιν, ἐπεὶ παντὸς τριγώνου ἡ ἐκτὸς γωνία τῆς ἐντὸς to both. Thus, (the sum of) CE and EB is greater than

καὶ ἀπεναντίον μείζων ἐστίν, τοῦ ΓΔΕ ἄρα τριγώνου ἡ (the sum of) CD and DB. But, (the sum of) BA and
ἐκτὸς γωνία ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΓΕΔ. διὰ AC was shown (to be) greater than (the sum of) BE and
ταὐτὰ τοίνυν καὶ τοῦ ΑΒΕ τριγώνου ἡ ἐκτὸς γωνία ἡ ὑπὸ EC. Thus, (the sum of) BA and AC is much greater than

STOIQEIWN aþ. ELEMENTS BOOK 1
ΓΕΒ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ. ἀλλὰ τῆς ὑπὸ ΓΕΒ μείζων (the sum of) BD and DC.
ἐδείχθη ἡ ὑπὸ ΒΔΓ· πολλῷ ἄρα ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ Again, since in any triangle the external angle is
τῆς ὑπὸ ΒΑΓ. greater than the internal and opposite (angles) [Prop.
᾿Εὰν ἄρα τριγώνου ἐπὶ μιᾶς τῶν πλευρῶν ἀπὸ τῶν 1.16], in triangle CDE the external angle BDC is thus

περάτων δύο εὐθεῖαι ἐντὸς συσταθῶσιν, αἱ συσταθεῖσαι τῶν greater than CED. Accordingly, for the same (reason),
λοιπῶν τοῦ τριγώνου δύο πλευρῶν ἐλάττονες μέν εἰσιν, the external angle CEB of the triangle ABE is also
μείζονα δὲ γωνίαν περιέχουσιν· ὅπερ ἔδει δεῖξαι. greater than BAC. But, BDC was shown (to be) greater

than CEB. Thus, BDC is much greater than BAC.
Thus, if two internal straight-lines are constructed on

one of the sides of a triangle, from its ends, the con-

structed (straight-lines) are less than the two remain-
ing sides of the triangle, but encompass a greater angle.

(Which is) the very thing it was required to show.kbþ. Proposition 22
᾿Εκ τριῶν εὐθειῶν, αἵ εἰσιν ἴσαι τρισὶ ταῖς δοθείσαις To construct a triangle from three straight-lines which

[εὐθείαις], τρίγωνον συστήσασθαι· δεῖ δὲ τὰς δύο τῆς λοιπῆς are equal to three given [straight-lines]. It is necessary
μείζονας εἶναι πάντῃ μεταλαμβανομένας [διὰ τὸ καὶ παντὸς for (the sum of) two (of the straight-lines) taken together
τριγώνου τὰς δύο πλευρὰς τῆς λοιπῆς μείζονας εἶναι πάντῃ in any (possible way) to be greater than the remaining
μεταλαμβανομένας]. (one), [on account of the (fact that) in any triangle (the

sum of) two sides taken together in any (possible way) is

greater than the remaining (one) [Prop. 1.20] ].

Ζ
∆ Ε

D
F

῎Εστωσαν αἱ δοθεῖσαι τρεῖς εὐθεῖαι αἱ Α, Β, Γ, ὧν αἱ Let A, B, and C be the three given straight-lines, of
δύο τῆς λοιπῆς μείζονες ἔστωσαν πάντῃ μεταλαμβανόμεναι, which let (the sum of) two taken together in any (possible
αἱ μὲν Α, Β τῆς Γ, αἱ δὲ Α, Γ τῆς Β, καὶ ἔτι αἱ Β, Γ τῆς Α· way) be greater than the remaining (one). (Thus), (the
δεῖ δὴ ἐκ τῶν ἴσων ταῖς Α, Β, Γ τρίγωνον συστήσασθαι. sum of) A and B (is greater) than C, (the sum of) A and
᾿Εκκείσθω τις εὐθεῖα ἡ ΔΕ πεπερασμένη μὲν κατὰ τὸ C than B, and also (the sum of) B and C than A. So

Δ ἄπειρος δὲ κατὰ τὸ Ε, καὶ κείσθω τῇ μὲν Α ἴση ἡ ΔΖ, it is required to construct a triangle from (straight-lines)
τῇ δὲ Β ἴση ἡ ΖΗ, τῇ δὲ Γ ἴση ἡ ΗΘ· καὶ κέντρῳ μὲν τῷ equal to A, B, and C.
Ζ, διαστήματι δὲ τῷ ΖΔ κύκλος γεγράφθω ὁ ΔΚΛ· πάλιν Let some straight-line DE be set out, terminated at D,
κέντρῳ μὲν τῷ Η, διαστήματι δὲ τῷ ΗΘ κύκλος γεγράφθω and infinite in the direction of E. And let DF made equal
ὁ ΚΛΘ, καὶ ἐπεζεύχθωσαν αἱ ΚΖ, ΚΗ· λέγω, ὅτι ἐκ τριῶν to A, and FG equal to B, and GH equal to C [Prop. 1.3].
εὐθειῶν τῶν ἴσων ταῖς Α, Β, Γ τρίγωνον συνέσταται τὸ And let the circle DKL have been drawn with center F
ΚΖΗ. and radius FD. Again, let the circle KLH have been
᾿Επεὶ γὰρ τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΔΚΛ κύκλου, drawn with center G and radius GH . And let KF and

ἴση ἐστὶν ἡ ΖΔ τῇ ΖΚ· ἀλλὰ ἡ ΖΔ τῇ Α ἐστιν ἴση. καὶ ἡ KG have been joined. I say that the triangle KFG has

STOIQEIWN aþ. ELEMENTS BOOK 1
ΚΖ ἄρα τῇ Α ἐστιν ἴση. πάλιν, ἐπεὶ τὸ Η σημεῖον κέντρον been constructed from three straight-lines equal to A, B,
ἐστὶ τοῦ ΛΚΘ κύκλου, ἴση ἐστὶν ἡ ΗΘ τῇ ΗΚ· ἀλλὰ ἡ ΗΘ and C.
τῇ Γ ἐστιν ἴση· καὶ ἡ ΚΗ ἄρα τῇ Γ ἐστιν ἴση. ἐστὶ δὲ καὶ ἡ For since point F is the center of the circle DKL, FD
ΖΗ τῇ Β ἴση· αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΚΖ, ΖΗ, ΗΚ τρισὶ ταῖς is equal to FK. But, FD is equal to A. Thus, KF is also
Α, Β, Γ ἴσαι εἰσίν. equal to A. Again, since point G is the center of the circle
᾿Εκ τριῶν ἄρα εὐθειῶν τῶν ΚΖ, ΖΗ, ΗΚ, αἵ εἰσιν LKH , GH is equal to GK. But, GH is equal to C. Thus,

ἴσαι τρισὶ ταῖς δοθείσαις εὐθείαις ταῖς Α, Β, Γ, τρίγωνον KG is also equal to C. And FG is also equal to B. Thus,
συνέσταται τὸ ΚΖΗ· ὅπερ ἔδει ποιῆσαι. the three straight-lines KF , FG, and GK are equal to A,

B, and C (respectively).

Thus, the triangle KFG has been constructed from

the three straight-lines KF , FG, and GK, which are
equal to the three given straight-lines A, B, and C (re-

spectively). (Which is) the very thing it was required to
do.kgþ. Proposition 23

Πρὸς τῇ δοθείσῃ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ To construct a rectilinear angle equal to a given recti-
τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ ἴσην γωνίαν εὐθύγραμμον linear angle at a (given) point on a given straight-line.
συστήσασθαι.

∆

Α
Η Β

G
A

῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ πρὸς αὐτῇ Let AB be the given straight-line, A the (given) point
σημεῖον τὸ Α, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ ΔΓΕ· on it, and DCE the given rectilinear angle. So it is re-
δεῖ δὴ πρὸς τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ quired to construct a rectilinear angle equal to the given
σημείῳ τῷ Α τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ ΔΓΕ rectilinear angle DCE at the (given) point A on the given
ἴσην γωνίαν εὐθύγραμμον συστήσασθαι. straight-line AB.
Εἰλήφθω ἐφ᾿ ἑκατέρας τῶν ΓΔ, ΓΕ τυχόντα σημεῖα τὰ Let the points D and E have been taken at random

Δ, Ε, καὶ ἐπεζεύχθω ἡ ΔΕ· καὶ ἐκ τριῶν εὐθειῶν, αἵ εἰσιν on each of the (straight-lines) CD and CE (respectively),
ἴσαι τρισὶ ταῖς ΓΔ, ΔΕ, ΓΕ, τρίγωνον συνεστάτω τὸ ΑΖΗ, and let DE have been joined. And let the triangle AFG
ὥστε ἴσην εἶναι τὴν μὲν ΓΔ τῇ ΑΖ, τὴν δὲ ΓΕ τῇ ΑΗ, καὶ have been constructed from three straight-lines which are
ἔτι τὴν ΔΕ τῇ ΖΗ. equal to CD, DE, and CE, such that CD is equal to AF ,
᾿Επεὶ οὖν δύο αἱ ΔΓ, ΓΕ δύο ταῖς ΖΑ, ΑΗ ἴσαι εἰσὶν CE to AG, and further DE to FG [Prop. 1.22].

ἑκατέρα ἑκατέρᾳ, καὶ βάσις ἡ ΔΕ βάσει τῇ ΖΗ ἴση, γωνία Therefore, since the two (straight-lines) DC, CE are
ἄρα ἡ ὑπὸ ΔΓΕ γωνίᾳ τῇ ὑπὸ ΖΑΗ ἐστιν ἴση. equal to the two (straight-lines) FA, AG, respectively,
Πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ and the base DE is equal to the base FG, the angle DCE

σημείῳ τῷ Α τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ ΔΓΕ is thus equal to the angle FAG [Prop. 1.8].
ἴση γωνία εὐθύγραμμος συνέσταται ἡ ὑπὸ ΖΑΗ· ὅπερ ἔδει Thus, the rectilinear angle FAG, equal to the given
ποιῆσαι. rectilinear angle DCE, has been constructed at the

(given) point A on the given straight-line AB. (Which

STOIQEIWN aþ. ELEMENTS BOOK 1
is) the very thing it was required to do.kdþ. Proposition 24

᾿Εὰν δύο τρίγωνα τὰς δύο πλευρὰς [ταῖς] δύο πλευραῖς If two triangles have two sides equal to two sides, re-
ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ γωνίαν τῆς γωνίας spectively, but (one) has the angle encompassed by the
μείζονα ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ equal straight-lines greater than the (corresponding) an-
τὴν βάσιν τῆς βάσεως μείζονα ἕξει. gle (in the other), then (the former triangle) will also

have a base greater than the base (of the latter).

Α ∆

Η
Ζ F

῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς Let ABC and DEF be two triangles having the two
τὰς ΑΒ, ΑΓ ταῖς δύο πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα sides AB and AC equal to the two sides DE and DF ,
ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ, ἡ respectively. (That is), AB (equal) to DE, and AC to
δὲ πρὸς τῷ Α γωνία τῆς πρὸς τῷ Δ γωνίας μείζων ἔστω· DF . Let them also have the angle at A greater than the
λέγω, ὅτι καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ μείζων ἐστίν. angle at D. I say that the base BC is also greater than
᾿Επεὶ γὰρ μείζων ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ the base EF .

γωνίας, συνεστάτω πρὸς τῇ ΔΕ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ For since angle BAC is greater than angle EDF ,
σημείῳ τῷ Δ τῇ ὑπὸ ΒΑΓ γωνίᾳ ἴση ἡ ὑπὸ ΕΔΗ, καὶ κείσθω let (angle) EDG, equal to angle BAC, have been
ὁποτέρᾳ τῶν ΑΓ, ΔΖ ἴση ἡ ΔΗ, καὶ ἐπεζεύχθωσαν αἱ ΕΗ, constructed at the point D on the straight-line DE
ΖΗ. [Prop. 1.23]. And let DG be made equal to either of
᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ, ἡ δὲ ΑΓ τῇ ΔΗ, AC or DF [Prop. 1.3], and let EG and FG have been

δύο δὴ αἱ ΒΑ, ΑΓ δυσὶ ταῖς ΕΔ, ΔΗ ἴσαι εἰσὶν ἑκατέρα joined.
ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΒΑΓ γωνίᾳ τῇ ὑπὸ ΕΔΗ ἴση· Therefore, since AB is equal to DE and AC to DG,
βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΗ ἐστιν ἴση. πάλιν, ἐπεὶ ἴση the two (straight-lines) BA, AC are equal to the two
ἐστὶν ἡ ΔΖ τῇ ΔΗ, ἴση ἐστὶ καὶ ἡ ὑπὸ ΔΗΖ γωνία τῇ ὑπὸ (straight-lines) ED, DG, respectively. Also the angle
ΔΖΗ· μείζων ἄρα ἡ ὑπὸ ΔΖΗ τῆς ὑπὸ ΕΗΖ· πολλῷ ἄρα BAC is equal to the angle EDG. Thus, the base BC
μείζων ἐστὶν ἡ ὑπὸ ΕΖΗ τῆς ὑπὸ ΕΗΖ. καὶ ἐπεὶ τρίγωνόν is equal to the base EG [Prop. 1.4]. Again, since DF
ἐστι τὸ ΕΖΗ μείζονα ἔχον τὴν ὑπὸ ΕΖΗ γωνίαν τῆς ὑπὸ is equal to DG, angle DGF is also equal to angle DFG
ΕΗΖ, ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει, [Prop. 1.5]. Thus, DFG (is) greater than EGF . Thus,
μείζων ἄρα καὶ πλευρὰ ἡ ΕΗ τῆς ΕΖ. ἴση δὲ ἡ ΕΗ τῇ ΒΓ· EFG is much greater than EGF . And since triangle
μείζων ἄρα καὶ ἡ ΒΓ τῆς ΕΖ. EFG has angle EFG greater than EGF , and the greater
᾿Εὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς δυσὶ πλευραῖς angle is subtended by the greater side [Prop. 1.19], side

ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ γωνίαν τῆς γωνίας EG (is) thus also greater than EF . But EG (is) equal to
μείζονα ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ BC. Thus, BC (is) also greater than EF .
τὴν βάσιν τῆς βάσεως μείζονα ἕξει· ὅπερ ἔδει δεῖξαι. Thus, if two triangles have two sides equal to two

sides, respectively, but (one) has the angle encompassed

by the equal straight-lines greater than the (correspond-
ing) angle (in the other), then (the former triangle) will

also have a base greater than the base (of the latter).

STOIQEIWN aþ. ELEMENTS BOOK 1
(Which is) the very thing it was required to show.keþ. Proposition 25

᾿Εὰν δύο τρίγωνα τὰς δύο πλευρὰς δυσὶ πλευραῖς ἴσας If two triangles have two sides equal to two sides,
ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ βάσιν τῆς βάσεως μείζονα respectively, but (one) has a base greater than the base
ἔχῃ, καὶ τὴν γωνίαν τῆς γωνίας μείζονα ἕξει τὴν ὑπὸ τῶν (of the other), then (the former triangle) will also have
ἴσων εὐθειῶν περιεχομένην. the angle encompassed by the equal straight-lines greater

than the (corresponding) angle (in the latter).

∆

Ε Ζ

῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς Let ABC and DEF be two triangles having the two
τὰς ΑΒ, ΑΓ ταῖς δύο πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα sides AB and AC equal to the two sides DE and DF ,
ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ, τὴν δὲ ΑΓ τῇ ΔΖ· respectively (That is), AB (equal) to DE, and AC to DF .
βάσις δὲ ἡ ΒΓ βάσεως τῆς ΕΖ μείζων ἔστω· λέγω, ὅτι καὶ And let the base BC be greater than the base EF . I say
γωνία ἡ ὑπὸ ΒΑΓ γωνίας τῆς ὑπὸ ΕΔΖ μείζων ἐστίν. that angle BAC is also greater than EDF .
Εἰ γὰρ μή, ἤτοι ἴση ἐστὶν αὐτῇ ἢ ἐλάσσων· ἴση μὲν οὖν For if not, (BAC) is certainly either equal to, or less

οὐκ ἔστιν ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ· ἴση γὰρ ἂν ἦν καὶ βάσις than, (EDF ). In fact, BAC is not equal to EDF . For
ἡ ΒΓ βάσει τῇ ΕΖ· οὐκ ἔστι δέ. οὐκ ἄρα ἴση ἐστὶ γωνία ἡ then the base BC would also have been equal to the base
ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ· οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ὑπὸ ΒΑΓ EF [Prop. 1.4]. But it is not. Thus, angle BAC is not
τῆς ὑπὸ ΕΔΖ· ἐλάσσων γὰρ ἂν ἦν καὶ βάσις ἡ ΒΓ βάσεως equal to EDF . Neither, indeed, is BAC less than EDF .
τῆς ΕΖ· οὐκ ἔστι δέ· οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ὑπὸ ΒΑΓ For then the base BC would also have been less than the
γωνία τῆς ὑπὸ ΕΔΖ. ἐδείχθη δέ, ὅτι οὐδὲ ἴση· μείζων ἄρα base EF [Prop. 1.24]. But it is not. Thus, angle BAC is
ἐστὶν ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ. not less than EDF . But it was shown that (BAC is) not
᾿Εὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς δυσὶ πλευραῖς equal (to EDF ) either. Thus, BAC is greater than EDF .

ἴσας ἔχῃ ἑκατέραν ἑκάτερᾳ, τὴν δὲ βασίν τῆς βάσεως Thus, if two triangles have two sides equal to two
μείζονα ἔχῃ, καὶ τὴν γωνίαν τῆς γωνίας μείζονα ἕξει τὴν sides, respectively, but (one) has a base greater than the
ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην· ὅπερ ἔδει δεῖξαι. base (of the other), then (the former triangle) will also

have the angle encompassed by the equal straight-lines

greater than the (corresponding) angle (in the latter).
(Which is) the very thing it was required to show.k�þ. Proposition 26

᾿Εὰν δύο τρίγωνα τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχῃ If two triangles have two angles equal to two angles,
ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ἤτοι τὴν respectively, and one side equal to one side—in fact, ei-
πρὸς ταῖς ἴσαις γωνίαις ἢ τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ther that by the equal angles, or that subtending one of
ἴσων γωνιῶν, καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς the equal angles—then (the triangles) will also have the
ἴσας ἕξει [ἑκατέραν ἑκατέρᾳ] καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ remaining sides equal to the [corresponding] remaining
γωνίᾳ. sides, and the remaining angle (equal) to the remaining
῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο γωνίας τὰς angle.

STOIQEIWN aþ. ELEMENTS BOOK 1
ὑπὸ ΑΒΓ, ΒΓΑ δυσὶ ταῖς ὑπὸ ΔΕΖ, ΕΖΔ ἴσας ἔχοντα Let ABC and DEF be two triangles having the two
ἑκατέραν ἑκατέρᾳ, τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ, τὴν angles ABC and BCA equal to the two (angles) DEF
δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ· ἐχέτω δὲ καὶ μίαν πλευρὰν μιᾷ and EFD, respectively. (That is) ABC (equal) to DEF ,
πλευρᾷ ἴσην, πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις τὴν and BCA to EFD. And let them also have one side equal
ΒΓ τῇ ΕΖ· λέγω, ὅτι καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς to one side. First of all, the (side) by the equal angles.
πλευραῖς ἴσας ἕξει ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ (That is) BC (equal) to EF . I say that they will have
τὴν δὲ ΑΓ τῇ ΔΖ, καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ, the remaining sides equal to the corresponding remain-
τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ. ing sides. (That is) AB (equal) to DE, and AC to DF .

And (they will have) the remaining angle (equal) to the

remaining angle. (That is) BAC (equal) to EDF .

∆

Θ
Γ

Ε Ζ

B
H

Εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΔΕ, μία αὐτῶν μείζων For if AB is unequal to DE then one of them is
ἐστίν. ἔστω μείζων ἡ ΑΒ, καὶ κείσθω τῇ ΔΕ ἴση ἡ ΒΗ, καὶ greater. Let AB be greater, and let BG be made equal
ἐπεζεύχθω ἡ ΗΓ. to DE [Prop. 1.3], and let GC have been joined.
᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΒΗ τῇ ΔΕ, ἡ δὲ ΒΓ τῇ ΕΖ, δύο Therefore, since BG is equal to DE, and BC to EF ,

δὴ αἱ ΒΗ, ΒΓ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· the two (straight-lines) GB, BC† are equal to the two
καὶ γωνία ἡ ὑπὸ ΗΒΓ γωνίᾳ τῇ ὑπὸ ΔΕΖ ἴση ἐστίν· βάσις (straight-lines) DE, EF , respectively. And angle GBC is
ἄρα ἡ ΗΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ ΗΒΓ τρίγωνον τῷ equal to angle DEF . Thus, the base GC is equal to the
ΔΕΖ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς base DF , and triangle GBC is equal to triangle DEF ,
γωνίαις ἴσαι ἔσονται, ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν· and the remaining angles subtended by the equal sides
ἴση ἄρα ἡ ὑπὸ ΗΓΒ γωνία τῇ ὑπὸ ΔΖΕ. ἀλλὰ ἡ ὑπὸ ΔΖΕ will be equal to the (corresponding) remaining angles
τῇ ὑπὸ ΒΓΑ ὑπόκειται ἴση· καὶ ἡ ὑπὸ ΒΓΗ ἄρα τῇ ὑπὸ ΒΓΑ [Prop. 1.4]. Thus, GCB (is equal) to DFE. But, DFE
ἴση ἐστίν, ἡ ἐλάσσων τῇ μείζονι· ὅπερ ἀδύνατον. οὐκ ἄρα was assumed (to be) equal to BCA. Thus, BCG is also
ἄνισός ἐστιν ἡ ΑΒ τῇ ΔΕ. ἴση ἄρα. ἔστι δὲ καὶ ἡ ΒΓ τῇ ΕΖ equal to BCA, the lesser to the greater. The very thing
ἴση· δύο δὴ αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα (is) impossible. Thus, AB is not unequal to DE. Thus,
ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΔΕΖ ἐστιν (it is) equal. And BC is also equal to EF . So the two
ἴση· βάσις ἄρα ἡ ΑΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ λοιπὴ γωνία (straight-lines) AB, BC are equal to the two (straight-
ἡ ὑπὸ ΒΑΓ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση ἐστίν. lines) DE, EF , respectively. And angle ABC is equal to
Ἀλλὰ δὴ πάλιν ἔστωσαν αἱ ὑπὸ τὰς ἴσας γωνίας πλευραὶ angle DEF . Thus, the base AC is equal to the base DF ,

ὑποτείνουσαι ἴσαι, ὡς ἡ ΑΒ τῇ ΔΕ· λέγω πάλιν, ὅτι καὶ αἱ and the remaining angle BAC is equal to the remaining
λοιπαὶ πλευραὶ ταῖς λοιπαῖς πλευραῖς ἴσαι ἔσονται, ἡ μὲν ΑΓ angle EDF [Prop. 1.4].
τῇ ΔΖ, ἡ δὲ ΒΓ τῇ ΕΖ καὶ ἔτι ἡ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ But, again, let the sides subtending the equal angles
τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση ἐστίν. be equal: for instance, (let) AB (be equal) to DE. Again,
Εἰ γὰρ ἄνισός ἐστιν ἡ ΒΓ τῇ ΕΖ, μία αὐτῶν μείζων ἐστίν. I say that the remaining sides will be equal to the remain-

ἔστω μείζων, εἰ δυνατόν, ἡ ΒΓ, καὶ κείσθω τῇ ΕΖ ἴση ἡ ΒΘ, ing sides. (That is) AC (equal) to DF , and BC to EF .
καὶ ἐπεζεύχθω ἡ ΑΘ. καὶ ἐπὲι ἴση ἐστὶν ἡ μὲν ΒΘ τῇ ΕΖ Furthermore, the remaining angle BAC is equal to the
ἡ δὲ ΑΒ τῇ ΔΕ, δύο δὴ αἱ ΑΒ, ΒΘ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι remaining angle EDF .
εἰσὶν ἑκατέρα ἑκαρέρᾳ· καὶ γωνίας ἴσας περιέχουσιν· βάσις For if BC is unequal to EF then one of them is
ἄρα ἡ ΑΘ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ ΑΒΘ τρίγωνον τῷ greater. If possible, let BC be greater. And let BH be
ΔΕΖ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς made equal to EF [Prop. 1.3], and let AH have been
γωνίαις ἴσαι ἔσονται, ὑφ᾿ ἃς αἱ ἴσας πλευραὶ ὑποτείνουσιν· joined. And since BH is equal to EF , and AB to DE,
ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΘΑ γωνία τῇ ὑπὸ ΕΖΔ. ἀλλὰ ἡ ὑπὸ the two (straight-lines) AB, BH are equal to the two

STOIQEIWN aþ. ELEMENTS BOOK 1
ΕΖΔ τῇ ὑπὸ ΒΓΑ ἐστιν ἴση· τριγώνου δὴ τοῦ ΑΘΓ ἡ ἐκτὸς (straight-lines) DE, EF , respectively. And the angles
γωνία ἡ ὑπὸ ΒΘΑ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ they encompass (are also equal). Thus, the base AH is
ΒΓΑ· ὅπερ ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ΒΓ τῇ ΕΖ· ἴση equal to the base DF , and the triangle ABH is equal to
ἄρα. ἐστὶ δὲ καὶ ἡ ΑΒ τῇ ΔΕ ἴση. δύο δὴ αἱ ΑΒ, ΒΓ δύο the triangle DEF , and the remaining angles subtended
ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνίας ἴσας by the equal sides will be equal to the (corresponding)
περιέχουσι· βάσις ἄρα ἡ ΑΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ remaining angles [Prop. 1.4]. Thus, angle BHA is equal
ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον καὶ λοιπὴ γωνία ἡ to EFD. But, EFD is equal to BCA. So, in triangle
ὑπὸ ΒΑΓ τῇ λοιπῂ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση. AHC, the external angle BHA is equal to the internal
᾿Εὰν ἄρα δύο τρίγωνα τὰς δύο γωνίας δυσὶ γωνίαις ἴσας and opposite angle BCA. The very thing (is) impossi-

ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ble [Prop. 1.16]. Thus, BC is not unequal to EF . Thus,
ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις, ἢ τὴν ὑποτείνουσαν ὑπὸ (it is) equal. And AB is also equal to DE. So the two
μίαν τῶν ἴσων γωνιῶν, καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς (straight-lines) AB, BC are equal to the two (straight-
πλευραῖς ἴσας ἕξει καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ· lines) DE, EF , respectively. And they encompass equal
ὅπερ ἔδει δεῖξαι. angles. Thus, the base AC is equal to the base DF , and

triangle ABC (is) equal to triangle DEF , and the re-

maining angle BAC (is) equal to the remaining angle
EDF [Prop. 1.4].

Thus, if two triangles have two angles equal to two

angles, respectively, and one side equal to one side—in
fact, either that by the equal angles, or that subtending

one of the equal angles—then (the triangles) will also
have the remaining sides equal to the (corresponding) re-

maining sides, and the remaining angle (equal) to the re-

maining angle. (Which is) the very thing it was required
to show.

† The Greek text has “BG, BC”, which is obviously a mistake.kzþ. Proposition 27
᾿Εὰν εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὰς ἐναλλὰξ If a straight-line falling across two straight-lines

γωνίας ἴσας ἀλλήλαις ποιῇ, παράλληλοι ἔσονται ἀλλήλαις αἱ makes the alternate angles equal to one another then
εὐθεῖαι. the (two) straight-lines will be parallel to one another.

ΖΓ

Α Ε Β

∆

E B

Εἰς γὰρ δύο εὐθείας τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπίπτουσα ἡ For let the straight-line EF , falling across the two
ΕΖ τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ ΑΕΖ, ΕΖΔ ἴσας ἀλλήλαις straight-lines AB and CD, make the alternate angles
ποιείτω· λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ. AEF and EFD equal to one another. I say that AB and
Εἰ γὰρ μή, ἐκβαλλόμεναι αἱ ΑΒ, ΓΔ συμπεσοῦνται ἤτοι CD are parallel.

ἐπὶ τὰ Β, Δ μέρη ἢ ἐπὶ τὰ Α, Γ. ἐκβεβλήσθωσαν καὶ συμ- For if not, being produced, AB and CD will certainly
πιπτέτωσαν ἐπὶ τὰ Β, Δ μέρη κατὰ τὸ Η. τριγώνου δὴ τοῦ meet together: either in the direction of B and D, or (in
ΗΕΖ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΕΖ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπε- the direction) of A and C [Def. 1.23]. Let them have
ναντίον τῇ ὑπὸ ΕΖΗ· ὅπερ ἐστὶν ἀδύνατον· οὐκ ἄρα αἱ ΑΒ, been produced, and let them meet together in the di-
ΔΓ ἐκβαλλόμεναι συμπεσοῦνται ἐπὶ τὰ Β, Δ μέρη. ὁμοίως rection of B and D at (point) G. So, for the triangle

STOIQEIWN aþ. ELEMENTS BOOK 1
δὴ δειχθήσεται, ὅτι οὐδὲ ἐπὶ τὰ Α, Γ· αἱ δὲ ἐπὶ μηδέτερα τὰ GEF , the external angle AEF is equal to the interior
μέρη συμπίπτουσαι παράλληλοί εἰσιν· παράλληλος ἄρα ἐστὶν and opposite (angle) EFG. The very thing is impossible
ἡ ΑΒ τῇ ΓΔ. [Prop. 1.16]. Thus, being produced, AB and CD will not
᾿Εὰν ἄρα εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὰς ἐναλλὰξ meet together in the direction of B and D. Similarly, it

γωνίας ἴσας ἀλλήλαις ποιῇ, παράλληλοι ἔσονται αἱ εὐθεῖαι· can be shown that neither (will they meet together) in
ὅπερ ἔδει δεῖξαι. (the direction of) A and C. But (straight-lines) meeting

in neither direction are parallel [Def. 1.23]. Thus, AB

and CD are parallel.
Thus, if a straight-line falling across two straight-lines

makes the alternate angles equal to one another then

the (two) straight-lines will be parallel (to one another).
(Which is) the very thing it was required to show.khþ. Proposition 28

᾿Εὰν εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὴν ἐκτὸς If a straight-line falling across two straight-lines
γωνίαν τῇ ἐντὸς καὶ ἀπεναντίον καὶ ἐπὶ τὰ αὐτὰ μέρη ἴσην makes the external angle equal to the internal and oppo-
ποιῇ ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας, site angle on the same side, or (makes) the (sum of the)
παράλληλοι ἔσονται ἀλλήλαις αἱ εὐθεῖαι. internal (angles) on the same side equal to two right-

angles, then the (two) straight-lines will be parallel to
one another.

Α Β

Γ ∆Θ

G B

Εἰς γὰρ δύο εὐθείας τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπίπτουσα ἡ For let EF , falling across the two straight-lines AB
ΕΖ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον and CD, make the external angle EGB equal to the in-
γωνίᾳ τῇ ὑπὸ ΗΘΔ ἴσην ποιείτω ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ ternal and opposite angle GHD, or the (sum of the) in-
αὐτὰ μέρη τὰς ὑπὸ ΒΗΘ, ΗΘΔ δυσὶν ὀρθαῖς ἴσας· λέγω, ternal (angles) on the same side, BGH and GHD, equal
ὅτι παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ. to two right-angles. I say that AB is parallel to CD.
᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΗΘΔ, ἀλλὰ ἡ ὑπὸ For since (in the first case) EGB is equal to GHD, but

ΕΗΒ τῇ ὑπὸ ΑΗΘ ἐστιν ἴση, καὶ ἡ ὑπὸ ΑΗΘ ἄρα τῇ ὑπὸ EGB is equal to AGH [Prop. 1.15], AGH is thus also
ΗΘΔ ἐστιν ἴση· καί εἰσιν ἐναλλάξ· παράλληλος ἄρα ἐστὶν ἡ equal to GHD. And they are alternate (angles). Thus,
ΑΒ τῇ ΓΔ. AB is parallel to CD [Prop. 1.27].
Πάλιν, ἐπεὶ αἱ ὑπὸ ΒΗΘ, ΗΘΔ δύο ὀρθαῖς ἴσαι εἰσίν, Again, since (in the second case, the sum of) BGH

εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΗΘ, ΒΗΘ δυσὶν ὀρθαῖς ἴσαι, αἱ ἄρα and GHD is equal to two right-angles, and (the sum
ὑπὸ ΑΗΘ, ΒΗΘ ταῖς ὑπὸ ΒΗΘ, ΗΘΔ ἴσαι εἰσίν· κοινὴ of) AGH and BGH is also equal to two right-angles
ἀφῃρήσθω ἡ ὑπὸ ΒΗΘ· λοιπὴ ἄρα ἡ ὑπὸ ΑΗΘ λοιπῇ τῇ [Prop. 1.13], (the sum of) AGH and BGH is thus equal
ὑπὸ ΗΘΔ ἐστιν ἴση· καί εἰσιν ἐναλλάξ· παράλληλος ἄρα to (the sum of) BGH and GHD. Let BGH have been
ἐστὶν ἡ ΑΒ τῇ ΓΔ. subtracted from both. Thus, the remainder AGH is equal
᾿Εὰν ἄρα εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὴν ἐκτὸς to the remainder GHD. And they are alternate (angles).

γωνίαν τῇ ἐντὸς καὶ ἀπεναντίον καὶ ἐπὶ τὰ αὐτὰ μέρη ἴσην Thus, AB is parallel to CD [Prop. 1.27].

STOIQEIWN aþ. ELEMENTS BOOK 1
ποιῇ ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας, Thus, if a straight-line falling across two straight-lines
παράλληλοι ἔσονται αἱ εὐθεῖαι· ὅπερ ἔδει δεῖξαι. makes the external angle equal to the internal and oppo-

site angle on the same side, or (makes) the (sum of the)
internal (angles) on the same side equal to two right-

angles, then the (two) straight-lines will be parallel (to
one another). (Which is) the very thing it was required

to show.kjþ. Proposition 29
῾Η εἰς τὰς παραλλήλους εὐθείας εὐθεῖα ἐμπίπτουσα τάς A straight-line falling across parallel straight-lines

τε ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιεῖ καὶ τὴν ἐκτὸς τῇ makes the alternate angles equal to one another, the ex-
ἐντὸς καὶ ἀπεναντίον ἴσην καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ ternal (angle) equal to the internal and opposite (angle),
μέρη δυσὶν ὀρθαῖς ἴσας. and the (sum of the) internal (angles) on the same side

equal to two right-angles.

Α Β

Γ ∆Θ

G B

Εἰς γὰρ παραλλήλους εὐθείας τὰς ΑΒ, ΓΔ εὐθεῖα For let the straight-line EF fall across the parallel
ἐμπιπτέτω ἡ ΕΖ· λέγω, ὅτι τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ straight-lines AB and CD. I say that it makes the alter-
ΑΗΘ, ΗΘΔ ἴσας ποιεῖ καὶ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ nate angles, AGH and GHD, equal, the external angle
τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΗΘΔ ἴσην καὶ τὰς ἐντὸς EGB equal to the internal and opposite (angle) GHD,
καὶ ἐπὶ τὰ αὐτὰ μέρη τὰς ὑπὸ ΒΗΘ, ΗΘΔ δυσὶν ὀρθαῖς and the (sum of the) internal (angles) on the same side,
ἴσας. BGH and GHD, equal to two right-angles.
Εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ, μία αὐτῶν For if AGH is unequal to GHD then one of them is

μείζων ἐστίν. ἔστω μείζων ἡ ὑπὸ ΑΗΘ· κοινὴ προσκείσθω greater. Let AGH be greater. Let BGH have been added
ἡ ὑπὸ ΒΗΘ· αἱ ἄρα ὑπὸ ΑΗΘ, ΒΗΘ τῶν ὑπὸ ΒΗΘ, ΗΘΔ to both. Thus, (the sum of) AGH and BGH is greater
μείζονές εἰσιν. ἀλλὰ αἱ ὑπὸ ΑΗΘ, ΒΗΘ δυσὶν ὀρθαῖς ἴσαι than (the sum of) BGH and GHD. But, (the sum of)
εἰσίν. [καὶ] αἱ ἄρα ὑπὸ ΒΗΘ, ΗΘΔ δύο ὀρθῶν ἐλάσσονές AGH and BGH is equal to two right-angles [Prop 1.13].
εἰσιν. αἱ δὲ ἀπ᾿ ἐλασσόνων ἢ δύο ὀρθῶν ἐκβαλλόμεναι Thus, (the sum of) BGH and GHD is [also] less than
εἰς ἄπειρον συμπίπτουσιν· αἱ ἄρα ΑΒ, ΓΔ ἐκβαλλόμεναι two right-angles. But (straight-lines) being produced to
εἰς ἄπειρον συμπεσοῦνται· οὐ συμπίπτουσι δὲ διὰ τὸ πα- infinity from (internal angles whose sum is) less than two
ραλλήλους αὐτὰς ὑποκεῖσθαι· οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ right-angles meet together [Post. 5]. Thus, AB and CD,
ΑΗΘ τῇ ὑπὸ ΗΘΔ· ἴση ἄρα. ἀλλὰ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΕΗΒ being produced to infinity, will meet together. But they do
ἐστιν ἴση· καὶ ἡ ὑπὸ ΕΗΒ ἄρα τῇ ὑπὸ ΗΘΔ ἐστιν ἴση· κοινὴ not meet, on account of them (initially) being assumed
προσκείσθω ἡ ὑπὸ ΒΗΘ· αἱ ἄρα ὑπὸ ΕΗΒ, ΒΗΘ ταῖς ὑπὸ parallel (to one another) [Def. 1.23]. Thus, AGH is not
ΒΗΘ, ΗΘΔ ἴσαι εἰσίν. ἀλλὰ αἱ ὑπὸ ΕΗΒ, ΒΗΘ δύο ὀρθαῖς unequal to GHD. Thus, (it is) equal. But, AGH is equal
ἴσαι εἰσίν· καὶ αἱ ὑπὸ ΒΗΘ, ΗΘΔ ἄρα δύο ὀρθαῖς ἴσαι εἰσίν. to EGB [Prop. 1.15]. And EGB is thus also equal to
῾Η ἄρα εἰς τὰς παραλλήλους εὐθείας εὐθεῖα ἐμπίπτουσα GHD. Let BGH be added to both. Thus, (the sum of)

τάς τε ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιεῖ καὶ τὴν ἐκτὸς EGB and BGH is equal to (the sum of) BGH and GHD.
τῇ ἐντὸς καὶ ἀπεναντίον ἴσην καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ But, (the sum of) EGB and BGH is equal to two right-

STOIQEIWN aþ. ELEMENTS BOOK 1
μέρη δυσὶν ὀρθαῖς ἴσας· ὅπερ ἔδει δεῖξαι. angles [Prop. 1.13]. Thus, (the sum of) BGH and GHD

is also equal to two right-angles.

Thus, a straight-line falling across parallel straight-
lines makes the alternate angles equal to one another, the

external (angle) equal to the internal and opposite (an-
gle), and the (sum of the) internal (angles) on the same

side equal to two right-angles. (Which is) the very thing

it was required to show.lþ. Proposition 30
Αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ ἀλλήλαις εἰσὶ παράλλη- (Straight-lines) parallel to the same straight-line are

λοι. also parallel to one another.

∆
Κ

῎Εστω ἑκατέρα τῶν ΑΒ, ΓΔ τῇ ΕΖ παράλληλος· λέγω, Let each of the (straight-lines) AB and CD be parallel
ὅτι καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος. to EF . I say that AB is also parallel to CD.
᾿Εμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ. For let the straight-line GK fall across (AB, CD, and
Καὶ ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΑΒ, ΕΖ εὐθεῖα EF ).

ἐμπέπτωκεν ἡ ΗΚ, ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ. And since the straight-line GK has fallen across the
πάλιν, ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΕΖ, ΓΔ εὐθεῖα parallel straight-lines AB and EF , (angle) AGK (is) thus
ἐμπέπτωκεν ἡ ΗΚ, ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ. equal to GHF [Prop. 1.29]. Again, since the straight-line
ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση. καὶ ἡ ὑπὸ ΑΗΚ GK has fallen across the parallel straight-lines EF and
ἄρα τῇ ὑπὸ ΗΚΔ ἐστιν ἴση· καί εἰσιν ἐναλλάξ. παράλληλος CD, (angle) GHF is equal to GKD [Prop. 1.29]. But
ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. AGK was also shown (to be) equal to GHF . Thus, AGK
[Αἱ ἄρα τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ ἀλλήλαις εἰσὶ is also equal to GKD. And they are alternate (angles).

παράλληλοι·] ὅπερ ἔδει δεῖξαι. Thus, AB is parallel to CD [Prop. 1.27].
[Thus, (straight-lines) parallel to the same straight-

line are also parallel to one another.] (Which is) the very

thing it was required to show.laþ. Proposition 31
Διὰ τοῦ δοθέντος σημείου τῇ δοθείσῃ εὐθείᾳ παράλληλον To draw a straight-line parallel to a given straight-line,

εὐθεῖαν γραμμὴν ἀγαγεῖν. through a given point.
῎Εστω τὸ μὲν δοθὲν σημεῖον τὸ Α, ἡ δὲ δοθεῖσα εὐθεῖα Let A be the given point, and BC the given straight-

ἡ ΒΓ· δεῖ δὴ διὰ τοῦ Α σημείου τῇ ΒΓ εὐθείᾳ παράλληλον line. So it is required to draw a straight-line parallel to
εὐθεῖαν γραμμὴν ἀγαγεῖν. the straight-line BC, through the point A.
Εἰλήφθω ἐπὶ τῆς ΒΓ τυχὸν σημεῖον τὸ Δ, καὶ ἐπεζεύχθω Let the point D have been taken a random on BC, and

ἡ ΑΔ· καὶ συνεστάτω πρὸς τῇ ΔΑ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ let AD have been joined. And let (angle) DAE, equal to
σημείῳ τῷ Α τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ἡ ὑπὸ ΔΑΕ· καὶ angle ADC, have been constructed on the straight-line

STOIQEIWN aþ. ELEMENTS BOOK 1
ἐκβεβλήσθω ἐπ᾿ εὐθείας τῇ ΕΑ εὐθεῖα ἡ ΑΖ. DA at the point A on it [Prop. 1.23]. And let the straight-

line AF have been produced in a straight-line with EA.

Β
∆

Α
Ζ

B
D

A
F

Καὶ ἐπεὶ εἰς δύο εὐθείας τὰς ΒΓ, ΕΖ εὐθεῖα ἐμπίπτουσα And since the straight-line AD, (in) falling across the
ἡ ΑΔ τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ ΕΑΔ, ΑΔΓ ἴσας two straight-lines BC and EF , has made the alternate
ἀλλήλαις πεποίηκεν, παράλληλος ἄρα ἐστὶν ἡ ΕΑΖ τῇ ΒΓ. angles EAD and ADC equal to one another, EAF is thus
Διὰ τοῦ δοθέντος ἄρα σημείου τοῦ Α τῇ δοθείσῃ εὐθείᾳ parallel to BC [Prop. 1.27].

τῇ ΒΓ παράλληλος εὐθεῖα γραμμὴ ἦκται ἡ ΕΑΖ· ὅπερ ἔδει Thus, the straight-line EAF has been drawn parallel
ποιῆσαι. to the given straight-line BC, through the given point A.

(Which is) the very thing it was required to do.lbþ. Proposition 32
Παντὸς τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης In any triangle, (if) one of the sides (is) produced

ἡ ἐκτὸς γωνία δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ἴση ἐστίν, καὶ (then) the external angle is equal to the (sum of the) two
αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. internal and opposite (angles), and the (sum of the) three

internal angles of the triangle is equal to two right-angles.

∆Γ

Α Ε

Β C

A E

῎Εστω τρίγωνον τὸ ΑΒΓ, καὶ προσεκβεβλήσθω αὐτοῦ Let ABC be a triangle, and let one of its sides BC
μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ· λέγω, ὅτι ἡ ἐκτὸς γωνία ἡ ὑπὸ have been produced to D. I say that the external angle
ΑΓΔ ἴση ἐστὶ δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ταῖς ὑπὸ ΓΑΒ, ACD is equal to the (sum of the) two internal and oppo-
ΑΒΓ, καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι αἱ ὑπὸ ΑΒΓ, site angles CAB and ABC, and the (sum of the) three
ΒΓΑ, ΓΑΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν. internal angles of the triangle—ABC, BCA, and CAB—
῎Ηχθω γὰρ διὰ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ παράλληλος is equal to two right-angles.

ἡ ΓΕ. For let CE have been drawn through point C parallel
Καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ, καὶ εἰς αὐτὰς to the straight-line AB [Prop. 1.31].

ἐμπέπτωκεν ἡ ΑΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ, ΑΓΕ ἴσαι And since AB is parallel to CE, and AC has fallen
ἀλλήλαις εἰσίν. πάλιν, ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ, across them, the alternate angles BAC and ACE are
καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΒΔ, ἡ ἐκτὸς γωνία ἡ equal to one another [Prop. 1.29]. Again, since AB is
ὑπὸ ΕΓΔ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΒΓ. parallel to CE, and the straight-line BD has fallen across
ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΓ ἴση· ὅλη ἄρα ἡ ὑπὸ them, the external angle ECD is equal to the internal
ΑΓΔ γωνία ἴση ἐστὶ δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ταῖς ὑπὸ and opposite (angle) ABC [Prop. 1.29]. But ACE was
ΒΑΓ, ΑΒΓ. also shown (to be) equal to BAC. Thus, the whole an-

STOIQEIWN aþ. ELEMENTS BOOK 1
Κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ· αἱ ἄρα ὑπὸ ΑΓΔ, ΑΓΒ gle ACD is equal to the (sum of the) two internal and

τρισὶ ταῖς ὑπὸ ΑΒΓ, ΒΓΑ, ΓΑΒ ἴσαι εἰσίν. ἀλλ᾿ αἱ ὑπὸ ΑΓΔ, opposite (angles) BAC and ABC.
ΑΓΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν· καὶ αἱ ὑπὸ ΑΓΒ, ΓΒΑ, ΓΑΒ Let ACB have been added to both. Thus, (the sum
ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν. of) ACD and ACB is equal to the (sum of the) three
Παντὸς ἄρα τριγώνου μιᾶς τῶν πλευρῶν προσεκ- (angles) ABC, BCA, and CAB. But, (the sum of) ACD

βληθείσης ἡ ἐκτὸς γωνία δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον and ACB is equal to two right-angles [Prop. 1.13]. Thus,
ἴση ἐστίν, καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι δυσὶν (the sum of) ACB, CBA, and CAB is also equal to two
ὀρθαῖς ἴσαι εἰσίν· ὅπερ ἔδει δεῖξαι. right-angles.

Thus, in any triangle, (if) one of the sides (is) pro-

duced (then) the external angle is equal to the (sum of

the) two internal and opposite (angles), and the (sum of
the) three internal angles of the triangle is equal to two

right-angles. (Which is) the very thing it was required to
show.lgþ. Proposition 33

Αἱ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ αὐτὰ μέρη ἐπι- Straight-lines joining equal and parallel (straight-
ζευγνύουσαι εὐθεῖαι καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν. lines) on the same sides are themselves also equal and

parallel.

ΑΒ

Γ∆ D C

B A

῎Εστωσαν ἴσαι τε καὶ παράλληλοι αἱ ΑΒ, ΓΔ, καὶ ἐπι- Let AB and CD be equal and parallel (straight-lines),
ζευγνύτωσαν αὐτὰς ἐπὶ τὰ αὐτὰ μέρη εὐθεῖαι αἱ ΑΓ, ΒΔ· and let the straight-lines AC and BD join them on the
λέγω, ὅτι καὶ αἱ ΑΓ, ΒΔ ἴσαι τε καὶ παράλληλοί εἰσιν. same sides. I say that AC and BD are also equal and
᾿Επεζεύχθω ἡ ΒΓ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ parallel.

ΓΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ, αἱ ἐναλλὰξ γωνίαι αἱ Let BC have been joined. And since AB is paral-
ὑπὸ ΑΒΓ, ΒΓΔ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ lel to CD, and BC has fallen across them, the alter-
τῇ ΓΔ κοινὴ δὲ ἡ ΒΓ, δύο δὴ αἱ ΑΒ, ΒΓ δύο ταῖς ΒΓ, ΓΔ ἴσαι nate angles ABC and BCD are equal to one another
εἰσίν· καὶ γωνία ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση· βάσις [Prop. 1.29]. And since AB is equal to CD, and BC
ἄρα ἡ ΑΓ βάσει τῇ ΒΔ ἐστιν ἴση, καὶ τὸ ΑΒΓ τρίγωνον τῷ is common, the two (straight-lines) AB, BC are equal

ΒΓΔ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς to the two (straight-lines) DC, CB.†And the angle ABC
γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ is equal to the angle BCD. Thus, the base AC is equal
ὑποτείνουσιν· ἴση ἄρα ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΓΒΔ. καὶ to the base BD, and triangle ABC is equal to triangle

ἐπεὶ εἰς δύο εὐθείας τὰς ΑΓ, ΒΔ εὐθεῖα ἐμπίπτουσα ἡ ΒΓ DCB‡, and the remaining angles will be equal to the
τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις πεποίηκεν, παράλληλος corresponding remaining angles subtended by the equal
ἄρα ἐστὶν ἡ ΑΓ τῇ ΒΔ. ἐδείχθη δὲ αὐτῇ καὶ ἴση. sides [Prop. 1.4]. Thus, angle ACB is equal to CBD.
Αἱ ἄρα τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ αὐτὰ μέρη Also, since the straight-line BC, (in) falling across the

ἐπιζευγνύουσαι εὐθεῖαι καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί two straight-lines AC and BD, has made the alternate
εἰσιν· ὅπερ ἔδει δεῖξαι. angles (ACB and CBD) equal to one another, AC is thus

parallel to BD [Prop. 1.27]. And (AC) was also shown
(to be) equal to (BD).

Thus, straight-lines joining equal and parallel (straight-

STOIQEIWN aþ. ELEMENTS BOOK 1
lines) on the same sides are themselves also equal and

parallel. (Which is) the very thing it was required to

show.

† The Greek text has “BC, CD”, which is obviously a mistake.
‡ The Greek text has “DCB”, which is obviously a mistake.ldþ. Proposition 34
Τῶν παραλληλογράμμων χωρίων αἱ ἀπεναντίον πλευραί In parallelogrammic figures the opposite sides and angles

τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν, καὶ ἡ διάμετρος αὐτὰ δίχα are equal to one another, and a diagonal cuts them in half.
τέμνει.

∆ C

A B

῎Εστω παραλληλόγραμμον χωρίον τὸ ΑΓΔΒ, διάμετρος Let ACDB be a parallelogrammic figure, and BC its
δὲ αὐτοῦ ἡ ΒΓ· λέγω, ὅτι τοῦ ΑΓΔΒ παραλληλογράμμου αἱ diagonal. I say that for parallelogram ACDB, the oppo-
ἀπεναντίον πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν, καὶ ἡ site sides and angles are equal to one another, and the
ΒΓ διάμετρος αὐτὸ δίχα τέμνει. diagonal BC cuts it in half.
᾿Επεὶ γὰρ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ, καὶ εἰς αὐτὰς For since AB is parallel to CD, and the straight-line

ἐμπέπτωκεν εὐθεῖα ἡ ΒΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ, BC has fallen across them, the alternate angles ABC and
ΒΓΔ ἴσαι ἀλλήλαις εἰσίν. πάλιν ἐπεὶ παράλληλός ἐστιν ἡ ΑΓ BCD are equal to one another [Prop. 1.29]. Again, since
τῇ ΒΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ, αἱ ἐναλλὰξ γωνίαι AC is parallel to BD, and BC has fallen across them,
αἱ ὑπὸ ΑΓΒ, ΓΒΔ ἴσαι ἀλλήλαις εἰσίν. δύο δὴ τρίγωνά ἐστι the alternate angles ACB and CBD are equal to one
τὰ ΑΒΓ, ΒΓΔ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ, ΒΓΑ δυσὶ another [Prop. 1.29]. So ABC and BCD are two tri-
ταῖς ὑπὸ ΒΓΔ, ΓΒΔ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν angles having the two angles ABC and BCA equal to
πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν πρὸς ταῖς ἴσαις γωνίαις κοινὴν the two (angles) BCD and CBD, respectively, and one
αὐτῶν τὴν ΒΓ· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς side equal to one side—the (one) by the equal angles and
ἴσας ἕξει ἑκατέραν ἑκατέρᾳ καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ common to them, (namely) BC. Thus, they will also
γωνίᾳ· ἴση ἄρα ἡ μὲν ΑΒ πλευρὰ τῇ ΓΔ, ἡ δὲ ΑΓ τῇ ΒΔ, have the remaining sides equal to the corresponding re-
καὶ ἔτι ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΓΔΒ. καὶ ἐπεὶ maining (sides), and the remaining angle (equal) to the
ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΒΓΔ, ἡ δὲ ὑπὸ ΓΒΔ remaining angle [Prop. 1.26]. Thus, side AB is equal to
τῇ ὑπὸ ΑΓΒ, ὅλη ἄρα ἡ ὑπὸ ΑΒΔ ὅλῃ τῇ ὑπὸ ΑΓΔ ἐστιν CD, and AC to BD. Furthermore, angle BAC is equal
ἴση. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΒ ἴση. to CDB. And since angle ABC is equal to BCD, and
Τῶν ἄρα παραλληλογράμμων χωρίων αἱ ἀπεναντίον CBD to ACB, the whole (angle) ABD is thus equal to

πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν. the whole (angle) ACD. And BAC was also shown (to
Λέγω δή, ὅτι καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει. ἐπεὶ γὰρ be) equal to CDB.

ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ, κοινὴ δὲ ἡ ΒΓ, δύο δὴ αἱ ΑΒ, ΒΓ Thus, in parallelogrammic figures the opposite sides
δυσὶ ταῖς ΓΔ, ΒΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνία ἡ and angles are equal to one another.
ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση. καὶ βάσις ἄρα ἡ ΑΓ τῇ And, I also say that a diagonal cuts them in half. For
ΔΒ ἴση. καὶ τὸ ΑΒΓ [ἄρα] τρίγωνον τῷ ΒΓΔ τριγώνῳ ἴσον since AB is equal to CD, and BC (is) common, the two
ἐστίν. (straight-lines) AB, BC are equal to the two (straight-

῾Η ἄρα ΒΓ διάμετρος δίχα τέμνει τὸ ΑΒΓΔ παραλ- lines) DC, CB†, respectively. And angle ABC is equal to
ληλόγραμμον· ὅπερ ἔδει δεῖξαι. angle BCD. Thus, the base AC (is) also equal to DB,

STOIQEIWN aþ. ELEMENTS BOOK 1
and triangle ABC is equal to triangle BCD [Prop. 1.4].

Thus, the diagonal BC cuts the parallelogram ACDB‡

in half. (Which is) the very thing it was required to show.

† The Greek text has “CD, BC”, which is obviously a mistake.
‡ The Greek text has “ABCD”, which is obviously a mistake.leþ. Proposition 35
Τὰ παραλληλόγραμμα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ Parallelograms which are on the same base and be-

ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν. tween the same parallels are equal† to one another.

∆

Β Γ

ΕΑ Ζ

B C

D E

A F

῎Εστω παραλληλόγραμμα τὰ ΑΒΓΔ, ΕΒΓΖ ἐπὶ τῆς Let ABCD and EBCF be parallelograms on the same
αὐτῆς βάσεως τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς base BC, and between the same parallels AF and BC. I
ΑΖ, ΒΓ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓΔ τῷ ΕΒΓΖ παραλλη- say that ABCD is equal to parallelogram EBCF .
λογράμμῳ. For since ABCD is a parallelogram, AD is equal to
᾿Επεὶ γὰρ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ, ἴση ἐστὶν BC [Prop. 1.34]. So, for the same (reasons), EF is also

ἡ ΑΔ τῇ ΒΓ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση· equal to BC. So AD is also equal to EF . And DE is
ὥστε καὶ ἡ ΑΔ τῇ ΕΖ ἐστιν ἴση· καὶ κοινὴ ἡ ΔΕ· ὅλη ἄρα common. Thus, the whole (straight-line) AE is equal to
ἡ ΑΕ ὅλῃ τῇ ΔΖ ἐστιν ἴση. ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΓ ἴση· the whole (straight-line) DF . And AB is also equal to
δύο δὴ αἱ ΕΑ, ΑΒ δύο ταῖς ΖΔ, ΔΓ ἴσαι εἰσὶν ἑκατέρα DC. So the two (straight-lines) EA, AB are equal to
ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΖΔΓ γωνίᾳ τῇ ὑπὸ ΕΑΒ ἐστιν the two (straight-lines) FD, DC, respectively. And angle
ἴση ἡ ἐκτὸς τῇ ἐντός· βάσις ἄρα ἡ ΕΒ βάσει τῇ ΖΓ ἴση ἐστίν, FDC is equal to angle EAB, the external to the inter-
καὶ τὸ ΕΑΒ τρίγωνον τῷ ΔΖΓ τριγώνῳ ἴσον ἔσται· κοινὸν nal [Prop. 1.29]. Thus, the base EB is equal to the base
ἀφῃρήσθω τὸ ΔΗΕ· λοιπὸν ἄρα τὸ ΑΒΗΔ τραπέζιον λοιπῷ FC, and triangle EAB will be equal to triangle DFC
τῷ ΕΗΓΖ τραπεζίῳ ἐστὶν ἴσον· κοινὸν προσκείσθω τὸ ΗΒΓ [Prop. 1.4]. Let DGE have been taken away from both.
τρίγωνον· ὅλον ἄρα τὸ ΑΒΓΔ παραλληλόγραμμον ὅλῳ τῷ Thus, the remaining trapezium ABGD is equal to the re-
ΕΒΓΖ παραλληλογράμμῳ ἴσον ἐστίν. maining trapezium EGCF . Let triangle GBC have been
Τὰ ἄρα παραλληλόγραμμα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα added to both. Thus, the whole parallelogram ABCD is

καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει equal to the whole parallelogram EBCF .
δεῖξαι. Thus, parallelograms which are on the same base and

between the same parallels are equal to one another.

(Which is) the very thing it was required to show.

† Here, for the first time, “equal” means “equal in area”, rather than “congruent”.l�þ. Proposition 36
Τὰ παραλληλόγραμμα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν Parallelograms which are on equal bases and between

ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν. the same parallels are equal to one another.
῎Εστω παραλληλόγραμμα τὰ ΑΒΓΔ, ΕΖΗΘ ἐπὶ ἴσων Let ABCD and EFGH be parallelograms which are

βάσεων ὄντα τῶν ΒΓ, ΖΗ καὶ ἐν ταῖς αὐταῖς παραλλήλοις on the equal bases BC and FG, and (are) between the
ταῖς ΑΘ, ΒΗ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓΔ παραλ- same parallels AH and BG. I say that the parallelogram

STOIQEIWN aþ. ELEMENTS BOOK 1
ληλόγραμμον τῷ ΕΖΗΘ. ABCD is equal to EFGH .

ΗΓ Ζ

ΘΕΑ ∆

Β G

A D E H

FCB

᾿Επεζεύχθωσαν γὰρ αἱ ΒΕ, ΓΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ For let BE and CH have been joined. And since BC is
ΒΓ τῇ ΖΗ, ἀλλὰ ἡ ΖΗ τῇ ΕΘ ἐστιν ἴση, καὶ ἡ ΒΓ ἄρα τῇ equal to FG, but FG is equal to EH [Prop. 1.34], BC is
ΕΘ ἐστιν ἴση. εἰσὶ δὲ καὶ παράλληλοι. καὶ ἐπιζευγνύουσιν thus equal to EH . And they are also parallel, and EB and
αὐτὰς αἱ ΕΒ, ΘΓ· αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ HC join them. But (straight-lines) joining equal and par-
τὰ αὐτὰ μέρη ἐπιζευγνύουσαι ἴσαι τε καὶ παράλληλοί εἰσι allel (straight-lines) on the same sides are (themselves)
[καὶ αἱ ΕΒ, ΘΓ ἄρα ἴσαι τέ εἰσι καὶ παράλληλοι]. παραλ- equal and parallel [Prop. 1.33] [thus, EB and HC are
ληλόγραμμον ἄρα ἐστὶ τὸ ΕΒΓΘ. καί ἐστιν ἴσον τῷ ΑΒΓΔ· also equal and parallel]. Thus, EBCH is a parallelogram
βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει τὴν ΒΓ, καὶ ἐν ταῖς αὐταῖς [Prop. 1.34], and is equal to ABCD. For it has the same
παραλλήλοις ἐστὶν αὐτῷ ταῖς ΒΓ, ΑΘ. δὶα τὰ αὐτὰ δὴ καὶ τὸ base, BC, as (ABCD), and is between the same paral-
ΕΖΗΘ τῷ αὐτῷ τῷ ΕΒΓΘ ἐστιν ἴσον· ὥστε καὶ τὸ ΑΒΓΔ lels, BC and AH , as (ABCD) [Prop. 1.35]. So, for the
παραλληλόγραμμον τῷ ΕΖΗΘ ἐστιν ἴσον. same (reasons), EFGH is also equal to the same (par-
Τὰ ἄρα παραλληλόγραμμα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ allelogram) EBCH [Prop. 1.34]. So that the parallelo-

ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει gram ABCD is also equal to EFGH .
δεῖξαι. Thus, parallelograms which are on equal bases and

between the same parallels are equal to one another.

(Which is) the very thing it was required to show.lzþ. Proposition 37
Τὰ τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς Triangles which are on the same base and between

αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν. the same parallels are equal to one another.

Α ∆

Ε Ζ

A
E

῎Εστω τρίγωνα τὰ ΑΒΓ, ΔΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς Let ABC and DBC be triangles on the same base BC,
ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΑΔ, ΒΓ· λέγω, ὅτι and between the same parallels AD and BC. I say that
ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ. triangle ABC is equal to triangle DBC.
᾿Εκβεβλήσθω ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε, Ζ, καὶ Let AD have been produced in both directions to E

διὰ μὲν τοῦ Β τῇ ΓΑ παράλληλος ἤχθω ἡ ΒΕ, δὶα δὲ τοῦ Γ τῇ and F , and let the (straight-line) BE have been drawn
ΒΔ παράλληλος ἤχθω ἡ ΓΖ. παραλληλόγραμμον ἄρα ἐστὶν through B parallel to CA [Prop. 1.31], and let the
ἑκάτερον τῶν ΕΒΓΑ, ΔΒΓΖ· καί εἰσιν ἴσα· ἐπί τε γὰρ τῆς (straight-line) CF have been drawn through C parallel
αὐτῆς βάσεώς εἰσι τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις to BD [Prop. 1.31]. Thus, EBCA and DBCF are both
ταῖς ΒΓ, ΕΖ· καί ἐστι τοῦ μὲν ΕΒΓΑ παραλληλογράμμου parallelograms, and are equal. For they are on the same
ἥμισυ τὸ ΑΒΓ τρίγωνον· ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα base BC, and between the same parallels BC and EF
τέμνει· τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ἥμισυ τὸ ΔΒΓ [Prop. 1.35]. And the triangle ABC is half of the paral-
τρίγωνον· ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει. [τὰ δὲ lelogram EBCA. For the diagonal AB cuts the latter in

STOIQEIWN aþ. ELEMENTS BOOK 1
τῶν ἴσων ἡμίση ἴσα ἀλλήλοις ἐστίν]. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ half [Prop. 1.34]. And the triangle DBC (is) half of the
τρίγωνον τῷ ΔΒΓ τριγώνῳ. parallelogram DBCF . For the diagonal DC cuts the lat-
Τὰ ἄρα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς ter in half [Prop. 1.34]. [And the halves of equal things

αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι. are equal to one another.]† Thus, triangle ABC is equal
to triangle DBC.

Thus, triangles which are on the same base and

between the same parallels are equal to one another.

(Which is) the very thing it was required to show.

† This is an additional common notion.lhþ. Proposition 38
Τὰ τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς Triangles which are on equal bases and between the

παραλλήλοις ἴσα ἀλλήλοις ἐστίν. same parallels are equal to one another.

ΖΕ

ΑΗ ∆ Θ

Β Γ FE

A DG H

B C

῎Εστω τρίγωνα τὰ ΑΒΓ, ΔΕΖ ἐπὶ ἴσων βάσεων τῶν ΒΓ, Let ABC and DEF be triangles on the equal bases
ΕΖ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΖ, ΑΔ· λέγω, ὅτι BC and EF , and between the same parallels BF and
ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. AD. I say that triangle ABC is equal to triangle DEF .
᾿Εκβεβλήσθω γὰρ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Η, For let AD have been produced in both directions

Θ, καὶ διὰ μὲν τοῦ Β τῇ ΓΑ παράλληλος ἤχθω ἡ ΒΗ, δὶα δὲ to G and H , and let the (straight-line) BG have been
τοῦ Ζ τῇ ΔΕ παράλληλος ἤχθω ἡ ΖΘ. παραλληλόγραμμον drawn through B parallel to CA [Prop. 1.31], and let the
ἄρα ἐστὶν ἑκάτερον τῶν ΗΒΓΑ, ΔΕΖΘ· καὶ ἴσον τὸ ΗΒΓΑ (straight-line) FH have been drawn through F parallel
τῷ ΔΕΖΘ· ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΓ, ΕΖ καὶ to DE [Prop. 1.31]. Thus, GBCA and DEFH are each
ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΖ, ΗΘ· καί ἐστι τοῦ μὲν parallelograms. And GBCA is equal to DEFH . For they
ΗΒΓΑ παραλληλογράμμου ἥμισυ τὸ ΑΒΓ τρίγωνον. ἡ γὰρ are on the equal bases BC and EF , and between the
ΑΒ διάμετρος αὐτὸ δίχα τέμνει· τοῦ δὲ ΔΕΖΘ παραλλη- same parallels BF and GH [Prop. 1.36]. And triangle
λογράμμου ἥμισυ τὸ ΖΕΔ τρίγωνον· ἡ γὰρ ΔΖ δίαμετρος ABC is half of the parallelogram GBCA. For the diago-
αὐτὸ δίχα τέμνει [τὰ δὲ τῶν ἴσων ἡμίση ἴσα ἀλλήλοις ἐστίν]. nal AB cuts the latter in half [Prop. 1.34]. And triangle
ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. FED (is) half of parallelogram DEFH . For the diagonal
Τὰ ἄρα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς DF cuts the latter in half. [And the halves of equal things

αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι. are equal to one another.] Thus, triangle ABC is equal
to triangle DEF .

Thus, triangles which are on equal bases and between

the same parallels are equal to one another. (Which is)

the very thing it was required to show.ljþ. Proposition 39
Τὰ ἴσα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐπὶ Equal triangles which are on the same base, and on

τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν. the same side, are also between the same parallels.
῎Εστω ἴσα τρίγωνα τὰ ΑΒΓ, ΔΒΓ ἐπὶ τῆς αὐτῆς βάσεως Let ABC and DBC be equal triangles which are on

ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη τῆς ΒΓ· λέγω, ὅτι καὶ ἐν ταῖς the same base BC, and on the same side (of it). I say that

STOIQEIWN aþ. ELEMENTS BOOK 1
αὐταῖς παραλλήλοις ἐστίν. they are also between the same parallels.

Β Γ

∆

᾿Επεζεύχθω γὰρ ἡ ΑΔ· λέγω, ὅτι παράλληλός ἐστιν ἡ For let AD have been joined. I say that AD and BC
ΑΔ τῇ ΒΓ. are parallel.
Εἰ γὰρ μή, ἤχθω διὰ τοῦ Α σημείου τῇ ΒΓ εὐθείᾳ For, if not, let AE have been drawn through point A

παράλληλος ἡ ΑΕ, καὶ ἐπεζεύχθω ἡ ΕΓ. ἴσον ἄρα ἐστὶ τὸ parallel to the straight-line BC [Prop. 1.31], and let EC
ΑΒΓ τρίγωνον τῷ ΕΒΓ τριγώνῳ· ἐπί τε γὰρ τῆς αὐτῆς have been joined. Thus, triangle ABC is equal to triangle
βάσεώς ἐστιν αὐτῷ τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις. EBC. For it is on the same base as it, BC, and between
ἀλλὰ τὸ ΑΒΓ τῷ ΔΒΓ ἐστιν ἴσον· καὶ τὸ ΔΒΓ ἄρα τῷ ΕΒΓ the same parallels [Prop. 1.37]. But ABC is equal to
ἴσον ἐστὶ τὸ μεῖζον τῷ ἐλάσσονι· ὅπερ ἐστὶν ἀδύνατον· οὐκ DBC. Thus, DBC is also equal to EBC, the greater to
ἄρα παράλληλός ἐστιν ἡ ΑΕ τῇ ΒΓ. ὁμοίως δὴ δείξομεν, the lesser. The very thing is impossible. Thus, AE is not
ὅτι οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ· ἡ ΑΔ ἄρα τῇ ΒΓ ἐστι parallel to BC. Similarly, we can show that neither (is)
παράλληλος. any other (straight-line) than AD. Thus, AD is parallel
Τὰ ἄρα ἴσα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ to BC.

ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν· ὅπερ Thus, equal triangles which are on the same base, and
ἔδει δεῖξαι. on the same side, are also between the same parallels.

(Which is) the very thing it was required to show.mþ. Proposition 40†
Τὰ ἴσα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐπὶ τὰ αὐτὰ Equal triangles which are on equal bases, and on the

μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν. same side, are also between the same parallels.

Α ∆

Β Ε
Γ

῎Εστω ἴσα τρίγωνα τὰ ΑΒΓ, ΓΔΕ ἐπὶ ἴσων βάσεων τῶν Let ABC and CDE be equal triangles on the equal
ΒΓ, ΓΕ καὶ ἐπὶ τὰ αὐτὰ μέρη. λέγω, ὅτι καὶ ἐν ταῖς αὐταῖς bases BC and CE (respectively), and on the same side
παραλλήλοις ἐστίν. (of BE). I say that they are also between the same par-
᾿Επεζεύχθω γὰρ ἡ ΑΔ· λέγω, ὅτι παράλληλός ἐστιν ἡ allels.

ΑΔ τῇ ΒΕ. For let AD have been joined. I say that AD is parallel
Εἰ γὰρ μή, ἤχθω διὰ τοῦ Α τῇ ΒΕ παράλληλος ἡ ΑΖ, to BE.

καὶ ἐπεζεύχθω ἡ ΖΕ. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον For if not, let AF have been drawn through A parallel
τῷ ΖΓΕ τριγώνῳ· ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΓ, to BE [Prop. 1.31], and let FE have been joined. Thus,
ΓΕ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΕ, ΑΖ. ἀλλὰ τὸ triangle ABC is equal to triangle FCE. For they are on
ΑΒΓ τρίγωνον ἴσον ἐστὶ τῷ ΔΓΕ [τρίγωνῳ]· καὶ τὸ ΔΓΕ equal bases, BC and CE, and between the same paral-
ἄρα [τρίγωνον] ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ τὸ μεῖζον τῷ lels, BE and AF [Prop. 1.38]. But, triangle ABC is equal

STOIQEIWN aþ. ELEMENTS BOOK 1
ἐλάσσονι· ὅπερ ἐστὶν ἀδύνατον· οὐκ ἄρα παράλληλος ἡ ΑΖ to [triangle] DCE. Thus, [triangle] DCE is also equal to
τῇ ΒΕ. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ· triangle FCE, the greater to the lesser. The very thing is
ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος. impossible. Thus, AF is not parallel to BE. Similarly, we
Τὰ ἄρα ἴσα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐπὶ τὰ can show that neither (is) any other (straight-line) than

αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν· ὅπερ ἔδει AD. Thus, AD is parallel to BE.
δεῖξαι. Thus, equal triangles which are on equal bases, and

on the same side, are also between the same parallels.

(Which is) the very thing it was required to show.

† This whole proposition is regarded by Heiberg as a relatively early interpolation to the original text.maþ. Proposition 41
᾿Εὰν παραλληλόγραμμον τριγώνῳ βάσιν τε ἔχῃ τὴν If a parallelogram has the same base as a triangle, and

αὐτὴν καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ, διπλάσιόν ἐστί is between the same parallels, then the parallelogram is
τὸ παραλληλόγραμμον τοῦ τριγώνου. double (the area) of the triangle.

∆Α

Γ B

A D E

Παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ τριγώνῳ τῷ ΕΒΓ For let parallelogram ABCD have the same base BC
βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ καὶ ἐν ταῖς αὐταῖς πα- as triangle EBC, and let it be between the same parallels,
ραλλήλοις ἔστω ταῖς ΒΓ, ΑΕ· λέγω, ὅτι διπλάσιόν ἐστι τὸ BC and AE. I say that parallelogram ABCD is double
ΑΒΓΔ παραλληλόγραμμον τοῦ ΒΕΓ τριγώνου. (the area) of triangle BEC.
᾿Επεζεύχθω γὰρ ἡ ΑΓ. ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον For let AC have been joined. So triangle ABC is equal

τῷ ΕΒΓ τριγώνῳ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν to triangle EBC. For it is on the same base, BC, as
αὐτῷ τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΑΕ. (EBC), and between the same parallels, BC and AE
ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον διπλάσιόν ἐστι τοῦ ΑΒΓ [Prop. 1.37]. But, parallelogram ABCD is double (the
τριγώνου· ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει· ὥστε area) of triangle ABC. For the diagonal AC cuts the for-
τὸ ΑΒΓΔ παραλληλόγραμμον καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ mer in half [Prop. 1.34]. So parallelogram ABCD is also
διπλάσιον. double (the area) of triangle EBC.
᾿Εὰν ἄρα παραλληλόγραμμον τριγώνῳ βάσιν τε ἔχῃ τὴν Thus, if a parallelogram has the same base as a trian-

αὐτὴν καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ, διπλάσιόν ἐστί τὸ gle, and is between the same parallels, then the parallel-
παραλληλόγραμμον τοῦ τριγώνου· ὅπερ ἔδει δεῖξαι. ogram is double (the area) of the triangle. (Which is) the

very thing it was required to show.mbþ. Proposition 42
Τῷ δοθέντι τριγώνῳ ἴσον παραλληλόγραμμον συστή- To construct a parallelogram equal to a given triangle

σασθαι ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ. in a given rectilinear angle.
῎Εστω τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ, ἡ δὲ δοθεῖσα Let ABC be the given triangle, and D the given recti-

γωνία εὐθύγραμμος ἡ Δ· δεῖ δὴ τῷ ΑΒΓ τριγώνῳ ἴσον πα- linear angle. So it is required to construct a parallelogram
ραλληλόγραμμον συστήσασθαι ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ. equal to triangle ABC in the rectilinear angle D.

STOIQEIWN aþ. ELEMENTS BOOK 1

∆

ΖΑ F

Τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε, καὶ ἐπεζεύχθω ἡ ΑΕ, Let BC have been cut in half at E [Prop. 1.10], and
καὶ συνεστάτω πρὸς τῇ ΕΓ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ let AE have been joined. And let (angle) CEF , equal to
τῷ Ε τῇ Δ γωνίᾳ ἴση ἡ ὑπὸ ΓΕΖ, καὶ διὰ μὲν τοῦ Α τῇ ΕΓ angle D, have been constructed at the point E on the
παράλληλος ἤχθω ἡ ΑΗ, διὰ δὲ τοῦ Γ τῇ ΕΖ παράλληλος straight-line EC [Prop. 1.23]. And let AG have been
ἤχθω ἡ ΓΗ· παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΕΓΗ. καὶ ἐπεὶ drawn through A parallel to EC [Prop. 1.31], and let CG
ἴση ἐστὶν ἡ ΒΕ τῇ ΕΓ, ἴσον ἐστὶ καὶ τὸ ΑΒΕ τρίγωνον τῷ have been drawn through C parallel to EF [Prop. 1.31].
ΑΕΓ τριγώνῳ· ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΕ, ΕΓ καὶ Thus, FECG is a parallelogram. And since BE is equal
ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΑΗ· διπλάσιον ἄρα ἐστὶ to EC, triangle ABE is also equal to triangle AEC. For
τὸ ΑΒΓ τρίγωνον τοῦ ΑΕΓ τριγώνου. ἔστι δὲ καὶ τὸ ΖΕΓΗ they are on the equal bases, BE and EC, and between
παραλληλόγραμμον διπλάσιον τοῦ ΑΕΓ τριγώνου· βάσιν τε the same parallels, BC and AG [Prop. 1.38]. Thus, tri-
γὰρ αὐτῷ τὴν αὐτὴν ἔχει καὶ ἐν ταῖς αὐταῖς ἐστιν αὐτῷ angle ABC is double (the area) of triangle AEC. And
παραλλήλοις· ἴσον ἄρα ἐστὶ τὸ ΖΕΓΗ παραλληλόγραμμον parallelogram FECG is also double (the area) of triangle
τῷ ΑΒΓ τριγώνῳ. καὶ ἔχει τὴν ὑπὸ ΓΕΖ γωνίαν ἴσην τῇ AEC. For it has the same base as (AEC), and is between
δοθείσῃ τῇ Δ. the same parallels as (AEC) [Prop. 1.41]. Thus, paral-
Τῷ ἄρα δοθέντι τριγώνῳ τῷ ΑΒΓ ἴσον παραλληλόγραμ- lelogram FECG is equal to triangle ABC. (FECG) also

μον συνέσταται τὸ ΖΕΓΗ ἐν γωνίᾳ τῇ ὑπὸ ΓΕΖ, ἥτις ἐστὶν has the angle CEF equal to the given (angle) D.
ἴση τῇ Δ· ὅπερ ἔδει ποιῆσαι. Thus, parallelogram FECG, equal to the given trian-

gle ABC, has been constructed in the angle CEF , which
is equal to D. (Which is) the very thing it was required

to do.mgþ. Proposition 43
Παντὸς παραλληλογράμμου τῶν περὶ τὴν διάμετρον πα- For any parallelogram, the complements of the paral-

ραλληλογράμμων τὰ παραπληρώματα ἴσα ἀλλήλοις ἐστίν. lelograms about the diagonal are equal to one another.
῎Εστω παραλληλόγραμμον τὸ ΑΒΓΔ, διάμετρος δὲ Let ABCD be a parallelogram, and AC its diagonal.

αὐτοῦ ἡ ΑΓ, περὶ δὲ τὴν ΑΓ παραλληλόγραμμα μὲν ἔστω And let EH and FG be the parallelograms about AC, and
τὰ ΕΘ, ΖΗ, τὰ δὲ λεγόμενα παραπληρώματα τὰ ΒΚ, ΚΔ· BK and KD the so-called complements (about AC). I
λέγω, ὅτι ἴσον ἐστὶ τὸ ΒΚ παραπλήρωμα τῷ ΚΔ παρα- say that the complement BK is equal to the complement
πληρώματι. KD.
᾿Επεὶ γὰρ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ, διάμετρος For since ABCD is a parallelogram, and AC its diago-

δὲ αὐτοῦ ἡ ΑΓ, ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ nal, triangle ABC is equal to triangle ACD [Prop. 1.34].
τριγώνῳ. πάλιν, ἐπεὶ παραλληλόγραμμόν ἐστι τὸ ΕΘ, Again, since EH is a parallelogram, and AK is its diago-
διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΑΚ, ἴσον ἐστὶ τὸ ΑΕΚ τρίγωνον nal, triangle AEK is equal to triangle AHK [Prop. 1.34].
τῷ ΑΘΚ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΚΖΓ τρίγωνον So, for the same (reasons), triangle KFC is also equal to
τῷ ΚΗΓ ἐστιν ἴσον. ἐπεὶ οὖν τὸ μὲν ΑΕΚ τρίγωνον τῷ (triangle) KGC. Therefore, since triangle AEK is equal
ΑΘΚ τριγώνῳ ἐστὶν ἴσον, τὸ δὲ ΚΖΓ τῷ ΚΗΓ, τὸ ΑΕΚ to triangle AHK, and KFC to KGC, triangle AEK plus
τρίγωνον μετὰ τοῦ ΚΗΓ ἴσον ἐστὶ τῷ ΑΘΚ τριγώνῳ μετὰ KGC is equal to triangle AHK plus KFC. And the
τοῦ ΚΖΓ· ἔστι δὲ καὶ ὅλον τὸ ΑΒΓ τρίγωνον ὅλῳ τῷ ΑΔΓ whole triangle ABC is also equal to the whole (triangle)
ἴσον· λοιπὸν ἄρα τὸ ΒΚ παραπλήρωμα λοιπῷ τῷ ΚΔ παρα- ADC. Thus, the remaining complement BK is equal to

STOIQEIWN aþ. ELEMENTS BOOK 1
πληρώματί ἐστιν ἴσον. the remaining complement KD.

Α Θ ∆

Β Η

Κ K

B G

Παντὸς ἄρα παραλληλογράμμου χωρίου τῶν περὶ τὴν Thus, for any parallelogramic figure, the comple-
διάμετρον παραλληλογράμμων τὰ παραπληρώματα ἴσα ἀλλή- ments of the parallelograms about the diagonal are equal
λοις ἐστίν· ὅπερ ἔδει δεῖξαι. to one another. (Which is) the very thing it was required

to show.mdþ. Proposition 44
Παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι τριγώνῳ ἴσον πα- To apply a parallelogram equal to a given triangle to

ραλληλόγραμμον παραβαλεῖν ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμ- a given straight-line in a given rectilinear angle.
μῳ.

∆

Ζ Ε

Η
Β

ΛΑΘ

Μ
B

F E K

LAH

῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν τρίγωνον Let AB be the given straight-line, C the given trian-
τὸ Γ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ· δεῖ δὴ παρὰ gle, and D the given rectilinear angle. So it is required to
τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ τῷ δοθέντι τριγώνῳ τῷ Γ apply a parallelogram equal to the given triangle C to the
ἴσον παραλληλόγραμμον παραβαλεῖν ἐν ἴσῃ τῇ Δ γωνίᾳ. given straight-line AB in an angle equal to (angle) D.
Συνεστάτω τῷ Γ τριγώνῳ ἴσον παραλληλόγραμμον τὸ Let the parallelogram BEFG, equal to the triangle C,

ΒΕΖΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ, ἥ ἐστιν ἴση τῇ Δ· καὶ κείσθω have been constructed in the angle EBG, which is equal
ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΒΕ τῇ ΑΒ, καὶ διήχθω ἡ ΖΗ to D [Prop. 1.42]. And let it have been placed so that

ἐπὶ τὸ Θ, καὶ διὰ τοῦ Α ὁποτέρᾳ τῶν ΒΗ, ΕΖ παράλληλος BE is straight-on to AB.† And let FG have been drawn
ἤχθω ἡ ΑΘ, καὶ ἐπεζεύχθω ἡ ΘΒ. καὶ ἐπεὶ εἰς παραλλήλους through to H , and let AH have been drawn through A
τὰς ΑΘ, ΕΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ, αἱ ἄρα ὑπὸ ΑΘΖ, ΘΖΕ parallel to either of BG or EF [Prop. 1.31], and let HB
γωνίαι δυσὶν ὀρθαῖς εἰσιν ἴσαι. αἱ ἄρα ὑπὸ ΒΘΗ, ΗΖΕ have been joined. And since the straight-line HF falls
δύο ὀρθῶν ἐλάσσονές εἰσιν· αἱ δὲ ἀπὸ ἐλασσόνων ἢ δύο across the parallels AH and EF , the (sum of the) an-
ὀρθῶν εἰς ἄπειρον ἐκβαλλόμεναι συμπίπτουσιν· αἱ ΘΒ, ΖΕ gles AHF and HFE is thus equal to two right-angles

STOIQEIWN aþ. ELEMENTS BOOK 1
ἄρα ἐκβαλλόμεναι συμπεσοῦνται. ἐκβεβλήσθωσαν καὶ συμ- [Prop. 1.29]. Thus, (the sum of) BHG and GFE is less
πιπτέτωσαν κατὰ τὸ Κ, καὶ διὰ τοῦ Κ σημείου ὁποτέρᾳ than two right-angles. And (straight-lines) produced to
τῶν ΕΑ, ΖΘ παράλληλος ἤχθω ἡ ΚΛ, καὶ ἐκβεβλήσθωσαν infinity from (internal angles whose sum is) less than two
αἱ ΘΑ, ΗΒ ἐπὶ τὰ Λ, Μ σημεῖα. παραλληλόγραμμον ἄρα right-angles meet together [Post. 5]. Thus, being pro-
ἐστὶ τὸ ΘΛΚΖ, διάμετρος δὲ αὐτοῦ ἡ ΘΚ, περὶ δὲ τὴν ΘΚ duced, HB and FE will meet together. Let them have
παραλληλόγραμμα μὲν τὰ ΑΗ, ΜΕ, τὰ δὲ λεγόμενα παρα- been produced, and let them meet together at K. And let
πληρώματα τὰ ΛΒ, ΒΖ· ἴσον ἄρα ἐστὶ τὸ ΛΒ τῷ ΒΖ. ἀλλὰ KL have been drawn through point K parallel to either
τὸ ΒΖ τῷ Γ τριγώνῳ ἐστὶν ἴσον· καὶ τὸ ΛΒ ἄρα τῷ Γ ἐστιν of EA or FH [Prop. 1.31]. And let HA and GB have
ἴσον. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΗΒΕ γωνία τῇ ὑπὸ ΑΒΜ, been produced to points L and M (respectively). Thus,
ἀλλὰ ἡ ὑπὸ ΗΒΕ τῇ Δ ἐστιν ἴση, καὶ ἡ ὑπὸ ΑΒΜ ἄρα τῇ Δ HLKF is a parallelogram, and HK its diagonal. And
γωνίᾳ ἐστὶν ἴση. AG and ME (are) parallelograms, and LB and BF the
Παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν τὴν ΑΒ τῷ δοθέντι so-called complements, about HK. Thus, LB is equal to

τριγώνῳ τῷ Γ ἴσον παραλληλόγραμμον παραβέβληται τὸ ΛΒ BF [Prop. 1.43]. But, BF is equal to triangle C. Thus,
ἐν γωνίᾳ τῇ ὑπὸ ΑΒΜ, ἥ ἐστιν ἴση τῇ Δ· ὅπερ ἔδει ποιῆσαι. LB is also equal to C. Also, since angle GBE is equal to

ABM [Prop. 1.15], but GBE is equal to D, ABM is thus

also equal to angle D.
Thus, the parallelogram LB, equal to the given trian-

gle C, has been applied to the given straight-line AB in

the angle ABM , which is equal to D. (Which is) the very
thing it was required to do.

† This can be achieved using Props. 1.3, 1.23, and 1.31.meþ. Proposition 45
Τῷ δοθέντι εὐθυγράμμῳ ἴσον παραλληλόγραμμον συστ- To construct a parallelogram equal to a given rectilin-

ήσασθαι ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ. ear figure in a given rectilinear angle.

῎Εστω τὸ μὲν δοθὲν εὐθύγραμμον τὸ ΑΒΓΔ, ἡ δὲ Let ABCD be the given rectilinear figure,† and E the
δοθεῖσα γωνία εὐθύγραμμος ἡ Ε· δεῖ δὴ τῷ ΑΒΓΔ εὐθυ- given rectilinear angle. So it is required to construct a
γράμμῳ ἴσον παραλληλόγραμμον συστήσασθαι ἐν τῇ δοθείσῃ parallelogram equal to the rectilinear figure ABCD in
γωνίᾳ τῇ Ε. the given angle E.
᾿Επεζεύχθω ἡ ΔΒ, καὶ συνεστάτω τῷ ΑΒΔ τριγώνῳ Let DB have been joined, and let the parallelogram

ἴσον παραλληλόγραμμον τὸ ΖΘ ἐν τῇ ὑπὸ ΘΚΖ γωνίᾳ, ἥ FH , equal to the triangle ABD, have been constructed
ἐστιν ἴση τῇ Ε· καὶ παραβεβλήσθω παρὰ τὴν ΗΘ εὐθεῖαν τῷ in the angle HKF , which is equal to E [Prop. 1.42]. And
ΔΒΓ τριγώνῳ ἴσον παραλληλόγραμμον τὸ ΗΜ ἐν τῇ ὑπὸ let the parallelogram GM , equal to the triangle DBC,
ΗΘΜ γωνίᾳ, ἥ ἐστιν ἴση τῇ Ε. καὶ ἐπεὶ ἡ Ε γωνία ἑκατέρᾳ have been applied to the straight-line GH in the angle
τῶν ὑπὸ ΘΚΖ, ΗΘΜ ἐστιν ἴση, καὶ ἡ ὑπὸ ΘΚΖ ἄρα τῇ ὑπὸ GHM , which is equal to E [Prop. 1.44]. And since angle
ΗΘΜ ἐστιν ἴση. κοινὴ προσκείσθω ἡ ὑπὸ ΚΘΗ· αἱ ἄρα E is equal to each of (angles) HKF and GHM , (an-
ὑπὸ ΖΚΘ, ΚΘΗ ταῖς ὑπὸ ΚΘΗ, ΗΘΜ ἴσαι εἰσίν. ἀλλ᾿ αἱ gle) HKF is thus also equal to GHM . Let KHG have
ὑπὸ ΖΚΘ, ΚΘΗ δυσὶν ὀρθαῖς ἴσαι εἰσίν· καὶ αἱ ὑπὸ ΚΘΗ, been added to both. Thus, (the sum of) FKH and KHG
ΗΘΜ ἄρα δύο ὀρθαῖς ἴσαι εἰσίν. πρὸς δή τινι εὐθεῖᾳ τῇ ΗΘ is equal to (the sum of) KHG and GHM . But, (the
καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ δύο εὐθεῖαι αἱ ΚΘ, ΘΜ μὴ sum of) FKH and KHG is equal to two right-angles
ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δύο ὀρθαῖς [Prop. 1.29]. Thus, (the sum of) KHG and GHM is
ἴσας ποιοῦσιν· ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΚΘ τῇ ΘΜ· καὶ also equal to two right-angles. So two straight-lines, KH
ἐπεὶ εἰς παραλλήλους τὰς ΚΜ, ΖΗ εὐθεῖα ἐνέπεσεν ἡ ΘΗ, and HM , not lying on the same side, make adjacent an-
αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΜΘΗ, ΘΗΖ ἴσαι ἀλλήλαις εἰσίν. gles with some straight-line GH , at the point H on it,
κοινὴ προσκείσθω ἡ ὑπὸ ΘΗΛ· αἱ ἄρα ὑπὸ ΜΘΗ, ΘΗΛ ταῖς (whose sum is) equal to two right-angles. Thus, KH is
ὑπὸ ΘΗΖ, ΘΗΛ ἴσαι εἰσιν. ἀλλ᾿ αἱ ὑπὸ ΜΘΗ, ΘΗΛ δύο straight-on to HM [Prop. 1.14]. And since the straight-
ὀρθαῖς ἴσαι εἰσίν· καὶ αἱ ὑπὸ ΘΗΖ, ΘΗΛ ἄρα δύο ὀρθαῖς line HG falls across the parallels KM and FG, the al-
ἴσαι εἰσίν· ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΛ. καὶ ἐπεὶ ἡ ternate angles MHG and HGF are equal to one another
ΖΚ τῇ ΘΗ ἴση τε καὶ παράλληλός ἐστιν, ἀλλὰ καὶ ἡ ΘΗ τῇ [Prop. 1.29]. Let HGL have been added to both. Thus,
ΜΛ, καὶ ἡ ΚΖ ἄρα τῇ ΜΛ ἴση τε καὶ παράλληλός ἐστιν· καὶ (the sum of) MHG and HGL is equal to (the sum of)

STOIQEIWN aþ. ELEMENTS BOOK 1
ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ΚΜ, ΖΛ· καὶ αἱ ΚΜ, ΖΛ HGF and HGL. But, (the sum of) MHG and HGL is
ἄρα ἴσαι τε καὶ παράλληλοί εἰσιν· παραλληλόγραμμον ἄρα equal to two right-angles [Prop. 1.29]. Thus, (the sum of)
ἐστὶ τὸ ΚΖΛΜ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ μὲν ΑΒΔ τρίγωνον HGF and HGL is also equal to two right-angles. Thus,
τῷ ΖΘ παραλληλογράμμῳ, τὸ δὲ ΔΒΓ τῷ ΗΜ, ὅλον ἄρα FG is straight-on to GL [Prop. 1.14]. And since FK is
τὸ ΑΒΓΔ εὐθύγραμμον ὅλῳ τῷ ΚΖΛΜ παραλληλογράμμῳ equal and parallel to HG [Prop. 1.34], but also HG to
ἐστὶν ἴσον. ML [Prop. 1.34], KF is thus also equal and parallel to

ML [Prop. 1.30]. And the straight-lines KM and FL

join them. Thus, KM and FL are equal and parallel as
well [Prop. 1.33]. Thus, KFLM is a parallelogram. And

since triangle ABD is equal to parallelogram FH , and

DBC to GM , the whole rectilinear figure ABCD is thus
equal to the whole parallelogram KFLM .

∆

Κ Θ Μ

Ζ F

K H

Τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓΔ ἴσον παραλ- Thus, the parallelogram KFLM , equal to the given
ληλόγραμμον συνέσταται τὸ ΚΖΛΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΚΜ, rectilinear figure ABCD, has been constructed in the an-
ἥ ἐστιν ἴση τῇ δοθείσῃ τῇ Ε· ὅπερ ἔδει ποιῆσαι. gle FKM , which is equal to the given (angle) E. (Which

is) the very thing it was required to do.

† The proof is only given for a four-sided figure. However, the extension to many-sided figures is trivial.m�þ. Proposition 46
Ἀπὸ τῆς δοθείσης εὐθείας τετράγωνον ἀναγράψαι. To describe a square on a given straight-line.
῎Εστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ· δεῖ δὴ ἀπὸ τῆς ΑΒ Let AB be the given straight-line. So it is required to

εὐθείας τετράγωνον ἀναγράψαι. describe a square on the straight-line AB.
῎Ηχθω τῇ ΑΒ εὐθείᾳ ἀπὸ τοῦ πρὸς αὐτῇ σημείου τοῦ Let AC have been drawn at right-angles to the

Α πρὸς ὀρθὰς ἡ ΑΓ, καὶ κείσθω τῇ ΑΒ ἴση ἡ ΑΔ· καὶ διὰ straight-line AB from the point A on it [Prop. 1.11],
μὲν τοῦ Δ σημείου τῇ ΑΒ παράλληλος ἤχθω ἡ ΔΕ, διὰ and let AD have been made equal to AB [Prop. 1.3].
δὲ τοῦ Β σημείου τῇ ΑΔ παράλληλος ἤχθω ἡ ΒΕ. παραλ- And let DE have been drawn through point D parallel to
ληλόγραμμον ἄρα ἐστὶ τὸ ΑΔΕΒ· ἴση ἄρα ἐστὶν ἡ μὲν ΑΒ AB [Prop. 1.31], and let BE have been drawn through
τῇ ΔΕ, ἡ δὲ ΑΔ τῇ ΒΕ. ἀλλὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση· point B parallel to AD [Prop. 1.31]. Thus, ADEB is a
αἱ τέσσαρες ἄρα αἱ ΒΑ, ΑΔ, ΔΕ, ΕΒ ἴσαι ἀλλήλαις εἰσίν· parallelogram. Therefore, AB is equal to DE, and AD to
ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΔΕΒ παραλληλόγραμμον. λέγω BE [Prop. 1.34]. But, AB is equal to AD. Thus, the four
δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ εἰς παραλλήλους τὰς ΑΒ, (sides) BA, AD, DE, and EB are equal to one another.
ΔΕ εὐθεῖα ἐνέπεσεν ἡ ΑΔ, αἱ ἄρα ὑπὸ ΒΑΔ, ΑΔΕ γωνίαι Thus, the parallelogram ADEB is equilateral. So I say
δύο ὀρθαῖς ἴσαι εἰσίν. ὀρθὴ δὲ ἡ ὑπὸ ΒΑΔ· ὀρθὴ ἄρα καὶ that (it is) also right-angled. For since the straight-line

STOIQEIWN aþ. ELEMENTS BOOK 1
ἡ ὑπὸ ΑΔΕ. τῶν δὲ παραλληλογράμμων χωρίων αἱ ἀπε- AD falls across the parallels AB and DE, the (sum of
ναντίον πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν· ὀρθὴ ἄρα the) angles BAD and ADE is equal to two right-angles
καὶ ἑκατέρα τῶν ἀπεναντίον τῶν ὑπὸ ΑΒΕ, ΒΕΔ γωνιῶν· [Prop. 1.29]. But BAD (is a) right-angle. Thus, ADE
ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΔΕΒ. ἐδείχθη δὲ καὶ ἰσόπλευρον. (is) also a right-angle. And for parallelogrammic figures,

the opposite sides and angles are equal to one another
[Prop. 1.34]. Thus, each of the opposite angles ABE

and BED (are) also right-angles. Thus, ADEB is right-

angled. And it was also shown (to be) equilateral.

ΒΑ

∆

Ε D

A B

Τετράγωνον ἄρα ἐστίν· καί ἐστιν ἀπὸ τῆς ΑΒ εὐθείας Thus, (ADEB) is a square [Def. 1.22]. And it is de-
ἀναγεγραμμένον· ὅπερ ἔδει ποιῆσαι. scribed on the straight-line AB. (Which is) the very thing

it was required to do.mzþ. Proposition 47
᾿Εν τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν In right-angled triangles, the square on the side sub-

γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἴσον ἐστὶ τοῖς tending the right-angle is equal to the (sum of the)
ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν τε- squares on the sides containing the right-angle.
τραγώνοις. Let ABC be a right-angled triangle having the angle
῎Εστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν BACa right-angle. I say that the square on BC is equal

ὑπὸ ΒΑΓ γωνίαν· λέγω, ὅτι τὸ ἀπὸ τῆς ΒΓ τετράγωνον to the (sum of the) squares on BA and AC.
ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ τετραγώνοις. For let the square BDEC have been described on
Ἀναγεγράφθω γὰρ ἀπὸ μὲν τῆς ΒΓ τετράγωνον τὸ BC, and (the squares) GB and HC on AB and AC

ΒΔΕΓ, ἀπὸ δὲ τῶν ΒΑ, ΑΓ τὰ ΗΒ, ΘΓ, καὶ διὰ τοῦ (respectively) [Prop. 1.46]. And let AL have been
Α ὁποτέρᾳ τῶν ΒΔ, ΓΕ παράλληλος ἤχθω ἡ ΑΛ· καὶ drawn through point A parallel to either of BD or CE
ἐπεζεύχθωσαν αἱ ΑΔ, ΖΓ. καὶ ἐπεὶ ὀρθή ἐστιν ἑκατέρα [Prop. 1.31]. And let AD and FC have been joined. And
τῶν ὑπὸ ΒΑΓ, ΒΑΗ γωνιῶν, πρὸς δή τινι εὐθείᾳ τῇ ΒΑ since angles BAC and BAG are each right-angles, then
καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α δύο εὐθεῖαι αἱ ΑΓ, ΑΗ μὴ two straight-lines AC and AG, not lying on the same
ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δυσὶν ὀρθαῖς side, make the adjacent angles with some straight-line
ἴσας ποιοῦσιν· ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΓΑ τῇ ΑΗ. διὰ τὰ BA, at the point A on it, (whose sum is) equal to two
αὐτὰ δὴ καὶ ἡ ΒΑ τῇ ΑΘ ἐστιν ἐπ᾿ εὐθείας. καὶ ἐπεὶ ἴση right-angles. Thus, CA is straight-on to AG [Prop. 1.14].
ἐστὶν ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΖΒΑ· ὀρθὴ γὰρ ἑκατέρα· So, for the same (reasons), BA is also straight-on to AH .
κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ· ὅλη ἄρα ἡ ὑπὸ ΔΒΑ ὅλῃ τῇ And since angle DBC is equal to FBA, for (they are)
ὑπὸ ΖΒΓ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΔΒ τῇ ΒΓ, ἡ both right-angles, let ABC have been added to both.
δὲ ΖΒ τῇ ΒΑ, δύο δὴ αἱ ΔΒ, ΒΑ δύο ταῖς ΖΒ, ΒΓ ἴσαι εἰσὶν Thus, the whole (angle) DBA is equal to the whole (an-
ἑκατέρα ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΔΒΑ γωνίᾳ τῇ ὑπὸ ΖΒΓ gle) FBC. And since DB is equal to BC, and FB to
ἴση· βάσις ἄρα ἡ ΑΔ βάσει τῇ ΖΓ [ἐστιν] ἴση, καὶ τὸ ΑΒΔ BA, the two (straight-lines) DB, BA are equal to the

STOIQEIWN aþ. ELEMENTS BOOK 1
τρίγωνον τῷ ΖΒΓ τριγώνῳ ἐστὶν ἴσον· καί [ἐστι] τοῦ μὲν two (straight-lines) CB, BF ,† respectively. And angle
ΑΒΔ τριγώνου διπλάσιον τὸ ΒΛ παραλληλόγραμμον· βάσιν DBA (is) equal to angle FBC. Thus, the base AD [is]
τε γὰρ τὴν αὐτὴν ἔχουσι τὴν ΒΔ καὶ ἐν ταῖς αὐταῖς εἰσι equal to the base FC, and the triangle ABD is equal to
παραλλήλοις ταῖς ΒΔ, ΑΛ· τοῦ δὲ ΖΒΓ τριγώνου διπλάσιον the triangle FBC [Prop. 1.4]. And parallelogram BL
τὸ ΗΒ τετράγωνον· βάσιν τε γὰρ πάλιν τὴν αὐτὴν ἔχουσι [is] double (the area) of triangle ABD. For they have
τὴν ΖΒ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ταῖς ΖΒ, ΗΓ. the same base, BD, and are between the same parallels,
[τὰ δὲ τῶν ἴσων διπλάσια ἴσα ἀλλήλοις ἐστίν·] ἴσον ἄρα ἐστὶ BD and AL [Prop. 1.41]. And square GB is double (the
καὶ τὸ ΒΛ παραλληλόγραμμον τῷ ΗΒ τετραγώνῳ. ὁμοίως area) of triangle FBC. For again they have the same
δὴ ἐπιζευγνυμένων τῶν ΑΕ, ΒΚ δειχθήσεται καὶ τὸ ΓΛ base, FB, and are between the same parallels, FB and
παραλληλόγραμμον ἴσον τῷ ΘΓ τετραγώνῳ· ὅλον ἄρα τὸ GC [Prop. 1.41]. [And the doubles of equal things are

ΒΔΕΓ τετράγωνον δυσὶ τοῖς ΗΒ, ΘΓ τετραγώνοις ἴσον equal to one another.]‡ Thus, the parallelogram BL is
ἐστίν. καί ἐστι τὸ μὲν ΒΔΕΓ τετράγωνον ἀπὸ τῆς ΒΓ ἀνα- also equal to the square GB. So, similarly, AE and BK
γραφέν, τὰ δὲ ΗΒ, ΘΓ ἀπὸ τῶν ΒΑ, ΑΓ. τὸ ἄρα ἀπὸ τῆς being joined, the parallelogram CL can be shown (to
ΒΓ πλευρᾶς τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ be) equal to the square HC. Thus, the whole square
πλευρῶν τετραγώνοις. BDEC is equal to the (sum of the) two squares GB and

HC. And the square BDEC is described on BC, and
the (squares) GB and HC on BA and AC (respectively).

Thus, the square on the side BC is equal to the (sum of

the) squares on the sides BA and AC.

∆ Λ Ε

D L E

᾿Εν ἄρα τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν Thus, in right-angled triangles, the square on the
ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἴσον ἐστὶ side subtending the right-angle is equal to the (sum of
τοῖς ἀπὸ τῶν τὴν ὀρθὴν [γωνίαν] περιεχουσῶν πλευρῶν τε- the) squares on the sides surrounding the right-[angle].
τραγώνοις· ὅπερ ἔδει δεῖξαι. (Which is) the very thing it was required to show.

† The Greek text has “FB, BC”, which is obviously a mistake.

‡ This is an additional common notion.

STOIQEIWN aþ. ELEMENTS BOOK 1mhþ. Proposition 48
᾿Εὰν τριγώνου τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον If the square on one of the sides of a triangle is equal

ἴσον ᾖ τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν to the (sum of the) squares on the two remaining sides of
τετραγώνοις, ἡ περιεχομένη γωνία ὑπὸ τῶν λοιπῶν τοῦ the triangle then the angle contained by the two remain-
τριγώνου δύο πλευρῶν ὀρθή ἐστιν. ing sides of the triangle is a right-angle.

Α Β

∆ A BD

Τριγώνου γὰρ τοῦ ΑΒΓ τὸ ἀπὸ μιᾶς τῆς ΒΓ πλευρᾶς For let the square on one of the sides, BC, of triangle
τετράγωνον ἴσον ἔστω τοῖς ἀπὸ τῶν ΒΑ, ΑΓ πλευρῶν τε- ABC be equal to the (sum of the) squares on the sides
τραγώνοις· λέγω, ὅτι ὀρθή ἐστιν ἡ ὑπὸ ΒΑΓ γωνία. BA and AC. I say that angle BAC is a right-angle.
῎Ηχθω γὰρ ἀπὸ τοῦ Α σημείου τῇ ΑΓ εὐθείᾳ πρὸς ὀρθὰς For let AD have been drawn from point A at right-

ἡ ΑΔ καὶ κείσθω τῇ ΒΑ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΓ. angles to the straight-line AC [Prop. 1.11], and let AD
ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς have been made equal to BA [Prop. 1.3], and let DC
ΔΑ τετράγωνον τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. κοινὸν προ- have been joined. Since DA is equal to AB, the square

σκείσθω τὸ ἀπὸ τῆς ΑΓ τετράγωνον· τὰ ἄρα ἀπὸ τῶν ΔΑ, on DA is thus also equal to the square on AB.† Let the
ΑΓ τετράγωνα ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ τετραγώνοις. square on AC have been added to both. Thus, the (sum
ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΔΑ, ΑΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΔΓ· of the) squares on DA and AC is equal to the (sum of
ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΔΑΓ γωνία· τοῖς δὲ ἀπὸ τῶν ΒΑ, the) squares on BA and AC. But, the (square) on DC is
ΑΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΒΓ· ὑπόκειται γάρ· τὸ ἄρα ἀπὸ equal to the (sum of the squares) on DA and AC. For an-
τῆς ΔΓ τετράγωνον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ· gle DAC is a right-angle [Prop. 1.47]. But, the (square)
ὥστε καὶ πλευρὰ ἡ ΔΓ τῇ ΒΓ ἐστιν ἴση· καὶ ἐπεὶ ἴση ἐστὶν on BC is equal to (sum of the squares) on BA and AC.
ἡ ΔΑ τῇ ΑΒ, κοινὴ δὲ ἡ ΑΓ, δύο δὴ αἱ ΔΑ, ΑΓ δύο ταῖς For (that) was assumed. Thus, the square on DC is equal
ΒΑ, ΑΓ ἴσαι εἰσίν· καὶ βάσις ἡ ΔΓ βάσει τῇ ΒΓ ἴση· γωνία to the square on BC. So side DC is also equal to (side)
ἄρα ἡ ὑπὸ ΔΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ [ἐστιν] ἴση. ὀρθὴ δὲ ἡ BC. And since DA is equal to AB, and AC (is) com-
ὑπὸ ΔΑΓ· ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΑΓ. mon, the two (straight-lines) DA, AC are equal to the
᾿Εὰν ἀρὰ τριγώνου τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον two (straight-lines) BA, AC. And the base DC is equal

ἴσον ᾖ τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν to the base BC. Thus, angle DAC [is] equal to angle
τετραγώνοις, ἡ περιεχομένη γωνία ὑπὸ τῶν λοιπῶν τοῦ BAC [Prop. 1.8]. But DAC is a right-angle. Thus, BAC
τριγώνου δύο πλευρῶν ὀρθή ἐστιν· ὅπερ ἔδει δεῖξαι. is also a right-angle.

Thus, if the square on one of the sides of a triangle is

equal to the (sum of the) squares on the remaining two
sides of the triangle then the angle contained by the re-

maining two sides of the triangle is a right-angle. (Which

is) the very thing it was required to show.

† Here, use is made of the additional common notion that the squares of equal things are themselves equal. Later on, the inverse notion is used.

ELEMENTS BOOK 2

Fundamentals of Geometric Algebra

STOIQEIWN bþ. ELEMENTS BOOK 2VOroi. Definitions
αʹ. Πᾶν παραλληλόγραμμον ὀρθογώνιον περιέχεσθαι 1. Any rectangular parallelogram is said to be con-

λέγεται ὑπὸ δύο τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν tained by the two straight-lines containing the right-
εὐθειῶν. angle.
βʹ. Παντὸς δὲ παραλληλογράμμου χωρίου τῶν περὶ τὴν 2. And in any parallelogrammic figure, let any one

διάμετρον αὐτοῦ παραλληλογράμμων ἓν ὁποιονοῦν σὺν τοῖς whatsoever of the parallelograms about its diagonal,
δυσὶ παραπληρώμασι γνώμων καλείσθω. (taken) with its two complements, be called a gnomon.aþ. Proposition 1†
᾿Εὰν ὦσι δύο εὐθεῖαι, τμηθῇ δὲ ἡ ἑτέρα αὐτῶν εἰς ὁσα- If there are two straight-lines, and one of them is cut

δηποτοῦν τμήματα, τὸ περιεχόμενον ὀρθογώνιον ὑπὸ τῶν into any number of pieces whatsoever, then the rectangle
δύο εὐθειῶν ἴσον ἐστὶ τοῖς ὑπό τε τῆς ἀτμήτου καὶ ἑκάστου contained by the two straight-lines is equal to the (sum
τῶν τμημάτων περιεχομένοις ὀρθογωνίοις. of the) rectangles contained by the uncut (straight-line),

and every one of the pieces (of the cut straight-line).

Β ∆

B D E C

F
K L

῎Εστωσαν δύο εὐθεῖαι αἱ Α, ΒΓ, καὶ τετμήσθω ἡ ΒΓ, Let A and BC be the two straight-lines, and let BC
ὡς ἔτυχεν, κατὰ τὰ Δ, Ε σημεῖα· λέγω, ὅτι τὸ ὑπὸ τῶν Α, be cut, at random, at points D and E. I say that the rect-
ΒΓ περιεχομένον ὀρθογώνιον ἴσον ἐστὶ τῷ τε ὑπὸ τῶν Α, angle contained by A and BC is equal to the rectangle(s)
ΒΔ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ὑπὸ τῶν Α, ΔΕ καὶ ἔτι contained by A and BD, by A and DE, and, finally, by A
τῷ ὑπὸ τῶν Α, ΕΓ. and EC.
῎Ηχθω γὰρ ἀπὸ τοῦ Β τῇ ΒΓ πρὸς ὀρθὰς ἡ ΒΖ, καὶ For let BF have been drawn from point B, at right-

κείσθω τῇ Α ἴση ἡ ΒΗ, καὶ διὰ μὲν τοῦ Η τῇ ΒΓ παράλληλος angles to BC [Prop. 1.11], and let BG be made equal
ἤχθω ἡ ΗΘ, διὰ δὲ τῶν Δ, Ε, Γ τῇ ΒΗ παράλληλοι ἤχθωσαν to A [Prop. 1.3], and let GH have been drawn through
αἱ ΔΚ, ΕΛ, ΓΘ. (point) G, parallel to BC [Prop. 1.31], and let DK, EL,
῎Ισον δή ἐστι τὸ ΒΘ τοῖς ΒΚ, ΔΛ, ΕΘ. καί ἐστι τὸ and CH have been drawn through (points) D, E, and C

μὲν ΒΘ τὸ ὑπὸ τῶν Α, ΒΓ· περιέχεται μὲν γὰρ ὑπὸ τῶν (respectively), parallel to BG [Prop. 1.31].
ΗΒ, ΒΓ, ἴση δὲ ἡ ΒΗ τῇ Α· τὸ δὲ ΒΚ τὸ ὑπὸ τῶν Α, ΒΔ· So the (rectangle) BH is equal to the (rectangles)
περιέχεται μὲν γὰρ ὑπὸ τῶν ΗΒ, ΒΔ, ἴση δὲ ἡ ΒΗ τῇ Α. τὸ BK, DL, and EH . And BH is the (rectangle contained)
δὲ ΔΛ τὸ ὑπὸ τῶν Α, ΔΕ· ἴση γὰρ ἡ ΔΚ, τουτέστιν ἡ ΒΗ, by A and BC. For it is contained by GB and BC, and BG
τῇ Α. καὶ ἔτι ὁμοίως τὸ ΕΘ τὸ ὑπὸ τῶν Α, ΕΓ· τὸ ἄρα ὑπὸ (is) equal to A. And BK (is) the (rectangle contained) by
τῶν Α, ΒΓ ἴσον ἐστὶ τῷ τε ὑπὸ Α, ΒΔ καὶ τῷ ὑπὸ Α, ΔΕ A and BD. For it is contained by GB and BD, and BG
καὶ ἔτι τῷ ὑπὸ Α, ΕΓ. (is) equal to A. And DL (is) the (rectangle contained) by
᾿Εὰν ἄρα ὦσι δύο εὐθεῖαι, τμηθῇ δὲ ἡ ἑτέρα αὐτῶν εἰς A and DE. For DK, that is to say BG [Prop. 1.34], (is)

ὁσαδηποτοῦν τμήματα, τὸ περιεχόμενον ὀρθογώνιον ὑπὸ equal to A. Similarly, EH (is) also the (rectangle con-
τῶν δύο εὐθειῶν ἴσον ἐστὶ τοῖς ὑπό τε τῆς ἀτμήτου καὶ tained) by A and EC. Thus, the (rectangle contained)
ἑκάστου τῶν τμημάτων περιεχομένοις ὀρθογωνίοις· ὅπερ by A and BC is equal to the (rectangles contained) by A

STOIQEIWN bþ. ELEMENTS BOOK 2
ἔδει δεῖξαι. and BD, by A and DE, and, finally, by A and EC.

Thus, if there are two straight-lines, and one of them

is cut into any number of pieces whatsoever, then the
rectangle contained by the two straight-lines is equal

to the (sum of the) rectangles contained by the uncut
(straight-line), and every one of the pieces (of the cut

straight-line). (Which is) the very thing it was required

to show.

† This proposition is a geometric version of the algebraic identity: a (b + c + d + · · · ) = a b + a c + a d + · · · .bþ. Proposition 2†
᾿Εὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς ὅλης If a straight-line is cut at random then the (sum of

καὶ ἑκατέρου τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον the) rectangle(s) contained by the whole (straight-line),
ἐστὶ τῷ ἀπὸ τῆς ὅλης τετραγώνῳ. and each of the pieces (of the straight-line), is equal to

the square on the whole.

ΓΑ

∆

ΕΖ

BA C

D F E
Εὐθεῖα γὰρ ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ For let the straight-line AB have been cut, at random,

σημεῖον· λέγω, ὅτι τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον at point C. I say that the rectangle contained by AB and
ὀρθογώνιον μετὰ τοῦ ὑπὸ ΒΑ, ΑΓ περιεχομένου ὀρθο- BC, plus the rectangle contained by BA and AC, is equal
γωνίου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. to the square on AB.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ, For let the square ADEB have been described on AB

καὶ ἤχθω διὰ τοῦ Γ ὁποτέρᾳ τῶν ΑΔ, ΒΕ παράλληλος ἡ [Prop. 1.46], and let CF have been drawn through C,
ΓΖ. parallel to either of AD or BE [Prop. 1.31].
῎Ισον δή ἐστὶ τὸ ΑΕ τοῖς ΑΖ, ΓΕ. καί ἐστι τὸ μὲν ΑΕ So the (square) AE is equal to the (rectangles) AF

τὸ ἀπὸ τῆς ΑΒ τετράγωνον, τὸ δὲ ΑΖ τὸ ὑπὸ τῶν ΒΑ, and CE. And AE is the square on AB. And AF (is) the
ΑΓ περιεχόμενον ὀρθογώνιον· περιέχεται μὲν γὰρ ὑπὸ τῶν rectangle contained by the (straight-lines) BA and AC.
ΔΑ, ΑΓ, ἴση δὲ ἡ ΑΔ τῇ ΑΒ· τὸ δὲ ΓΕ τὸ ὑπὸ τῶν ΑΒ, For it is contained by DA and AC, and AD (is) equal to
ΒΓ· ἴση γὰρ ἡ ΒΕ τῇ ΑΒ. τὸ ἄρα ὑπὸ τῶν ΒΑ, ΑΓ μετὰ AB. And CE (is) the (rectangle contained) by AB and
τοῦ ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. BC. For BE (is) equal to AB. Thus, the (rectangle con-
᾿Εὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς tained) by BA and AC, plus the (rectangle contained) by

ὅλης καὶ ἑκατέρου τῶν τμημάτων περιεχόμενον ὀρθογώνιον AB and BC, is equal to the square on AB.
ἴσον ἐστὶ τῷ ἀπὸ τῆς ὅλης τετραγώνῳ· ὅπερ ἔδει δεῖξαι. Thus, if a straight-line is cut at random then the (sum

of the) rectangle(s) contained by the whole (straight-

line), and each of the pieces (of the straight-line), is equal
to the square on the whole. (Which is) the very thing it

was required to show.

STOIQEIWN bþ. ELEMENTS BOOK 2
† This proposition is a geometric version of the algebraic identity: a b + a c = a2 if a = b + c.gþ. Proposition 3†
᾿Εὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς ὅλης If a straight-line is cut at random then the rectangle

καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ contained by the whole (straight-line), and one of the
τῷ τε ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ καὶ τῷ pieces (of the straight-line), is equal to the rectangle con-
ἀπὸ τοῦ προειρημένου τμήματος τετραγώνῳ. tained by (both of) the pieces, and the square on the

aforementioned piece.

ΕΖ

Α Γ

∆

Β BA C

F D E
Εὐθεῖα γὰρ ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ· For let the straight-line AB have been cut, at random,

λέγω, ὅτι τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον at (point) C. I say that the rectangle contained by AB
ἴσον ἐστὶ τῷ τε ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ and BC is equal to the rectangle contained by AC and
μετὰ τοῦ ἀπὸ τῆς ΒΓ τετραγώνου. CB, plus the square on BC.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΓΒ τετράγωνον τὸ ΓΔΕΒ, For let the square CDEB have been described on CB

καὶ διήχθω ἡ ΕΔ ἐπὶ τὸ Ζ, καὶ διὰ τοῦ Α ὁποτέρᾳ τῶν ΓΔ, [Prop. 1.46], and let ED have been drawn through to
ΒΕ παράλληλος ἤχθω ἡ ΑΖ. ἴσον δή ἐστι τὸ ΑΕ τοῖς ΑΔ, F , and let AF have been drawn through A, parallel to
ΓΕ· καί ἐστι τὸ μὲν ΑΕ τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον either of CD or BE [Prop. 1.31]. So the (rectangle) AE
ὀρθογώνιον· περιέχεται μὲν γὰρ ὑπὸ τῶν ΑΒ, ΒΕ, ἴση δὲ ἡ is equal to the (rectangle) AD and the (square) CE. And
ΒΕ τῇ ΒΓ· τὸ δὲ ΑΔ τὸ ὑπὸ τῶν ΑΓ, ΓΒ· ἴση γὰρ ἡ ΔΓ AE is the rectangle contained by AB and BC. For it is
τῇ ΓΒ· τὸ δὲ ΔΒ τὸ ἀπὸ τῆς ΓΒ τετράγωνον· τὸ ἄρα ὑπὸ contained by AB and BE, and BE (is) equal to BC. And
τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ AD (is) the (rectangle contained) by AC and CB. For
τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ μετὰ τοῦ ἀπὸ τῆς ΒΓ DC (is) equal to CB. And DB (is) the square on CB.
τετραγώνου. Thus, the rectangle contained by AB and BC is equal to
᾿Εὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς the rectangle contained by AC and CB, plus the square

ὅλης καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον on BC.
ἐστὶ τῷ τε ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ καὶ Thus, if a straight-line is cut at random then the rect-
τῷ ἀπὸ τοῦ προειρημένου τμήματος τετραγώνῳ· ὅπερ ἔδει angle contained by the whole (straight-line), and one of
δεῖξαι. the pieces (of the straight-line), is equal to the rectangle

contained by (both of) the pieces, and the square on the
aforementioned piece. (Which is) the very thing it was

required to show.

† This proposition is a geometric version of the algebraic identity: (a + b) a = a b + a2.dþ. Proposition 4†
᾿Εὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς If a straight-line is cut at random then the square

ὅλης τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν τμημάτων τε- on the whole (straight-line) is equal to the (sum of the)
τραγώνοις καὶ τῷ δὶς ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθο- squares on the pieces (of the straight-line), and twice the

STOIQEIWN bþ. ELEMENTS BOOK 2
γωνίῳ. rectangle contained by the pieces.

∆

Κ
Η

A C B

G
K

D F E

Εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ For let the straight-line AB have been cut, at random,
τὸ Γ. λέγω, ὅτι τὸ ἀπὸ τῆς ΑΒ τετράγωνον ἴσον ἐστὶ τοῖς at (point) C. I say that the square on AB is equal to
τε ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, the (sum of the) squares on AC and CB, and twice the
ΓΒ περιεχομένῳ ὀρθογωνίῳ. rectangle contained by AC and CB.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ, For let the square ADEB have been described on AB

καὶ ἐπεζεύχθω ἡ ΒΔ, καὶ διὰ μὲν τοῦ Γ ὁποτέρᾳ τῶν ΑΔ, [Prop. 1.46], and let BD have been joined, and let CF
ΕΒ παράλληλος ἤχθω ἡ ΓΖ, διὰ δὲ τοῦ Η ὁποτέρᾳ τῶν ΑΒ, have been drawn through C, parallel to either of AD or
ΔΕ παράλληλος ἤχθω ἡ ΘΚ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ EB [Prop. 1.31], and let HK have been drawn through
ΓΖ τῇ ΑΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΔ, ἡ ἐκτὸς γωνία G, parallel to either of AB or DE [Prop. 1.31]. And since
ἡ ὑπὸ ΓΗΒ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΔΒ. CF is parallel to AD, and BD has fallen across them, the
ἀλλ᾿ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση, ἐπεὶ καὶ πλευρὰ ἡ external angle CGB is equal to the internal and opposite
ΒΑ τῇ ΑΔ ἐστιν ἴση· καὶ ἡ ὑπὸ ΓΗΒ ἄρα γωνιά τῇ ὑπὸ ΗΒΓ (angle) ADB [Prop. 1.29]. But, ADB is equal to ABD,
ἐστιν ἴση· ὥστε καὶ πλευρὰ ἡ ΒΓ πλευρᾷ τῇ ΓΗ ἐστιν ἴση· since the side BA is also equal to AD [Prop. 1.5]. Thus,
ἀλλ᾿ ἡ μὲν ΓΒ τῇ ΗΚ ἐστιν ἴση. ἡ δὲ ΓΗ τῇ ΚΒ· καὶ ἡ ΗΚ angle CGB is also equal to GBC. So the side BC is
ἄρα τῇ ΚΒ ἐστιν ἴση· ἰσόπλευρον ἄρα ἐστὶ τὸ ΓΗΚΒ. λέγω equal to the side CG [Prop. 1.6]. But, CB is equal to
δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ παράλληλός ἐστιν ἡ ΓΗ GK, and CG to KB [Prop. 1.34]. Thus, GK is also equal
τῇ ΒΚ [καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΓΒ], αἱ ἄρα ὑπὸ to KB. Thus, CGKB is equilateral. So I say that (it is)
ΚΒΓ, ΗΓΒ γωνίαι δύο ὀρθαῖς εἰσιν ἴσαι. ὀρθὴ δὲ ἡ ὑπὸ also right-angled. For since CG is parallel to BK [and the
ΚΒΓ· ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΓΗ· ὥστε καὶ αἱ ἀπεναντίον straight-line CB has fallen across them], the angles KBC
αἱ ὑπὸ ΓΗΚ, ΗΚΒ ὀρθαί εἰσιν. ὀρθογώνιον ἄρα ἐστὶ τὸ and GCB are thus equal to two right-angles [Prop. 1.29].
ΓΗΚΒ· ἐδείχθη δὲ καὶ ἰσόπλευρον· τετράγωνον ἄρα ἐστίν· But KBC (is) a right-angle. Thus, BCG (is) also a right-
καί ἐστιν ἀπὸ τῆς ΓΒ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΘΖ τετράγωνόν angle. So the opposite (angles) CGK and GKB are also
ἐστιν· καί ἐστιν ἀπὸ τῆς ΘΗ, τουτέστιν [ἀπὸ] τῆς ΑΓ· τὰ right-angles [Prop. 1.34]. Thus, CGKB is right-angled.
ἄρα ΘΖ, ΚΓ τετράγωνα ἀπὸ τῶν ΑΓ, ΓΒ εἰσιν. καὶ ἐπεὶ And it was also shown (to be) equilateral. Thus, it is a
ἴσον ἐστὶ τὸ ΑΗ τῷ ΗΕ, καί ἐστι τὸ ΑΗ τὸ ὑπὸ τῶν ΑΓ, square. And it is on CB. So, for the same (reasons),
ΓΒ· ἴση γὰρ ἡ ΗΓ τῇ ΓΒ· καὶ τὸ ΗΕ ἄρα ἴσον ἐστὶ τῷ HF is also a square. And it is on HG, that is to say [on]
ὑπὸ ΑΓ, ΓΒ· τὰ ἄρα ΑΗ, ΗΕ ἴσα ἐστὶ τῷ δὶς ὑπὸ τῶν AC [Prop. 1.34]. Thus, the squares HF and KC are
ΑΓ, ΓΒ. ἔστι δὲ καὶ τὰ ΘΖ, ΓΚ τετράγωνα ἀπὸ τῶν ΑΓ, on AC and CB (respectively). And the (rectangle) AG
ΓΒ· τὰ ἄρα τέσσαρα τὰ ΘΖ, ΓΚ, ΑΗ, ΗΕ ἴσα ἐστὶ τοῖς τε is equal to the (rectangle) GE [Prop. 1.43]. And AG is
ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ the (rectangle contained) by AC and CB. For GC (is)
περιεχομένῳ ὀρθογωνίῳ. ἀλλὰ τὰ ΘΖ, ΓΚ, ΑΗ, ΗΕ ὅλον equal to CB. Thus, GE is also equal to the (rectangle
ἐστὶ τὸ ΑΔΕΒ, ὅ ἐστιν ἀπὸ τῆς ΑΒ τετράγωνον· τὸ ἄρα contained) by AC and CB. Thus, the (rectangles) AG
ἀπὸ τῆς ΑΒ τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΑΓ, and GE are equal to twice the (rectangle contained) by
ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ AC and CB. And HF and CK are the squares on AC
ὀρθογωνίῳ. and CB (respectively). Thus, the four (figures) HF , CK,
᾿Εὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς AG, and GE are equal to the (sum of the) squares on

STOIQEIWN bþ. ELEMENTS BOOK 2
ὅλης τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν τμημάτων τε- AC and BC, and twice the rectangle contained by AC
τραγώνοις καὶ τῷ δὶς ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθο- and CB. But, the (figures) HF , CK, AG, and GE are
γωνίῳ· ὅπερ ἔδει δεῖξαι. (equivalent to) the whole of ADEB, which is the square

on AB. Thus, the square on AB is equal to the (sum

of the) squares on AC and CB, and twice the rectangle
contained by AC and CB.

Thus, if a straight-line is cut at random then the

square on the whole (straight-line) is equal to the (sum
of the) squares on the pieces (of the straight-line), and

twice the rectangle contained by the pieces. (Which is)

the very thing it was required to show.

† This proposition is a geometric version of the algebraic identity: (a + b)2 = a2 + b2 + 2 a b.eþ. Proposition 5‡
᾿Εὰν εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὸ ὑπὸ τῶν If a straight-line is cut into equal and unequal (pieces)

ἀνίσων τῆς ὅλης τμημάτων περιεχόμενον ὀρθογώνιον μετὰ then the rectangle contained by the unequal pieces of the
τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν τετραγώνου ἴσον ἐστὶ τῷ whole (straight-line), plus the square on the (difference)
ἀπὸ τῆς ἡμισείας τετραγώνῳ. between the (equal and unequal) pieces, is equal to the

square on half (of the straight-line).

Α Γ

∆

Η Ζ

A BC D

E G

K
N

P
L

Εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω εἰς μὲν ἴσα κατὰ τὸ For let any straight-line AB have been cut—equally at
Γ, εἰς δὲ ἄνισα κατὰ τὸ Δ· λέγω, ὅτι τὸ ὑπὸ τῶν ΑΔ, ΔΒ C, and unequally at D. I say that the rectangle contained
περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΔ τετραγώνου by AD and DB, plus the square on CD, is equal to the
ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΒ τετραγώνῳ. square on CB.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΓΒ τετράγωνον τὸ ΓΕΖΒ, For let the square CEFB have been described on CB

καὶ ἐπεζεύχθω ἡ ΒΕ, καὶ διὰ μὲν τοῦ Δ ὁποτέρᾳ τῶν ΓΕ, [Prop. 1.46], and let BE have been joined, and let DG
ΒΖ παράλληλος ἤχθω ἡ ΔΗ, διὰ δὲ τοῦ Θ ὁποτέρᾳ τῶν have been drawn through D, parallel to either of CE or
ΑΒ, ΕΖ παράλληλος πάλιν ἤχθω ἡ ΚΜ, καὶ πάλιν διὰ τοῦ Α BF [Prop. 1.31], and again let KM have been drawn
ὁποτέρᾳ τῶν ΓΛ, ΒΜ παράλληλος ἤχθω ἡ ΑΚ. καὶ ἐπεὶ ἴσον through H , parallel to either of AB or EF [Prop. 1.31],
ἐστὶ τὸ ΓΘ παραπλήρωμα τῷ ΘΖ παραπληρώματι, κοινὸν and again let AK have been drawn through A, parallel to
προσκείσθω τὸ ΔΜ· ὅλον ἄρα τὸ ΓΜ ὅλῳ τῷ ΔΖ ἴσον either of CL or BM [Prop. 1.31]. And since the comple-
ἐστίν. ἀλλὰ τὸ ΓΜ τῷ ΑΛ ἴσον ἐστίν, ἐπεὶ καὶ ἡ ΑΓ τῇ ment CH is equal to the complement HF [Prop. 1.43],
ΓΒ ἐστιν ἴση· καὶ τὸ ΑΛ ἄρα τῷ ΔΖ ἴσον ἐστίν. κοινὸν let the (square) DM have been added to both. Thus,

προσκείσθω τὸ ΓΘ· ὅλον ἄρα τὸ ΑΘ τῷ ΜΝΞ† γνώμονι the whole (rectangle) CM is equal to the whole (rect-
ἴσον ἐστίν. ἀλλὰ τὸ ΑΘ τὸ ὑπὸ τῶν ΑΔ, ΔΒ ἐστιν· ἴση angle) DF . But, (rectangle) CM is equal to (rectangle)
γὰρ ἡ ΔΘ τῇ ΔΒ· καὶ ὁ ΜΝΞ ἄρα γνώμων ἴσος ἐστὶ τῷ AL, since AC is also equal to CB [Prop. 1.36]. Thus,
ὑπὸ ΑΔ, ΔΒ. κοινὸν προσκείσθω τὸ ΛΗ, ὅ ἐστιν ἴσον τῷ (rectangle) AL is also equal to (rectangle) DF . Let (rect-
ἀπὸ τῆς ΓΔ· ὁ ἄρα ΜΝΞ γνώμων καὶ τὸ ΛΗ ἴσα ἐστὶ τῷ angle) CH have been added to both. Thus, the whole
ὑπὸ τῶν ΑΔ, ΔΒ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς (rectangle) AH is equal to the gnomon NOP . But, AH

STOIQEIWN bþ. ELEMENTS BOOK 2
ΓΔ τετραγώνῳ. ἀλλὰ ὁ ΜΝΞ γνώμων καὶ τὸ ΛΗ ὅλον ἐστὶ is the (rectangle contained) by AD and DB. For DH
τὸ ΓΕΖΒ τετράγωνον, ὅ ἐστιν ἀπὸ τῆς ΓΒ· τὸ ἄρα ὑπὸ τῶν (is) equal to DB. Thus, the gnomon NOP is also equal
ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΔ to the (rectangle contained) by AD and DB. Let LG,
τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΒ τετραγώνῳ. which is equal to the (square) on CD, have been added to
᾿Εὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὸ ὑπὸ both. Thus, the gnomon NOP and the (square) LG are

τῶν ἀνίσων τῆς ὅλης τμημάτων περιεχόμενον ὀρθογώνιον equal to the rectangle contained by AD and DB, and the
μετὰ τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν τετραγώνου ἴσον ἐστὶ square on CD. But, the gnomon NOP and the (square)
τῷ ἀπὸ τῆς ἡμισείας τετραγώνῳ. ὅπερ ἔδει δεῖξαι. LG is (equivalent to) the whole square CEFB, which is

on CB. Thus, the rectangle contained by AD and DB,

plus the square on CD, is equal to the square on CB.

Thus, if a straight-line is cut into equal and unequal
(pieces) then the rectangle contained by the unequal

pieces of the whole (straight-line), plus the square on the
(difference) between the (equal and unequal) pieces, is

equal to the square on half (of the straight-line). (Which

is) the very thing it was required to show.

† Note the (presumably mistaken) double use of the label M in the Greek text.

‡ This proposition is a geometric version of the algebraic identity: a b + [(a + b)/2 − b]2 = [(a + b)/2]2 .�þ. Proposition 6†
᾿Εὰν εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις αὐτῇ If a straight-line is cut in half, and any straight-line

εὐθεῖα ἐπ᾿ εὐθείας, τὸ ὑπὸ τῆς ὅλης σὺν τῇ προσκειμένῃ καὶ added to it straight-on, then the rectangle contained by
τῆς προσκειμένης περιεχόμενον ὀρθόγώνιον μετὰ τοῦ ἀπὸ the whole (straight-line) with the (straight-line) having
τῆς ἡμισείας τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς συγκειμένης being added, and the (straight-line) having being added,
ἔκ τε τῆς ἡμισείας καὶ τῆς προσκειμένης τετραγώνῳ. plus the square on half (of the original straight-line), is

equal to the square on the sum of half (of the original

straight-line) and the (straight-line) having been added.

Γ Β

∆

Λ
Ο

Ξ O

C B D

E G F

M
N

K L

Εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω δίχα κατὰ τὸ Γ σημεῖον, For let any straight-line AB have been cut in half at
προσκείσθω δέ τις αὐτῇ εὐθεῖα ἐπ᾿ εὐθείας ἡ ΒΔ· λέγω, point C, and let any straight-line BD have been added to
ὅτι τὸ ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ it straight-on. I say that the rectangle contained by AD
τοῦ ἀπὸ τῆς ΓΒ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΔ τε- and DB, plus the square on CB, is equal to the square
τραγώνῳ. on CD.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΓΔ τετράγωνον τὸ ΓΕΖΔ, For let the square CEFD have been described on

καὶ ἐπεζεύχθω ἡ ΔΕ, καὶ διὰ μὲν τοῦ Β σημείου ὁποτέρᾳ CD [Prop. 1.46], and let DE have been joined, and
τῶν ΕΓ, ΔΖ παράλληλος ἤχθω ἡ ΒΗ, διὰ δὲ τοῦ Θ σημείου let BG have been drawn through point B, parallel to
ὁποτέρᾳ τῶν ΑΒ, ΕΖ παράλληλος ἤχθω ἡ ΚΜ, καὶ ἔτι διὰ either of EC or DF [Prop. 1.31], and let KM have
τοῦ Α ὁποτέρᾳ τῶν ΓΛ, ΔΜ παράλληλος ἤχθω ἡ ΑΚ. been drawn through point H , parallel to either of AB
᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, ἴσον ἐστὶ καὶ τὸ ΑΛ or EF [Prop. 1.31], and finally let AK have been drawn

STOIQEIWN bþ. ELEMENTS BOOK 2
τῷ ΓΘ. ἀλλὰ τὸ ΓΘ τῷ ΘΖ ἴσον ἐστίν. καὶ τὸ ΑΛ ἄρα τῷ through A, parallel to either of CL or DM [Prop. 1.31].
ΘΖ ἐστιν ἴσον. κοινὸν προσκείσθω τὸ ΓΜ· ὅλον ἄρα τὸ Therefore, since AC is equal to CB, (rectangle) AL is
ΑΜ τῷ ΝΞΟ γνώμονί ἐστιν ἴσον. ἀλλὰ τὸ ΑΜ ἐστι τὸ ὑπὸ also equal to (rectangle) CH [Prop. 1.36]. But, (rectan-
τῶν ΑΔ, ΔΒ· ἴση γάρ ἐστιν ἡ ΔΜ τῇ ΔΒ· καὶ ὁ ΝΞΟ ἄρα gle) CH is equal to (rectangle) HF [Prop. 1.43]. Thus,
γνώμων ἴσος ἐστὶ τῷ ὑπὸ τῶν ΑΔ, ΔΒ [περιεχομένῳ ὀρθο- (rectangle) AL is also equal to (rectangle) HF . Let (rect-
γωνίῳ]. κοινὸν προσκείσθω τὸ ΛΗ, ὅ ἐστιν ἴσον τῷ ἀπὸ angle) CM have been added to both. Thus, the whole
τῆς ΒΓ τετραγώνῳ· τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον (rectangle) AM is equal to the gnomon NOP . But, AM
ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΒ τετραγώνου ἴσον ἐστὶ is the (rectangle contained) by AD and DB. For DM is
τῷ ΝΞΟ γνώμονι καὶ τῷ ΛΗ. ἀλλὰ ὁ ΝΞΟ γνώμων καὶ equal to DB. Thus, gnomon NOP is also equal to the
τὸ ΛΗ ὅλον ἐστὶ τὸ ΓΕΖΔ τετράγωνον, ὅ ἐστιν ἀπὸ τῆς [rectangle contained] by AD and DB. Let LG, which
ΓΔ· τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον is equal to the square on BC, have been added to both.
μετὰ τοῦ ἀπὸ τῆς ΓΒ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΔ Thus, the rectangle contained by AD and DB, plus the
τετραγώνῳ. square on CB, is equal to the gnomon NOP and the
᾿Εὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις (square) LG. But the gnomon NOP and the (square)

αὐτῇ εὐθεῖα ἐπ᾿ εὐθείας, τὸ ὑπὸ τῆς ὅλης σὺν τῇ προ- LG is (equivalent to) the whole square CEFD, which is
σκειμένῃ καὶ τῆς προσκειμένης περιεχόμενον ὀρθόγώνιον on CD. Thus, the rectangle contained by AD and DB,
μετὰ τοῦ ἀπὸ τῆς ἡμισείας τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ plus the square on CB, is equal to the square on CD.
τῆς συγκειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προσκειμένης Thus, if a straight-line is cut in half, and any straight-
τετραγώνῳ· ὅπερ ἔδει δεῖξαι. line added to it straight-on, then the rectangle contained

by the whole (straight-line) with the (straight-line) hav-

ing being added, and the (straight-line) having being
added, plus the square on half (of the original straight-

line), is equal to the square on the sum of half (of the

original straight-line) and the (straight-line) having been
added. (Which is) the very thing it was required to show.

† This proposition is a geometric version of the algebraic identity: (2 a + b) b + a2 = (a + b)2.zþ. Proposition 7†
᾿Εὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης If a straight-line is cut at random then the sum of

καὶ τὸ ἀφ᾿ ἑνὸς τῶν τμημάτων τὰ συναμφότερα τετράγωνα the squares on the whole (straight-line), and one of the
ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος pieces (of the straight-line), is equal to twice the rectan-
περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ λοιποῦ τμήματος gle contained by the whole, and the said piece, and the
τετραγώνῳ. square on the remaining piece.

ΓΑ

Ζ
Η

∆

A C B

D N E

F
GK

Εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ For let any straight-line AB have been cut, at random,
σημεῖον· λέγω, ὅτι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ at point C. I say that the (sum of the) squares on AB and
τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ BC is equal to twice the rectangle contained by AB and

STOIQEIWN bþ. ELEMENTS BOOK 2
ἀπὸ τῆς ΓΑ τετραγώνῳ. BC, and the square on CA.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ· For let the square ADEB have been described on AB

καὶ καταγεγράφθω τὸ σχῆμα. [Prop. 1.46], and let the (rest of) the figure have been
᾿Επεὶ οὖν ἴσον ἐστὶ τὸ ΑΗ τῷ ΗΕ, κοινὸν προσκείσθω drawn.

τὸ ΓΖ· ὅλον ἄρα τὸ ΑΖ ὅλῳ τῷ ΓΕ ἴσον ἐστίν· τὰ ἄρα Therefore, since (rectangle) AG is equal to (rectan-
ΑΖ, ΓΕ διπλάσιά ἐστι τοῦ ΑΖ. ἀλλὰ τὰ ΑΖ, ΓΕ ὁ ΚΛΜ gle) GE [Prop. 1.43], let the (square) CF have been
ἐστι γνώμων καὶ τὸ ΓΖ τετράγωνον· ὁ ΚΛΜ ἄρα γνώμων added to both. Thus, the whole (rectangle) AF is equal
καὶ τὸ ΓΖ διπλάσιά ἐστι τοῦ ΑΖ. ἔστι δὲ τοῦ ΑΖ διπλάσιον to the whole (rectangle) CE. Thus, (rectangle) AF plus
καὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ· ἴση γὰρ ἡ ΒΖ τῇ ΒΓ· ὁ ἄρα (rectangle) CE is double (rectangle) AF . But, (rectan-
ΚΛΜ γνώμων καὶ τὸ ΓΖ τετράγωνον ἴσον ἐστὶ τῷ δὶς ὑπὸ gle) AF plus (rectangle) CE is the gnomon KLM , and
τῶν ΑΒ, ΒΓ. κοινὸν προσκείσθω τὸ ΔΗ, ὅ ἐστιν ἀπὸ τῆς the square CF . Thus, the gnomon KLM , and the square
ΑΓ τετράγωνον· ὁ ἄρα ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ CF , is double the (rectangle) AF . But double the (rect-
τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ angle) AF is also twice the (rectangle contained) by AB
ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΑΓ τετραγώνῳ. ἀλλὰ ὁ ΚΛΜ and BC. For BF (is) equal to BC. Thus, the gnomon
γνώμων καὶ τὰ ΒΗ, ΗΔ τετράγωνα ὅλον ἐστὶ τὸ ΑΔΕΒ καὶ KLM , and the square CF , are equal to twice the (rect-
τὸ ΓΖ, ἅ ἐστιν ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα· τὰ ἄρα ἀπὸ τῶν angle contained) by AB and BC. Let DG, which is
ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ τῷ [τε] δὶς ὑπὸ τῶν ΑΒ, ΒΓ the square on AC, have been added to both. Thus, the
περιεχομένῳ ὀρθογωνίῳ μετὰ τοῦ ἀπὸ τῆς ΑΓ τετραγώνου. gnomon KLM , and the squares BG and GD, are equal
᾿Εὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ to twice the rectangle contained by AB and BC, and the

τῆς ὅλης καὶ τὸ ἀφ᾿ ἑνὸς τῶν τμημάτων τὰ συναμφότερα square on AC. But, the gnomon KLM and the squares
τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ BG and GD is (equivalent to) the whole of ADEB and
εἰρημένου τμήματος περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ CF , which are the squares on AB and BC (respectively).
τοῦ λοιποῦ τμήματος τετραγώνῳ· ὅπερ ἔδει δεῖξαι. Thus, the (sum of the) squares on AB and BC is equal

to twice the rectangle contained by AB and BC, and the
square on AC.

Thus, if a straight-line is cut at random then the sum

of the squares on the whole (straight-line), and one of
the pieces (of the straight-line), is equal to twice the rect-

angle contained by the whole, and the said piece, and the
square on the remaining piece. (Which is) the very thing

it was required to show.

† This proposition is a geometric version of the algebraic identity: (a + b)2 + a2 = 2 (a + b) a + b2.hþ. Proposition 8†
᾿Εὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ τετράκις ὑπὸ If a straight-line is cut at random then four times the

τῆς ὅλης καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώνιον rectangle contained by the whole (straight-line), and one
μετὰ τοῦ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνου ἴσον ἐστὶ of the pieces (of the straight-line), plus the square on the
τῷ ἀπό τε τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος ὡς ἀπὸ remaining piece, is equal to the square described on the
μιᾶς ἀναγραφέντι τετραγώνῳ. whole and the former piece, as on one (complete straight-
Εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ line).

Γ σημεῖον· λέγω, ὅτι τὸ τετράκις ὑπὸ τῶν ΑΒ, ΒΓ πε- For let any straight-line AB have been cut, at random,
ριεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΑΓ τετραγώνου at point C. I say that four times the rectangle contained
ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ, ΒΓ ὡς ἀπὸ μιᾶς ἀναγραφέντι by AB and BC, plus the square on AC, is equal to the
τετραγώνῳ. square described on AB and BC, as on one (complete
᾿Εκβεβλήσθω γὰρ ἐπ᾿ εὐθείας [τῇ ΑΒ εὐθεῖα] ἡ ΒΔ, straight-line).

καὶ κείσθω τῇ ΓΒ ἴση ἡ ΒΔ, καὶ ἀναγεγράφθω ἀπὸ τῆς For let BD have been produced in a straight-line
ΑΔ τετράγωνον τὸ ΑΕΖΔ, καὶ καταγεγράφθω διπλοῦν τὸ [with the straight-line AB], and let BD be made equal
σχῆμα. to CB [Prop. 1.3], and let the square AEFD have been

described on AD [Prop. 1.46], and let the (rest of the)
figure have been drawn double.

STOIQEIWN bþ. ELEMENTS BOOK 2
Υ

Α Γ Β

Ν
Η

Π Ρ

Ε Θ Λ

Ξ
Σ

∆

A C B D

P
Q

G
K

E H L F

᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΓΒ τῇ ΒΔ, ἀλλὰ ἡ μὲν ΓΒ τῇ ΗΚ Therefore, since CB is equal to BD, but CB is equal
ἐστιν ἴση, ἡ δὲ ΒΔ τῇ ΚΝ, καὶ ἡ ΗΚ ἄρα τῇ ΚΝ ἐστιν ἴση. to GK [Prop. 1.34], and BD to KN [Prop. 1.34], GK is
διὰ τὰ αὐτὰ δὴ καὶ ἡ ΠΡ τῇ ΡΟ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν thus also equal to KN . So, for the same (reasons), QR is
ἡ ΒΓ τῂ ΒΔ, ἡ δὲ ΗΚ τῇ ΚΝ, ἴσον ἄρα ἐστὶ καὶ τὸ μὲν equal to RP . And since BC is equal to BD, and GK to
ΓΚ τῷ ΚΔ, τὸ δὲ ΗΡ τῷ ΡΝ. ἀλλὰ τὸ ΓΚ τῷ ΡΝ ἐστιν KN , (square) CK is thus also equal to (square) KD, and
ἴσον· παραπληρώματα γὰρ τοῦ ΓΟ παραλληλογράμμου· καὶ (square) GR to (square) RN [Prop. 1.36]. But, (square)
τὸ ΚΔ ἄρα τῷ ΗΡ ἴσον ἐστίν· τὰ τέσσαρα ἄρα τὰ ΔΚ, ΓΚ, CK is equal to (square) RN . For (they are) comple-
ΗΡ, ΡΝ ἴσα ἀλλήλοις ἐστίν. τὰ τέσσαρα ἄρα τετραπλάσιά ments in the parallelogram CP [Prop. 1.43]. Thus,
ἐστι τοῦ ΓΚ. πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΒΔ, ἀλλὰ ἡ μὲν (square) KD is also equal to (square) GR. Thus, the
ΒΔ τῇ ΒΚ, τουτέστι τῇ ΓΗ ἴση, ἡ δὲ ΓΒ τῇ ΗΚ, τουτέστι four (squares) DK, CK, GR, and RN are equal to one
τῇ ΗΠ, ἐστιν ἴση, καὶ ἡ ΓΗ ἄρα τῇ ΗΠ ἴση ἐστίν. καὶ ἐπεὶ another. Thus, the four (taken together) are quadruple
ἴση ἐστὶν ἡ μὲν ΓΗ τῇ ΗΠ, ἡ δὲ ΠΡ τῇ ΡΟ, ἴσον ἐστὶ καὶ τὸ (square) CK. Again, since CB is equal to BD, but BD
μὲν ΑΗ τῷ ΜΠ, τὸ δὲ ΠΛ τῷ ΡΖ. ἀλλὰ τὸ ΜΠ τῷ ΠΛ ἐστιν (is) equal to BK—that is to say, CG—and CB is equal
ἴσον· παραπληρώματα γὰρ τοῦ ΜΛ παραλληλογράμμου· καὶ to GK—that is to say, GQ—CG is thus also equal to GQ.
τὸ ΑΗ ἄρα τῷ ΡΖ ἴσον ἐστίν· τὰ τέσσαρα ἄρα τὰ ΑΗ, ΜΠ, And since CG is equal to GQ, and QR to RP , (rectan-
ΠΛ, ΡΖ ἴσα ἀλλήλοις ἐστίν· τὰ τέσσαρα ἄρα τοῦ ΑΗ ἐστι gle) AG is also equal to (rectangle) MQ, and (rectangle)
τετραπλάσια. ἐδείχθη δὲ καὶ τὰ τέσσαρα τὰ ΓΚ, ΚΔ, ΗΡ, QL to (rectangle) RF [Prop. 1.36]. But, (rectangle) MQ
ΡΝ τοῦ ΓΚ τετραπλάσια· τὰ ἄρα ὀκτώ, ἃ περιέχει τὸν ΣΤΥ is equal to (rectangle) QL. For (they are) complements
γνώμονα, τετραπλάσιά ἐστι τοῦ ΑΚ. καὶ ἐπεὶ τὸ ΑΚ τὸ ὑπὸ in the parallelogram ML [Prop. 1.43]. Thus, (rectangle)
τῶν ΑΒ, ΒΔ ἐστιν· ἴση γὰρ ἡ ΒΚ τῇ ΒΔ· τὸ ἄρα τετράκις AG is also equal to (rectangle) RF . Thus, the four (rect-
ὑπὸ τῶν ΑΒ, ΒΔ τετραπλάσιόν ἐστι τοῦ ΑΚ. ἐδείχθη δὲ angles) AG, MQ, QL, and RF are equal to one another.
τοῦ ΑΚ τετραπλάσιος καὶ ὁ ΣΤΥ γνώμων· τὸ ἄρα τετράκις Thus, the four (taken together) are quadruple (rectan-
ὑπὸ τῶν ΑΒ, ΒΔ ἴσον ἐστὶ τῷ ΣΤΥ γνώμονι. κοινὸν προ- gle) AG. And it was also shown that the four (squares)
σκείσθω τὸ ΞΘ, ὅ ἐστιν ἴσον τῷ ἀπὸ τῆς ΑΓ τετραγώνῳ· τὸ CK, KD, GR, and RN (taken together are) quadruple
ἄρα τετράκις ὑπὸ τῶν ΑΒ, ΒΔ περιεχόμενον ὀρθογώνιον (square) CK. Thus, the eight (figures taken together),
μετὰ τοῦ ἀπὸ ΑΓ τετραγώνου ἴσον ἐστὶ τῷ ΣΤΥ γνώμονι which comprise the gnomon STU , are quadruple (rect-
καὶ τῷ ΞΘ. ἀλλὰ ὁ ΣΤΥ γνώμων καὶ τὸ ΞΘ ὅλον ἐστὶ τὸ angle) AK. And since AK is the (rectangle contained)
ΑΕΖΔ τετράγωνον, ὅ ἐστιν ἀπὸ τῆς ΑΔ· τὸ ἄρα τετράκις by AB and BD, for BK (is) equal to BD, four times the
ὑπὸ τῶν ΑΒ, ΒΔ μετὰ τοῦ ἀπὸ ΑΓ ἴσον ἐστὶ τῷ ἀπὸ ΑΔ (rectangle contained) by AB and BD is quadruple (rect-
τετραγώνῳ· ἴση δὲ ἡ ΒΔ τῇ ΒΓ. τὸ ἄρα τετράκις ὑπὸ τῶν angle) AK. But the gnomon STU was also shown (to
ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ ΑΓ τε- be equal to) quadruple (rectangle) AK. Thus, four times
τραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΔ, τουτέστι τῷ ἀπὸ τῆς the (rectangle contained) by AB and BD is equal to the
ΑΒ καὶ ΒΓ ὡς ἀπὸ μιᾶς ἀναγραφέντι τετραγώνῳ. gnomon STU . Let OH , which is equal to the square on
᾿Εὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ τετράκις AC, have been added to both. Thus, four times the rect-

ὑπὸ τῆς ὅλης καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώ- angle contained by AB and BD, plus the square on AC,
νιον μετὰ τοῦ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνου ἴσου is equal to the gnomon STU , and the (square) OH . But,

STOIQEIWN bþ. ELEMENTS BOOK 2
ἐστὶ τῷ ἀπό τε τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος ὡς the gnomon STU and the (square) OH is (equivalent to)
ἀπὸ μιᾶς ἀναγραφέντι τετραγώνῳ· ὅπερ ἔδει δεῖξαι. the whole square AEFD, which is on AD. Thus, four

times the (rectangle contained) by AB and BD, plus the
(square) on AC, is equal to the square on AD. And BD

(is) equal to BC. Thus, four times the rectangle con-
tained by AB and BC, plus the square on AC, is equal to

the (square) on AD, that is to say the square described

on AB and BC, as on one (complete straight-line).
Thus, if a straight-line is cut at random then four times

the rectangle contained by the whole (straight-line), and

one of the pieces (of the straight-line), plus the square
on the remaining piece, is equal to the square described

on the whole and the former piece, as on one (complete
straight-line). (Which is) the very thing it was required

to show.

† This proposition is a geometric version of the algebraic identity: 4 (a + b) a + b2 = [(a + b) + a]2.jþ. Proposition 9†
᾿Εὰν εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὰ ἀπὸ If a straight-line is cut into equal and unequal (pieces)

τῶν ἀνίσων τῆς ὅλης τμημάτων τετράγωνα διπλάσιά ἐστι then the (sum of the) squares on the unequal pieces of the
τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν whole (straight-line) is double the (sum of the) square
τετραγώνου. on half (the straight-line) and (the square) on the (dif-

ference) between the (equal and unequal) pieces.

Α Γ ∆

Η F

A C BD

Εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω εἰς μὲν ἴσα κατὰ τὸ For let any straight-line AB have been cut—equally at
Γ, εἱς δὲ ἄνισα κατὰ τὸ Δ· λέγω, ὅτι τὰ ἀπὸ τῶν ΑΔ, ΔΒ C, and unequally at D. I say that the (sum of the) squares
τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. on AD and DB is double the (sum of the squares) on AC
῎Ηχθω γὰρ ἀπὸ τοῦ Γ τῇ ΑΒ πρὸς ὀρθὰς ἡ ΓΕ, καὶ and CD.

κείσθω ἴση ἑκατέρᾳ τῶν ΑΓ, ΓΒ, καὶ ἐπεζεύχθωσαν αἱ ΕΑ, For let CE have been drawn from (point) C, at right-
ΕΒ, καὶ διὰ μὲν τοῦ Δ τῇ ΕΓ παράλληλος ἤχθω ἡ ΔΖ, διὰ angles to AB [Prop. 1.11], and let it be made equal to
δὲ τοῦ Ζ τῇ ΑΒ ἡ ΖΗ, καὶ ἐπεζεύχθω ἡ ΑΖ. καὶ ἐπεὶ ἴση each of AC and CB [Prop. 1.3], and let EA and EB
ἐστὶν ἡ ΑΓ τῇ ΓΕ, ἴση ἐστὶ καὶ ἡ ὑπὸ ΕΑΓ γωνία τῇ ὑπὸ have been joined. And let DF have been drawn through
ΑΕΓ. καὶ ἐπεὶ ὀρθή ἐστιν ἡ πρὸς τῷ Γ, λοιπαὶ ἄρα αἱ ὑπὸ (point) D, parallel to EC [Prop. 1.31], and (let) FG
ΕΑΓ, ΑΕΓ μιᾷ ὀρθῇ ἴσαι εἰσίν· καί εἰσιν ἴσαι· ἡμίσεια ἄρα (have been drawn) through (point) F , (parallel) to AB
ὀρθῆς ἐστιν ἑκατέρα τῶν ὑπὸ ΓΕΑ, ΓΑΕ. δὶα τὰ αὐτὰ δὴ [Prop. 1.31]. And let AF have been joined. And since
καὶ ἑκατέρα τῶν ὑπὸ ΓΕΒ, ΕΒΓ ἡμίσειά ἐστιν ὀρθῆς· ὅλη AC is equal to CE, the angle EAC is also equal to the
ἄρα ἡ ὑπὸ ΑΕΒ ὀρθή ἐστιν. καὶ ἐπεὶ ἡ ὑπὸ ΗΕΖ ἡμίσειά (angle) AEC [Prop. 1.5]. And since the (angle) at C is
ἐστιν ὀρθῆς, ὀρθὴ δὲ ἡ ὑπὸ ΕΗΖ· ἴση γάρ ἐστι τῇ ἐντὸς καὶ a right-angle, the (sum of the) remaining angles (of tri-
ἀπεναντίον τῇ ὑπὸ ΕΓΒ· λοιπὴ ἄρα ἡ ὑπὸ ΕΖΗ ἡμίσειά ἐστιν angle AEC), EAC and AEC, is thus equal to one right-

STOIQEIWN bþ. ELEMENTS BOOK 2
ὀρθῆς· ἴση ἄρα [ἐστὶν] ἡ ὑπὸ ΗΕΖ γωνία τῇ ὑπὸ ΕΖΗ· ὥστε angle [Prop. 1.32]. And they are equal. Thus, (angles)
καὶ πλευρὰ ἡ ΕΗ τῇ ΗΖ ἐστιν ἴση. πάλιν ἐπεὶ ἡ πρὸς τῷ Β CEA and CAE are each half a right-angle. So, for the
γωνία ἡμίσειά ἐστιν ὀρθῆς, ὀρθὴ δὲ ἡ ὑπὸ ΖΔΒ· ἴση γὰρ same (reasons), (angles) CEB and EBC are also each
πάλιν ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΕΓΒ· λοιπὴ ἄρα half a right-angle. Thus, the whole (angle) AEB is a
ἡ ὑπὸ ΒΖΔ ἡμίσειά ἐστιν ὀρθῆς· ἴση ἄρα ἡ πρὸς τῷ Β γωνία right-angle. And since GEF is half a right-angle, and
τῇ ὑπὸ ΔΖΒ· ὥστε καὶ πλευρὰ ἡ ΖΔ πλευρᾷ τῇ ΔΒ ἐστιν EGF (is) a right-angle—for it is equal to the internal and
ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΕ, ἴσον ἐστὶ καὶ τὸ ἀπὸ ΑΓ opposite (angle) ECB [Prop. 1.29]—the remaining (an-
τῷ ἀπὸ ΓΕ· τὰ ἄρα ἀπὸ τῶν ΑΓ, ΓΕ τετράγωνα διπλάσιά gle) EFG is thus half a right-angle [Prop. 1.32]. Thus,
ἐστι τοῦ ἀπὸ ΑΓ. τοῖς δὲ ἀπὸ τῶν ΑΓ, ΓΕ ἴσον ἐστὶ τὸ ἀπὸ angle GEF [is] equal to EFG. So the side EG is also
τῆς ΕΑ τετράγωνον· ὀρθὴ γὰρ ἡ ὑπὸ ΑΓΕ γωνία· τὸ ἄρα equal to the (side) GF [Prop. 1.6]. Again, since the an-
ἀπὸ τῆς ΕΑ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΑΓ. πάλιν, ἐπεὶ ἴση gle at B is half a right-angle, and (angle) FDB (is) a
ἐστὶν ἡ ΕΗ τῇ ΗΖ, ἴσον καὶ τὸ ἀπὸ τῆς ΕΗ τῷ ἀπὸ τῆς ΗΖ· right-angle—for again it is equal to the internal and op-
τὰ ἄρα ἀπὸ τῶν ΕΗ, ΗΖ τετράγωνα διπλάσιά ἐστι τοῦ ἀπὸ posite (angle) ECB [Prop. 1.29]—the remaining (angle)
τῆς ΗΖ τετραγώνου. τοῖς δὲ ἀπὸ τῶν ΕΗ, ΗΖ τετραγώνοις BFD is half a right-angle [Prop. 1.32]. Thus, the angle at
ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΖ τετράγωνον· τὸ ἄρα ἀπὸ τῆς ΕΖ B (is) equal to DFB. So the side FD is also equal to the
τετράγωνον διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΗΖ. ἴση δὲ ἡ ΗΖ side DB [Prop. 1.6]. And since AC is equal to CE, the
τῇ ΓΔ· τὸ ἄρα ἀπὸ τῆς ΕΖ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΓΔ. (square) on AC (is) also equal to the (square) on CE.
ἔστι δὲ καὶ τὸ ἀπὸ τῆς ΕΑ διπλάσιον τοῦ ἀπὸ τῆς ΑΓ· τὰ Thus, the (sum of the) squares on AC and CE is dou-
ἄρα ἀπὸ τῶν ΑΕ, ΕΖ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ble the (square) on AC. And the square on EA is equal
ΑΓ, ΓΔ τετραγώνων. τοῖς δὲ ἀπὸ τῶν ΑΕ, ΕΖ ἴσον ἐστὶ to the (sum of the) squares on AC and CE. For angle
τὸ ἀπὸ τῆς ΑΖ τετράγωνον· ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΑΕΖ ACE (is) a right-angle [Prop. 1.47]. Thus, the (square)
γωνία· τὸ ἄρα ἀπὸ τῆς ΑΖ τετράγωνον διπλάσιόν ἐστι τῶν on EA is double the (square) on AC. Again, since EG
ἀπὸ τῶν ΑΓ, ΓΔ. τῷ δὲ ἀπὸ τῆς ΑΖ ἴσα τὰ ἀπὸ τῶν ΑΔ, is equal to GF , the (square) on EG (is) also equal to
ΔΖ· ὀρθὴ γὰρ ἡ πρὸς τῷ Δ γωνία· τὰ ἄρα ἀπὸ τῶν ΑΔ, ΔΖ the (square) on GF . Thus, the (sum of the squares) on
διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἴση δὲ ἡ EG and GF is double the square on GF . And the square
ΔΖ τῇ ΔΒ· τὰ ἄρα ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα διπλάσιά on EF is equal to the (sum of the) squares on EG and
ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετράγώνων. GF [Prop. 1.47]. Thus, the square on EF is double the
᾿Εὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὰ ἀπὸ (square) on GF . And GF (is) equal to CD [Prop. 1.34].

τῶν ἀνίσων τῆς ὅλης τμημάτων τετράγωνα διπλάσιά ἐστι Thus, the (square) on EF is double the (square) on CD.
τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν And the (square) on EA is also double the (square) on
τετραγώνου· ὅπερ ἔδει δεῖξαι. AC. Thus, the (sum of the) squares on AE and EF is

double the (sum of the) squares on AC and CD. And
the square on AF is equal to the (sum of the squares)

on AE and EF . For the angle AEF is a right-angle

[Prop. 1.47]. Thus, the square on AF is double the (sum
of the squares) on AC and CD. And the (sum of the

squares) on AD and DF (is) equal to the (square) on
AF . For the angle at D is a right-angle [Prop. 1.47].

Thus, the (sum of the squares) on AD and DF is double

the (sum of the) squares on AC and CD. And DF (is)
equal to DB. Thus, the (sum of the) squares on AD and

DB is double the (sum of the) squares on AC and CD.

Thus, if a straight-line is cut into equal and unequal
(pieces) then the (sum of the) squares on the unequal

pieces of the whole (straight-line) is double the (sum of
the) square on half (the straight-line) and (the square) on

the (difference) between the (equal and unequal) pieces.

(Which is) the very thing it was required to show.

† This proposition is a geometric version of the algebraic identity: a2 + b2 = 2[([a + b]/2)2 + ([a + b]/2 − b)2].

STOIQEIWN bþ. ELEMENTS BOOK 2iþ. Proposition 10†
᾿Εὰν εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις αὐτῇ If a straight-line is cut in half, and any straight-line

εὐθεῖα ἐπ᾿ εὐθείας, τὸ ἀπὸ τῆς ὅλης σὺν τῇ προσκειμένῃ added to it straight-on, then the sum of the square on
καὶ τὸ ἀπὸ τῆς προσκειμένης τὰ συναμφότερα τετράγωνα the whole (straight-line) with the (straight-line) having
διπλάσιά ἐστι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ ἀπὸ τῆς συγ- been added, and the (square) on the (straight-line) hav-
κειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προσκειμένης ὡς ἀπὸ ing been added, is double the (sum of the square) on half
μιᾶς ἀναγραφέντος τετραγώνου. (the straight-line), and the square described on the sum

of half (the straight-line) and (straight-line) having been

added, as on one (complete straight-line).

∆

Α Γ Β

A C B

Εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω δίχα κατὰ τὸ Γ, προ- For let any straight-line AB have been cut in half at
σκείσθω δέ τις αὐτῇ εὐθεῖα ἐπ᾿ εὐθείας ἡ ΒΔ· λέγω, ὅτι τὰ (point) C, and let any straight-line BD have been added
ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν to it straight-on. I say that the (sum of the) squares on
ΑΓ, ΓΔ τετραγώνων. AD and DB is double the (sum of the) squares on AC
῎Ηχθω γὰρ ἀπὸ τοῦ Γ σημείου τῇ ΑΒ πρὸς ὀρθὰς ἡ ΓΕ, and CD.

καὶ κείσθω ἴση ἑκατέρᾳ τῶν ΑΓ, ΓΒ, καὶ ἐπεζεύχθωσαν For let CE have been drawn from point C, at right-
αἱ ΕΑ, ΕΒ· καὶ διὰ μὲν τοῦ Ε τῇ ΑΔ παράλληλος ἤχθω ἡ angles to AB [Prop. 1.11], and let it be made equal to
ΕΖ, διὰ δὲ τοῦ Δ τῇ ΓΕ παράλληλος ἤχθω ἡ ΖΔ. καὶ ἐπεὶ each of AC and CB [Prop. 1.3], and let EA and EB have
εἰς παραλλήλους εὐθείας τὰς ΕΓ, ΖΔ εὐθεῖά τις ἐνέπεσεν been joined. And let EF have been drawn through E,
ἡ ΕΖ, αἱ ὑπὸ ΓΕΖ, ΕΖΔ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν· αἱ parallel to AD [Prop. 1.31], and let FD have been drawn
ἄρα ὑπὸ ΖΕΒ, ΕΖΔ δύο ὀρθῶν ἐλάσσονές εἰσιν· αἱ δὲ through D, parallel to CE [Prop. 1.31]. And since some
ἀπ᾿ ἐλασσόνων ἢ δύο ὀρθῶν ἐκβαλλόμεναι συμπίπτουσιν· straight-line EF falls across the parallel straight-lines EC
αἱ ἄρα ΕΒ, ΖΔ ἐκβαλλόμεναι ἐπὶ τὰ Β, Δ μέρη συμ- and FD, the (internal angles) CEF and EFD are thus
πεσοῦνται. ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν κατὰ τὸ Η, equal to two right-angles [Prop. 1.29]. Thus, FEB and
καὶ ἐπεζεύχθω ἡ ΑΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΕ, ἴση EFD are less than two right-angles. And (straight-lines)
ἐστὶ καὶ γωνία ἡ ὑπὸ ΕΑΓ τῇ ὑπὸ ΑΕΓ· καὶ ὀρθὴ ἡ πρὸς τῷ produced from (internal angles whose sum is) less than
Γ· ἡμίσεια ἄρα ὀρθῆς [ἐστιν] ἑκατέρα τῶν ὑπὸ ΕΑΓ, ΑΕΓ. two right-angles meet together [Post. 5]. Thus, being pro-
διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ὑπὸ ΓΕΒ, ΕΒΓ ἡμίσειά ἐστιν duced in the direction of B and D, the (straight-lines)
ὀρθῆς· ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΑΕΒ. καὶ ἐπεὶ ἡμίσεια ὀρθῆς EB and FD will meet. Let them have been produced,
ἐστιν ἡ ὑπὸ ΕΒΓ, ἡμίσεια ἄρα ὀρθῆς καὶ ἡ ὑπὸ ΔΒΗ. ἔστι and let them meet together at G, and let AG have been
δὲ καὶ ἡ ὑπὸ ΒΔΗ ὀρθή· ἴση γάρ ἐστι τῇ ὑπὸ ΔΓΕ· ἐναλλὰξ joined. And since AC is equal to CE, angle EAC is also
γάρ· λοιπὴ ἅρα ἡ ὑπὸ ΔΗΒ ἡμίσειά ἐστιν ὀρθῆς· ἡ ἄρα ὑπὸ equal to (angle) AEC [Prop. 1.5]. And the (angle) at
ΔΗΒ τῇ ὑπὸ ΔΒΗ ἐστιν ἴση· ὥστε καὶ πλευρὰ ἡ ΒΔ πλευρᾷ C (is) a right-angle. Thus, EAC and AEC [are] each
τῇ ΗΔ ἐστιν ἴση. πάλιν, ἐπεὶ ἡ ὑπὸ ΕΗΖ ἡμίσειά ἐστιν half a right-angle [Prop. 1.32]. So, for the same (rea-
ὀρθῆς, ὀρθὴ δὲ ἡ πρὸς τῷ Ζ· ἴση γάρ ἐστι τῇ ἀπεναντίον τῇ sons), CEB and EBC are also each half a right-angle.
πρὸς τῷ Γ· λοιπὴ ἄρα ἡ ὑπὸ ΖΕΗ ἡμίσειά ἐστιν ὀρθῆς· ἴση Thus, (angle) AEB is a right-angle. And since EBC
ἄρα ἡ ὑπὸ ΕΗΖ γωνία τῇ ὑπὸ ΖΕΗ· ὥστε καὶ πλευρὰ ἡ ΗΖ is half a right-angle, DBG (is) thus also half a right-
πλευρᾷ τῇ ΕΖ ἐστιν ἴση. καὶ ἐπεὶ [ἴση ἐστὶν ἡ ΕΓ τῇ ΓΑ], angle [Prop. 1.15]. And BDG is also a right-angle. For
ἴσον ἐστὶ [καὶ] τὸ ἀπὸ τῆς ΕΓ τετράγωνον τῷ ἀπὸ τῆς ΓΑ it is equal to DCE. For (they are) alternate (angles)

STOIQEIWN bþ. ELEMENTS BOOK 2
τετραγώνῳ· τὰ ἄρα ἀπὸ τῶν ΕΓ, ΓΑ τετράγωνα διπλάσιά [Prop. 1.29]. Thus, the remaining (angle) DGB is half
ἐστι τοῦ ἀπὸ τῆς ΓΑ τετραγώνου. τοῖς δὲ ἀπὸ τῶν ΕΓ, ΓΑ a right-angle. Thus, DGB is equal to DBG. So side BD
ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΑ· τὸ ἄρα ἀπὸ τῆς ΕΑ τετράγωνον is also equal to side GD [Prop. 1.6]. Again, since EGF is
διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΑΓ τετραγώνου. πάλιν, ἐπεὶ ἴση half a right-angle, and the (angle) at F (is) a right-angle,
ἐστὶν ἡ ΖΗ τῇ ΕΖ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΖΗ τῷ ἀπὸ τῆς for it is equal to the opposite (angle) at C [Prop. 1.34],
ΖΕ· τὰ ἄρα ἀπὸ τῶν ΗΖ, ΖΕ διπλάσιά ἐστι τοῦ ἀπὸ τῆς ΕΖ. the remaining (angle) FEG is thus half a right-angle.
τοῖς δὲ ἀπὸ τῶν ΗΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΗ· τὸ ἄρα Thus, angle EGF (is) equal to FEG. So the side GF
ἀπὸ τῆς ΕΗ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΕΖ. ἴση δὲ ἡ ΕΖ τῇ is also equal to the side EF [Prop. 1.6]. And since [EC
ΓΔ· τὸ ἄρα ἀπὸ τῆς ΕΗ τετράγωνον διπλάσιόν ἐστι τοῦ ἀπὸ is equal to CA] the square on EC is [also] equal to the
τῆς ΓΔ. ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς ΕΑ διπλάσιον τοῦ ἀπὸ square on CA. Thus, the (sum of the) squares on EC
τῆς ΑΓ· τὰ ἄρα ἀπὸ τῶν ΑΕ, ΕΗ τετράγωνα διπλάσιά ἐστι and CA is double the square on CA. And the (square)
τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. τοῖς δὲ ἀπὸ τῶν ΑΕ, on EA is equal to the (sum of the squares) on EC and
ΕΗ τετραγώνοις ἴσον ἐστὶ τὸ ἀπὸ τῆς ΑΗ τετράγωνον· τὸ CA [Prop. 1.47]. Thus, the square on EA is double the
ἄρα ἀπὸ τῆς ΑΗ διπλάσιόν ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ. τῷ square on AC. Again, since FG is equal to EF , the
δὲ ἀπὸ τῆς ΑΗ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΔ, ΔΗ· τὰ ἄρα ἀπὸ (square) on FG is also equal to the (square) on FE.
τῶν ΑΔ, ΔΗ [τετράγωνα] διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, Thus, the (sum of the squares) on GF and FE is dou-
ΓΔ [τετραγώνων]. ἴση δὲ ἡ ΔΗ τῇ ΔΒ· τὰ ἄρα ἀπὸ τῶν ble the (square) on EF . And the (square) on EG is equal
ΑΔ, ΔΒ [τετράγωνα] διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ to the (sum of the squares) on GF and FE [Prop. 1.47].
τετραγώνων. Thus, the (square) on EG is double the (square) on EF .
᾿Εὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις And EF (is) equal to CD [Prop. 1.34]. Thus, the square

αὐτῇ εὐθεῖα ἐπ᾿ εὐθείας, τὸ ἀπὸ τῆς ὅλης σὺν τῇ προ- on EG is double the (square) on CD. But it was also
σκειμένῃ καὶ τὸ ἀπὸ τῆς προσκειμένης τὰ συναμφότερα shown that the (square) on EA (is) double the (square)
τετράγωνα διπλάσιά ἐστι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ on AC. Thus, the (sum of the) squares on AE and EG is
ἀπὸ τῆς συγκειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προ- double the (sum of the) squares on AC and CD. And the
σκειμένης ὡς ἀπὸ μιᾶς ἀναγραφέντος τετραγώνου· ὅπερ square on AG is equal to the (sum of the) squares on AE
ἔδει δεῖξαι. and EG [Prop. 1.47]. Thus, the (square) on AG is double

the (sum of the squares) on AC and CD. And the (sum
of the squares) on AD and DG is equal to the (square)

on AG [Prop. 1.47]. Thus, the (sum of the) [squares] on
AD and DG is double the (sum of the) [squares] on AC

and CD. And DG (is) equal to DB. Thus, the (sum of

the) [squares] on AD and DB is double the (sum of the)
squares on AC and CD.

Thus, if a straight-line is cut in half, and any straight-

line added to it straight-on, then the sum of the square
on the whole (straight-line) with the (straight-line) hav-

ing been added, and the (square) on the (straight-line)
having been added, is double the (sum of the square) on

half (the straight-line), and the square described on the

sum of half (the straight-line) and (straight-line) having
been added, as on one (complete straight-line). (Which

is) the very thing it was required to show.

† This proposition is a geometric version of the algebraic identity: (2 a + b)2 + b2 = 2 [a2 + (a + b)2].iaþ. Proposition 11†
Τὴν δοθεῖσαν εὐθεῖαν τεμεῖν ὥστε τὸ ὑπὸ τῆς ὅλης καὶ To cut a given straight-line such that the rectangle

τοῦ ἑτέρου τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον contained by the whole (straight-line), and one of the
εἶναι τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ. pieces (of the straight-line), is equal to the square on the

remaining piece.

STOIQEIWN bþ. ELEMENTS BOOK 2

∆

F G

C K

H
B

῎Εστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ· δεῖ δὴ τὴν ΑΒ τεμεῖν Let AB be the given straight-line. So it is required to
ὥστε τὸ ὑπὸ τῆς ὅλης καὶ τοῦ ἑτέρου τῶν τμημάτων cut AB such that the rectangle contained by the whole
περιεχόμενον ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τοῦ λοιποῦ (straight-line), and one of the pieces (of the straight-
τμήματος τετραγώνῳ. line), is equal to the square on the remaining piece.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΒΔΓ, For let the square ABDC have been described on AB

καὶ τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε σημεῖον, καὶ ἐπεζεύχθω [Prop. 1.46], and let AC have been cut in half at point
ἡ ΒΕ, καὶ διήχθω ἡ ΓΑ ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΒΕ ἴση ἡ E [Prop. 1.10], and let BE have been joined. And let
ΕΖ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΑΖ τετράγωνον τὸ ΖΘ, καὶ CA have been drawn through to (point) F , and let EF
διήχθω ἡ ΗΘ ἐπὶ τὸ Κ· λέγω, ὅτι ἡ ΑΒ τέτμηται κατὰ τὸ Θ, be made equal to BE [Prop. 1.3]. And let the square
ὥστε τὸ ὑπὸ τῶν ΑΒ, ΒΘ περιεχόμενον ὀρθογώνιον ἴσον FH have been described on AF [Prop. 1.46], and let GH
ποιεῖν τῷ ἀπὸ τῆς ΑΘ τετραγώνῳ. have been drawn through to (point) K. I say that AB has
᾿Επεὶ γὰρ εὐθεῖα ἡ ΑΓ τέτμηται δίχα κατὰ τὸ Ε, been cut at H such as to make the rectangle contained by

πρόσκειται δὲ αὐτῇ ἡ ΖΑ, τὸ ἄρα ὑπὸ τῶν ΓΖ, ΖΑ πε- AB and BH equal to the square on AH .
ριεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΑΕ τετραγώνου For since the straight-line AC has been cut in half at
ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΖ τετραγώνῳ. ἴση δὲ ἡ ΕΖ τῇ ΕΒ· E, and FA has been added to it, the rectangle contained
τὸ ἄρα ὑπὸ τῶν ΓΖ, ΖΑ μετὰ τοῦ ἀπὸ τῆς ΑΕ ἴσον ἐστὶ by CF and FA, plus the square on AE, is thus equal to
τῷ ἀπὸ ΕΒ. ἀλλὰ τῷ ἀπὸ ΕΒ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΒΑ, the square on EF [Prop. 2.6]. And EF (is) equal to EB.
ΑΕ· ὀρθὴ γὰρ ἡ πρὸς τῷ Α γωνία· τὸ ἄρα ὑπὸ τῶν ΓΖ, Thus, the (rectangle contained) by CF and FA, plus the
ΖΑ μετὰ τοῦ ἀπὸ τῆς ΑΕ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΕ. (square) on AE, is equal to the (square) on EB. But,
κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΑΕ· λοιπὸν ἄρα τὸ ὑπὸ τῶν the (sum of the squares) on BA and AE is equal to the
ΓΖ, ΖΑ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ (square) on EB. For the angle at A (is) a right-angle
τετραγώνῳ. καί ἐστι τὸ μὲν ὑπὸ τῶν ΓΖ, ΖΑ τὸ ΖΚ· ἴση [Prop. 1.47]. Thus, the (rectangle contained) by CF and
γὰρ ἡ ΑΖ τῇ ΖΗ· τὸ δὲ ἀπὸ τῆς ΑΒ τὸ ΑΔ· τὸ ἄρα ΖΚ ἴσον FA, plus the (square) on AE, is equal to the (sum of
ἐστὶ τῷ ΑΔ. κοινὸν ἀρῃρήσθω τὸ ΑΚ· λοιπὸν ἄρα τὸ ΖΘ the squares) on BA and AE. Let the square on AE have
τῷ ΘΔ ἴσον ἐστίν. καί ἐστι τὸ μὲν ΘΔ τὸ ὑπὸ τῶν ΑΒ, been subtracted from both. Thus, the remaining rectan-
ΒΘ· ἴση γὰρ ἡ ΑΒ τῇ ΒΔ· τὸ δὲ ΖΘ τὸ ἀπὸ τῆς ΑΘ· τὸ gle contained by CF and FA is equal to the square on
ἄρα ὑπὸ τῶν ΑΒ, ΒΘ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ AB. And FK is the (rectangle contained) by CF and
τῷ ἀπὸ ΘΑ τετραγώνῳ. FA. For AF (is) equal to FG. And AD (is) the (square)
῾Η ἄρα δοθεῖσα εὐθεῖα ἡ ΑΒ τέτμηται κατὰ τὸ Θ ὥστε on AB. Thus, the (rectangle) FK is equal to the (square)

τὸ ὑπὸ τῶν ΑΒ, ΒΘ περιεχόμενον ὀρθογώνιον ἴσον ποιεῖν AD. Let (rectangle) AK have been subtracted from both.
τῷ ἀπὸ τῆς ΘΑ τετραγώνῳ· ὅπερ ἔδει ποιῆσαι. Thus, the remaining (square) FH is equal to the (rectan-

gle) HD. And HD is the (rectangle contained) by AB

and BH . For AB (is) equal to BD. And FH (is) the

(square) on AH . Thus, the rectangle contained by AB

STOIQEIWN bþ. ELEMENTS BOOK 2
and BH is equal to the square on HA.

Thus, the given straight-line AB has been cut at

(point) H such as to make the rectangle contained by
AB and BH equal to the square on HA. (Which is) the

very thing it was required to do.

† This manner of cutting a straight-line—so that the ratio of the whole to the larger piece is equal to the ratio of the larger to the smaller piece—is

sometimes called the “Golden Section”.ibþ. Proposition 12†
᾿Εν τοῖς ἀμβλυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν In obtuse-angled triangles, the square on the side sub-

ἀμβλεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον μεῖζόν tending the obtuse angle is greater than the (sum of the)
ἐστι τῶν ἀπὸ τῶν τὴν ἀμβλεῖαν γωνίαν περιεχουσῶν squares on the sides containing the obtuse angle by twice
πλευρῶν τετραγώνων τῷ περιεχομένῳ δὶς ὑπὸ τε μιᾶς τῶν the (rectangle) contained by one of the sides around the
περὶ τὴν ἀμβλεῖαν γωνίαν, ἐφ᾿ ἣν ἡ κάθετος πίπτει, καὶ τῆς obtuse angle, to which a perpendicular (straight-line)
ἀπολαμβανομένης ἐκτὸς ὑπὸ τῆς καθέτου πρὸς τῇ ἀμβλείᾳ falls, and the (straight-line) cut off outside (the triangle)
γωνίᾳ. by the perpendicular (straight-line) towards the obtuse

angle.

∆ Α Γ

D A C
῎Εστω ἀμβλυγώνιον τρίγωνον τὸ ΑΒΓ ἀμβλεῖαν ἔχον Let ABC be an obtuse-angled triangle, having the an-

τὴν ὑπὸ ΒΑΓ, καὶ ἤχθω ἀπὸ τοῦ Β σημείου ἐπὶ τὴν ΓΑ gle BAC obtuse. And let BD be drawn from point B,
ἐκβληθεῖσαν κάθετος ἡ ΒΔ. λέγω, ὅτι τὸ ἀπὸ τῆς ΒΓ perpendicular to CA produced [Prop. 1.12]. I say that
τετράγωνον μεῖζόν ἐστι τῶν ἀπὸ τῶν ΒΑ, ΑΓ τετραγώνων the square on BC is greater than the (sum of the) squares
τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ. on BA and AC, by twice the rectangle contained by CA
᾿Επεὶ γὰρ εὐθεῖα ἡ ΓΔ τέτμηται, ὡς ἔτυχεν, κατὰ τὸ Α and AD.

σημεῖον, τὸ ἄρα ἀπὸ τῆς ΔΓ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΓΑ, For since the straight-line CD has been cut, at ran-
ΑΔ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ dom, at point A, the (square) on DC is thus equal to
ὀρθογωνίῳ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΔΒ· τὰ ἄρα the (sum of the) squares on CA and AD, and twice the
ἀπὸ τῶν ΓΔ, ΔΒ ἴσα ἐστὶ τοῖς τε ἀπὸ τῶν ΓΑ, ΑΔ, ΔΒ τε- rectangle contained by CA and AD [Prop. 2.4]. Let the
τραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ [περιεχομένῳ ὀρθο- (square) on DB have been added to both. Thus, the (sum
γωνίῳ]. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΓΔ, ΔΒ ἴσον ἐστὶ τὸ ἀπὸ of the squares) on CD and DB is equal to the (sum of
τῆς ΓΒ· ὀρθὴ γὰρ ἡ προς τῷ Δ γωνία· τοῖς δὲ ἀπὸ τῶν ΑΔ, the) squares on CA, AD, and DB, and twice the [rect-
ΔΒ ἴσον τὸ ἀπὸ τῆς ΑΒ· τὸ ἄρα ἀπὸ τῆς ΓΒ τετράγωνον angle contained] by CA and AD. But, the (square) on
ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΓΑ, ΑΒ τετραγώνοις καὶ τῷ δὶς CB is equal to the (sum of the squares) on CD and DB.
ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ· ὥστε τὸ ἀπὸ τῆς For the angle at D (is) a right-angle [Prop. 1.47]. And
ΓΒ τετράγωνον τῶν ἀπὸ τῶν ΓΑ, ΑΒ τετραγώνων μεῖζόν the (square) on AB (is) equal to the (sum of the squares)
ἐστι τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ. on AD and DB [Prop. 1.47]. Thus, the square on CB
᾿Εν ἄρα τοῖς ἀμβλυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν is equal to the (sum of the) squares on CA and AB, and

ἀμβλεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον μεῖζόν twice the rectangle contained by CA and AD. So the
ἐστι τῶν ἀπὸ τῶν τὴν ἀμβλεῖαν γωνίαν περιεχουσῶν square on CB is greater than the (sum of the) squares on

STOIQEIWN bþ. ELEMENTS BOOK 2
πλευρῶν τετραγώνων τῷ περιχομένῳ δὶς ὑπό τε μιᾶς τῶν CA and AB by twice the rectangle contained by CA and
περὶ τὴν ἀμβλεῖαν γωνίαν, ἐφ᾿ ἣν ἡ κάθετος πίπτει, καὶ τῆς AD.
ἀπολαμβανομένης ἐκτὸς ὑπὸ τῆς καθέτου πρὸς τῇ ἀμβλείᾳ Thus, in obtuse-angled triangles, the square on the
γωνίᾳ· ὅπερ ἔδει δεῖξαι. side subtending the obtuse angle is greater than the (sum

of the) squares on the sides containing the obtuse an-
gle by twice the (rectangle) contained by one of the

sides around the obtuse angle, to which a perpendicu-

lar (straight-line) falls, and the (straight-line) cut off out-
side (the triangle) by the perpendicular (straight-line) to-

wards the obtuse angle. (Which is) the very thing it was

required to show.

† This proposition is equivalent to the well-known cosine formula: BC 2 = AB 2 + AC 2 − 2 AB AC cos BAC, since cos BAC = −AD/AB.igþ. Proposition 13†
᾿Εν τοῖς ὀξυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀξεῖαν In acute-angled triangles, the square on the side sub-

γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἔλαττόν ἐστι tending the acute angle is less than the (sum of the)
τῶν ἀπὸ τῶν τὴν ὀξεῖαν γωνίαν περιεχουσῶν πλευρῶν τε- squares on the sides containing the acute angle by twice
τραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν the (rectangle) contained by one of the sides around the
ὀξεῖαν γωνίαν, ἐφ᾿ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβα- acute angle, to which a perpendicular (straight-line) falls,
νομένης ἐντὸς ὑπὸ τῆς καθέτου πρὸς τῇ ὀξείᾳ γωνίᾳ. and the (straight-line) cut off inside (the triangle) by the

perpendicular (straight-line) towards the acute angle.

∆ Γ B D C

῎Εστω ὀξυγώνιον τρίγωνον τὸ ΑΒΓ ὀξεῖαν ἔχον τὴν Let ABC be an acute-angled triangle, having the an-
πρὸς τῷ Β γωνίαν, καὶ ἤχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὴν gle at (point) B acute. And let AD have been drawn from
ΒΓ κάθετος ἡ ΑΔ· λέγω, ὅτι τὸ ἀπὸ τῆς ΑΓ τετράγωνον point A, perpendicular to BC [Prop. 1.12]. I say that the
ἔλαττόν ἐστι τῶν ἀπὸ τῶν ΓΒ, ΒΑ τετραγώνων τῷ δὶς ὑπὸ square on AC is less than the (sum of the) squares on
τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ. CB and BA, by twice the rectangle contained by CB and
᾿Επεὶ γὰρ εὐθεῖα ἡ ΓΒ τέτμηται, ὡς ἔτυχεν, κατὰ τὸ BD.

Δ, τὰ ἄρα ἀπὸ τῶν ΓΒ, ΒΔ τετράγωνα ἴσα ἐστὶ τῷ τε For since the straight-line CB has been cut, at ran-
δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ dom, at (point) D, the (sum of the) squares on CB and
τῆς ΔΓ τετραγώνῳ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΔΑ BD is thus equal to twice the rectangle contained by CB
τετράγωνον· τὰ ἄρα ἀπὸ τῶν ΓΒ, ΒΔ, ΔΑ τετράγωνα ἴσα and BD, and the square on DC [Prop. 2.7]. Let the
ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ square on DA have been added to both. Thus, the (sum
καὶ τοῖς ἀπὸ τῶν ΑΔ, ΔΓ τετραγώνιος. ἀλλὰ τοῖς μὲν ἀπὸ of the) squares on CB, BD, and DA is equal to twice
τῶν ΒΔ, ΔΑ ἴσον τὸ ἀπὸ τῆς ΑΒ· ὀρθὴ γὰρ ἡ πρὸς τῷ Δ the rectangle contained by CB and BD, and the (sum of
γωνίᾳ· τοῖς δὲ ἀπὸ τῶν ΑΔ, ΔΓ ἴσον τὸ ἀπὸ τῆς ΑΓ· τὰ the) squares on AD and DC. But, the (square) on AB
ἄρα ἀπὸ τῶν ΓΒ, ΒΑ ἴσα ἐστὶ τῷ τε ἀπὸ τῆς ΑΓ καὶ τῷ δὶς (is) equal to the (sum of the squares) on BD and DA.
ὑπὸ τῶν ΓΒ, ΒΔ· ὥστε μόνον τὸ ἀπὸ τῆς ΑΓ ἔλαττόν ἐστι For the angle at (point) D is a right-angle [Prop. 1.47].

STOIQEIWN bþ. ELEMENTS BOOK 2
τῶν ἀπὸ τῶν ΓΒ, ΒΑ τετραγώνων τῷ δὶς ὑπὸ τῶν ΓΒ, ΒΔ And the (square) on AC (is) equal to the (sum of the
περιεχομένῳ ὀρθογωνίῳ. squares) on AD and DC [Prop. 1.47]. Thus, the (sum of
᾿Εν ἄρα τοῖς ὀξυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀξεῖαν the squares) on CB and BA is equal to the (square) on

γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἔλαττόν ἐστι AC, and twice the (rectangle contained) by CB and BD.
τῶν ἀπὸ τῶν τὴν ὀξεῖαν γωνίαν περιεχουσῶν πλευρῶν τε- So the (square) on AC alone is less than the (sum of the)
τραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν squares on CB and BA by twice the rectangle contained
ὀξεῖαν γωνίαν, ἐφ᾿ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβα- by CB and BD.
νομένης ἐντὸς ὑπὸ τῆς καθέτου πρὸς τῇ ὀξείᾳ γωνίᾳ· ὅπερ Thus, in acute-angled triangles, the square on the side
ἔδει δεῖξαι. subtending the acute angle is less than the (sum of the)

squares on the sides containing the acute angle by twice

the (rectangle) contained by one of the sides around the
acute angle, to which a perpendicular (straight-line) falls,

and the (straight-line) cut off inside (the triangle) by
the perpendicular (straight-line) towards the acute angle.

(Which is) the very thing it was required to show.

† This proposition is equivalent to the well-known cosine formula: AC 2 = AB 2 + BC 2 − 2 AB BC cos ABC, since cos ABC = BD/AB.idþ. Proposition 14
Τῷ δοθέντι εὐθυγράμμῳ ἴσον τετράγωνον συστήσας- To construct a square equal to a given rectilinear fig-

θαι. ure.

Ε
Β

Γ ∆

Ζ
Η

E
F

῎Εστω τὸ δοθὲν εὐθύγραμμον τὸ Α· δεῖ δὴ τῷ Α Let A be the given rectilinear figure. So it is required
εὐθυγράμμῳ ἴσον τετράγωνον συστήσασθαι. to construct a square equal to the rectilinear figure A.
Συνεστάτω γὰρ τῷ Α ἐυθυγράμμῳ ἴσον παραλληλό- For let the right-angled parallelogram BD, equal to

γραμμον ὀρθογώνιον τὸ ΒΔ· εἰ μὲν οὖν ἴση ἐστὶν ἡ ΒΕ the rectilinear figure A, have been constructed [Prop.
τῇ ΕΔ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. συνέσταται γὰρ τῷ 1.45]. Therefore, if BE is equal to ED then that (which)
Α εὐθυγράμμῳ ἴσον τετράγωνον τὸ ΒΔ· εἰ δὲ οὔ, μία τῶν was prescribed has taken place. For the square BD, equal
ΒΕ, ΕΔ μείζων ἐστίν. ἔστω μείζων ἡ ΒΕ, καὶ ἐκβεβλήσθω to the rectilinear figure A, has been constructed. And if
ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΕΔ ἴση ἡ ΕΖ, καὶ τετμήσθω ἡ ΒΖ not, then one of the (straight-lines) BE or ED is greater
δίχα κατὰ τὸ Η, καὶ κέντρῳ τῷ Η, διαστήματι δὲ ἑνὶ τῶν (than the other). Let BE be greater, and let it have
ΗΒ, ΗΖ ἡμικύκλιον γεγράφθω τὸ ΒΘΖ, καὶ ἐκβεβλήσθω ἡ been produced to F , and let EF be made equal to ED
ΔΕ ἐπὶ τὸ Θ, καὶ ἐπεζεύχθω ἡ ΗΘ. [Prop. 1.3]. And let BF have been cut in half at (point)
᾿Επεὶ οὖν εὐθεῖα ἡ ΒΖ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Η, εἰς G [Prop. 1.10]. And, with center G, and radius one of

δὲ ἄνισα κατὰ τὸ Ε, τὸ ἄρα ὑπὸ τῶν ΒΕ, ΕΖ περιεχόμενον the (straight-lines) GB or GF , let the semi-circle BHF
ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΕΗ τετραγώνου ἴσον ἐστὶ have been drawn. And let DE have been produced to H ,
τῷ ἀπὸ τῆς ΗΖ τετραγώνῳ. ἴση δὲ ἡ ΗΖ τῇ ΗΘ· τὸ ἄρα and let GH have been joined.
ὑπὸ τῶν ΒΕ, ΕΖ μετὰ τοῦ ἀπὸ τῆς ΗΕ ἴσον ἐστὶ τῷ ἀπὸ Therefore, since the straight-line BF has been cut—
τῆς ΗΘ. τῷ δὲ ἀπὸ τῆς ΗΘ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΘΕ, ΕΗ equally at G, and unequally at E—the rectangle con-

STOIQEIWN bþ. ELEMENTS BOOK 2
τετράγωνα· τὸ ἄρα ὑπὸ τῶν ΒΕ, ΕΖ μετὰ τοῦ ἀπὸ ΗΕ ἴσα tained by BE and EF , plus the square on EG, is thus
ἐστὶ τοῖς ἀπὸ τῶν ΘΕ, ΕΗ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΗΕ equal to the square on GF [Prop. 2.5]. And GF (is) equal
τετράγωνον· λοιπὸν ἄρα τὸ ὑπὸ τῶν ΒΕ, ΕΖ περιεχόμενον to GH . Thus, the (rectangle contained) by BE and EF ,
ὄρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ τετραγώνῳ. ἀλλὰ τὸ plus the (square) on GE, is equal to the (square) on GH .
ὑπὸ τῶν ΒΕ, ΕΖ τὸ ΒΔ ἐστιν· ἴση γὰρ ἡ ΕΖ τῇ ΕΔ· τὸ And the (sum of the) squares on HE and EG is equal to
ἄρα ΒΔ παραλληλόγραμμον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΘΕ τε- the (square) on GH [Prop. 1.47]. Thus, the (rectangle
τραγώνῳ. ἴσον δὲ τὸ ΒΔ τῷ Α εὐθυγράμμῳ. καὶ τὸ Α ἄρα contained) by BE and EF , plus the (square) on GE, is
εὐθύγραμμον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ ἀναγραφησομένῳ equal to the (sum of the squares) on HE and EG. Let
τετραγώνῳ. the square on GE have been taken from both. Thus, the
Τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ Α ἴσον τετράγωνον remaining rectangle contained by BE and EF is equal to

συνέσταται τὸ ἀπὸ τῆς ΕΘ ἀναγραφησόμενον· ὅπερ ἔδει the square on EH . But, BD is the (rectangle contained)
ποιῆσαι. by BE and EF . For EF (is) equal to ED. Thus, the par-

allelogram BD is equal to the square on HE. And BD
(is) equal to the rectilinear figure A. Thus, the rectilin-

ear figure A is also equal to the square (which) can be

described on EH .
Thus, a square—(namely), that (which) can be de-

scribed on EH—has been constructed, equal to the given

rectilinear figure A. (Which is) the very thing it was re-
quired to do.

ELEMENTS BOOK 3

Fundamentals of Plane Geometry Involving
Circles

STOIQEIWN gþ. ELEMENTS BOOK 3VOroi. Definitions
αʹ. ῎Ισοι κύκλοι εἰσίν, ὧν αἱ διάμετροι ἴσαι εἰσίν, ἢ ὧν αἱ 1. Equal circles are (circles) whose diameters are

ἐκ τῶν κέντρων ἴσαι εἰσίν. equal, or whose (distances) from the centers (to the cir-
βʹ. Εὐθεῖα κύκλου ἐφάπτεσθαι λέγεται, ἥτις ἁπτομένη cumferences) are equal (i.e., whose radii are equal).

τοῦ κύκλου καὶ ἐκβαλλομένη οὐ τέμνει τὸν κύκλον. 2. A straight-line said to touch a circle is any (straight-
γʹ. Κύκλοι ἐφάπτεσθαι ἀλλήλων λέγονται οἵτινες ἁπτό- line) which, meeting the circle and being produced, does

μενοι ἀλλήλων οὐ τέμνουσιν ἀλλήλους. not cut the circle.
δʹ. ᾿Εν κύκλῳ ἴσον ἀπέχειν ἀπὸ τοῦ κέντρου εὐθεῖαι 3. Circles said to touch one another are any (circles)

λέγονται, ὅταν αἱ ἀπὸ τοῦ κέντρου ἐπ᾿ αὐτὰς κάθετοι which, meeting one another, do not cut one another.
ἀγόμεναι ἴσαι ὦσιν. 4. In a circle, straight-lines are said to be equally far
εʹ. Μεῖζον δὲ ἀπέχειν λέγεται, ἐφ᾿ ἣν ἡ μείζων κάθετος from the center when the perpendiculars drawn to them

πίπτει. from the center are equal.
ϛʹ. Τμῆμα κύκλου ἐστὶ τὸ περιεχόμενον σχῆμα ὑπό τε 5. And (that straight-line) is said to be further (from

εὐθείας καὶ κύκλου περιφερείας. the center) on which the greater perpendicular falls
ζʹ. Τμήματος δὲ γωνία ἐστὶν ἡ περιεχομένη ὑπό τε (from the center).

εὐθείας καὶ κύκλου περιφερείας. 6. A segment of a circle is the figure contained by a
ηʹ. ᾿Εν τμήματι δὲ γωνία ἐστίν, ὅταν ἐπὶ τῆς περι- straight-line and a circumference of a circle.

φερείας τοῦ τμήματος ληφθῇ τι σημεῖον καὶ ἀπ᾿ αὐτοῦ ἐπὶ 7. And the angle of a segment is that contained by a
τὰ πέρατα τῆς εὐθείας, ἥ ἐστι βάσις τοῦ τμήματος, ἐπι- straight-line and a circumference of a circle.
ζευχθῶσιν εὐθεῖαι, ἡ περιεχομένη γωνία ὑπὸ τῶν ἐπιζευ- 8. And the angle in a segment is the angle contained
χθεισῶν εὐθειῶν. by the joined straight-lines, when any point is taken on
θʹ. ῞Οταν δὲ αἱ περιέχουσαι τὴν γωνίαν εὐθεῖαι ἀπο- the circumference of a segment, and straight-lines are

λαμβάνωσί τινα περιφέρειαν, ἐπ᾿ ἐκείνης λέγεται βεβηκέναι joined from it to the ends of the straight-line which is
ἡ γωνία. the base of the segment.
ιʹ. Τομεὺς δὲ κύκλου ἐστίν, ὅταν πρὸς τῷ κέντρῷ τοῦ 9. And when the straight-lines containing an angle

κύκλου συσταθῇ γωνία, τὸ περιεχόμενον σχῆμα ὑπό τε τῶν cut off some circumference, the angle is said to stand
τὴν γωνίαν περιεχουσῶν εὐθειῶν καὶ τῆς ἀπολαμβανομένης upon that (circumference).
ὑπ᾿ αὐτῶν περιφερείας. 10. And a sector of a circle is the figure contained by
ιαʹ. ῞Ομοία τμήματα κύκλων ἐστὶ τὰ δεχόμενα γωνίας the straight-lines surrounding an angle, and the circum-

ἴσας, ἤ ἐν οἷς αἱ γωνίαι ἴσαι ἀλλήλαις εἰσίν. ference cut off by them, when the angle is constructed at
the center of a circle.

11. Similar segments of circles are those accepting

equal angles, or in which the angles are equal to one an-
other.aþ. Proposition 1

Τοῦ δοθέντος κύκλου τὸ κέντρον εὑρεῖν. To find the center of a given circle.
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ· δεῖ δὴ τοῦ ΑΒΓ κύκλου Let ABC be the given circle. So it is required to find

τὸ κέντρον εὑρεῖν. the center of circle ABC.
Διήχθω τις εἰς αὐτόν, ὡς ἔτυχεν, εὐθεῖα ἡ ΑΒ, καὶ Let some straight-line AB have been drawn through

τετμήσθω δίχα κατὰ τὸ Δ σημεῖον, καὶ ἀπὸ τοῦ Δ τῇ ΑΒ (ABC), at random, and let (AB) have been cut in half at
πρὸς ὀρθὰς ἤχθω ἡ ΔΓ καὶ διήχθω ἐπὶ τὸ Ε, καὶ τετμήσθω point D [Prop. 1.9]. And let DC have been drawn from
ἡ ΓΕ δίχα κατὰ τὸ Ζ· λέγω, ὅτι τὸ Ζ κέντρον ἐστὶ τοῦ ΑΒΓ D, at right-angles to AB [Prop. 1.11]. And let (CD) have
[κύκλου]. been drawn through to E. And let CE have been cut in
Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, ἔστω τὸ Η, καὶ ἐπεζεύχθωσαν half at F [Prop. 1.9]. I say that (point) F is the center of

αἱ ΗΑ, ΗΔ, ΗΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ ἡ the [circle] ABC.
ΔΗ, δύο δὴ αἱ ΑΔ, ΔΗ δύο ταῖς ΗΔ, ΔΒ ἴσαι εἰσὶν ἑκατέρα For (if) not then, if possible, let G (be the center of the
ἑκατέρᾳ· καὶ βάσις ἡ ΗΑ βάσει τῇ ΗΒ ἐστιν ἴση· ἐκ κέντρου circle), and let GA, GD, and GB have been joined. And
γάρ· γωνία ἄρα ἡ ὑπὸ ΑΔΗ γωνίᾳ τῇ ὑπὸ ΗΔΒ ἴση ἐστίν. since AD is equal to DB, and DG (is) common, the two

STOIQEIWN gþ. ELEMENTS BOOK 3
ὅταν δὲ εὐθεῖα ἐπ᾿ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας (straight-lines) AD, DG are equal to the two (straight-

ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν· lines) BD, DG,† respectively. And the base GA is equal
ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΗΔΒ. ἐστὶ δὲ καὶ ἡ ὑπὸ ΖΔΒ ὀρθή· to the base GB. For (they are both) radii. Thus, angle
ἴση ἄρα ἡ ὑπὸ ΖΔΒ τῇ ὑπὸ ΗΔΒ, ἡ μείζων τῇ ἐλάττονι· ADG is equal to angle GDB [Prop. 1.8]. And when a
ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Η κέντρον ἐστὶ τοῦ ΑΒΓ straight-line stood upon (another) straight-line make ad-
κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾿ ἄλλο τι πλὴν τοῦ Ζ. jacent angles (which are) equal to one another, each of

the equal angles is a right-angle [Def. 1.10]. Thus, GDB

is a right-angle. And FDB is also a right-angle. Thus,
FDB (is) equal to GDB, the greater to the lesser. The

very thing is impossible. Thus, (point) G is not the center

of the circle ABC. So, similarly, we can show that neither
is any other (point) except F .

∆
Β

Η
Ζ

Α
D

Τὸ Ζ ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ [κύκλου]. Thus, point F is the center of the [circle] ABC.Pìrisma. Corollary
᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν ἐν κύκλῳ εὐθεῖά τις So, from this, (it is) manifest that if any straight-line

εὐθεῖάν τινα δίχα καὶ πρὸς ὀρθὰς τέμνῃ, ἐπὶ τῆς τεμνούσης in a circle cuts any (other) straight-line in half, and at
ἐστὶ τὸ κέντρον τοῦ κύκλου. — ὅπερ ἔδει ποιῆσαι. right-angles, then the center of the circle is on the for-

mer (straight-line). — (Which is) the very thing it was
required to do.

† The Greek text has “GD, DB”, which is obviously a mistake.bþ. Proposition 2
᾿Εὰν κύκλου ἐπὶ τῆς περιφερείας ληφθῇ δύο τυχόντα If two points are taken at random on the circumfer-

σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται ence of a circle then the straight-line joining the points
τοῦ κύκλου. will fall inside the circle.
῎Εστω κύκλος ὁ ΑΒΓ, καὶ ἐπὶ τῆς περιφερείας αὐτοῦ Let ABC be a circle, and let two points A and B have

εἰλήφθω δύο τυχόντα σημεῖα τὰ Α, Β· λέγω, ὅτι ἡ ἀπὸ been taken at random on its circumference. I say that the
τοῦ Α ἐπὶ τὸ Β ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται τοῦ straight-line joining A to B will fall inside the circle.
κύκλου. For (if) not then, if possible, let it fall outside (the
Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, πιπτέτω ἐκτὸς ὡς ἡ ΑΕΒ, καὶ circle), like AEB (in the figure). And let the center of

εἰλήφθω τὸ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ Δ, καὶ the circle ABC have been found [Prop. 3.1], and let it be
ἐπεζεύχθωσαν αἱ ΔΑ, ΔΒ, καὶ διήχθω ἡ ΔΖΕ. (at point) D. And let DA and DB have been joined, and
᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΔΒ, ἴση ἄρα καὶ γωνία ἡ let DFE have been drawn through.

ὑπὸ ΔΑΕ τῇ ὑπὸ ΔΒΕ· καὶ ἐπεὶ τριγώνου τοῦ ΔΑΕ μία Therefore, since DA is equal to DB, the angle DAE

STOIQEIWN gþ. ELEMENTS BOOK 3
πλευρὰ προσεκβέβληται ἡ ΑΕΒ, μείζων ἄρα ἡ ὑπὸ ΔΕΒ (is) thus also equal to DBE [Prop. 1.5]. And since in tri-
γωνία τῆς ὑπὸ ΔΑΕ. ἴση δὲ ἡ ὑπὸ ΔΑΕ τῇ ὑπὸ ΔΒΕ· angle DAE the one side, AEB, has been produced, an-
μείζων ἄρα ἡ ὑπὸ ΔΕΒ τῆς ὑπὸ ΔΒΕ. ὑπὸ δὲ τὴν μείζονα gle DEB (is) thus greater than DAE [Prop. 1.16]. And
γωνίαν ἡ μείζων πλευρὰ ὑποτείνει· μείζων ἄρα ἡ ΔΒ τῆς DAE (is) equal to DBE [Prop. 1.5]. Thus, DEB (is)
ΔΕ. ἴση δὲ ἡ ΔΒ τῇ ΔΖ. μείζων ἄρα ἡ ΔΖ τῆς ΔΕ ἡ greater than DBE. And the greater angle is subtended
ἐλάττων τῆς μείζονος· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ by the greater side [Prop. 1.19]. Thus, DB (is) greater
ἀπὸ τοῦ Α ἐπὶ τὸ Β ἐπιζευγνυμένη εὐθεῖα ἐκτὸς πεσεῖται than DE. And DB (is) equal to DF . Thus, DF (is)
τοῦ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἐπ᾿ αὐτῆς τῆς greater than DE, the lesser than the greater. The very
περιφερείας· ἐντὸς ἄρα. thing is impossible. Thus, the straight-line joining A to

B will not fall outside the circle. So, similarly, we can

show that neither (will it fall) on the circumference itself.
Thus, (it will fall) inside (the circle).

∆

Β
Ε

᾿Εὰν ἄρα κύκλου ἐπὶ τῆς περιφερείας ληφθῇ δύο τυχόντα Thus, if two points are taken at random on the cir-
σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται cumference of a circle then the straight-line joining the
τοῦ κύκλου· ὅπερ ἔδει δεῖξαι. points will fall inside the circle. (Which is) the very thing

it was required to show.gþ. Proposition 3
᾿Εὰν ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν τινα In a circle, if any straight-line through the center cuts

μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει· in half any straight-line not through the center then it
καὶ ἐὰν πρὸς ὀρθὰς αὐτὴν τέμνῃ, καὶ δίχα αὐτὴν τέμνει. also cuts it at right-angles. And (conversely) if it cuts it
῎Εστω κύκλος ὁ ΑΒΓ, καὶ ἐν αὐτῷ εὐθεῖά τις διὰ τοῦ at right-angles then it also cuts it in half.

κέντρου ἡ ΓΔ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΒ δίχα Let ABC be a circle, and, within it, let some straight-
τεμνέτω κατὰ τὸ Ζ σημεῖον· λέγω, ὅτι καὶ πρὸς ὀρθὰς αὐτὴν line through the center, CD, cut in half some straight-line
τέμνει. not through the center, AB, at the point F . I say that
Εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω (CD) also cuts (AB) at right-angles.

τὸ Ε, καὶ ἐπεζεύχθωσαν αἱ ΕΑ, ΕΒ. For let the center of the circle ABC have been found
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, κοινὴ δὲ ἡ ΖΕ, δύο δυσὶν [Prop. 3.1], and let it be (at point) E, and let EA and

ἴσαι [εἰσίν]· καὶ βάσις ἡ ΕΑ βάσει τῇ ΕΒ ἴση· γωνία ἄρα ἡ EB have been joined.
ὑπὸ ΑΖΕ γωνίᾳ τῇ ὑπὸ ΒΖΕ ἴση ἐστίν. ὅταν δὲ εὐθεῖα ἐπ᾿ And since AF is equal to FB, and FE (is) common,
εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, two (sides of triangle AFE) [are] equal to two (sides of
ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν· ἑκατέρα ἄρα τῶν triangle BFE). And the base EA (is) equal to the base
ὑπὸ ΑΖΕ, ΒΖΕ ὀρθή ἐστιν. ἡ ΓΔ ἄρα διὰ τοῦ κέντρου EB. Thus, angle AFE is equal to angle BFE [Prop. 1.8].
οὖσα τὴν ΑΒ μὴ διὰ τοῦ κέντρου οὖσαν δίχα τέμνουσα καὶ And when a straight-line stood upon (another) straight-
πρὸς ὀρθὰς τέμνει. line makes adjacent angles (which are) equal to one an-

other, each of the equal angles is a right-angle [Def. 1.10].

Thus, AFE and BFE are each right-angles. Thus, the

STOIQEIWN gþ. ELEMENTS BOOK 3
(straight-line) CD, which is through the center and cuts

in half the (straight-line) AB, which is not through the

center, also cuts (AB) at right-angles.

Α Β

∆ D

F
B

Ἀλλὰ δὴ ἡ ΓΔ τὴν ΑΒ πρὸς ὀρθὰς τεμνέτω· λέγω, ὅτι And so let CD cut AB at right-angles. I say that it
καὶ δίχα αὐτὴν τέμνει, τουτέστιν, ὅτι ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ. also cuts (AB) in half. That is to say, that AF is equal to
Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ ἴση ἐστὶν ἡ FB.

ΕΑ τῇ ΕΒ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΕΑΖ τῇ ὑπὸ ΕΒΖ. For, with the same construction, since EA is equal
ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΑΖΕ ὀρθῇ τῇ ὑπὸ ΒΖΕ ἴση· δύο to EB, angle EAF is also equal to EBF [Prop. 1.5].
ἄρα τρίγωνά ἐστι ΕΑΖ, ΕΖΒ τὰς δύο γωνίας δυσὶ γωνίαις And the right-angle AFE is also equal to the right-angle
ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν BFE. Thus, EAF and EFB are two triangles having
τὴν ΕΖ ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν· καὶ τὰς two angles equal to two angles, and one side equal to
λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει· ἴση ἄρα one side—(namely), their common (side) EF , subtend-
ἡ ΑΖ τῇ ΖΒ. ing one of the equal angles. Thus, they will also have the
᾿Εὰν ἄρα ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν remaining sides equal to the (corresponding) remaining

τινα μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν sides [Prop. 1.26]. Thus, AF (is) equal to FB.
τέμνει· καὶ ἐὰν πρὸς ὀρθὰς αὐτὴν τέμνῃ, καὶ δίχα αὐτὴν Thus, in a circle, if any straight-line through the cen-
τέμνει· ὅπερ ἔδει δεῖξαι. ter cuts in half any straight-line not through the center

then it also cuts it at right-angles. And (conversely) if it

cuts it at right-angles then it also cuts it in half. (Which
is) the very thing it was required to show.dþ. Proposition 4

᾿Εὰν ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας μὴ δὶα τοῦ In a circle, if two straight-lines, which are not through
κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα. the center, cut one another then they do not cut one an-
῎Εστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ δύο εὐθεῖαι αἱ ΑΓ, other in half.

ΒΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε μὴ διὰ τοῦ κέντρου Let ABCD be a circle, and within it, let two straight-
οὖσαι· λέγω, ὅτι οὐ τέμνουσιν ἀλλήλας δίχα. lines, AC and BD, which are not through the center, cut
Εἰ γὰρ δυνατόν, τεμνέτωσαν ἀλλήλας δίχα ὥστε ἴσην one another at (point) E. I say that they do not cut one

εἶναι τὴν μὲν ΑΕ τῇ ΕΓ, τὴν δὲ ΒΕ τῇ ΕΔ· καὶ εἰλήφθω τὸ another in half.
κέντρον τοῦ ΑΒΓΔ κύκλου, καὶ ἔστω τὸ Ζ, καὶ ἐπεζεύχθω For, if possible, let them cut one another in half, such
ἡ ΖΕ. that AE is equal to EC, and BE to ED. And let the
᾿Επεὶ οὖν εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΖΕ εὐθεῖάν τινα center of the circle ABCD have been found [Prop. 3.1],

μὴ διὰ τοῦ κέντρου τὴν ΑΓ δίχα τέμνει, καὶ πρὸς ὀρθὰς and let it be (at point) F , and let FE have been joined.
αὐτὴν τέμνει· ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΕΑ· πάλιν, ἐπεὶ εὐθεῖά Therefore, since some straight-line through the center,
τις ἡ ΖΕ εὐθεῖάν τινα τὴν ΒΔ δίχα τέμνει, καὶ πρὸς ὀρθὰς FE, cuts in half some straight-line not through the cen-
αὐτὴν τέμνει· ὀρθὴ ἄρα ἡ ὑπὸ ΖΕΒ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ter, AC, it also cuts it at right-angles [Prop. 3.3]. Thus,
ΖΕΑ ὀρθή· ἴση ἄρα ἡ ὑπὸ ΖΕΑ τῇ ὑπὸ ΖΕΒ ἡ ἐλάττων τῇ FEA is a right-angle. Again, since some straight-line FE

STOIQEIWN gþ. ELEMENTS BOOK 3
μείζονι· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα αἱ ΑΓ, ΒΔ τέμνουσιν cuts in half some straight-line BD, it also cuts it at right-
ἀλλήλας δίχα. angles [Prop. 3.3]. Thus, FEB (is) a right-angle. But

FEA was also shown (to be) a right-angle. Thus, FEA
(is) equal to FEB, the lesser to the greater. The very

thing is impossible. Thus, AC and BD do not cut one
another in half.

∆

ΖΑ

C
B

D
A

᾿Εὰν ἄρα ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας μὴ δὶα Thus, in a circle, if two straight-lines, which are not
τοῦ κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα· ὅπερ ἔδει through the center, cut one another then they do not cut
δεῖξαι. one another in half. (Which is) the very thing it was re-

quired to show.eþ. Proposition 5
᾿Εὰν δύο κύκλοι τέμνωσιν ἀλλήλους, οὐκ ἔσται αὐτῶν If two circles cut one another then they will not have

τὸ αὐτὸ κέντρον. the same center.

∆

ΖΒ

A C

Δύο γὰρ κύκλοι οἱ ΑΒΓ, ΓΔΗ τεμνέτωσαν ἀλλήλους For let the two circles ABC and CDG cut one another
κατὰ τὰ Β, Γ σημεῖα. λέγω, ὅτι οὐκ ἔσται αὐτῶν τὸ αὐτὸ at points B and C. I say that they will not have the same
κέντρον. center.
Εἰ γὰρ δυνατόν, ἔστω τὸ Ε, καὶ ἐπεζεύχθω ἡ ΕΓ, καὶ For, if possible, let E be (the common center), and

διήχθω ἡ ΕΖΗ, ὡς ἔτυχεν. καὶ ἐπεὶ τὸ Ε σημεῖον κέντρον let EC have been joined, and let EFG have been drawn
ἐστὶ τοῦ ΑΒΓ κύκλου, ἵση ἐστὶν ἡ ΕΓ τῇ ΕΖ. πάλιν, ἐπεὶ τὸ through (the two circles), at random. And since point
Ε σημεῖον κέντρον ἐστὶ τοῦ ΓΔΗ κύκλου, ἴση ἐστὶν ἡ ΕΓ E is the center of the circle ABC, EC is equal to EF .
τῇ ΕΗ· ἐδείχθη δὲ ἡ ΕΓ καὶ τῇ ΕΖ ἴση· καὶ ἡ ΕΖ ἄρα τῇ ΕΗ Again, since point E is the center of the circle CDG, EC
ἐστιν ἴση ἡ ἐλάσσων τῇ μείζονι· ὅπερ ἐστὶν ἀδύνατον. οὐκ is equal to EG. But EC was also shown (to be) equal
ἄρα τὸ Ε σημεῖον κέντρον ἐστὶ τῶν ΑΒΓ, ΓΔΗ κύκλων. to EF . Thus, EF is also equal to EG, the lesser to the
᾿Εὰν ἄρα δύο κύκλοι τέμνωσιν ἀλλήλους, οὐκ ἔστιν greater. The very thing is impossible. Thus, point E is not

STOIQEIWN gþ. ELEMENTS BOOK 3
αὐτῶν τὸ αὐτὸ κέντρον· ὅπερ ἔδει δεῖξαι. the (common) center of the circles ABC and CDG.

Thus, if two circles cut one another then they will not

have the same center. (Which is) the very thing it was
required to show.�þ. Proposition 6

᾿Εὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων, οὐκ ἔσται αὐτῶν If two circles touch one another then they will not
τὸ αὐτὸ κέντρον. have the same center.

∆

Γ C

Δύο γὰρ κύκλοι οἱ ΑΒΓ, ΓΔΕ ἐφαπτέσθωσαν ἀλλήλων For let the two circles ABC and CDE touch one an-
κατὰ τὸ Γ σημεῖον· λέγω, ὅτι οὐκ ἔσται αὐτῶν τὸ αὐτὸ other at point C. I say that they will not have the same
κέντρον. center.
Εἰ γὰρ δυνατόν, ἔστω τὸ Ζ, καὶ ἐπεζεύχθω ἡ ΖΓ, καὶ For, if possible, let F be (the common center), and

διήχθω, ὡς ἔτυχεν, ἡ ΖΕΒ. let FC have been joined, and let FEB have been drawn
᾿Επεὶ οὖν τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου, through (the two circles), at random.

ἴση ἐστὶν ἡ ΖΓ τῇ ΖΒ. πάλιν, ἐπεὶ τὸ Ζ σημεῖον κέντρον Therefore, since point F is the center of the circle
ἐστὶ τοῦ ΓΔΕ κύκλου, ἴση ἐστὶν ἡ ΖΓ τῇ ΖΕ. ἐδείχθη δὲ ἡ ABC, FC is equal to FB. Again, since point F is the
ΖΓ τῇ ΖΒ ἴση· καὶ ἡ ΖΕ ἄρα τῇ ΖΒ ἐστιν ἴση, ἡ ἐλάττων center of the circle CDE, FC is equal to FE. But FC
τῇ μείζονι· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Ζ σημεῖον was shown (to be) equal to FB. Thus, FE is also equal
κέντρον ἐστὶ τῶν ΑΒΓ, ΓΔΕ κύκλων. to FB, the lesser to the greater. The very thing is impos-
᾿Εὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων, οὐκ ἔσται sible. Thus, point F is not the (common) center of the

αὐτῶν τὸ αὐτὸ κέντρον· ὅπερ ἔδει δεῖξαι. circles ABC and CDE.
Thus, if two circles touch one another then they will

not have the same center. (Which is) the very thing it was

required to show.zþ. Proposition 7
᾿Εὰν κύκλου ἐπὶ τῆς διαμέτρου ληφθῇ τι σημεῖον, ὃ μή If some point, which is not the center of the circle,

ἐστι κέντρον τοῦ κύκλου, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν is taken on the diameter of a circle, and some straight-
κύκλον προσπίπτωσιν εὐθεῖαί τινες, μεγίστη μὲν ἔσται, ἐφ᾿ lines radiate from the point towards the (circumference
ἧς τὸ κέντρον, ἐλαχίστη δὲ ἡ λοιπή, τῶν δὲ ἄλλων ἀεὶ ἡ of the) circle, then the greatest (straight-line) will be that
ἔγγιον τῆς δὶα τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, on which the center (lies), and the least the remainder
δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται πρὸς (of the same diameter). And for the others, a (straight-

τὸν κύκλον ἐφ᾿ ἑκάτερα τῆς ἐλαχίστης. line) nearer† to the (straight-line) through the center is
always greater than a (straight-line) further away. And

only two equal (straight-lines) will radiate from the point
towards the (circumference of the) circle, (one) on each

STOIQEIWN gþ. ELEMENTS BOOK 3
(side) of the least (straight-line).

Γ Η

∆
Ζ

Κ
Θ

F
D

῎Εστω κύκλος ὁ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἔστω ἡ ΑΔ, Let ABCD be a circle, and let AD be its diameter, and
καὶ ἐπὶ τῆς ΑΔ εἰλήφθω τι σημεῖον τὸ Ζ, ὃ μή ἐστι κέντρον let some point F , which is not the center of the circle,
τοῦ κύκλου, κέντρον δὲ τοῦ κύκλου ἔστω τὸ Ε, καὶ ἀπὸ τοῦ have been taken on AD. Let E be the center of the circle.
Ζ πρὸς τὸν ΑΒΓΔ κύκλον προσπιπτέτωσαν εὐθεῖαί τινες αἱ And let some straight-lines, FB, FC, and FG, radiate
ΖΒ, ΖΓ, ΖΗ· λέγω, ὅτι μεγίστη μέν ἐστιν ἡ ΖΑ, ἐλαχίστη from F towards (the circumference of) circle ABCD. I
δὲ ἡ ΖΔ, τῶν δὲ ἄλλων ἡ μὲν ΖΒ τῆς ΖΓ μείζων, ἡ δὲ ΖΓ say that FA is the greatest (straight-line), FD the least,
τῆς ΖΗ. and of the others, FB (is) greater than FC, and FC than
᾿Επεζεύχθωσαν γὰρ αἱ ΒΕ, ΓΕ, ΗΕ. καὶ ἐπεὶ παντὸς FG.

τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσιν, αἱ ἄρα For let BE, CE, and GE have been joined. And since
ΕΒ, ΕΖ τῆς ΒΖ μείζονές εἰσιν. ἴση δὲ ἡ ΑΕ τῇ ΒΕ [αἱ ἄρα for every triangle (any) two sides are greater than the
ΒΕ, ΕΖ ἴσαι εἰσὶ τῇ ΑΖ]· μείζων ἄρα ἡ ΑΖ τῆς ΒΖ. πάλιν, remaining (side) [Prop. 1.20], EB and EF is thus greater
ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΓΕ, κοινὴ δὲ ἡ ΖΕ, δύο δὴ αἱ ΒΕ, than BF . And AE (is) equal to BE [thus, BE and EF
ΕΖ δυσὶ ταῖς ΓΕ, ΕΖ ἴσαι εἰσίν. ἀλλὰ καὶ γωνία ἡ ὑπὸ ΒΕΖ is equal to AF ]. Thus, AF (is) greater than BF . Again,
γωνίας τῆς ὑπὸ ΓΕΖ μείζων· βάσις ἄρα ἡ ΒΖ βάσεως τῆς since BE is equal to CE, and FE (is) common, the two
ΓΖ μείζων ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΓΖ τῆς ΖΗ μείζων (straight-lines) BE, EF are equal to the two (straight-
ἐστίν. lines) CE, EF (respectively). But, angle BEF (is) also

Πάλιν, ἐπεὶ αἱ ΗΖ, ΖΕ τῆς ΕΗ μείζονές εἰσιν, ἴση δὲ ἡ greater than angle CEF .‡ Thus, the base BF is greater
ΕΗ τῇ ΕΔ, αἱ ἄρα ΗΖ, ΖΕ τῆς ΕΔ μείζονές εἰσιν. κοινὴ than the base CF . Thus, the base BF is greater than the
ἀφῃρήσθω ἡ ΕΖ· λοιπὴ ἄρα ἡ ΗΖ λοιπῆς τῆς ΖΔ μείζων base CF [Prop. 1.24]. So, for the same (reasons), CF is
ἐστίν. μεγίστη μὲν ἄρα ἡ ΖΑ, ἐλαχίστη δὲ ἡ ΖΔ, μείζων δὲ also greater than FG.
ἡ μὲν ΖΒ τῆς ΖΓ, ἡ δὲ ΖΓ τῆς ΖΗ. Again, since GF and FE are greater than EG
Λέγω, ὅτι καὶ ἀπὸ τοῦ Ζ σημείου δύο μόνον ἴσαι προ- [Prop. 1.20], and EG (is) equal to ED, GF and FE

σπεσοῦνται πρὸς τὸν ΑΒΓΔ κύκλον ἐφ᾿ ἑκάτερα τῆς ΖΔ are thus greater than ED. Let EF have been taken from
ἐλαχίστης. συνεστάτω γὰρ πρὸς τῇ ΕΖ εὐθείᾳ καὶ τῷ πρὸς both. Thus, the remainder GF is greater than the re-
αὐτῇ σημείῳ τῷ Ε τῇ ὑπὸ ΗΕΖ γωνίᾳ ἴση ἡ ὑπὸ ΖΕΘ, καὶ mainder FD. Thus, FA (is) the greatest (straight-line),
ἐπεζεύχθω ἡ ΖΘ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΗΕ τῇ ΕΘ, κοινὴ FD the least, and FB (is) greater than FC, and FC than
δὲ ἡ ΕΖ, δύο δὴ αἱ ΗΕ, ΕΖ δυσὶ ταῖς ΘΕ, ΕΖ ἴσαι εἰσίν· FG.
καὶ γωνία ἡ ὑπὸ ΗΕΖ γωνίᾳ τῇ ὑπὸ ΘΕΖ ἴση· βάσις ἄρα I also say that from point F only two equal (straight-
ἡ ΖΗ βάσει τῇ ΖΘ ἴση ἐστίν. λέγω δή, ὅτι τῇ ΖΗ ἄλλη lines) will radiate towards (the circumference of) circle
ἴση οὐ προσπεσεῖται πρὸς τὸν κύκλον ἀπὸ τοῦ Ζ σημείου. ABCD, (one) on each (side) of the least (straight-line)
εἰ γὰρ δυνατόν, προσπιπτέτω ἡ ΖΚ. καὶ ἐπεὶ ἡ ΖΚ τῇ ΖΗ FD. For let the (angle) FEH , equal to angle GEF , have
ἴση ἐστίν, ἀλλὰ ἡ ΖΘ τῇ ΖΗ [ἴση ἐστίν], καὶ ἡ ΖΚ ἄρα τῇ been constructed on the straight-line EF , at the point E
ΖΘ ἐστιν ἴση, ἡ ἔγγιον τῆς διὰ τοῦ κέντρου τῇ ἀπώτερον on it [Prop. 1.23], and let FH have been joined. There-
ἴση· ὅπερ ἀδύνατον. οὐκ ἄρα ἀπὸ τοῦ Ζ σημείου ἑτέρα τις fore, since GE is equal to EH , and EF (is) common,

STOIQEIWN gþ. ELEMENTS BOOK 3
προσπεσεῖται πρὸς τὸν κύκλον ἴση τῇ ΗΖ· μία ἄρα μόνη. the two (straight-lines) GE, EF are equal to the two
᾿Εὰν ἄρα κύκλου ἐπὶ τῆς διαμέτρου ληφθῇ τι σημεῖον, (straight-lines) HE, EF (respectively). And angle GEF

ὃ μή ἐστι κέντρον τοῦ κύκλου, ἀπὸ δὲ τοῦ σημείου πρὸς (is) equal to angle HEF . Thus, the base FG is equal to
τὸν κύκλον προσπίπτωσιν εὐθεῖαί τινες, μεγίστη μὲν ἔσται, the base FH [Prop. 1.4]. So I say that another (straight-
ἐφ᾿ ἧς τὸ κέντρον, ἐλαχίστη δὲ ἡ λοιπή, τῶν δὲ ἄλλων ἀεὶ ἡ line) equal to FG will not radiate towards (the circumfer-
ἔγγιον τῆς δὶα τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, δύο ence of) the circle from point F . For, if possible, let FK
δὲ μόνον ἴσαι ἀπὸ τοῦ αὐτοῦ σημείου προσπεσοῦνται πρὸς (so) radiate. And since FK is equal to FG, but FH [is
τὸν κύκλον ἐφ᾿ ἑκάτερα τῆς ἐλαχίστης· ὅπερ ἔδει δεῖξαι. equal] to FG, FK is thus also equal to FH , the nearer

to the (straight-line) through the center equal to the fur-

ther away. The very thing (is) impossible. Thus, another

(straight-line) equal to GF will not radiate from the point
F towards (the circumference of) the circle. Thus, (there

is) only one (such straight-line).
Thus, if some point, which is not the center of the

circle, is taken on the diameter of a circle, and some

straight-lines radiate from the point towards the (circum-
ference of the) circle, then the greatest (straight-line)

will be that on which the center (lies), and the least

the remainder (of the same diameter). And for the oth-
ers, a (straight-line) nearer to the (straight-line) through

the center is always greater than a (straight-line) further
away. And only two equal (straight-lines) will radiate

from the same point towards the (circumference of the)

circle, (one) on each (side) of the least (straight-line).
(Which is) the very thing it was required to show.

† Presumably, in an angular sense.
‡ This is not proved, except by reference to the figure.hþ. Proposition 8
᾿Εὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ If some point is taken outside a circle, and some

σημείου πρὸς τὸν κύκλον διαχθῶσιν εὐθεῖαί τινες, ὧν μία straight-lines are drawn from the point to the (circum-
μὲν διὰ τοῦ κέντρου, αἱ δὲ λοιπαί, ὡς ἔτυχεν, τῶν μὲν πρὸς ference of the) circle, one of which (passes) through
τὴν κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη the center, the remainder (being) random, then for the
μέν ἐστιν ἡ διὰ τοῦ κέντρου, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς straight-lines radiating towards the concave (part of the)
διὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, τῶν δὲ πρὸς circumference, the greatest is that (passing) through the

τὴν κυρτὴν περιφέρειαν προσπιπτουσῶν εὐθειῶν ἐλαχίστη center. For the others, a (straight-line) nearer† to the
μέν ἐστιν ἡ μεταξὺ τοῦ τε σημείου καὶ τῆς διαμέτρου, τῶν (straight-line) through the center is always greater than
δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς ἐλαχίστης τῆς ἀπώτερόν ἐστιν one further away. For the straight-lines radiating towards
ἐλάττων, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται the convex (part of the) circumference, the least is that
πρὸς τὸν κύκλον ἐφ᾿ ἑκάτερα τῆς ἐλαχίστης. between the point and the diameter. For the others, a
῎Εστω κύκλος ὁ ΑΒΓ, καὶ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον (straight-line) nearer to the least (straight-line) is always

ἐκτὸς τὸ Δ, καὶ ἀπ᾿ αὐτοῦ διήχθωσαν εὐθεῖαί τινες αἱ ΔΑ, less than one further away. And only two equal (straight-
ΔΕ, ΔΖ, ΔΓ, ἔστω δὲ ἡ ΔΑ διὰ τοῦ κέντρου. λέγω, lines) will radiate from the point towards the (circum-
ὅτι τῶν μὲν πρὸς τὴν ΑΕΖΓ κοίλην περιφέρειαν προσπι- ference of the) circle, (one) on each (side) of the least
πτουσῶν εὐθειῶν μεγίστη μέν ἐστιν ἡ διὰ τοῦ κέντρου ἡ (straight-line).
ΔΑ, μείζων δὲ ἡ μὲν ΔΕ τῆς ΔΖ ἡ δὲ ΔΖ τῆς ΔΓ, τῶν Let ABC be a circle, and let some point D have been
δὲ πρὸς τὴν ΘΛΚΗ κυρτὴν περιφέρειαν προσπιπτουσῶν taken outside ABC, and from it let some straight-lines,
εὐθειῶν ἐλαχίστη μέν ἐστιν ἡ ΔΗ ἡ μεταξὺ τοῦ σημείου καὶ DA, DE, DF , and DC, have been drawn through (the
τῆς διαμέτρου τῆς ΑΗ, ἀεὶ δὲ ἡ ἔγγιον τῆς ΔΗ ἐλαχίστης circle), and let DA be through the center. I say that for
ἐλάττων ἐστὶ τῆς ἀπώτερον, ἡ μὲν ΔΚ τῆς ΔΛ, ἡ δὲ ΔΛ the straight-lines radiating towards the concave (part of

STOIQEIWN gþ. ELEMENTS BOOK 3
τῆς ΔΘ. the) circumference, AEFC, the greatest is the one (pass-

ing) through the center, (namely) AD, and (that) DE (is)

greater than DF , and DF than DC. For the straight-lines
radiating towards the convex (part of the) circumference,

HLKG, the least is the one between the point and the di-
ameter AG, (namely) DG, and a (straight-line) nearer to

the least (straight-line) DG is always less than one far-

ther away, (so that) DK (is less) than DL, and DL than
than DH .

Λ
Κ
Η

∆

Γ C

H L
K

Εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓ κύκλου καὶ ἔστω τὸ For let the center of the circle have been found
Μ· καὶ ἐπεζεύχθωσαν αἱ ΜΕ, ΜΖ, ΜΓ, ΜΚ, ΜΛ, ΜΘ. [Prop. 3.1], and let it be (at point) M [Prop. 3.1]. And let
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΜ τῇ ΕΜ, κοινὴ προσκείσθω ἡ ME, MF , MC, MK, ML, and MH have been joined.

ΜΔ· ἡ ἄρα ΑΔ ἴση ἐστὶ ταῖς ΕΜ, ΜΔ. ἀλλ᾿ αἱ ΕΜ, ΜΔ And since AM is equal to EM , let MD have been
τῆς ΕΔ μείζονές εἰσιν· καὶ ἡ ΑΔ ἄρα τῆς ΕΔ μείζων ἐστίν. added to both. Thus, AD is equal to EM and MD. But,
πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΜΕ τῇ ΜΖ, κοινὴ δὲ ἡ ΜΔ, αἱ ΕΜ, EM and MD is greater than ED [Prop. 1.20]. Thus,
ΜΔ ἄρα ταῖς ΖΜ, ΜΔ ἴσαι εἰσίν· καὶ γωνία ἡ ὑπὸ ΕΜΔ AD is also greater than ED. Again, since ME is equal
γωνίας τῆς ὑπὸ ΖΜΔ μείζων ἐστίν. βάσις ἄρα ἡ ΕΔ βάσεως to MF , and MD (is) common, the (straight-lines) EM ,
τῆς ΖΔ μείζων ἐστίν· ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΖΔ τῆς MD are thus equal to FM , MD. And angle EMD is

ΓΔ μείζων ἐστίν· μεγίστη μὲν ἄρα ἡ ΔΑ, μείζων δὲ ἡ μὲν greater than angle FMD.‡ Thus, the base ED is greater
ΔΕ τῆς ΔΖ, ἡ δὲ ΔΖ τῆς ΔΓ. than the base FD [Prop. 1.24]. So, similarly, we can
Καὶ ἐπεὶ αἱ ΜΚ, ΚΔ τῆς ΜΔ μείζονές εἰσιν, ἴση δὲ ἡ show that FD is also greater than CD. Thus, AD (is) the

ΜΗ τῇ ΜΚ, λοιπὴ ἄρα ἡ ΚΔ λοιπῆς τῆς ΗΔ μείζων ἐστίν· greatest (straight-line), and DE (is) greater than DF ,
ὥστε ἡ ΗΔ τῆς ΚΔ ἐλάττων ἐστίν· καὶ ἐπεὶ τριγώνου τοῦ and DF than DC.
ΜΛΔ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΜΔ δύο εὐθεῖαι ἐντὸς And since MK and KD is greater than MD [Prop.
συνεστάθησαν αἱ ΜΚ, ΚΔ, αἱ ἄρα ΜΚ, ΚΔ τῶν ΜΛ, ΛΔ 1.20], and MG (is) equal to MK, the remainder KD
ἐλάττονές εἰσιν· ἴση δὲ ἡ ΜΚ τῇ ΜΛ· λοιπὴ ἄρα ἡ ΔΚ is thus greater than the remainder GD. So GD is less
λοιπῆς τῆς ΔΛ ἐλάττων ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι than KD. And since in triangle MLD, the two inter-
καὶ ἡ ΔΛ τῆς ΔΘ ἐλάττων ἐστίν· ἐλαχίστη μὲν ἄρα ἡ ΔΗ, nal straight-lines MK and KD were constructed on one
ἐλάττων δὲ ἡ μὲν ΔΚ τῆς ΔΛ ἡ δὲ ΔΛ τῆς ΔΘ. of the sides, MD, then MK and KD are thus less than
Λέγω, ὅτι καὶ δύο μόνον ἴσαι ἀπὸ τοῦ Δ σημείου ML and LD [Prop. 1.21]. And MK (is) equal to ML.

προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾿ ἑκάτερα τῆς ΔΗ Thus, the remainder DK is less than the remainder DL.
ἐλαχίστης· συνεστάτω πρὸς τῇ ΜΔ εὐθείᾳ καὶ τῷ πρὸς So, similarly, we can show that DL is also less than DH .
αὐτῇ σημείῳ τῷ Μ τῇ ὑπὸ ΚΜΔ γωνίᾳ ἴση γωνία ἡ ὑπὸ Thus, DG (is) the least (straight-line), and DK (is) less
ΔΜΒ, καὶ ἐπεζεύχθω ἡ ΔΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΜΚ τῇ than DL, and DL than DH .
ΜΒ, κοινὴ δὲ ἡ ΜΔ, δύο δὴ αἱ ΚΜ, ΜΔ δύο ταῖς ΒΜ, ΜΔ I also say that only two equal (straight-lines) will radi-

STOIQEIWN gþ. ELEMENTS BOOK 3
ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΚΜΔ γωνίᾳ ate from point D towards (the circumference of) the cir-
τῇ ὑπὸ ΒΜΔ ἴση· βάσις ἄρα ἡ ΔΚ βάσει τῇ ΔΒ ἴση ἐστίν. cle, (one) on each (side) on the least (straight-line), DG.
λέγω [δή], ὅτι τῇ ΔΚ εὐθείᾳ ἄλλη ἴση οὐ προσπεσεῖται Let the angle DMB, equal to angle KMD, have been
πρὸς τὸν κύκλον ἀπὸ τοῦ Δ σημείου. εἰ γὰρ δυνατόν, προ- constructed on the straight-line MD, at the point M on
σπιπτέτω καὶ ἔστω ἡ ΔΝ. ἐπεὶ οὖν ἡ ΔΚ τῇ ΔΝ ἐστιν ἴση, it [Prop. 1.23], and let DB have been joined. And since
ἀλλ᾿ ἡ ΔΚ τῇ ΔΒ ἐστιν ἴση, καὶ ἡ ΔΒ ἄρα τῇ ΔΝ ἐστιν MK is equal to MB, and MD (is) common, the two
ἴση, ἡ ἔγγιον τῆς ΔΗ ἐλαχίστης τῇ ἀπώτερον [ἐστιν] ἴση· (straight-lines) KM , MD are equal to the two (straight-
ὅπερ ἀδύνατον ἐδείχθη. οὐκ ἄρα πλείους ἢ δύο ἴσαι πρὸς lines) BM , MD, respectively. And angle KMD (is)
τὸν ΑΒΓ κύκλον ἀπὸ τοῦ Δ σημείου ἐφ᾿ ἑκάτερα τῆς ΔΗ equal to angle BMD. Thus, the base DK is equal to the
ἐλαχίστης προσπεσοῦνται. base DB [Prop. 1.4]. [So] I say that another (straight-
᾿Εὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ line) equal to DK will not radiate towards the (circum-

σημείου πρὸς τὸν κύκλον διαχθῶσιν εὐθεῖαί τινες, ὧν μία ference of the) circle from point D. For, if possible, let
μὲν διὰ τοῦ κέντρου αἱ δὲ λοιπαί, ὡς ἔτυχεν, τῶν μὲν πρὸς (such a straight-line) radiate, and let it be DN . There-
τὴν κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη fore, since DK is equal to DN , but DK is equal to DB,
μέν ἐστιν ἡ διὰ τοῦ κέντου, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς then DB is thus also equal to DN , (so that) a (straight-
διὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, τῶν δὲ πρὸς line) nearer to the least (straight-line) DG [is] equal to
τὴν κυρτὴν περιφέρειαν προσπιπτουσῶν εὐθειῶν ἐλαχίστη one further away. The very thing was shown (to be) im-
μέν ἐστιν ἡ μεταξὺ τοῦ τε σημείου καὶ τῆς διαμέτρου, τῶν possible. Thus, not more than two equal (straight-lines)
δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς ἐλαχίστης τῆς ἀπώτερόν ἐστιν will radiate towards (the circumference of) circle ABC
ἐλάττων, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται from point D, (one) on each side of the least (straight-
πρὸς τὸν κύκλον ἐφ᾿ ἑκάτερα τῆς ἐλαχίστης· ὅπερ ἔδει line) DG.
δεῖξαι. Thus, if some point is taken outside a circle, and some

straight-lines are drawn from the point to the (circumfer-

ence of the) circle, one of which (passes) through the cen-
ter, the remainder (being) random, then for the straight-

lines radiating towards the concave (part of the) circum-

ference, the greatest is that (passing) through the center.
For the others, a (straight-line) nearer to the (straight-

line) through the center is always greater than one fur-
ther away. For the straight-lines radiating towards the

convex (part of the) circumference, the least is that be-

tween the point and the diameter. For the others, a
(straight-line) nearer to the least (straight-line) is always

less than one further away. And only two equal (straight-

lines) will radiate from the point towards the (circum-
ference of the) circle, (one) on each (side) of the least

(straight-line). (Which is) the very thing it was required
to show.

† Presumably, in an angular sense.

‡ This is not proved, except by reference to the figure.jþ. Proposition 9
᾿Εὰν κύκλου ληφθῇ τι σημεῖον ἐντός, ἀπο δὲ τοῦ If some point is taken inside a circle, and more than

σημείου πρὸς τὸν κύκλον προσπίπτωσι πλείους ἢ δύο ἴσαι two equal straight-lines radiate from the point towards
εὐθεῖαι, τὸ ληφθὲν σημεῖον κέντρον ἐστὶ τοῦ κύκλου. the (circumference of the) circle, then the point taken is
῎Εστω κύκλος ὁ ΑΒΓ, ἐντὸς δὲ αὐτοῦ σημεῖον τὸ Δ, καὶ the center of the circle.

ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν πλείους Let ABC be a circle, and D a point inside it, and let
ἢ δύο ἴσαι εὐθεῖαι αἱ ΔΑ, ΔΒ, ΔΓ· λέγω, ὅτι τὸ Δ σημεῖον more than two equal straight-lines, DA, DB, and DC, ra-
κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. diate from D towards (the circumference of) circle ABC.

STOIQEIWN gþ. ELEMENTS BOOK 3
I say that point D is the center of circle ABC.

Κ
Ε

Γ
Ζ

∆

D
E

B
C

᾿Επεζεύχθωσαν γὰρ αἱ ΑΒ, ΒΓ καὶ τετμήσθωσαν For let AB and BC have been joined, and (then)
δίχα κατὰ τὰ Ε, Ζ σημεῖα, καὶ ἐπιζευχθεῖσαι αἱ ΕΔ, ΖΔ have been cut in half at points E and F (respectively)
διήχθωσαν ἐπὶ τὰ Η, Κ, Θ, Λ σημεῖα. [Prop. 1.10]. And ED and FD being joined, let them
᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΑΕ τῇ ΕΒ, κοινὴ δὲ ἡ ΕΔ, δύο have been drawn through to points G, K, H , and L.

δὴ αἱ ΑΕ, ΕΔ δύο ταῖς ΒΕ, ΕΔ ἴσαι εἰσίν· καὶ βάσις ἡ ΔΑ Therefore, since AE is equal to EB, and ED (is) com-
βάσει τῇ ΔΒ ἴση· γωνία ἄρα ἡ ὑπὸ ΑΕΔ γωνίᾳ τῇ ὑπὸ ΒΕΔ mon, the two (straight-lines) AE, ED are equal to the
ἴση ἐστίν· ὀρθὴ ἄρα ἑκατέρα τῶν ὑπὸ ΑΕΔ, ΒΕΔ γωνιῶν· ἡ two (straight-lines) BE, ED (respectively). And the base
ΗΚ ἄρα τὴν ΑΒ τέμνει δίχα καὶ πρὸς ὀρθάς. καὶ ἐπεί, ἐὰν DA (is) equal to the base DB. Thus, angle AED is equal
ἐν κύκλῳ εὐθεῖά τις εὐθεῖάν τινα δίχα τε καὶ πρὸς ὀρθὰς to angle BED [Prop. 1.8]. Thus, angles AED and BED
τέμνῃ, ἐπὶ τῆς τεμνούσης ἐστὶ τὸ κέντρον τοῦ κύκλου, ἐπὶ (are) each right-angles [Def. 1.10]. Thus, GK cuts AB in
τῆς ΗΚ ἄρα ἐστὶ τὸ κέντρον τοῦ κύκλου. διὰ τὰ αὐτὰ δὴ καὶ half, and at right-angles. And since, if some straight-line
ἐπὶ τῆς ΘΛ ἐστι τὸ κέντρον τοῦ ΑΒΓ κύκλου. καὶ οὐδὲν in a circle cuts some (other) straight-line in half, and at
ἕτερον κοινὸν ἔχουσιν αἱ ΗΚ, ΘΛ εὐθεῖαι ἢ τὸ Δ σημεῖον· right-angles, then the center of the circle is on the former
τὸ Δ ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. (straight-line) [Prop. 3.1 corr.], the center of the circle is
᾿Εὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐντός, ἀπὸ δὲ τοῦ thus on GK. So, for the same (reasons), the center of

σημείου πρὸς τὸν κύκλον προσπίπτωσι πλείους ἢ δύο ἴσαι circle ABC is also on HL. And the straight-lines GK and
εὐθεῖαι, τὸ ληφθὲν σημεῖον κέντρον ἐστὶ τοῦ κύκλου· ὅπερ HL have no common (point) other than point D. Thus,
ἔδει δεῖξαι. point D is the center of circle ABC.

Thus, if some point is taken inside a circle, and more

than two equal straight-lines radiate from the point to-

wards the (circumference of the) circle, then the point
taken is the center of the circle. (Which is) the very thing

it was required to show.iþ. Proposition 10
Κύκλος κύκλον οὐ τέμνει κατὰ πλείονα σημεῖα ἢ δύο. A circle does not cut a(nother) circle at more than two
Εἰ γὰρ δυνατόν, κύκλος ὁ ΑΒΓ κύκλον τὸν ΔΕΖ points.

τεμνέτω κατὰ πλείονα σημεῖα ἢ δύο τὰ Β, Η, Ζ, Θ, καὶ For, if possible, let the circle ABC cut the circle DEF
ἐπιζευχθεῖσαι αἱ ΒΘ, ΒΗ δίχα τεμνέσθωσαν κατὰ τὰ Κ, Λ at more than two points, B, G, F , and H . And BH and
σημεῖα· καὶ ἀπὸ τῶν Κ, Λ ταῖς ΒΘ, ΒΗ πρὸς ὀρθὰς ἀχθεῖσαι BG being joined, let them (then) have been cut in half
αἱ ΚΓ, ΛΜ διήχθωσαν ἐπὶ τὰ Α, Ε σημεῖα. at points K and L (respectively). And KC and LM be-

ing drawn at right-angles to BH and BG from K and

L (respectively) [Prop. 1.11], let them (then) have been

drawn through to points A and E (respectively).

STOIQEIWN gþ. ELEMENTS BOOK 3
Β∆

Ξ Ε

Ο
Μ

EL
O

M
P

᾿Επεὶ οὖν ἐν κύκλῳ τῷ ΑΒΓ εὐθεῖά τις ἡ ΑΓ εὐθεῖάν Therefore, since in circle ABC some straight-line
τινα τὴν ΒΘ δίχα καὶ πρὸς ὀρθὰς τέμνει, ἐπὶ τῆς ΑΓ ἄρα AC cuts some (other) straight-line BH in half, and at
ἐστὶ τὸ κέντρον τοῦ ΑΒΓ κύκλου. πάλιν, ἐπεὶ ἐν κύκλῳ τῷ right-angles, the center of circle ABC is thus on AC
αὐτῷ τῷ ΑΒΓ εὐθεῖά τις ἡ ΝΞ εὐθεῖάν τινα τὴν ΒΗ δίχα [Prop. 3.1 corr.]. Again, since in the same circle ABC
καὶ πρὸς ὀρθὰς τέμνει, ἐπὶ τῆς ΝΞ ἄρα ἐστὶ τὸ κέντρον some straight-line NO cuts some (other straight-line) BG
τοῦ ΑΒΓ κύκλου. ἐδείχθη δὲ καὶ ἐπὶ τῆς ΑΓ, καὶ κατ᾿ in half, and at right-angles, the center of circle ABC is
οὐδὲν συμβάλλουσιν αἱ ΑΓ, ΝΞ εὐθεῖαι ἢ κατὰ τὸ Ο· τὸ thus on NO [Prop. 3.1 corr.]. And it was also shown (to
Ο ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ be) on AC. And the straight-lines AC and NO meet at
δείξομεν, ὅτι καὶ τοῦ ΔΕΖ κύκλου κέντρον ἐστὶ τὸ Ο· δύο no other (point) than P . Thus, point P is the center of
ἄρα κύκλων τεμνόντων ἀλλήλους τῶν ΑΒΓ, ΔΕΖ τὸ αὐτό circle ABC. So, similarly, we can show that P is also the
ἐστι κέντρον τὸ Ο· ὅπερ ἐστὶν ἀδύνατον. center of circle DEF . Thus, two circles cutting one an-
Οὐκ ἄρα κύκλος κύκλον τέμνει κατὰ πλείονα σημεῖα ἢ other, ABC and DEF , have the same center P . The very

δύο· ὅπερ ἔδει δεῖξαι. thing is impossible [Prop. 3.5].
Thus, a circle does not cut a(nother) circle at more

than two points. (Which is) the very thing it was required
to show.iaþ. Proposition 11

᾿Εὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐντός, καὶ ληφθῇ If two circles touch one another internally, and their
αὐτῶν τὰ κέντρα, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη centers are found, then the straight-line joining their cen-
εὐθεῖα καὶ ἐκβαλλομένη ἐπὶ τὴν συναφὴν πεσεῖται τῶν ters, being produced, will fall upon the point of union of
κύκλων. the circles.
Δύο γὰρ κύκλοι οἱ ΑΒΓ, ΑΔΕ ἐφαπτέσθωσαν ἀλλήλων For let two circles, ABC and ADE, touch one another

ἐντὸς κατὰ τὸ Α σημεῖον, καὶ εἰλήφθω τοῦ μὲν ΑΒΓ κύκλου internally at point A, and let the center F of circle ABC
κέντρον τὸ Ζ, τοῦ δὲ ΑΔΕ τὸ Η· λέγω, ὅτι ἡ ἀπὸ τοῦ Η ἐπὶ have been found [Prop. 3.1], and (the center) G of (cir-
τὸ Ζ ἐπιζευγνυμένη εὐθεῖα ἐκβαλλομένη ἐπὶ τὸ Α πεσεῖται. cle) ADE [Prop. 3.1]. I say that the straight-line joining
Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, πιπτέτω ὡς ἡ ΖΗΘ, καὶ G to F , being produced, will fall on A.

ἐπεζεύχθωσαν αἱ ΑΖ, ΑΗ. For (if) not then, if possible, let it fall like FGH (in
᾿Επεὶ οὖν αἱ ΑΗ, ΗΖ τῆς ΖΑ, τουτέστι τῆς ΖΘ, μείζονές the figure), and let AF and AG have been joined.

εἰσιν, κοινὴ ἀφῃρήσθω ἡ ΖΗ· λοιπὴ ἄρα ἡ ΑΗ λοιπῆς τῆς Therefore, since AG and GF is greater than FA, that
ΗΘ μείζων ἐστίν. ἴση δὲ ἡ ΑΗ τῇ ΗΔ· καὶ ἡ ΗΔ ἄρα is to say FH [Prop. 1.20], let FG have been taken from
τῆς ΗΘ μείζων ἐστὶν ἡ ἐλάττων τῆς μείζονος· ὅπερ ἐστὶν both. Thus, the remainder AG is greater than the re-
ἀδύνατον· οὐκ ἄρα ἡ ἀπὸ τοῦ Ζ ἐπὶ τὸ Η ἐπιζευγνυμένη mainder GH . And AG (is) equal to GD. Thus, GD is
εὐθεὶα ἐκτὸς πεσεῖται· κατὰ τὸ Α ἄρα ἐπὶ τῆς συναφῆς also greater than GH , the lesser than the greater. The
πεσεῖται. very thing is impossible. Thus, the straight-line joining F

to G will not fall outside (one circle but inside the other).
Thus, it will fall upon the point of union (of the circles)

STOIQEIWN gþ. ELEMENTS BOOK 3
at point A.

∆

᾿Εὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐντός, [καὶ Thus, if two circles touch one another internally, [and
ληφθῇ αὐτῶν τὰ κέντρα], ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευ- their centers are found], then the straight-line joining
γνυμένη εὐθεῖα [καὶ ἐκβαλλομένη] ἐπὶ τὴν συναφὴν πεσεῖται their centers, [being produced], will fall upon the point
τῶν κύκλων· ὅπερ ἔδει δεῖξαι. of union of the circles. (Which is) the very thing it was

required to show.ibþ. Proposition 12
᾿Εὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐκτός, ἡ ἐπὶ τὰ If two circles touch one another externally then the

κέντρα αὐτῶν ἐπιζευγνυμένη διὰ τῆς ἐπαφῆς ἐλεύσεται. (straight-line) joining their centers will go through the
point of union.

∆
Γ

D
C

Δύο γὰρ κύκλοι οἱ ΑΒΓ, ΑΔΕ ἐφαπτέσθωσαν ἀλλήλων For let two circles, ABC and ADE, touch one an-
ἐκτὸς κατὰ τὸ Α σημεῖον, καὶ εἰλήφθω τοῦ μὲν ΑΒΓ other externally at point A, and let the center F of ABC
κέντρον τὸ Ζ, τοῦ δὲ ΑΔΕ τὸ Η· λέγω, ὅτι ἡ ἀπὸ τοῦ have been found [Prop. 3.1], and (the center) G of ADE
Ζ ἐπὶ τὸ Η ἐπιζευγνυμένη εὐθεῖα διὰ τῆς κατὰ τὸ Α ἐπαφῆς [Prop. 3.1]. I say that the straight-line joining F to G will
ἐλεύσεται. go through the point of union at A.
Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, ἐρχέσθω ὡς ἡ ΖΓΔΗ, καὶ For (if) not then, if possible, let it go like FCDG (in

ἐπεζεύχθωσαν αἱ ΑΖ, ΑΗ. the figure), and let AF and AG have been joined.
᾿Επεὶ οὖν τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου, Therefore, since point F is the center of circle ABC,

ἴση ἐστὶν ἡ ΖΑ τῇ ΖΓ. πάλιν, ἐπεὶ τὸ Η σημεῖον κέντρον FA is equal to FC. Again, since point G is the center of
ἐστὶ τοῦ ΑΔΕ κύκλου, ἴση ἐστὶν ἡ ΗΑ τῇ ΗΔ. ἐδείχθη circle ADE, GA is equal to GD. And FA was also shown

STOIQEIWN gþ. ELEMENTS BOOK 3
δὲ καὶ ἡ ΖΑ τῇ ΖΓ ἴση· αἱ ἄρα ΖΑ, ΑΗ ταῖς ΖΓ, ΗΔ ἴσαι (to be) equal to FC. Thus, the (straight-lines) FA and
εἰσίν· ὥστε ὅλη ἡ ΖΗ τῶν ΖΑ, ΑΗ μείζων ἐστίν· ἀλλὰ καὶ AG are equal to the (straight-lines) FC and GD. So the
ἐλάττων· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ Ζ ἐπὶ whole of FG is greater than FA and AG. But, (it is) also
τὸ Η ἐπιζευγνυμένη εὐθεῖα διὰ τῆς κατὰ τὸ Α ἐπαφῆς οὐκ less [Prop. 1.20]. The very thing is impossible. Thus, the
ἐλεύσεται· δι᾿ αὐτῆς ἄρα. straight-line joining F to G cannot not go through the
᾿Εὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐκτός, ἡ ἐπὶ point of union at A. Thus, (it will go) through it.

τὰ κέντρα αὐτῶν ἐπιζευγνυμένη [εὐθεῖα] διὰ τῆς ἐπαφῆς Thus, if two circles touch one another externally then
ἐλεύσεται· ὅπερ ἔδει δεῖξαι. the [straight-line] joining their centers will go through

the point of union. (Which is) the very thing it was re-

quired to show.igþ. Proposition 13
Κύκλος κύκλου οὐκ ἐφάπτεται κατὰ πλείονα σημεῖα ἢ A circle does not touch a(nother) circle at more than

καθ᾿ ἕν, ἐάν τε ἐντὸς ἐάν τε ἐκτὸς ἐφάπτηται. one point, whether they touch internally or externally.

ΘΗ

Β ∆ D

G H

Εἰ γὰρ δυνατόν, κύκλος ὁ ΑΒΓΔ κύκλου τοῦ ΕΒΖΔ For, if possible, let circle ABDC† touch circle EBFD—
ἐφαπτέσθω πρότερον ἐντὸς κατὰ πλείονα σημεῖα ἢ ἓν τὰ Δ, first of all, internally—at more than one point, D and B.
Β. And let the center G of circle ABDC have been found
Καὶ εἰλήφθω τοῦ μὲν ΑΒΓΔ κύκλου κέντρον τὸ Η, τοῦ [Prop. 3.1], and (the center) H of EBFD [Prop. 3.1].

δὲ ΕΒΖΔ τὸ Θ. Thus, the (straight-line) joining G and H will fall on
῾Η ἄρα ἀπὸ τοῦ Η ἐπὶ τὸ Θ ἐπιζευγνυμένη ἐπὶ τὰ Β, B and D [Prop. 3.11]. Let it fall like BGHD (in the

Δ πεσεῖται. πιπτέτω ὡς ἡ ΒΗΘΔ. καὶ ἐπεὶ τὸ Η σημεῖον figure). And since point G is the center of circle ABDC,
κέντρον ἐστὶ τοῦ ΑΒΓΔ κύκλου, ἴση ἐστὶν ἡ ΒΗ τῇ ΗΔ· BG is equal to GD. Thus, BG (is) greater than HD.
μείζων ἄρα ἡ ΒΗ τῆς ΘΔ· πολλῷ ἄρα μείζων ἡ ΒΘ τῆς ΘΔ. Thus, BH (is) much greater than HD. Again, since point
πάλιν, ἐπεὶ τὸ Θ σημεῖον κέντρον ἐστὶ τοῦ ΕΒΖΔ κύκλου, H is the center of circle EBFD, BH is equal to HD.
ἴση ἐστὶν ἡ ΒΘ τῇ ΘΔ· ἐδείχθη δὲ αὐτῆς καὶ πολλῷ μείζων· But it was also shown (to be) much greater than it. The
ὅπερ ἀδύνατον· οὐκ ἄρα κύκλος κύκλου ἐφάπτεται ἐντὸς very thing (is) impossible. Thus, a circle does not touch
κατὰ πλείονα σημεῖα ἢ ἕν. a(nother) circle internally at more than one point.
Λέγω δή, ὅτι οὐδὲ ἐκτός. So, I say that neither (does it touch) externally (at
Εἰ γὰρ δυνατόν, κύκλος ὁ ΑΓΚ κύκλου τοῦ ΑΒΓΔ more than one point).

ἐφαπτέσθω ἐκτὸς κατὰ πλείονα σημεῖα ἢ ἓν τὰ Α, Γ, καὶ For, if possible, let circle ACK touch circle ABDC
ἐπεζεύχθω ἡ ΑΓ. externally at more than one point, A and C. And let AC
῎Επεὶ οὖν κύκλων τῶν ΑΒΓΔ, ΑΓΚ εἴληπται ἐπὶ τῆς have been joined.

περιφερείας ἑκατέρου δύο τυχόντα σημεῖα τὰ Α, Γ, ἡ ἐπὶ Therefore, since two points, A and C, have been taken
τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς ἑκατέρου πεσεῖται· at random on the circumference of each of the circles
ἀλλὰ τοῦ μὲν ΑΒΓΔ ἐντὸς ἔπεσεν, τοῦ δὲ ΑΓΚ ἐκτός· ABDC and ACK, the straight-line joining the points will
ὅπερ ἄτοπον· οὐκ ἄρα κύκλος κύκλου ἐφάπτεται ἐκτὸς κατὰ fall inside each (circle) [Prop. 3.2]. But, it fell inside
πλείονα σημεῖα ἢ ἕν. ἐδείχθη δέ, ὅτι οὐδὲ ἐντός. ABDC, and outside ACK [Def. 3.3]. The very thing

STOIQEIWN gþ. ELEMENTS BOOK 3
Κύκλος ἄρα κύκλου οὐκ ἐφάπτεται κατὰ πλείονα σημεῖα (is) absurd. Thus, a circle does not touch a(nother) circle

ἢ [καθ᾿] ἕν, ἐάν τε ἐντὸς ἐάν τε ἐκτὸς ἐφάπτηται· ὅπερ ἔδει externally at more than one point. And it was shown that
δεῖξαι. neither (does it) internally.

Thus, a circle does not touch a(nother) circle at more

than one point, whether they touch internally or exter-
nally. (Which is) the very thing it was required to show.

† The Greek text has “ABCD”, which is obviously a mistake.idþ. Proposition 14
᾿Εν κύκλῳ αἱ ἴσαι εὐθεῖαι ἴσον ἀπέχουσιν ἀπὸ τοῦ In a circle, equal straight-lines are equally far from the

κέντρου, καὶ αἱ ἴσον ἀπέχουσαι ἀπὸ τοῦ κέντρου ἴσαι center, and (straight-lines) which are equally far from the
ἀλλήλαις εἰσίν. center are equal to one another.

ΗΕ

∆

῎Εστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ ἴσαι εὐθεῖαι ἔστω- Let ABDC† be a circle, and let AB and CD be equal
σαν αἱ ΑΒ, ΓΔ· λέγω, ὅτι αἱ ΑΒ, ΓΔ ἴσον ἀπέχουσιν ἀπὸ straight-lines within it. I say that AB and CD are equally
τοῦ κέντρου. far from the center.
Εἰλήφθω γὰρ τὸ κέντον τοῦ ΑΒΓΔ κύκλου καὶ ἔστω τὸ For let the center of circle ABDC have been found

Ε, καὶ ἀπὸ τοῦ Ε ἐπὶ τὰς ΑΒ, ΓΔ κάθετοι ἤχθωσαν αἱ ΕΖ, [Prop. 3.1], and let it be (at) E. And let EF and EG
ΕΗ, καὶ ἐπεζεύχθωσαν αἱ ΑΕ, ΕΓ. have been drawn from (point) E, perpendicular to AB
᾿Επεὶ οὖν εὐθεῖά τις δὶα τοῦ κέντρου ἡ ΕΖ εὐθεῖάν τινα and CD (respectively) [Prop. 1.12]. And let AE and EC

μὴ διὰ τοῦ κέντρου τὴν ΑΒ πρὸς ὀρθὰς τέμνει, καὶ δίχα have been joined.
αὐτὴν τέμνει. ἴση ἄρα ἡ ΑΖ τῇ ΖΒ· διπλῆ ἄρα ἡ ΑΒ τῆς Therefore, since some straight-line, EF , through the
ΑΖ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΓΔ τῆς ΓΗ ἐστι διπλῆ· καί ἐστιν center (of the circle), cuts some (other) straight-line, AB,
ἴση ἡ ΑΒ τῇ ΓΔ· ἴση ἄρα καὶ ἡ ΑΖ τῇ ΓΗ. καὶ ἐπεὶ ἴση ἐστὶν not through the center, at right-angles, it also cuts it in
ἡ ΑΕ τῇ ΕΓ, ἴσον καὶ τὸ ἀπὸ τῆς ΑΕ τῷ ἀπὸ τῆς ΕΓ. ἀλλὰ half [Prop. 3.3]. Thus, AF (is) equal to FB. Thus, AB
τῷ μὲν ἀπὸ τῆς ΑΕ ἴσα τὰ ἀπὸ τῶν ΑΖ, ΕΖ· ὀρθὴ γὰρ ἡ (is) double AF . So, for the same (reasons), CD is also
πρὸς τῷ Ζ γωνία· τῷ δὲ ἀπὸ τῆς ΕΓ ἴσα τὰ ἀπὸ τῶν ΕΗ, ΗΓ· double CG. And AB is equal to CD. Thus, AF (is)
ὀρθὴ γὰρ ἡ πρὸς τῷ Η γωνία· τὰ ἄρα ἀπὸ τῶν ΑΖ, ΖΕ ἴσα also equal to CG. And since AE is equal to EC, the
ἐστὶ τοῖς ἀπὸ τῶν ΓΗ, ΗΕ, ὧν τὸ ἀπὸ τῆς ΑΖ ἴσον ἐστὶ τῷ (square) on AE (is) also equal to the (square) on EC.
ἀπὸ τῆς ΓΗ· ἴση γάρ ἐστιν ἡ ΑΖ τῇ ΓΗ· λοιπὸν ἄρα τὸ ἀπὸ But, the (sum of the squares) on AF and EF (is) equal
τῆς ΖΕ τῷ ἀπὸ τῆς ΕΗ ἴσον ἐστίν· ἴση ἄρα ἡ ΕΖ τῇ ΕΗ. ἐν to the (square) on AE. For the angle at F (is) a right-
δὲ κύκλῳ ἴσον ἀπέχειν ἀπὸ τοῦ κέντρου εὐθεῖαι λέγονται, angle [Prop. 1.47]. And the (sum of the squares) on EG
ὅταν αἱ ἀπὸ τοῦ κέντρου ἐπ᾿ αὐτὰς κάθετοι ἀγόμεναι ἴσαι and GC (is) equal to the (square) on EC. For the angle
ὦσιν· αἱ ἄρα ΑΒ, ΓΔ ἴσον ἀπέχουσιν ἀπὸ τοῦ κέντρου. at G (is) a right-angle [Prop. 1.47]. Thus, the (sum of
Ἀλλὰ δὴ αἱ ΑΒ, ΓΔ εὐθεῖαι ἴσον ἀπεχέτωσαν ἀπὸ τοῦ the squares) on AF and FE is equal to the (sum of the

κέντρου, τουτέστιν ἴση ἔστω ἡ ΕΖ τῇ ΕΗ. λέγω, ὅτι ἴση squares) on CG and GE, of which the (square) on AF
ἐστὶ καὶ ἡ ΑΒ τῇ ΓΔ. is equal to the (square) on CG. For AF is equal to CG.

STOIQEIWN gþ. ELEMENTS BOOK 3
Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, Thus, the remaining (square) on FE is equal to the (re-

ὅτι διπλῆ ἐστιν ἡ μὲν ΑΒ τῆς ΑΖ, ἡ δὲ ΓΔ τῆς ΓΗ· καὶ ἐπεὶ maining square) on EG. Thus, EF (is) equal to EG. And
ἴση ἐστὶν ἡ ΑΕ τῇ ΓΕ, ἴσον ἐστὶ τὸ ἀπὸ τῆς ΑΕ τῷ ἀπὸ straight-lines in a circle are said to be equally far from
τῆς ΓΕ· ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΕ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΕΖ, the center when perpendicular (straight-lines) which are
ΖΑ, τῷ δὲ ἀπὸ τῆς ΓΕ ἴσα τὰ ἀπὸ τῶν ΕΗ, ΗΓ. τὰ ἄρα ἀπὸ drawn to them from the center are equal [Def. 3.4]. Thus,
τῶν ΕΖ, ΖΑ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΕΗ, ΗΓ· ὧν τὸ ἀπὸ τῆς AB and CD are equally far from the center.
ΕΖ τῷ ἀπὸ τῆς ΕΗ ἐστιν ἴσον· ἴση γὰρ ἡ ΕΖ τῇ ΕΗ· λοιπὸν So, let the straight-lines AB and CD be equally far
ἄρα τὸ ἀπὸ τῆς ΑΖ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΗ· ἴση ἄρα ἡ ΑΖ from the center. That is to say, let EF be equal to EG. I
τῇ ΓΗ· καί ἐστι τῆς μὲν ΑΖ διπλῆ ἡ ΑΒ, τῆς δὲ ΓΗ διπλῆ say that AB is also equal to CD.
ἡ ΓΔ· ἴση ἄρα ἡ ΑΒ τῇ ΓΔ. For, with the same construction, we can, similarly,
᾿Εν κύκλῳ ἄρα αἱ ἴσαι εὐθεῖαι ἴσον ἀπέχουσιν ἀπὸ show that AB is double AF , and CD (double) CG. And

τοῦ κέντρου, καὶ αἱ ἴσον ἀπέχουσαι ἀπὸ τοῦ κέντρου ἴσαι since AE is equal to CE, the (square) on AE is equal to
ἀλλήλαις εἰσίν· ὅπερ ἔδει δεῖξαι. the (square) on CE. But, the (sum of the squares) on

EF and FA is equal to the (square) on AE [Prop. 1.47].

And the (sum of the squares) on EG and GC (is) equal

to the (square) on CE [Prop. 1.47]. Thus, the (sum of
the squares) on EF and FA is equal to the (sum of the

squares) on EG and GC, of which the (square) on EF is

equal to the (square) on EG. For EF (is) equal to EG.
Thus, the remaining (square) on AF is equal to the (re-

maining square) on CG. Thus, AF (is) equal to CG. And
AB is double AF , and CD double CG. Thus, AB (is)

equal to CD.

Thus, in a circle, equal straight-lines are equally far
from the center, and (straight-lines) which are equally far

from the center are equal to one another. (Which is) the

very thing it was required to show.

† The Greek text has “ABCD”, which is obviously a mistake.ieþ. Proposition 15
᾿Εν κύκλῳ μεγίστη μὲν ἡ διάμετρος, τῶν δὲ ἄλλων ἀεὶ In a circle, a diameter (is) the greatest (straight-line),

ἡ ἔγγιον τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν. and for the others, a (straight-line) nearer to the center
῎Εστω κύκλος ὁ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἔστω ἡ ΑΔ, is always greater than one further away.

κέντρον δὲ τὸ Ε, καὶ ἔγγιον μὲν τῆς ΑΔ διαμέτρου ἔστω ἡ Let ABCD be a circle, and let AD be its diameter,
ΒΓ, ἀπώτερον δὲ ἡ ΖΗ· λέγω, ὅτι μεγίστη μέν ἐστιν ἡ ΑΔ, and E (its) center. And let BC be nearer to the diameter

μείζων δὲ ἡ ΒΓ τῆς ΖΗ. AD,† and FG further away. I say that AD is the greatest
῎Ηχθωσαν γὰρ ἀπὸ τοῦ Ε κέντρου ἐπὶ τὰς ΒΓ, ΖΗ (straight-line), and BC (is) greater than FG.

κάθετοι αἱ ΕΘ, ΕΚ. καὶ ἐπεὶ ἔγγιον μὲν τοῦ κέντρου ἐστὶν For let EH and EK have been drawn from the cen-
ἡ ΒΓ, ἀπώτερον δὲ ἡ ΖΗ, μείζων ἄρα ἡ ΕΚ τῆς ΕΘ. κείσθω ter E, at right-angles to BC and FG (respectively)
τῇ ΕΘ ἴση ἡ ΕΛ, καὶ διὰ τοῦ Λ τῇ ΕΚ πρὸς ὀρθὰς ἀχθεῖσα [Prop. 1.12]. And since BC is nearer to the center,
ἡ ΛΜ διήχθω ἐπὶ τὸ Ν, καὶ ἐπεζεύχθωσαν αἱ ΜΕ, ΕΝ, ΖΕ, and FG further away, EK (is) thus greater than EH
ΕΗ. [Def. 3.5]. Let EL be made equal to EH [Prop. 1.3].
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΕΘ τῇ ΕΛ, ἴση ἐστὶ καὶ ἡ ΒΓ τῇ And LM being drawn through L, at right-angles to EK

ΜΝ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ μὲν ΑΕ τῇ ΕΜ, ἡ δὲ ΕΔ τῇ [Prop. 1.11], let it have been drawn through to N . And
ΕΝ, ἡ ἄρα ΑΔ ταῖς ΜΕ, ΕΝ ἴση ἐστίν. ἀλλ᾿ αἱ μὲν ΜΕ, ΕΝ let ME, EN , FE, and EG have been joined.
τῆς ΜΝ μείζονές εἰσιν [καὶ ἡ ΑΔ τῆς ΜΝ μείζων ἐστίν], And since EH is equal to EL, BC is also equal to
ἴση δὲ ἡ ΜΝ τῇ ΒΓ· ἡ ΑΔ ἄρα τῆς ΒΓ μείζων ἐστίν. καὶ MN [Prop. 3.14]. Again, since AE is equal to EM , and
ἐπεὶ δύο αἱ ΜΕ, ΕΝ δύο ταῖς ΖΕ, ΕΗ ἴσαι εἰσίν, καὶ γωνία ED to EN , AD is thus equal to ME and EN . But, ME
ἡ ὑπὸ ΜΕΝ γωνίας τῆς ὑπὸ ΖΕΗ μείζων [ἐστίν], βάσις ἄρα and EN is greater than MN [Prop. 1.20] [also AD is

STOIQEIWN gþ. ELEMENTS BOOK 3
ἡ ΜΝ βάσεως τῆς ΖΗ μείζων ἐστίν. ἀλλὰ ἡ ΜΝ τῇ ΒΓ greater than MN], and MN (is) equal to BC. Thus, AD
ἐδείχθη ἴση [καὶ ἡ ΒΓ τῆς ΖΗ μείζων ἐστίν]. μεγίστη μὲν is greater than BC. And since the two (straight-lines)
ἄρα ἡ ΑΔ διάμετρος, μείζων δὲ ἡ ΒΓ τῆς ΖΗ. ME, EN are equal to the two (straight-lines) FE, EG

(respectively), and angle MEN [is] greater than angle

FEG,‡ the base MN is thus greater than the base FG
[Prop. 1.24]. But, MN was shown (to be) equal to BC

[(so) BC is also greater than FG]. Thus, the diameter

AD (is) the greatest (straight-line), and BC (is) greater
than FG.

Ν
∆

Κ
Λ Ε

N
D

L E

᾿Εν κύκλῳ ἄρα μεγίστη μὲν έστιν ἡ διάμετρος, τῶν δὲ Thus, in a circle, a diameter (is) the greatest (straight-
ἄλλων ἀεὶ ἡ ἔγγιον τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν· line), and for the others, a (straight-line) nearer to the
ὅπερ ἔδει δεῖξαι. center is always greater than one further away. (Which

is) the very thing it was required to show.

† Euclid should have said “to the center”, rather than ”to the diameter AD”, since BC, AD and FG are not necessarily parallel.
‡ This is not proved, except by reference to the figure.i�þ. Proposition 16
῾Η τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾿ ἄκρας A (straight-line) drawn at right-angles to the diameter

ἀγομένη ἐκτὸς πεσεῖται τοῦ κύκλου, καὶ εἰς τὸν μεταξὺ of a circle, from its end, will fall outside the circle. And
τόπον τῆς τε εὐθείας καὶ τῆς περιφερείας ἑτέρα εὐθεῖα οὐ another straight-line cannot be inserted into the space be-
παρεμπεσεῖται, καὶ ἡ μὲν τοῦ ἡμικυκλίου γωνία ἁπάσης tween the (aforementioned) straight-line and the circum-
γωνίας ὀξείας εὐθυγράμμου μείζων ἐστίν, ἡ δὲ λοιπὴ ference. And the angle of the semi-circle is greater than
ἐλάττων. any acute rectilinear angle whatsoever, and the remain-
῎Εστω κύκλος ὁ ΑΒΓ περὶ κέντρον τὸ Δ καὶ διάμετρον ing (angle is) less (than any acute rectilinear angle).

τὴν ΑΒ· λέγω, ὅτι ἡ ἀπὸ τοῦ Α τῇ ΑΒ πρὸς ὀρθὰς ἀπ᾿ Let ABC be a circle around the center D and the di-
ἄκρας ἀγομένη ἐκτὸς πεσεῖται τοῦ κύκλου. ameter AB. I say that the (straight-line) drawn from A,
Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, πιπτέτω ἐντὸς ὡς ἡ ΓΑ, καὶ at right-angles to AB [Prop 1.11], from its end, will fall

ἐπεζεύχθω ἡ ΔΓ. outside the circle.
᾿Επεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΔΓ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ For (if) not then, if possible, let it fall inside, like CA

ΔΑΓ γωνίᾳ τῇ ὑπὸ ΑΓΔ. ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ· ὀρθὴ ἄρα (in the figure), and let DC have been joined.
καὶ ἡ ὑπὸ ΑΓΔ· τριγώνου δὴ τοῦ ΑΓΔ αἱ δύο γωνίαι αἱ Since DA is equal to DC, angle DAC is also equal
ὑπὸ ΔΑΓ, ΑΓΔ δύο ὀρθαῖς ἴσαι εἰσίν· ὅπερ ἐστὶν ἀδύνατον. to angle ACD [Prop. 1.5]. And DAC (is) a right-angle.
οὐκ ἄρα ἡ ἀπὸ τοῦ Α σημείου τῇ ΒΑ πρὸς ὀρθὰς ἀγομένη Thus, ACD (is) also a right-angle. So, in triangle ACD,
ἐντὸς πεσεῖται τοῦ κύκλου. ὁμοίως δὴ δεῖξομεν, ὅτι οὐδ᾿ the two angles DAC and ACD are equal to two right-
ἐπὶ τῆς περιφερείας· ἐκτὸς ἄρα. angles. The very thing is impossible [Prop. 1.17]. Thus,

the (straight-line) drawn from point A, at right-angles

STOIQEIWN gþ. ELEMENTS BOOK 3
to BA, will not fall inside the circle. So, similarly, we

can show that neither (will it fall) on the circumference.

Thus, (it will fall) outside (the circle).

∆
Γ

Θ H
F

Πιπτέτω ὡς ἡ ΑΕ· λέγω δή, ὅτι εἰς τὸν μεταξὺ τόπον Let it fall like AE (in the figure). So, I say that another
τῆς τε ΑΕ εὐθείας καὶ τῆς ΓΘΑ περιφερείας ἑτέρα εὐθεῖα straight-line cannot be inserted into the space between
οὐ παρεμπεσεῖται. the straight-line AE and the circumference CHA.
Εἰ γὰρ δυνατόν, παρεμπιπτέτω ὡς ἡ ΖΑ, καὶ ἤχθω ἀπὸ For, if possible, let it be inserted like FA (in the fig-

τοῦ Δ σημείου ἐπὶ τῆν ΖΑ κάθετος ἡ ΔΗ. καὶ ἐπεὶ ὀρθή ure), and let DG have been drawn from point D, perpen-
ἐστιν ἡ ὑπὸ ΑΗΔ, ἐλάττων δὲ ὀρθῆς ἡ ὑπὸ ΔΑΗ, μείζων dicular to FA [Prop. 1.12]. And since AGD is a right-
ἄρα ἡ ΑΔ τῆς ΔΗ. ἴση δὲ ἡ ΔΑ τῇ ΔΘ· μείζων ἄρα ἡ ΔΘ angle, and DAG (is) less than a right-angle, AD (is)
τῆς ΔΗ, ἡ ἐλάττων τῆς μείζονος· ὅπερ ἐστὶν ἀδύνατον. οὐκ thus greater than DG [Prop. 1.19]. And DA (is) equal
ἄρα εἰς τὸν μεταξὺ τόπον τῆς τε εὐθείας καὶ τῆς περιφερείας to DH . Thus, DH (is) greater than DG, the lesser than
ἑτέρα εὐθεῖα παρεμπεσεῖται. the greater. The very thing is impossible. Thus, another
Λέγω, ὅτι καὶ ἡ μὲν τοῦ ἡμικυκλίου γωνία ἡ περιεχομένη straight-line cannot be inserted into the space between

ὑπό τε τῆς ΒΑ εὐθείας καὶ τῆς ΓΘΑ περιφερείας ἁπάσης the straight-line (AE) and the circumference.
γωνίας ὀξείας εὐθυγράμμου μείζων ἐστίν, ἡ δὲ λοιπὴ ἡ πε- And I also say that the semi-circular angle contained
ριεχομένη ὑπό τε τῆς ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας by the straight-line BA and the circumference CHA is
ἁπάσης γωνίας ὀξείας εὐθυγράμμου ἐλάττων ἐστίν. greater than any acute rectilinear angle whatsoever, and
Εἰ γὰρ ἐστί τις γωνία εὐθύγραμμος μείζων μὲν τῆς the remaining (angle) contained by the circumference

περιεχομένης ὑπό τε τῆς ΒΑ εὐθείας καὶ τῆς ΓΘΑ περι- CHA and the straight-line AE is less than any acute rec-
φερείας, ἐλάττων δὲ τῆς περιεχομένης ὑπό τε τῆς ΓΘΑ tilinear angle whatsoever.
περιφερείας καὶ τὴς ΑΕ εὐθείας, εἰς τὸν μεταξὺ τόπον τῆς For if any rectilinear angle is greater than the (an-
τε ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας εὐθεῖα παρεμ- gle) contained by the straight-line BA and the circum-
πεσεῖται, ἥτις ποιήσει μείζονα μὲν τῆς περιεχομένης ὑπὸ ference CHA, or less than the (angle) contained by the
τε τῆς ΒΑ εὐθείας καὶ τῆς ΓΘΑ περιφερείας ὑπὸ εὐθειῶν circumference CHA and the straight-line AE, then a
περιεχομένην, ἐλάττονα δὲ τῆς περιεχομένης ὑπό τε τῆς straight-line can be inserted into the space between the
ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας. οὐ παρεμπίπτει δέ· circumference CHA and the straight-line AE—anything
οὐκ ἄρα τῆς περιεχομένης γωνίας ὑπό τε τῆς ΒΑ εὐθείας which will make (an angle) contained by straight-lines
καὶ τῆς ΓΘΑ περιφερείας ἔσται μείζων ὀξεῖα ὑπὸ εὐθειῶν greater than the angle contained by the straight-line BA
περιεχομένη, οὐδὲ μὴν ἐλάττων τῆς περιεχομένης ὑπό τε and the circumference CHA, or less than the (angle)
τῆς ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας. contained by the circumference CHA and the straight-

line AE. But (such a straight-line) cannot be inserted.

Thus, an acute (angle) contained by straight-lines cannot
be greater than the angle contained by the straight-line

BA and the circumference CHA, neither (can it be) less
than the (angle) contained by the circumference CHA

and the straight-line AE.

STOIQEIWN gþ. ELEMENTS BOOK 3Pìrisma. Corollary
᾿Εκ δὴ τούτου φανερόν, ὅτι ἡ τῇ διαμέτρῳ τοῦ κύκλου So, from this, (it is) manifest that a (straight-line)

πρὸς ὀρθὰς ἀπ᾿ ἄκρας ἀγομένη ἐφάπτεται τοῦ κύκλου drawn at right-angles to the diameter of a circle, from
[καὶ ὅτι εὐθεῖα κύκλου καθ᾿ ἓν μόνον ἐφάπτεται σημεῖον, its extremity, touches the circle [and that the straight-line
ἐπειδήπερ καὶ ἡ κατὰ δύο αὐτῷ συμβάλλουσα ἐντὸς αὐτοῦ touches the circle at a single point, inasmuch as it was
πίπτουσα ἐδείχθη]· ὅπερ ἔδει δεῖξαι. also shown that a (straight-line) meeting (the circle) at

two (points) falls inside it [Prop. 3.2] ]. (Which is) the

very thing it was required to show.izþ. Proposition 17
Ἀπὸ τοῦ δοθέντος σημείου τοῦ δοθέντος κύκλου ἐφα- To draw a straight-line touching a given circle from a

πτομένην εὐθεῖαν γραμμὴν ἀγαγεῖν. given point.

∆

ΗΓ

Ζ
D

C G
E

῎Εστω τὸ μὲν δοθὲν σημεῖον τὸ Α, ὁ δὲ δοθεὶς κύκλος Let A be the given point, and BCD the given circle.
ὁ ΒΓΔ· δεῖ δὴ ἀπὸ τοῦ Α σημείου τοῦ ΒΓΔ κύκλου ἐφα- So it is required to draw a straight-line touching circle
πτομένην εὐθεῖαν γραμμὴν ἀγαγεῖν. BCD from point A.
Εἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου τὸ Ε, καὶ For let the center E of the circle have been found

ἐπεζεύχθω ἡ ΑΕ, καὶ κέντρῳ μὲν τῷ Ε διαστήματι δὲ τῷ [Prop. 3.1], and let AE have been joined. And let (the
ΕΑ κύκλος γεγράφθω ὁ ΑΖΗ, καὶ ἀπὸ τοῦ Δ τῇ ΕΑ πρὸς circle) AFG have been drawn with center E and radius
ὀρθὰς ἤχθω ἡ ΔΖ, καὶ ἐπεζεύχθωσαν αἱ ΕΖ, ΑΒ· λέγω, EA. And let DF have been drawn from from (point) D,
ὅτι ἀπὸ τοῦ Α σημείου τοῦ ΒΓΔ κύκλου ἐφαπτομένη ἦκται at right-angles to EA [Prop. 1.11]. And let EF and AB
ἡ ΑΒ. have been joined. I say that the (straight-line) AB has
᾿Επεὶ γὰρ τὸ Ε κέντρον ἐστὶ τῶν ΒΓΔ, ΑΖΗ κύκλων, been drawn from point A touching circle BCD.

ἴση ἄρα ἐστὶν ἡ μὲν ΕΑ τῇ ΕΖ, ἡ δὲ ΕΔ τῇ ΕΒ· δύο δὴ For since E is the center of circles BCD and AFG,
αἱ ΑΕ, ΕΒ δύο ταῖς ΖΕ, ΕΔ ἴσαι εἰσίν· καὶ γωνίαν κοινὴν EA is thus equal to EF , and ED to EB. So the two
περιέχουσι τὴν πρὸς τῷ Ε· βάσις ἄρα ἡ ΔΖ βάσει τῇ ΑΒ (straight-lines) AE, EB are equal to the two (straight-
ἴση ἐστίν, καὶ τὸ ΔΕΖ τρίγωνον τῷ ΕΒΑ τριγώνῳ ἴσον lines) FE, ED (respectively). And they contain a com-
ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις· ἴση ἄρα ἡ mon angle at E. Thus, the base DF is equal to the
ὑπὸ ΕΔΖ τῇ ὑπὸ ΕΒΑ. ὀρθὴ δὲ ἡ ὑπὸ ΕΔΖ· ὀρθὴ ἄρα καὶ ἡ base AB, and triangle DEF is equal to triangle EBA,
ὑπὸ ΕΒΑ. καί ἐστιν ἡ ΕΒ ἐκ τοῦ κέντρου· ἡ δὲ τῇ διαμέτρῳ and the remaining angles (are equal) to the (corre-
τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾿ ἄκρας ἀγομένη ἐφάπτεται τοῦ sponding) remaining angles [Prop. 1.4]. Thus, (angle)
κύκλου· ἡ ΑΒ ἄρα ἐφάπτεται τοῦ ΒΓΔ κύκλου. EDF (is) equal to EBA. And EDF (is) a right-angle.
Ἀπὸ τοῦ ἄρα δοθέντος σημείου τοῦ Α τοῦ δοθέντος Thus, EBA (is) also a right-angle. And EB is a ra-

κύκλου τοῦ ΒΓΔ ἐφαπτομένη εὐθεῖα γραμμὴ ἦκται ἡ ΑΒ· dius. And a (straight-line) drawn at right-angles to the
ὅπερ ἔδει ποιῆσαι. diameter of a circle, from its extremity, touches the circle

[Prop. 3.16 corr.]. Thus, AB touches circle BCD.

Thus, the straight-line AB has been drawn touching

STOIQEIWN gþ. ELEMENTS BOOK 3
the given circle BCD from the given point A. (Which is)

the very thing it was required to do.ihþ. Proposition 18
᾿Εὰν κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τοῦ κέντρου If some straight-line touches a circle, and some

ἐπὶ τὴν ἁφὴν ἐπιζευχθῇ τις εὐθεῖα, ἡ ἐπιζευχθεῖσα κάθετος (other) straight-line is joined from the center (of the cir-
ἔσται ἐπὶ τὴν ἐφαπτομένην. cle) to the point of contact, then the (straight-line) so

joined will be perpendicular to the tangent.

∆

Ζ
Β

F
B

Κύκλου γὰρ τοῦ ΑΒΓ ἐφαπτέσθω τις εὐθεῖα ἡ ΔΕ κατὰ For let some straight-line DE touch the circle ABC at
τὸ Γ σημεῖον, καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓ κύκλου τὸ point C, and let the center F of circle ABC have been
Ζ, καὶ ἀπὸ τοῦ Ζ ἐπὶ τὸ Γ ἐπεζεύχθω ἡ ΖΓ· λέγω, ὅτι ἡ ΖΓ found [Prop. 3.1], and let FC have been joined from F
κάθετός ἐστιν ἐπὶ τὴν ΔΕ. to C. I say that FC is perpendicular to DE.
Εἰ γὰρ μή, ἤχθω ἀπὸ τοῦ Ζ ἐπὶ τὴν ΔΕ κάθετος ἡ ΖΗ. For if not, let FG have been drawn from F , perpen-
᾿Επεὶ οὖν ἡ ὑπὸ ΖΗΓ γωνία ὀρθή ἐστιν, ὀξεῖα ἄρα ἐστὶν dicular to DE [Prop. 1.12].

ἡ ὑπὸ ΖΓΗ· ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ Therefore, since angle FGC is a right-angle, (angle)
ὑποτείνει· μείζων ἄρα ἡ ΖΓ τῆς ΖΗ· ἴση δὲ ἡ ΖΓ τῇ ΖΒ· FCG is thus acute [Prop. 1.17]. And the greater angle is
μείζων ἄρα καὶ ἡ ΖΒ τῆς ΖΗ ἡ ἐλάττων τῆς μείζονος· ὅπερ subtended by the greater side [Prop. 1.19]. Thus, FC (is)
ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ΖΗ κάθετός ἐστιν ἐπὶ τὴν ΔΕ. greater than FG. And FC (is) equal to FB. Thus, FB
ὁμοίως δὴ δεῖξομεν, ὅτι οὐδ᾿ ἄλλη τις πλὴν τῆς ΖΓ· ἡ ΖΓ (is) also greater than FG, the lesser than the greater. The
ἄρα κάθετός ἐστιν ἐπὶ τὴν ΔΕ. very thing is impossible. Thus, FG is not perpendicular to
᾿Εὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τοῦ DE. So, similarly, we can show that neither (is) any other

κέντρου ἐπὶ τὴν ἁφὴν ἐπιζευχθῇ τις εὐθεῖα, ἡ ἐπιζευχθεῖσα (straight-line) except FC. Thus, FC is perpendicular to
κάθετος ἔσται ἐπὶ τὴν ἐφαπτομένην· ὅπερ ἔδει δεῖξαι. DE.

Thus, if some straight-line touches a circle, and some

(other) straight-line is joined from the center (of the cir-
cle) to the point of contact, then the (straight-line) so

joined will be perpendicular to the tangent. (Which is)
the very thing it was required to show.ijþ. Proposition 19

᾿Εὰν κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς τῇ If some straight-line touches a circle, and a straight-
ἐφαπτομένῃ πρὸς ὀρθὰς [γωνίας] εὐθεῖα γραμμὴ ἀχθῇ, ἐπὶ line is drawn from the point of contact, at right-[angles]
τῆς ἀχθείσης ἔσται τὸ κέντρον τοῦ κύκλου. to the tangent, then the center (of the circle) will be on
Κύκλου γὰρ τοῦ ΑΒΓ ἐφαπτέσθω τις εὐθεῖα ἡ ΔΕ κατὰ the (straight-line) so drawn.

τὸ Γ σημεῖον, καὶ ἀπὸ τοῦ Γ τῇ ΔΕ πρὸς ὀρθὰς ἤχθω ἡ For let some straight-line DE touch the circle ABC at
ΓΑ· λέγω, ὅτι ἐπὶ τῆς ΑΓ ἐστι τὸ κέντρον τοῦ κύκλου. point C. And let CA have been drawn from C, at right-

STOIQEIWN gþ. ELEMENTS BOOK 3
angles to DE [Prop. 1.11]. I say that the center of the

circle is on AC.

∆

D E
C

Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, ἔστω τὸ Ζ, καὶ ἐπεζεύχθω ἡ For (if) not, if possible, let F be (the center of the
ΓΖ. circle), and let CF have been joined.
᾿Επεὶ [οὖν] κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΔΕ, [Therefore], since some straight-line DE touches the

ἀπὸ δὲ τοῦ κέντρου ἐπὶ τὴν ἁφὴν ἐπέζευκται ἡ ΖΓ, ἡ ΖΓ ἄρα circle ABC, and FC has been joined from the center to
κάθετός ἐστιν ἐπὶ τὴν ΔΕ· ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΓΕ. ἐστὶ the point of contact, FC is thus perpendicular to DE
δὲ καὶ ἡ ὑπὸ ΑΓΕ ὀρθή· ἴση ἄρα ἐστὶν ἡ ὑπὸ ΖΓΕ τῇ ὑπὸ [Prop. 3.18]. Thus, FCE is a right-angle. And ACE
ΑΓΕ ἡ ἐλάττων τῇ μείζονι· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα is also a right-angle. Thus, FCE is equal to ACE, the
τὸ Ζ κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δείξομεν, lesser to the greater. The very thing is impossible. Thus,
ὅτι οὐδ᾿ ἄλλο τι πλὴν ἐπὶ τῆς ΑΓ. F is not the center of circle ABC. So, similarly, we can
᾿Εὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς show that neither is any (point) other (than one) on AC.

τῇ ἐφαπτομένῃ πρὸς ὀρθὰς εὐθεῖα γραμμὴ ἀχθῇ, ἐπὶ τῆς
ἀχθείσης ἔσται τὸ κέντρον τοῦ κύκλου· ὅπερ ἔδει δεῖξαι. Thus, if some straight-line touches a circle, and a straight-

line is drawn from the point of contact, at right-angles to

the tangent, then the center (of the circle) will be on the

(straight-line) so drawn. (Which is) the very thing it was
required to show.kþ. Proposition 20

᾿Εν κύκλῳ ἡ πρὸς τῷ κέντρῳ γωνία διπλασίων ἐστὶ τῆς In a circle, the angle at the center is double that at the
πρὸς τῇ περιφερείᾳ, ὅταν τὴν αὐτὴν περιφέρειαν βάσιν ἔχω- circumference, when the angles have the same circumfer-
σιν αἱ γωνίαι. ence base.
῎Εστω κύκλος ὁ ΑΒΓ, καὶ πρὸς μὲν τῷ κέντρῳ αὐτοῦ Let ABC be a circle, and let BEC be an angle at its

γωνία ἔστω ἡ ὑπὸ ΒΕΓ, πρὸς δὲ τῇ περιφερείᾳ ἡ ὑπὸ ΒΑΓ, center, and BAC (one) at (its) circumference. And let
ἐχέτωσαν δὲ τὴν αὐτὴν περιφέρειαν βάσιν τὴν ΒΓ· λέγω, them have the same circumference base BC. I say that
ὅτι διπλασίων ἐστὶν ἡ ὑπὸ ΒΕΓ γωνία τῆς ὑπὸ ΒΑΓ. angle BEC is double (angle) BAC.
᾿Επιζευχθεῖσα γὰρ ἡ ΑΕ διήχθω ἐπὶ τὸ Ζ. For being joined, let AE have been drawn through to
᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΕΑ τῇ ΕΒ, ἴση καὶ γωνία ἡ ὑπὸ F .

ΕΑΒ τῇ ὑπὸ ΕΒΑ· αἱ ἄρα ὑπὸ ΕΑΒ, ΕΒΑ γωνίαι τῆς ὑπὸ Therefore, since EA is equal to EB, angle EAB (is)
ΕΑΒ διπλασίους εἰσίν. ἴση δὲ ἡ ὑπὸ ΒΕΖ ταῖς ὑπὸ ΕΑΒ, also equal to EBA [Prop. 1.5]. Thus, angle EAB and
ΕΒΑ· καὶ ἡ ὑπὸ ΒΕΖ ἄρα τῆς ὑπὸ ΕΑΒ ἐστι διπλῆ. διὰ τὰ EBA is double (angle) EAB. And BEF (is) equal to
αὐτὰ δὴ καὶ ἡ ὑπὸ ΖΕΓ τῆς ὑπὸ ΕΑΓ ἐστι διπλῆ. ὅλη ἄρα EAB and EBA [Prop. 1.32]. Thus, BEF is also double
ἡ ὑπὸ ΒΕΓ ὅλης τῆς ὑπὸ ΒΑΓ ἐστι διπλῆ. EAB. So, for the same (reasons), FEC is also double

EAC. Thus, the whole (angle) BEC is double the whole
(angle) BAC.

STOIQEIWN gþ. ELEMENTS BOOK 3
Ε

∆

Κεκλάσθω δὴ πάλιν, καὶ ἔστω ἑτέρα γωνία ἡ ὑπὸ ΒΔΓ, So let another (straight-line) have been inflected, and
καὶ ἐπιζευχθεῖσα ἡ ΔΕ ἐκβεβλήσθω ἐπὶ τὸ Η. ὁμοίως δὴ let there be another angle, BDC. And DE being joined,
δείξομεν, ὅτι διπλῆ ἐστιν ἡ ὑπὸ ΗΕΓ γωνία τῆς ὑπὸ ΕΔΓ, let it have been produced to G. So, similarly, we can show
ὧν ἡ ὑπὸ ΗΕΒ διπλῆ ἐστι τῆς ὑπὸ ΕΔΒ· λοιπὴ ἄρα ἡ ὑπὸ that angle GEC is double EDC, of which GEB is double
ΒΕΓ διπλῆ ἐστι τῆς ὑπὸ ΒΔΓ. EDB. Thus, the remaining (angle) BEC is double the
᾿Εν κύκλῳ ἄρα ἡ πρὸς τῷ κέντρῳ γωνία διπλασίων ἐστὶ (remaining angle) BDC.

τῆς πρὸς τῇ περιφερείᾳ, ὅταν τὴν αὐτὴν περιφέρειαν βάσιν Thus, in a circle, the angle at the center is double that
ἔχωσιν [αἱ γωνίαι]· ὅπερ ἔδει δεῖξαι. at the circumference, when [the angles] have the same

circumference base. (Which is) the very thing it was re-

quired to show.kaþ. Proposition 21
᾿Εν κύκλῳ αἱ ἐν τῷ αὐτῷ τμήματι γωνίαι ἴσαι ἀλλήλαις In a circle, angles in the same segment are equal to

εἰσίν. one another.

Β ∆ D

F E

C
῎Εστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν τῷ αὐτῷ τμήματι τῷ Let ABCD be a circle, and let BAD and BED be

ΒΑΕΔ γωνίαι ἔστωσαν αἱ ὑπὸ ΒΑΔ, ΒΕΔ· λέγω, ὅτι αἱ angles in the same segment BAED. I say that angles
ὑπὸ ΒΑΔ, ΒΕΔ γωνίαι ἴσαι ἀλλήλαις εἰσίν. BAD and BED are equal to one another.
Εἰλήφθω γὰρ τοῦ ΑΒΓΔ κύκλου τὸ κέντρον, καὶ ἔστω For let the center of circle ABCD have been found

τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΒΖ, ΖΔ. [Prop. 3.1], and let it be (at point) F . And let BF and
Καὶ ἐπεὶ ἡ μὲν ὑπὸ ΒΖΔ γωνία πρὸς τῷ κέντρῳ ἐστίν, ἡ FD have been joined.

δὲ ὑπὸ ΒΑΔ πρὸς τῇ περιφερείᾳ, καὶ ἔχουσι τὴν αὐτὴν πε- And since angle BFD is at the center, and BAD at
ριφέρειαν βάσιν τὴν ΒΓΔ, ἡ ἄρα ὑπὸ ΒΖΔ γωνία διπλασίων the circumference, and they have the same circumference
ἐστὶ τῆς ὑπὸ ΒΑΔ. διὰ τὰ αὐτὰ δὴ ἡ ὑπὸ ΒΖΔ καὶ τῆς ὑπὸ base BCD, angle BFD is thus double BAD [Prop. 3.20].

STOIQEIWN gþ. ELEMENTS BOOK 3
ΒΕΔ ἐστι διπλσίων· ἴση ἄρα ἡ ὑπὸ ΒΑΔ τῇ ὑπὸ ΒΕΔ. So, for the same (reasons), BFD is also double BED.
᾿Εν κύκλῳ ἄρα αἱ ἐν τῷ αὐτῷ τμήματι γωνίαι ἴσαι Thus, BAD (is) equal to BED.

ἀλλήλαις εἰσίν· ὅπερ ἔδει δεῖξαι. Thus, in a circle, angles in the same segment are equal
to one another. (Which is) the very thing it was required

to show.kbþ. Proposition 22
Τῶν ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον γωνίαι For quadrilaterals within circles, the (sum of the) op-

δυσὶν ὀρθαῖς ἴσαι εἰσίν. posite angles is equal to two right-angles.

∆

῎Εστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ τετράπλευρον ἔστω Let ABCD be a circle, and let ABCD be a quadrilat-
τὸ ΑΒΓΔ· λέγω, ὅτι αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι eral within it. I say that the (sum of the) opposite angles
εἰσίν. is equal to two right-angles.
᾿Επεζεύχθωσαν αἱ ΑΓ, ΒΔ. Let AC and BD have been joined.
᾿Επεὶ οὖν παντὸς τριγώνου αἱ τρεῖς γωνίαι δυσὶν ὀρθαῖς Therefore, since the three angles of any triangle are

ἴσαι εἰσίν, τοῦ ΑΒΓ ἄρα τριγώνου αἱ τρεῖς γωνίαι αἱ ὑπὸ equal to two right-angles [Prop. 1.32], the three angles
ΓΑΒ, ΑΒΓ, ΒΓΑ δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἴση δὲ ἡ μὲν ὑπὸ CAB, ABC, and BCA of triangle ABC are thus equal
ΓΑΒ τῇ ὑπὸ ΒΔΓ· ἐν γὰρ τῷ αὐτῷ τμήματί εἰσι τῷ ΒΑΔΓ· to two right-angles. And CAB (is) equal to BDC. For
ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΑΔΒ· ἐν γὰρ τῷ αὐτῷ τμήματί εἰσι they are in the same segment BADC [Prop. 3.21]. And
τῷ ΑΔΓΒ· ὅλη ἄρα ἡ ὑπὸ ΑΔΓ ταῖς ὑπὸ ΒΑΓ, ΑΓΒ ἴση ACB (is equal) to ADB. For they are in the same seg-
ἐστίν. κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ· αἱ ἄρα ὑπὸ ΑΒΓ, ment ADCB [Prop. 3.21]. Thus, the whole of ADC is
ΒΑΓ, ΑΓΒ ταῖς ὑπὸ ΑΒΓ, ΑΔΓ ἴσαι εἰσίν. ἀλλ᾿ αἱ ὑπὸ equal to BAC and ACB. Let ABC have been added to
ΑΒΓ, ΒΑΓ, ΑΓΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν. καὶ αἱ ὑπὸ ΑΒΓ, both. Thus, ABC, BAC, and ACB are equal to ABC
ΑΔΓ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν. ὁμοίως δὴ δείξομεν, ὅτι and ADC. But, ABC, BAC, and ACB are equal to two
καὶ αἱ ὑπὸ ΒΑΔ, ΔΓΒ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. right-angles. Thus, ABC and ADC are also equal to two
Τῶν ἄρα ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον right-angles. Similarly, we can show that angles BAD

γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν· ὅπερ ἔδει δεῖξαι. and DCB are also equal to two right-angles.
Thus, for quadrilaterals within circles, the (sum of

the) opposite angles is equal to two right-angles. (Which

is) the very thing it was required to show.kgþ. Proposition 23
᾿Επὶ τῆς αὐτῆς εὐθείας δύο τμήματα κύκλων ὅμοια καὶ Two similar and unequal segments of circles cannot be

ἄνισα οὐ συσταθήσεται ἐπὶ τὰ αὐτὰ μέρη. constructed on the same side of the same straight-line.
Εἰ γὰρ δυνατόν, ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ δύο For, if possible, let the two similar and unequal seg-

τμήματα κύκλων ὅμοια καὶ ἄνισα συνεστάτω ἐπὶ τὰ αὐτὰ ments of circles, ACB and ADB, have been constructed
μέρη τὰ ΑΓΒ, ΑΔΒ, καὶ διήχθω ἡ ΑΓΔ, καὶ ἐπεζεύχθωσαν on the same side of the same straight-line AB. And let

STOIQEIWN gþ. ELEMENTS BOOK 3
αἱ ΓΒ, ΔΒ. ACD have been drawn through (the segments), and let

CB and DB have been joined.

Γ
∆

Α Β

A B

᾿Επεὶ οὖν ὅμοιόν ἐστι τὸ ΑΓΒ τμῆμα τῷ ΑΔΒ τμήματι, Therefore, since segment ACB is similar to segment
ὅμοια δὲ τμήματα κύκλων ἐστὶ τὰ δεχόμενα γωνίας ἴσας, ADB, and similar segments of circles are those accept-
ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΑΔΒ ἡ ἐκτὸς τῇ ing equal angles [Def. 3.11], angle ACB is thus equal
ἐντός· ὅπερ ἐστὶν ἀδύνατον. to ADB, the external to the internal. The very thing is
Οὐκ ἄρα ἐπὶ τῆς αὐτῆς εὐθείας δύο τμήματα κύκλων impossible [Prop. 1.16].

ὅμοια καὶ ἄνισα συσταθήσεται ἐπὶ τὰ αὐτὰ μέρη· ὅπερ ἔδει Thus, two similar and unequal segments of circles
δεῖξαι. cannot be constructed on the same side of the same

straight-line.kdþ. Proposition 24
Τὰ ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύλων ἴσα ἀλλήλοις Similar segments of circles on equal straight-lines are

ἐστίν. equal to one another.

Α Β

∆

A B

C D

῎Εστωσαν γὰρ ἐπὶ ἴσων εὐθειῶν τῶν ΑΒ, ΓΔ ὅμοια For let AEB and CFD be similar segments of circles
τμήματα κύκλων τὰ ΑΕΒ, ΓΖΔ· λέγω, ὅτι ἴσον ἐστὶ τὸ on the equal straight-lines AB and CD (respectively). I
ΑΕΒ τμῆμα τῷ ΓΖΔ τμήματι. say that segment AEB is equal to segment CFD.
᾿Εφαρμοζομένου γὰρ τοῦ ΑΕΒ τμήματος ἐπὶ τὸ ΓΖΔ καὶ For if the segment AEB is applied to the segment

τιθεμένου τοῦ μὲν Α σημείου ἐπὶ τὸ Γ τῆς δὲ ΑΒ εὐθείας CFD, and point A is placed on (point) C, and the
ἐπὶ τὴν ΓΔ, ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Δ σημεῖον straight-line AB on CD, then point B will also coincide
διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΓΔ· τῆς δὲ ΑΒ ἐπὶ τὴν ΓΔ ἐφαρ- with point D, on account of AB being equal to CD. And
μοσάσης ἐφαρμόσει καὶ τὸ ΑΕΒ τμῆμα ἐπὶ τὸ ΓΖΔ. εἰ γὰρ if AB coincides with CD then the segment AEB will also
ἡ ΑΒ εὐθεῖα ἐπὶ τὴν ΓΔ ἐφαρμόσει, τὸ δὲ ΑΕΒ τμῆμα ἐπὶ coincide with CFD. For if the straight-line AB coincides
τὸ ΓΖΔ μὴ ἐφαρμόσει, ἤτοι ἐντὸς αὐτοῦ πεσεῖται ἢ ἐκτὸς with CD, and the segment AEB does not coincide with

ἢ παραλλάξει, ὡς τὸ ΓΗΔ, καὶ κύκλος κύκλον τέμνει κατὰ CFD, then it will surely either fall inside it, outside (it),†

πλείονα σημεῖα ἢ δύο· ὅπερ ἐστίν ἀδύνατον. οὐκ ἄρα ἐφαρ- or it will miss like CGD (in the figure), and a circle (will)
μοζομένης τῆς ΑΒ εὐθείας ἐπὶ τὴν ΓΔ οὐκ ἐφαρμόσει καὶ cut (another) circle at more than two points. The very

STOIQEIWN gþ. ELEMENTS BOOK 3
τὸ ΑΕΒ τμῆμα ἐπὶ τὸ ΓΖΔ· ἐφαρμόσει ἄρα, καὶ ἴσον αὐτῷ thing is impossible [Prop. 3.10]. Thus, if the straight-line
ἔσται. AB is applied to CD, the segment AEB cannot not also
Τὰ ἄρα ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύκλων ἴσα coincide with CFD. Thus, it will coincide, and will be

ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι. equal to it [C.N. 4].
Thus, similar segments of circles on equal straight-

lines are equal to one another. (Which is) the very thing

it was required to show.

† Both this possibility, and the previous one, are precluded by Prop. 3.23.keþ. Proposition 25
Κύκλου τμήματος δοθέντος προσαναγράψαι τὸν κύκλον, For a given segment of a circle, to complete the circle,

οὗπέρ ἐστι τμῆμα. the very one of which it is a segment.

∆ ∆

∆

Β Ε

C C

A A

E
B BB

D
E

῎Εστω τὸ δοθὲν τμῆμα κύκλου τὸ ΑΒΓ· δεῖ δὴ τοῦ ΑΒΓ Let ABC be the given segment of a circle. So it is re-
τμήματος προσαναγράψαι τὸν κύκλον, οὖπέρ ἐστι τμῆμα. quired to complete the circle for segment ABC, the very
Τετμήσθω γὰρ ἡ ΑΓ δίχα κατὰ τὸ Δ, καὶ ἤχθω ἀπὸ τοῦ one of which it is a segment.

Δ σημείου τῇ ΑΓ πρὸς ὀρθὰς ἡ ΔΒ, καὶ ἐπεζεύχθω ἡ ΑΒ· For let AC have been cut in half at (point) D
ἡ ὑπὸ ΑΒΔ γωνία ἄρα τῆς ὑπὸ ΒΑΔ ἤτοι μείζων ἐστὶν ἢ [Prop. 1.10], and let DB have been drawn from point
ἴση ἢ ἐλάττων. D, at right-angles to AC [Prop. 1.11]. And let AB have
῎Εστω πρότερον μείζων, καὶ συνεστάτω πρὸς τῇ ΒΑ been joined. Thus, angle ABD is surely either greater

εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΑΒΔ γωνίᾳ than, equal to, or less than (angle) BAD.
ἴση ἡ ὑπὸ ΒΑΕ, καὶ διήχθω ἡ ΔΒ ἐπὶ τὸ Ε, καὶ ἐπεζεύχθω First of all, let it be greater. And let (angle) BAE,
ἡ ΕΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ὑπὸ ΑΒΕ γωνία τῇ ὑπὸ ΒΑΕ, equal to angle ABD, have been constructed on the
ἴση ἄρα ἐστὶ καὶ ἡ ΕΒ εὐθεῖα τῇ ΕΑ. καὶ ἐπεὶ ἴση ἐστὶν ἡ straight-line BA, at the point A on it [Prop. 1.23]. And
ΑΔ τῇ ΔΓ, κοινὴ δὲ ἡ ΔΕ, δύο δὴ αἱ ΑΔ, ΔΕ δύο ταῖς let DB have been drawn through to E, and let EC have
ΓΔ, ΔΕ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΑΔΕ been joined. Therefore, since angle ABE is equal to
γωνίᾳ τῇ ὑπὸ ΓΔΕ ἐστιν ἴση· ὀρθὴ γὰρ ἑκατέρα· βάσις ἄρα BAE, the straight-line EB is thus also equal to EA
ἡ ΑΕ βάσει τῇ ΓΕ ἐστιν ἴση. ἀλλὰ ἡ ΑΕ τῇ ΒΕ ἐδείχθη [Prop. 1.6]. And since AD is equal to DC, and DE (is)
ἴση· καὶ ἡ ΒΕ ἄρα τῇ ΓΕ ἐστιν ἴση· αἱ τρεῖς ἄρα αἱ ΑΕ, ΕΒ, common, the two (straight-lines) AD, DE are equal to
ΕΓ ἴσαι ἀλλήλαις εἰσίν· ὁ ἄρα κέντρῷ τῷ Ε διαστήματι δὲ the two (straight-lines) CD, DE, respectively. And angle
ἑνὶ τῶν ΑΕ, ΕΒ, ΕΓ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν ADE is equal to angle CDE. For each (is) a right-angle.
λοιπῶν σημείων καὶ ἔσται προσαναγεγραμμένος. κύκλου Thus, the base AE is equal to the base CE [Prop. 1.4].
ἄρα τμήματος δοθέντος προσαναγέγραπται ὁ κύκλος. καὶ But, AE was shown (to be) equal to BE. Thus, BE is
δῆλον, ὡς τὸ ΑΒΓ τμῆμα ἔλαττόν ἐστιν ἡμικυκλίου διὰ τὸ also equal to CE. Thus, the three (straight-lines) AE,
τὸ Ε κέντρον ἐκτὸς αὐτοῦ τυγχάνειν. EB, and EC are equal to one another. Thus, if a cir-
῾Ομοίως [δὲ] κἂν ᾖ ἡ ὑπὸ ΑΒΔ γωνία ἴση τῇ ὑπὸ ΒΑΔ, cle is drawn with center E, and radius one of AE, EB,

τῆς ΑΔ ἴσης γενομένης ἑκατέρᾳ τῶν ΒΔ, ΔΓ αἱ τρεῖς αἱ or EC, it will also go through the remaining points (of
ΔΑ, ΔΒ, ΔΓ ἴσαι ἀλλήλαις ἔσονται, καὶ ἔσται τὸ Δ κέντρον the segment), and the (associated circle) will have been
τοῦ προσαναπεπληρωμένου κύκλου, καὶ δηλαδὴ ἔσται τὸ completed [Prop. 3.9]. Thus, a circle has been completed
ΑΒΓ ἡμικύκλιον. from the given segment of a circle. And (it is) clear that
᾿Εὰν δὲ ἡ ὑπὸ ΑΒΔ ἐλάττων ᾖ τῆς ὑπὸ ΒΑΔ, καὶ συ- the segment ABC is less than a semi-circle, because the

στησώμεθα πρὸς τῇ ΒΑ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ center E happens to lie outside it.

STOIQEIWN gþ. ELEMENTS BOOK 3
τῷ Α τῇ ὑπὸ ΑΒΔ γωνίᾳ ἴσην, ἐντὸς τοῦ ΑΒΓ τμήματος [And], similarly, even if angle ABD is equal to BAD,
πεσεῖται τὸ κέντρον ἐπὶ τῆς ΔΒ, καὶ ἔσται δηλαδὴ τὸ ΑΒΓ (since) AD becomes equal to each of BD [Prop. 1.6] and
τμῆμα μεῖζον ἡμικυκλίου. DC, the three (straight-lines) DA, DB, and DC will be
Κύκλου ἄρα τμήματος δοθέντος προσαναγέγραπται ὁ equal to one another. And point D will be the center

κύκλος· ὅπερ ἔδει ποιῆσαι. of the completed circle. And ABC will manifestly be a
semi-circle.

And if ABD is less than BAD, and we construct (an-

gle BAE), equal to angle ABD, on the straight-line BA,
at the point A on it [Prop. 1.23], then the center will fall

on DB, inside the segment ABC. And segment ABC will

manifestly be greater than a semi-circle.
Thus, a circle has been completed from the given seg-

ment of a circle. (Which is) the very thing it was required
to do.k�þ. Proposition 26

᾿Εν τοῖς ἴσοις κύκλοις αἱ ἴσαι γωνίαι ἐπὶ ἴσων περιφε- In equal circles, equal angles stand upon equal cir-
ρειῶν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς cumferences whether they are standing at the center or
ταῖς περιφερείαις ὦσι βεβηκυῖαι. at the circumference.

∆

Κ Λ

B E

῎Εστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ καὶ ἐν αὐτοῖς ἴσαι Let ABC and DEF be equal circles, and within them
γωνίαι ἔστωσαν πρὸς μὲν τοῖς κέντροις αἱ ὑπὸ ΒΗΓ, ΕΘΖ, let BGC and EHF be equal angles at the center, and
πρὸς δὲ ταῖς περιφερείαις αἱ ὑπὸ ΒΑΓ, ΕΔΖ· λέγω, ὅτι ἴση BAC and EDF (equal angles) at the circumference. I
ἐστὶν ἡ ΒΚΓ περιφέρεια τῇ ΕΛΖ περιφερείᾳ. say that circumference BKC is equal to circumference
᾿Επεζεύχθωσαν γὰρ αἱ ΒΓ, ΕΖ. ELF .
Καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΑΒΓ, ΔΕΖ κύκλοι, ἴσαι εἰσὶν αἱ For let BC and EF have been joined.

ἐκ τῶν κέντρων· δύο δὴ αἱ ΒΗ, ΗΓ δύο ταῖς ΕΘ, ΘΖ ἴσαι· And since circles ABC and DEF are equal, their radii
καὶ γωνία ἡ πρὸς τῷ Η γωνίᾳ τῇ πρὸς τῷ Θ ἴση· βάσις ἄρα are equal. So the two (straight-lines) BG, GC (are) equal
ἡ ΒΓ βάσει τῇ ΕΖ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ πρὸς τῷ to the two (straight-lines) EH , HF (respectively). And
Α γωνία τῇ πρὸς τῷ Δ, ὅμοιον ἄρα ἐστὶ τὸ ΒΑΓ τμῆμα τῷ the angle at G (is) equal to the angle at H . Thus, the base
ΕΔΖ τμήματι· καί εἰσιν ἐπὶ ἴσων εὐθειῶν [τῶν ΒΓ, ΕΖ]· τὰ BC is equal to the base EF [Prop. 1.4]. And since the
δὲ ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύκλων ἴσα ἀλλήλοις angle at A is equal to the (angle) at D, the segment BAC
ἐστίν· ἴσον ἄρα τὸ ΒΑΓ τμῆμα τῷ ΕΔΖ. ἔστι δὲ καὶ ὅλος ὁ is thus similar to the segment EDF [Def. 3.11]. And
ΑΒΓ κύκλος ὅλῳ τῷ ΔΕΖ κύκλῳ ἴσος· λοιπὴ ἄρα ἡ ΒΚΓ they are on equal straight-lines [BC and EF ]. And simi-
περιφέρεια τῇ ΕΛΖ περιφερείᾳ ἐστὶν ἴση. lar segments of circles on equal straight-lines are equal to
᾿Εν ἄρα τοῖς ἴσοις κύκλοις αἱ ἴσαι γωνίαι ἐπὶ ἴσων περι- one another [Prop. 3.24]. Thus, segment BAC is equal to

φερειῶν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς (segment) EDF . And the whole circle ABC is also equal
ταῖς περιφερείας ὦσι βεβηκυῖαι· ὅπερ ἔδει δεῖξαι. to the whole circle DEF . Thus, the remaining circum-

ference BKC is equal to the (remaining) circumference
ELF .

Thus, in equal circles, equal angles stand upon equal
circumferences, whether they are standing at the center

STOIQEIWN gþ. ELEMENTS BOOK 3
or at the circumference. (Which is) the very thing which

it was required to show.kzþ. Proposition 27
᾿Εν τοῖς ἴσοις κύκλοις αἱ ἐπὶ ἴσων περιφερειῶν βεβηκυῖαι In equal circles, angles standing upon equal circum-

γωνίαι ἴσαι ἀλλήλαις εἰσίν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε ferences are equal to one another, whether they are
πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι. standing at the center or at the circumference.

∆

Γ Ζ

Β Ε

Α A

FB E

᾿Εν γὰρ ἴσοις κύκλοις τοῖς ΑΒΓ, ΔΕΖ ἐπὶ ἴσων περι- For let the angles BGC and EHF at the centers G
φερειῶν τῶν ΒΓ, ΕΖ πρὸς μὲν τοῖς Η, Θ κέντροις γωνίαι and H , and the (angles) BAC and EDF at the circum-
βεβηκέτωσαν αἱ ὑπὸ ΒΗΓ, ΕΘΖ, πρὸς δὲ ταῖς περιφερείαις ferences, stand upon the equal circumferences BC and
αἱ ὑπὸ ΒΑΓ, ΕΔΖ· λέγω, ὅτι ἡ μὲν ὑπὸ ΒΗΓ γωνία τῇ ὑπὸ EF , in the equal circles ABC and DEF (respectively). I
ΕΘΖ ἐστιν ἴση, ἡ δὲ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση. say that angle BGC is equal to (angle) EHF , and BAC
Εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΒΗΓ τῇ ὑπὸ ΕΘΖ, μία αὐτῶν is equal to EDF .

μείζων ἐστίν. ἔστω μείζων ἡ ὑπὸ ΒΗΓ, καὶ συνεστάτω For if BGC is unequal to EHF , one of them is greater.
πρὸς τῇ ΒΗ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Η τῇ ὑπὸ Let BGC be greater, and let the (angle) BGK, equal to
ΕΘΖ γωνίᾳ ἴση ἡ ὑπὸ ΒΗΚ· αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων angle EHF , have been constructed on the straight-line
περιφερειῶν βεβήκασιν, ὅταν πρὸς τοῖς κέντροις ὦσιν· ἴση BG, at the point G on it [Prop. 1.23]. But equal angles
ἄρα ἡ ΒΚ περιφέρεια τῇ ΕΖ περιφερείᾳ. ἀλλὰ ἡ ΕΖ τῇ ΒΓ (in equal circles) stand upon equal circumferences, when
ἐστιν ἴση· καὶ ἡ ΒΚ ἄρα τῇ ΒΓ ἐστιν ἴση ἡ ἐλάττων τῇ they are at the centers [Prop. 3.26]. Thus, circumference
μείζονι· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ BK (is) equal to circumference EF . But, EF is equal
ΒΗΓ γωνία τῇ ὑπὸ ΕΘΖ· ἴση ἄρα. καί ἐστι τῆς μὲν ὑπὸ to BC. Thus, BK is also equal to BC, the lesser to the
ΒΗΓ ἡμίσεια ἡ πρὸς τῷ Α, τῆς δὲ ὑπὸ ΕΘΖ ἡμίσεια ἡ πρὸς greater. The very thing is impossible. Thus, angle BGC
τῷ Δ· ἴση ἄρα καὶ ἡ πρὸς τῷ Α γωνία τῇ πρὸς τῷ Δ. is not unequal to EHF . Thus, (it is) equal. And the
᾿Εν ἄρα τοῖς ἴσοις κύκλοις αἱ ἐπὶ ἴσων περιφερειῶν βε- (angle) at A is half BGC, and the (angle) at D half EHF

βηκυῖαι γωνίαι ἴσαι ἀλλήλαις εἰσίν, ἐάν τε πρὸς τοῖς κέντροις [Prop. 3.20]. Thus, the angle at A (is) also equal to the
ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι· ὅπερ ἔδει δεῖξαι. (angle) at D.

Thus, in equal circles, angles standing upon equal cir-
cumferences are equal to one another, whether they are

standing at the center or at the circumference. (Which is)

the very thing it was required to show.khþ. Proposition 28
᾿Εν τοῖς ἴσοις κύκλοις αἱ ἴσαι εὐθεῖαι ἴσας περιφερείας In equal circles, equal straight-lines cut off equal cir-

ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ ἐλάττονα τῇ cumferences, the greater (circumference being equal) to
ἐλάττονι. the greater, and the lesser to the lesser.
῎Εστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ, καὶ ἐν τοῖς κύκλοις Let ABC and DEF be equal circles, and let AB

ἴσαι εὐθεῖαι ἔστωσαν αἱ ΑΒ, ΔΕ τὰς μὲν ΑΓΒ, ΑΖΕ περι- and DE be equal straight-lines in these circles, cutting
φερείας μείζονας ἀφαιροῦσαι τὰς δὲ ΑΗΒ, ΔΘΕ ἐλάττονας· off the greater circumferences ACB and DFE, and the
λέγω, ὅτι ἡ μὲν ΑΓΒ μείζων περιφέρεια ἴση ἐστὶ τῇ ΔΖΕ lesser (circumferences) AGB and DHE (respectively). I
μείζονι περιφερείᾳ ἡ δὲ ΑΗΒ ἐλάττων περιφέρεια τῇ ΔΘΕ. say that the greater circumference ACB is equal to the

greater circumference DFE, and the lesser circumfer-

STOIQEIWN gþ. ELEMENTS BOOK 3
ence AGB to (the lesser) DHE.

Η Θ

Β ∆ Ε

Εἰλήφθω γὰρ τὰ κέντρα τῶν κύκλων τὰ Κ, Λ, καὶ For let the centers of the circles, K and L, have been
ἐπεζεύχθωσαν αἱ ΑΚ, ΚΒ, ΔΛ, ΛΕ. found [Prop. 3.1], and let AK, KB, DL, and LE have
Καὶ ἐπεὶ ἴσοι κύκλοι εἰσίν, ἴσαι εἰσὶ καὶ αἱ ἐκ τῶν been joined.

κέντρων· δύο δὴ αἱ ΑΚ, ΚΒ δυσὶ ταῖς ΔΛ, ΛΕ ἴσαι εἰσίν· And since (ABC and DEF ) are equal circles, their
καὶ βάσις ἡ ΑΒ βάσει τῇ ΔΕ ἴση· γωνία ἄρα ἡ ὑπὸ ΑΚΒ radii are also equal [Def. 3.1]. So the two (straight-
γωνίᾳ τῇ ὑπὸ ΔΛΕ ἴση ἐστίν. αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων lines) AK, KB are equal to the two (straight-lines) DL,
περιφερειῶν βεβήκασιν, ὅταν πρὸς τοῖς κέντροις ὦσιν· ἴση LE (respectively). And the base AB (is) equal to the
ἄρα ἡ ΑΗΒ περιφέρεια τῇ ΔΘΕ. ἐστὶ δὲ καὶ ὅλος ὁ ΑΒΓ base DE. Thus, angle AKB is equal to angle DLE
κύκλος ὅλῳ τῷ ΔΕΖ κύκλῳ ἴσος· καὶ λοιπὴ ἄρα ἡ ΑΓΒ [Prop. 1.8]. And equal angles stand upon equal circum-
περιφέρεια λοιπῇ τῇ ΔΖΕ περιφερείᾳ ἴση ἐστίν. ferences, when they are at the centers [Prop. 3.26]. Thus,
᾿Εν ἄρα τοῖς ἴσοις κύκλοις αἱ ἴσαι εὐθεῖαι ἴσας πε- circumference AGB (is) equal to DHE. And the whole

ριφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ circle ABC is also equal to the whole circle DEF . Thus,
ἐλάττονα τῇ ἐλάττονι· ὅπερ ἔδει δεῖξαι. the remaining circumference ACB is also equal to the

remaining circumference DFE.

Thus, in equal circles, equal straight-lines cut off

equal circumferences, the greater (circumference being
equal) to the greater, and the lesser to the lesser. (Which

is) the very thing it was required to show.kjþ. Proposition 29
᾿Εν τοῖς ἴσοις κύκλοις τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι In equal circles, equal straight-lines subtend equal cir-

ὑποτείνουσιν. cumferences.

Κ Λ

Α ∆

Η Θ

Β Ε Ζ E

C F

῎Εστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ, καὶ ἐν αὐτοῖς ἴσαι Let ABC and DEF be equal circles, and within them
περιφέρειαι ἀπειλήφθωσαν αἱ ΒΗΓ, ΕΘΖ, καὶ ἐπεζεύχθωσαν let the equal circumferences BGC and EHF have been
αἱ ΒΓ, ΕΖ εὐθεῖαι· λέγω, ὅτι ἴση ἐστὶν ἡ ΒΓ τῇ ΕΖ. cut off. And let the straight-lines BC and EF have been
Εἰλήφθω γὰρ τὰ κέντρα τῶν κύκλων, καὶ ἔστω τὰ Κ, joined. I say that BC is equal to EF .

Λ, καὶ ἐπεζεύχθωσαν αἱ ΒΚ, ΚΓ, ΕΛ, ΛΖ. For let the centers of the circles have been found
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΗΓ περιφέρεια τῇ ΕΘΖ περιφερείᾳ, [Prop. 3.1], and let them be (at) K and L. And let BK,

STOIQEIWN gþ. ELEMENTS BOOK 3
ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΒΚΓ τῇ ὑπὸ ΕΛΖ. καὶ ἐπεὶ ἴσοι KC, EL, and LF have been joined.
εἰσὶν οἱ ΑΒΓ, ΔΕΖ κύκλοι, ἴσαι εἰσὶ καὶ αἱ ἐκ τῶν κέντρων· And since the circumference BGC is equal to the cir-
δύο δὴ αἱ ΒΚ, ΚΓ δυσὶ ταῖς ΕΛ, ΛΖ ἴσαι εἰσίν· καὶ γωνίας cumference EHF , the angle BKC is also equal to (an-
ἴσας περιέχουσιν· βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν· gle) ELF [Prop. 3.27]. And since the circles ABC and
᾿Εν ἄρα τοῖς ἴσοις κύκλοις τὰς ἴσας περιφερείας ἴσαι DEF are equal, their radii are also equal [Def. 3.1]. So

εὐθεῖαι ὑποτείνουσιν· ὅπερ ἔδει δεῖξαι. the two (straight-lines) BK, KC are equal to the two
(straight-lines) EL, LF (respectively). And they contain

equal angles. Thus, the base BC is equal to the base EF
[Prop. 1.4].

Thus, in equal circles, equal straight-lines subtend

equal circumferences. (Which is) the very thing it was
required to show.lþ. Proposition 30

Τὴν δοθεῖσαν περιφέρειαν δίχα τεμεῖν. To cut a given circumference in half.

Α Β

∆

A BC
῎Εστω ἡ δοθεῖσα περιφέρεια ἡ ΑΔΒ· δεῖ δὴ τὴν ΑΔΒ Let ADB be the given circumference. So it is required

περιφέρειαν δίχα τεμεῖν. to cut circumference ADB in half.
᾿Επεζεύχθω ἡ ΑΒ, καὶ τετμήσθω δίχα κατὰ τὸ Γ, καὶ Let AB have been joined, and let it have been cut in

ἀπὸ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ πρὸς ὀρθὰς ἤχθω ἡ ΓΔ, half at (point) C [Prop. 1.10]. And let CD have been
καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΒ. drawn from point C, at right-angles to AB [Prop. 1.11].
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, κοινὴ δὲ ἡ ΓΔ, δύο And let AD, and DB have been joined.

δὴ αἱ ΑΓ, ΓΔ δυσὶ ταῖς ΒΓ, ΓΔ ἴσαι εἰσίν· καὶ γωνία ἡ And since AC is equal to CB, and CD (is) com-
ὑπὸ ΑΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση· ὀρθὴ γὰρ ἑκατέρα· βάσις mon, the two (straight-lines) AC, CD are equal to the
ἄρα ἡ ΑΔ βάσει τῇ ΔΒ ἴση ἐστίν. αἱ δὲ ἴσαι εὐθεῖαι ἴσας two (straight-lines) BC, CD (respectively). And angle
περιφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ ACD (is) equal to angle BCD. For (they are) each right-
ἐλάττονα τῇ ἐλάττονι· κάι ἐστιν ἑκατέρα τῶν ΑΔ, ΔΒ πε- angles. Thus, the base AD is equal to the base DB
ριφερειῶν ἐλάττων ἡμικυκλίου· ἴση ἄρα ἡ ΑΔ περιφέρεια τῇ [Prop. 1.4]. And equal straight-lines cut off equal circum-
ΔΒ περιφερείᾳ. ferences, the greater (circumference being equal) to the
῾Η ἄρα δοθεῖσα περιφέρεια δίχα τέτμηται κατὰ τὸ Δ greater, and the lesser to the lesser [Prop. 1.28]. And the

σημεῖον· ὅπερ ἔδει ποιῆσαι. circumferences AD and DB are each less than a semi-
circle. Thus, circumference AD (is) equal to circumfer-

ence DB.
Thus, the given circumference has been cut in half at

point D. (Which is) the very thing it was required to do.laþ. Proposition 31
᾿Εν κύκλῳ ἡ μὲν ἐν τῷ ἡμικυκλίῳ γωνία ὀρθή ἐστιν, ἡ δὲ In a circle, the angle in a semi-circle is a right-angle,

ἐν τῷ μείζονι τμήματι ἐλάττων ὀρθῆς, ἡ δὲ ἐν τῷ ἐλάττονι and that in a greater segment (is) less than a right-angle,
τμήματι μείζων ὀρθῆς· καὶ ἔπι ἡ μὲν τοῦ μείζονος τμήματος and that in a lesser segment (is) greater than a right-
γωνία μείζων ἐστὶν ὀρθῆς, ἡ δὲ τοῦ ἐλάττονος τμήματος angle. And, further, the angle of a segment greater (than
γωνία ἐλάττων ὀρθῆς. a semi-circle) is greater than a right-angle, and the an-

STOIQEIWN gþ. ELEMENTS BOOK 3
gle of a segment less (than a semi-circle) is less than a

right-angle.

∆

῎Εστω κύκλος ὁ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἔστω ἡ ΒΓ, Let ABCD be a circle, and let BC be its diameter, and
κέντρον δὲ τὸ Ε, καὶ ἐπεζεύχθωσαν αἱ ΒΑ, ΑΓ, ΑΔ, ΔΓ· E its center. And let BA, AC, AD, and DC have been
λέγω, ὅτι ἡ μὲν ἐν τῷ ΒΑΓ ἡμικυκλίῳ γωνία ἡ ὑπὸ ΒΑΓ joined. I say that the angle BAC in the semi-circle BAC
ὀρθή ἐστιν, ἡ δὲ ἐν τῷ ΑΒΓ μείζονι τοῦ ἡμικυκλίου τμήματι is a right-angle, and the angle ABC in the segment ABC,
γωνία ἡ ὑπὸ ΑΒΓ ἐλάττων ἐστὶν ὀρθῆς, ἡ δὲ ἐν τῷ ΑΔΓ (which is) greater than a semi-circle, is less than a right-
ἐλάττονι τοῦ ἡμικυκλίου τμήματι γωνία ἡ ὑπὸ ΑΔΓ μείζων angle, and the angle ADC in the segment ADC, (which
ἐστὶν ὀρθῆς. is) less than a semi-circle, is greater than a right-angle.
᾿Επεζεύχθω ἡ ΑΕ, καὶ διήχθω ἡ ΒΑ ἐπὶ τὸ Ζ. Let AE have been joined, and let BA have been
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΑ, ἴση ἐστὶ καὶ γωνία ἡ drawn through to F .

ὑπὸ ΑΒΕ τῇ ὑπὸ ΒΑΕ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΓΕ τῇ ΕΑ, And since BE is equal to EA, angle ABE is also
ἴση ἐστὶ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΓΑΕ· ὅλη ἄρα ἡ ὑπὸ ΒΑΓ equal to BAE [Prop. 1.5]. Again, since CE is equal to
δυσὶ ταῖς ὑπὸ ΑΒΓ, ΑΓΒ ἴση ἐστίν. ἐστὶ δὲ καὶ ἡ ὑπὸ ΖΑΓ EA, ACE is also equal to CAE [Prop. 1.5]. Thus, the
ἐκτὸς τοῦ ΑΒΓ τριγώνου δυσὶ ταῖς ὑπὸ ΑΒΓ, ΑΓΒ γωνίαις whole (angle) BAC is equal to the two (angles) ABC
ἴση· ἴση ἄρα καὶ ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΖΑΓ· ὀρθὴ ἄρα and ACB. And FAC, (which is) external to triangle
ἑκατέρα· ἡ ἄρα ἐν τῷ ΒΑΓ ἡμικυκλίῳ γωνία ἡ ὑπὸ ΒΑΓ ABC, is also equal to the two angles ABC and ACB
ὀρθή ἐστιν. [Prop. 1.32]. Thus, angle BAC (is) also equal to FAC.
Καὶ ἐπεὶ τοῦ ΑΒΓ τρίγωνου δύο γωνίαι αἱ ὑπὸ ΑΒΓ, Thus, (they are) each right-angles. [Def. 1.10]. Thus, the

ΒΑΓ δύο ὀρθῶν ἐλάττονές εἰσιν, ὀρθὴ δὲ ἡ ὑπὸ ΒΑΓ, angle BAC in the semi-circle BAC is a right-angle.
ἐλάττων ἄρα ὀρθῆς ἐστιν ἡ ὑπὸ ΑΒΓ γωνία· καί ἐστιν ἐν And since the two angles ABC and BAC of trian-
τῷ ΑΒΓ μείζονι τοῦ ἡμικυκλίου τμήματι. gle ABC are less than two right-angles [Prop. 1.17], and
Καὶ ἐπεὶ ἐν κύκλῳ τετράπλευρόν ἐστι τὸ ΑΒΓΔ, τῶν δὲ BAC is a right-angle, angle ABC is thus less than a right-

ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον γωνίαι δυσὶν angle. And it is in segment ABC, (which is) greater than
ὀρθαῖς ἴσαι εἰσίν [αἱ ἄρα ὑπὸ ΑΒΓ, ΑΔΓ γωνίαι δυσὶν ὀρθαῖς a semi-circle.
ἴσας εἰσίν], καί ἐστιν ἡ ὑπὸ ΑΒΓ ἐλάττων ὀρθῆς· λοιπὴ ἄρα And since ABCD is a quadrilateral within a circle,
ἡ ὑπὸ ΑΔΓ γωνία μείζων ὀρθῆς ἐστιν· καί ἐστιν ἐν τῷ ΑΔΓ and for quadrilaterals within circles the (sum of the) op-
ἐλάττονι τοῦ ἡμικυκλίου τμήματι. posite angles is equal to two right-angles [Prop. 3.22]
Λέγω, ὅτι καὶ ἡ μὲν τοῦ μείζονος τμήματος γωνία ἡ πε- [angles ABC and ADC are thus equal to two right-

ριεχομένη ὑπό [τε] τῆς ΑΒΓ περιφερείας καὶ τῆς ΑΓ εὐθείας angles], and (angle) ABC is less than a right-angle. The
μείζων ἐστὶν ὀρθῆς, ἡ δὲ τοῦ ἐλάττονος τμήματος γωνία ἡ remaining angle ADC is thus greater than a right-angle.
περιεχομένη ὑπό [τε] τῆς ΑΔ[Γ] περιφερείας καὶ τῆς ΑΓ And it is in segment ADC, (which is) less than a semi-
εὐθείας ἐλάττων ἐστὶν ὀρθῆς. καί ἐστιν αὐτόθεν φανερόν. circle.
ἐπεὶ γὰρ ἡ ὑπὸ τῶν ΒΑ, ΑΓ εὐθειῶν ὀρθή ἐστιν, ἡ ἄρα I also say that the angle of the greater segment,
ὑπὸ τῆς ΑΒΓ περιφερείας καὶ τῆς ΑΓ εὐθείας περιεχομένη (namely) that contained by the circumference ABC and
μείζων ἐστὶν ὀρθῆς. πάλιν, ἐπεὶ ἡ ὑπὸ τῶν ΑΓ, ΑΖ εὐθειῶν the straight-line AC, is greater than a right-angle. And
ὀρθή ἐστιν, ἡ ἄρα ὑπὸ τῆς ΓΑ εὐθείας καὶ τῆς ΑΔ[Γ] περι- the angle of the lesser segment, (namely) that contained

STOIQEIWN gþ. ELEMENTS BOOK 3
φερείας περιεχομένη ἐλάττων ἐστὶν ὀρθῆς. by the circumference AD[C] and the straight-line AC, is
᾿Εν κύκλῳ ἄρα ἡ μὲν ἐν τῷ ἡμικυκλίῳ γωνία ὀρθή ἐστιν, less than a right-angle. And this is immediately apparent.

ἡ δὲ ἐν τῷ μείζονι τμήματι ἐλάττων ὀρθῆς, ἡ δὲ ἐν τῷ For since the (angle contained by) the two straight-lines
ἐλάττονι [τμήματι] μείζων ὀρθῆς· καὶ ἔπι ἡ μὲν τοῦ μείζονος BA and AC is a right-angle, the (angle) contained by
τμήματος [γωνία] μείζων [ἐστὶν] ὀρθῆς, ἡ δὲ τοῦ ἐλάττονος the circumference ABC and the straight-line AC is thus
τμήματος [γωνία] ἐλάττων ὀρθῆς· ὅπερ ἔδει δεῖξαι. greater than a right-angle. Again, since the (angle con-

tained by) the straight-lines AC and AF is a right-angle,

the (angle) contained by the circumference AD[C] and
the straight-line CA is thus less than a right-angle.

Thus, in a circle, the angle in a semi-circle is a right-

angle, and that in a greater segment (is) less than a
right-angle, and that in a lesser [segment] (is) greater

than a right-angle. And, further, the [angle] of a seg-
ment greater (than a semi-circle) [is] greater than a right-

angle, and the [angle] of a segment less (than a semi-

circle) is less than a right-angle. (Which is) the very thing
it was required to show.lbþ. Proposition 32

᾿Εὰν κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς εἰς If some straight-line touches a circle, and some
τὸν κύκλον διαχθῇ τις εὐθεῖα τέμνουσα τὸν κύκλον, ἃς (other) straight-line is drawn across, from the point of
ποιεῖ γωνίας πρὸς τῇ ἐφαπτομένῃ, ἴσαι ἔσονται ταῖς ἐν τοῖς contact into the circle, cutting the circle (in two), then
ἐναλλὰξ τοῦ κύκλου τμήμασι γωνίαις. those angles the (straight-line) makes with the tangent

will be equal to the angles in the alternate segments of
the circle.

E
B

Κύκλου γὰρ τοῦ ΑΒΓΔ ἐφαπτέσθω τις εὐθεῖα ἡ ΕΖ For let some straight-line EF touch the circle ABCD
κατὰ τὸ Β σημεῖον, καὶ ἀπὸ τοῦ Β σημείου διήχθω τις at the point B, and let some (other) straight-line BD
εὐθεῖα εἰς τὸν ΑΒΓΔ κύκλον τέμνουσα αὐτὸν ἡ ΒΔ. λέγω, have been drawn from point B into the circle ABCD,
ὅτι ἃς ποιεῖ γωνίας ἡ ΒΔ μετὰ τῆς ΕΖ ἐφαπτομένης, ἴσας cutting it (in two). I say that the angles BD makes with
ἔσονται ταῖς ἐν τοῖς ἐναλλὰξ τμήμασι τοῦ κύκλου γωνίαις, the tangent EF will be equal to the angles in the alter-
τουτέστιν, ὅτι ἡ μὲν ὑπὸ ΖΒΔ γωνία ἴση ἐστὶ τῇ ἐν τῷ ΒΑΔ nate segments of the circle. That is to say, that angle
τμήματι συνισταμένῃ γωνίᾳ, ἡ δὲ ὑπὸ ΕΒΔ γωνία ἴση ἐστὶ FBD is equal to the angle constructed in segment BAD,
τῇ ἐν τῷ ΔΓΒ τμήματι συνισταμένῃ γωνίᾳ. and angle EBD is equal to the angle constructed in seg-
῎Ηχθω γὰρ ἀπὸ τοῦ Β τῇ ΕΖ πρὸς ὀρθὰς ἡ ΒΑ, καὶ ment DCB.

εἰλήφθω ἐπὶ τῆς ΒΔ περιφερείας τυχὸν σημεῖον τὸ Γ, καὶ For let BA have been drawn from B, at right-angles
ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ, ΓΒ. to EF [Prop. 1.11]. And let the point C have been taken
Καὶ ἐπεὶ κύκλου τοῦ ΑΒΓΔ ἐφάπτεταί τις εὐθεῖα ἡ ΕΖ at random on the circumference BD. And let AD, DC,

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STOIQEIWN gþ. ELEMENTS BOOK 3
κατὰ τὸ Β, καὶ ἀπὸ τῆς ἁφῆς ἦκται τῇ ἐφαπτομένῃ πρὸς and CB have been joined.
ὀρθὰς ἡ ΒΑ, ἐπὶ τῆς ΒΑ ἄρα τὸ κέντρον ἐστὶ τοῦ ΑΒΓΔ And since some straight-line EF touches the circle
κύκλου. ἡ ΒΑ ἄρα διάμετός ἐστι τοῦ ΑΒΓΔ κύκλου· ἡ ἄρα ABCD at point B, and BA has been drawn from the
ὑπὸ ΑΔΒ γωνία ἐν ἡμικυκλίῳ οὖσα ὀρθή ἐστιν. λοιπαὶ ἄρα point of contact, at right-angles to the tangent, the center
αἱ ὑπὸ ΒΑΔ, ΑΒΔ μιᾷ ὀρθῇ ἴσαι εἰσίν. ἐστὶ δὲ καὶ ἡ ὑπὸ of circle ABCD is thus on BA [Prop. 3.19]. Thus, BA
ΑΒΖ ὀρθή· ἡ ἄρα ὑπὸ ΑΒΖ ἴση ἐστὶ ταῖς ὑπὸ ΒΑΔ, ΑΒΔ. is a diameter of circle ABCD. Thus, angle ADB, being
κοινὴ ἀφῃρήσθω ἡ ὑπὸ ΑΒΔ· λοιπὴ ἄρα ἡ ὑπὸ ΔΒΖ γωνία in a semi-circle, is a right-angle [Prop. 3.31]. Thus, the
ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τμήματι τοῦ κύκλου γωνίᾳ τῇ remaining angles (of triangle ADB) BAD and ABD are
ὑπὸ ΒΑΔ. καὶ ἐπεὶ ἐν κύκλῳ τετράπλευρόν ἐστι τὸ ΑΒΓΔ, equal to one right-angle [Prop. 1.32]. And ABF is also a
αἱ ἀπεναντίον αὐτοῦ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. εἰσὶ δὲ right-angle. Thus, ABF is equal to BAD and ABD. Let
καὶ αἱ ὑπὸ ΔΒΖ, ΔΒΕ δυσὶν ὀρθαῖς ἴσαι· αἱ ἄρα ὑπὸ ΔΒΖ, ABD have been subtracted from both. Thus, the remain-
ΔΒΕ ταῖς ὑπὸ ΒΑΔ, ΒΓΔ ἴσαι εἰσίν, ὧν ἡ ὑπὸ ΒΑΔ τῇ ὑπὸ ing angle DBF is equal to the angle BAD in the alternate
ΔΒΖ ἐδείχθη ἴση· λοιπὴ ἄρα ἡ ὑπὸ ΔΒΕ τῇ ἐν τῷ ἐναλλὰξ segment of the circle. And since ABCD is a quadrilateral
τοῦ κύκλου τμήματι τῷ ΔΓΒ τῇ ὑπὸ ΔΓΒ γωνίᾳ ἐστὶν ἴση. in a circle, (the sum of) its opposite angles is equal to
᾿Εὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς two right-angles [Prop. 3.22]. And DBF and DBE is

εἰς τὸν κύκλον διαχθῇ τις εὐθεῖα τέμνουσα τὸν κύκλον, ἃς also equal to two right-angles [Prop. 1.13]. Thus, DBF
ποιεῖ γωνίας πρὸς τῇ ἐφαπτομένῃ, ἴσαι ἔσονται ταῖς ἐν τοῖς and DBE is equal to BAD and BCD, of which BAD
ἐναλλὰξ τοῦ κύκλου τμήμασι γωνίαις· ὅπερ ἔδει δεῖξαι. was shown (to be) equal to DBF . Thus, the remaining

(angle) DBE is equal to the angle DCB in the alternate
segment DCB of the circle.

Thus, if some straight-line touches a circle, and some
(other) straight-line is drawn across, from the point of

contact into the circle, cutting the circle (in two), then

those angles the (straight-line) makes with the tangent
will be equal to the angles in the alternate segments of

the circle. (Which is) the very thing it was required to

show.lgþ. Proposition 33
᾿Επὶ τῆς δοθείσης εὐθείας γράψαι τμῆμα κύκλου δεχόμε- To draw a segment of a circle, accepting an angle

νον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ. equal to a given rectilinear angle, on a given straight-line.

Α Α

Ζ
Β

∆
Α

Γ Γ

∆∆

Θ
H

G
F

E
F

A
D

C C

E E

῎Εστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ, ἡ δὲ δοθεῖσα γωνία Let AB be the given straight-line, and C the given
εὐθύγραμμος ἡ πρὸς τῷ Γ· δεῖ δὴ ἐπὶ τῆς δοθείσης εὐθείας rectilinear angle. So it is required to draw a segment
τῆς ΑΒ γράψαι τμῆμα κύκλου δεχόμενον γωνίαν ἴσην τῇ of a circle, accepting an angle equal to C, on the given
πρὸς τῷ Γ. straight-line AB.
῾Η δὴ πρὸς τῷ Γ [γωνία] ἤτοι ὀξεῖά ἐστιν ἢ ὀρθὴ ἢ So the [angle] C is surely either acute, a right-angle,

ἀμβλεῖα· ἔστω πρότερον ὀξεῖα, καὶ ὡς ἐπὶ τῆς πρώτης κα- or obtuse. First of all, let it be acute. And, as in the first
ταγραφῆς συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ Α σημείῳ diagram (from the left), let (angle) BAD, equal to angle
τῇ πρὸς τῷ Γ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΔ· ὀξεῖα ἄρα ἐστὶ καὶ ἡ C, have been constructed on the straight-line AB, at the
ὑπὸ ΒΑΔ. ἤχθω τῇ ΔΑ πρὸς ὀρθὰς ἡ ΑΕ, καὶ τετμήσθω point A (on it) [Prop. 1.23]. Thus, BAD is also acute. Let
ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ ἤχθω ἀπὸ τοῦ Ζ σημείου τῇ ΑΒ AE have been drawn, at right-angles to DA [Prop. 1.11].

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STOIQEIWN gþ. ELEMENTS BOOK 3
πρὸς ὀρθὰς ἡ ΖΗ, καὶ ἐπεζεύχθω ἡ ΗΒ. And let AB have been cut in half at F [Prop. 1.10]. And
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, κοινὴ δὲ ἡ ΖΗ, δύο δὴ let FG have been drawn from point F , at right-angles to

αἱ ΑΖ, ΖΗ δύο ταῖς ΒΖ, ΖΗ ἴσαι εἰσίν· καὶ γωνία ἡ ὑπὸ AB [Prop. 1.11]. And let GB have been joined.
ΑΖΗ [γωνίᾳ] τῇ ὑπὸ ΒΖΗ ἴση· βάσις ἄρα ἡ ΑΗ βάσει τῇ And since AF is equal to FB, and FG (is) common,
ΒΗ ἴση ἐστίν. ὁ ἄρα κέντρῳ μὲν τῷ Η διαστήματι δὲ τῷ the two (straight-lines) AF , FG are equal to the two
ΗΑ κύκλος γραφόμενος ἥξει καὶ διὰ τοῦ Β. γεγράφθω καὶ (straight-lines) BF , FG (respectively). And angle AFG
ἔστω ὁ ΑΒΕ, καὶ ἐπεζεύχθω ἡ ΕΒ. ἐπεὶ οὖν ἀπ᾿ ἄκρας τῆς (is) equal to [angle] BFG. Thus, the base AG is equal to
ΑΕ διαμέτρου ἀπὸ τοῦ Α τῇ ΑΕ πρὸς ὀρθάς ἐστιν ἡ ΑΔ, the base BG [Prop. 1.4]. Thus, the circle drawn with
ἡ ΑΔ ἄρα ἐφάπτεται τοῦ ΑΒΕ κύκλου· ἐπεὶ οὖν κύκλου center G, and radius GA, will also go through B (as
τοῦ ΑΒΕ ἐφάπτεταί τις εὐθεῖα ἡ ΑΔ, καὶ ἀπὸ τῆς κατὰ τὸ well as A). Let it have been drawn, and let it be (de-
Α ἁφῆς εἰς τὸν ΑΒΕ κύκλον διῆκταί τις εὐθεῖα ἡ ΑΒ, ἡ noted) ABE. And let EB have been joined. Therefore,
ἄρα ὑπὸ ΔΑΒ γωνία ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου since AD is at the extremity of diameter AE, (namely,
τμήματι γωνίᾳ τῇ ὑπὸ ΑΕΒ. ἀλλ᾿ ἡ ὑπὸ ΔΑΒ τῇ πρὸς τῷ Γ point) A, at right-angles to AE, the (straight-line) AD
ἐστιν ἴση· καὶ ἡ πρὸς τῷ Γ ἄρα γωνία ἴση ἐστὶ τῇ ὑπὸ ΑΕΒ. thus touches the circle ABE [Prop. 3.16 corr.]. There-
᾿Επὶ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τμῆμα κύκλου fore, since some straight-line AD touches the circle ABE,

γέγραπται τὸ ΑΕΒ δεχόμενον γωνίαν τὴν ὑπὸ ΑΕΒ ἴσην and some (other) straight-line AB has been drawn across
τῇ δοθείσῃ τῇ πρὸς τῷ Γ. from the point of contact A into circle ABE, angle DAB
Ἀλλὰ δὴ ὀρθὴ ἔστω ἡ πρὸς τῷ Γ· καὶ δέον πάλιν ἔστω is thus equal to the angle AEB in the alternate segment

ἐπὶ τῆς ΑΒ γράψαι τμῆμα κύκλου δεχόμενον γωνίαν ἴσην τῇ of the circle [Prop. 3.32]. But, DAB is equal to C. Thus,
πρὸς τῷ Γ ὀρθῇ [γωνίᾳ]. συνεστάτω [πάλιν] τῇ πρὸς τῷ Γ angle C is also equal to AEB.
ὀρθῇ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΔ, ὡς ἔχει ἐπὶ τῆς δευτέρας κατα- Thus, a segment AEB of a circle, accepting the angle
γραφῆς, καὶ τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ κέντρῳ τῷ AEB (which is) equal to the given (angle) C, has been
Ζ, διαστήματι δὲ ὁποτέρῳ τῶν ΖΑ, ΖΒ, κύκλος γεγράφθω drawn on the given straight-line AB.
ὁ ΑΕΒ. And so let C be a right-angle. And let it again be
᾿Εφάπτεται ἄρα ἡ ΑΔ εὐθεῖα τοῦ ΑΒΕ κύκλου διὰ τὸ necessary to draw a segment of a circle on AB, accepting

ὀρθὴν εἶναι τὴν πρὸς τῷ Α γωνίαν. καὶ ἴση ἐστὶν ἡ ὑπὸ an angle equal to the right-[angle] C. Let the (angle)
ΒΑΔ γωνία τῇ ἐν τῷ ΑΕΒ τμήματι· ὀρθὴ γὰρ καὶ αὐτὴ ἐν BAD [again] have been constructed, equal to the right-
ἡμικυκλίῳ οὖσα. ἀλλὰ καὶ ἡ ὑπὸ ΒΑΔ τῇ πρὸς τῷ Γ ἴση angle C [Prop. 1.23], as in the second diagram (from the
ἐστίν. καὶ ἡ ἐν τῷ ΑΕΒ ἄρα ἴση ἐστὶ τῇ πρὸς τῷ Γ. left). And let AB have been cut in half at F [Prop. 1.10].
Γέγραπται ἄρα πάλιν ἐπὶ τῆς ΑΒ τμῆμα κύκλου τὸ ΑΕΒ And let the circle AEB have been drawn with center F ,

δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ. and radius either FA or FB.
Ἀλλὰ δὴ ἡ πρὸς τῷ Γ ἀμβλεῖα ἔστω· καὶ συνεστάτω Thus, the straight-line AD touches the circle ABE, on

αὐτῇ ἴση πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ Α σημείῳ ἡ ὑπὸ ΒΑΔ, account of the angle at A being a right-angle [Prop. 3.16
ὡς ἔχει ἐπὶ τῆς τρίτης καταγραφῆς, καὶ τῇ ΑΔ πρὸς ὀρθὰς corr.]. And angle BAD is equal to the angle in segment
ἤχθω ἡ ΑΕ, καὶ τετμήσθω πάλιν ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ AEB. For (the latter angle), being in a semi-circle, is also
τῇ ΑΒ πρὸς ὀρθὰς ἤχθω ἡ ΖΗ, καὶ ἐπεζεύχθω ἡ ΗΒ. a right-angle [Prop. 3.31]. But, BAD is also equal to C.
Καὶ ἐπεὶ πάλιν ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, καὶ κοινὴ ἡ ΖΗ, Thus, the (angle) in (segment) AEB is also equal to C.

δύο δὴ αἱ ΑΖ, ΖΗ δύο ταῖς ΒΖ, ΖΗ ἴσαι εἰσίν· καὶ γωνία ἡ Thus, a segment AEB of a circle, accepting an angle
ὑπὸ ΑΖΗ γωνίᾳ τῇ ὑπὸ ΒΖΗ ἴση· βάσις ἄρα ἡ ΑΗ βάσει equal to C, has again been drawn on AB.
τῇ ΒΗ ἴση ἐστίν· ὁ ἄρα κέντρῳ μὲν τῷ Η διαστήματι δὲ τῷ And so let (angle) C be obtuse. And let (angle) BAD,
ΗΑ κύκλος γραφόμενος ἥξει καὶ διὰ τοῦ Β. ἐρχέσθω ὡς ὁ equal to (C), have been constructed on the straight-line
ΑΕΒ. καὶ ἐπεὶ τῇ ΑΕ διαμέτρῳ ἀπ᾿ ἄκρας πρὸς ὀρθάς ἐστιν AB, at the point A (on it) [Prop. 1.23], as in the third
ἡ ΑΔ, ἡ ΑΔ ἄρα ἐφάπτεται τοῦ ΑΕΒ κύκλου. καὶ ἀπὸ τῆς diagram (from the left). And let AE have been drawn, at
κατὰ τὸ Α ἐπαφῆς διῆκται ἡ ΑΒ· ἡ ἄρα ὑπὸ ΒΑΔ γωνία right-angles to AD [Prop. 1.11]. And let AB have again
ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι τῷ ΑΘΒ been cut in half at F [Prop. 1.10]. And let FG have been
συνισταμένῃ γωνίᾳ. ἀλλ᾿ ἡ ὑπὸ ΒΑΔ γωνία τῇ πρὸς τῷ Γ drawn, at right-angles to AB [Prop. 1.10]. And let GB
ἴση ἐστίν. καὶ ἡ ἐν τῷ ΑΘΒ ἄρα τμήματι γωνία ἴση ἐστὶ τῇ have been joined.
πρὸς τῷ Γ. And again, since AF is equal to FB, and FG (is)
᾿Επὶ τῆς ἄρα δοθείσης εὐθείας τῆς ΑΒ γέγραπται τμῆμα common, the two (straight-lines) AF , FG are equal to

κύκλου τὸ ΑΘΒ δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ· ὅπερ the two (straight-lines) BF , FG (respectively). And an-
ἔδει ποιῆσαι. gle AFG (is) equal to angle BFG. Thus, the base AG is

102

STOIQEIWN gþ. ELEMENTS BOOK 3
equal to the base BG [Prop. 1.4]. Thus, a circle of center

G, and radius GA, being drawn, will also go through B

(as well as A). Let it go like AEB (in the third diagram
from the left). And since AD is at right-angles to the di-

ameter AE, at its extremity, AD thus touches circle AEB
[Prop. 3.16 corr.]. And AB has been drawn across (the

circle) from the point of contact A. Thus, angle BAD is

equal to the angle constructed in the alternate segment
AHB of the circle [Prop. 3.32]. But, angle BAD is equal

to C. Thus, the angle in segment AHB is also equal to

C.
Thus, a segment AHB of a circle, accepting an angle

equal to C, has been drawn on the given straight-line AB.
(Which is) the very thing it was required to do.ldþ. Proposition 34

Ἀπὸ τοῦ δοθέντος κύκλου τμῆμα ἀφελεῖν δεχόμενον To cut off a segment, accepting an angle equal to a
γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ. given rectilinear angle, from a given circle.

ΑΕ

∆

A
E

F
C

῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, ἡ δὲ δοθεῖσα γωνία Let ABC be the given circle, and D the given rectilin-
εὐθύγραμμος ἡ πρὸς τῷ Δ· δεῖ δὴ ἀπὸ τοῦ ΑΒΓ κύκλου ear angle. So it is required to cut off a segment, accepting
τμῆμα ἀφελεῖν δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ an angle equal to the given rectilinear angle D, from the
εὐθυγράμμῳ τῇ πρὸς τῷ Δ. given circle ABC.

῎Ηχθω τοῦ ΑΒΓ ἐφαπτομένη ἡ ΕΖ κατὰ τὸ Β σημεῖον, Let EF have been drawn touching ABC at point B.†

καὶ συνεστάτω πρὸς τῇ ΖΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ And let (angle) FBC, equal to angle D, have been con-
τῷ Β τῇ πρὸς τῷ Δ γωνίᾳ ἴση ἡ ὑπὸ ΖΒΓ. structed on the straight-line FB, at the point B on it
᾿Επεὶ οὖν κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΕΖ, [Prop. 1.23].

καὶ ἀπὸ τῆς κατὰ τὸ Β ἐπαφῆς διῆκται ἡ ΒΓ, ἡ ὑπὸ ΖΒΓ ἄρα Therefore, since some straight-line EF touches the
γωνία ἴση ἐστὶ τῇ ἐν τῷ ΒΑΓ ἐναλλὰξ τμήματι συνισταμένῃ circle ABC, and BC has been drawn across (the circle)
γωνίᾳ. ἀλλ᾿ ἡ ὑπὸ ΖΒΓ τῇ πρὸς τῷ Δ ἐστιν ἴση· καὶ ἡ ἐν from the point of contact B, angle FBC is thus equal
τῷ ΒΑΓ ἄρα τμήματι ἴση ἐστὶ τῇ πρὸς τῷ Δ [γωνίᾳ]. to the angle constructed in the alternate segment BAC
Ἀπὸ τοῦ δοθέντος ἄρα κύκλου τοῦ ΑΒΓ τμῆμα ἀφῄρηται [Prop. 1.32]. But, FBC is equal to D. Thus, the (angle)

τὸ ΒΑΓ δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμ- in the segment BAC is also equal to [angle] D.
μῳ τῇ πρὸς τῷ Δ· ὅπερ ἔδει ποιῆσαι. Thus, the segment BAC, accepting an angle equal to

the given rectilinear angle D, has been cut off from the

given circle ABC. (Which is) the very thing it was re-
quired to do.

† Presumably, by finding the center of ABC [Prop. 3.1], drawing a straight-line between the center and point B, and then drawing EF through

103

STOIQEIWN gþ. ELEMENTS BOOK 3
point B, at right-angles to the aforementioned straight-line [Prop. 1.11].leþ. Proposition 35
᾿Εὰν ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὸ ὑπὸ If two straight-lines in a circle cut one another then

τῶν τῆς μιᾶς τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ the rectangle contained by the pieces of one is equal to
τῷ ὑπὸ τῶν τῆς ἑτέρας τμημάτων περιεχομένῳ ὀρθογωνίῳ. the rectangle contained by the pieces of the other.

∆

∆Β

Β
Γ

B C

B D

᾿Εν γὰρ κύκλῳ τῷ ΑΒΓΔ δύο εὐθεῖαι αἱ ΑΓ, ΒΔ For let the two straight-lines AC and BD, in the circle
τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε σημεῖον· λέγω, ὅτι τὸ ὑπὸ ABCD, cut one another at point E. I say that the rect-
τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ angle contained by AE and EC is equal to the rectangle
τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ. contained by DE and EB.
Εἰ μὲν οὖν αἱ ΑΓ, ΒΔ διὰ τοῦ κέντρου εἰσὶν ὥστε τὸ Ε In fact, if AC and BD are through the center (as in

κέντρον εἶναι τοῦ ΑΒΓΔ κύκλου, φανερόν, ὅτι ἴσων οὐσῶν the first diagram from the left), so that E is the center of
τῶν ΑΕ, ΕΓ, ΔΕ, ΕΒ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον circle ABCD, then (it is) clear that, AE, EC, DE, and
ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ EB being equal, the rectangle contained by AE and EC
ὀρθογωνίῳ. is also equal to the rectangle contained by DE and EB.
Μὴ ἔστωσαν δὴ αἱ ΑΓ, ΔΒ διὰ τοῦ κέντρου, καὶ So let AC and DB not be though the center (as in

εἰλήφθω τὸ κέντρον τοῦ ΑΒΓΔ, καὶ ἔστω τὸ Ζ, καὶ ἀπὸ the second diagram from the left), and let the center of
τοῦ Ζ ἐπὶ τὰς ΑΓ, ΔΒ εὐθείας κάθετοι ἤχθωσαν αἱ ΖΗ, ABCD have been found [Prop. 3.1], and let it be (at) F .
ΖΘ, καὶ ἐπεζεύχθωσαν αἱ ΖΒ, ΖΓ, ΖΕ. And let FG and FH have been drawn from F , perpen-
Καὶ ἐπεὶ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΗΖ εὐθεῖάν τινα dicular to the straight-lines AC and DB (respectively)

μὴ διὰ τοῦ κέντρου τὴν ΑΓ πρὸς ὀρθὰς τέμνει, καὶ δίχα [Prop. 1.12]. And let FB, FC, and FE have been joined.
αὐτὴν τέμνει· ἴση ἄρα ἡ ΑΗ τῇ ΗΓ. ἐπεὶ οὖν εὐθεῖα ἡ ΑΓ And since some straight-line, GF , through the center,
τέτμηται εἰς μὲν ἴσα κατὰ τὸ Η, εἰς δὲ ἄνισα κατὰ τὸ Ε, τὸ cuts at right-angles some (other) straight-line, AC, not
ἄρα ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ through the center, then it also cuts it in half [Prop. 3.3].
ἀπὸ τῆς ΕΗ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΗΓ· [κοινὸν] Thus, AG (is) equal to GC. Therefore, since the straight-
προσκείσθω τὸ ἀπὸ τῆς ΗΖ· τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ line AC is cut equally at G, and unequally at E, the
τῶν ἀπὸ τῶν ΗΕ, ΗΖ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΓΗ, ΗΖ. ἀλλὰ rectangle contained by AE and EC plus the square on
τοῖς μὲν ἀπὸ τῶν ΕΗ, ΗΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΕ, τοὶς EG is thus equal to the (square) on GC [Prop. 2.5]. Let
δὲ ἀπὸ τῶν ΓΗ, ΗΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΓ· τὸ ἄρα ὑπὸ the (square) on GF have been added [to both]. Thus,
τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς the (rectangle contained) by AE and EC plus the (sum
ΖΓ. ἴση δὲ ἡ ΖΓ τῇ ΖΒ· τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ of the squares) on GE and GF is equal to the (sum of
ἀπὸ τῆς ΕΖ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΒ. διὰ τὰ αὐτὰ δὴ καὶ the squares) on CG and GF . But, the (square) on FE
τὸ ὑπὸ τῶν ΔΕ, ΕΒ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἰσον ἐστὶ τῷ ἀπὸ is equal to the (sum of the squares) on EG and GF
τῆς ΖΒ. ἐδείχθη δὲ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ [Prop. 1.47], and the (square) on FC is equal to the (sum
τῆς ΖΕ ἴσον τῷ ἀπὸ τῆς ΖΒ· τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ of the squares) on CG and GF [Prop. 1.47]. Thus, the
τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ μετὰ τοῦ (rectangle contained) by AE and EC plus the (square)
ἀπὸ τῆς ΖΕ. κοινὸν ἀφῇρήσθω τὸ ἀπὸ τῆς ΖΕ· λοιπὸν ἄρα on FE is equal to the (square) on FC. And FC (is)
τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ equal to FB. Thus, the (rectangle contained) by AE
ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ. and EC plus the (square) on FE is equal to the (square)
᾿Εὰν ἄρα ἐν κύκλῳ εὐθεῖαι δύο τέμνωσιν ἀλλήλας, τὸ on FB. So, for the same (reasons), the (rectangle con-

ὑπὸ τῶν τῆς μιᾶς τμημάτων περιεχόμενον ὀρθογώνιον ἴσον tained) by DE and EB plus the (square) on FE is equal

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STOIQEIWN gþ. ELEMENTS BOOK 3
ἐστὶ τῷ ὑπὸ τῶν τῆς ἑτέρας τμημάτων περιεχομένῳ ὀρθο- to the (square) on FB. And the (rectangle contained)
γωνίῳ· ὅπερ ἔδει δεῖξαι. by AE and EC plus the (square) on FE was also shown

(to be) equal to the (square) on FB. Thus, the (rect-
angle contained) by AE and EC plus the (square) on

FE is equal to the (rectangle contained) by DE and EB
plus the (square) on FE. Let the (square) on FE have

been taken from both. Thus, the remaining rectangle con-

tained by AE and EC is equal to the rectangle contained
by DE and EB.

Thus, if two straight-lines in a circle cut one another

then the rectangle contained by the pieces of one is equal
to the rectangle contained by the pieces of the other.

(Which is) the very thing it was required to show.l�þ. Proposition 36
᾿Εὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, καὶ ἀπ᾿ αὐτοῦ If some point is taken outside a circle, and two

πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν straight-lines radiate from it towards the circle, and (one)
τέμνῃ τὸν κύκλον, ἡ δὲ ἐφάπτηται, ἔσται τὸ ὑπὸ ὅλης τῆς of them cuts the circle, and the (other) touches (it), then
τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε the (rectangle contained) by the whole (straight-line)
σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς ἐφα- cutting (the circle), and the (part of it) cut off outside
πτομένης τετραγώνῳ. (the circle), between the point and the convex circumfer-

ence, will be equal to the square on the tangent (line).

∆

Ζ
Γ

B F

BC
D

Κύκλου γὰρ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον ἐκτὸς τὸ Δ, For let some point D have been taken outside circle
καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν δύο ABC, and let two straight-lines, DC[A] and DB, radi-
εὐθεῖαι αἱ ΔΓ[Α], ΔΒ· καὶ ἡ μὲν ΔΓΑ τεμνέτω τὸν ΑΒΓ ate from D towards circle ABC. And let DCA cut circle
κύκλον, ἡ δὲ ΒΔ ἐφαπτέσθω· λέγω, ὅτι τὸ ὑπὸ τῶν ΑΔ, ABC, and let BD touch (it). I say that the rectangle
ΔΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ contained by AD and DC is equal to the square on DB.
τετραγώνῳ. [D]CA is surely either through the center, or not. Let
῾Η ἄρα [Δ]ΓΑ ἤτοι διὰ τοῦ κέντρου ἐστὶν ἢ οὔ. ἔστω it first of all be through the center, and let F be the cen-

πρότερον διὰ τοῦ κέντρου, καὶ ἔστω τὸ Ζ κέντρον τοῦ ΑΒΓ ter of circle ABC, and let FB have been joined. Thus,
κύκλου, καὶ ἐπεζεύχθω ἡ ΖΒ· ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΒΔ. (angle) FBD is a right-angle [Prop. 3.18]. And since
καὶ ἐπεὶ εὐθεῖα ἡ ΑΓ δίχα τέτμηται κατὰ τὸ Ζ, πρόσκειται straight-line AC is cut in half at F , let CD have been
δὲ αὐτῇ ἡ ΓΔ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς added to it. Thus, the (rectangle contained) by AD and
ΖΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. ἴση δὲ ἡ ΖΓ τῇ ΖΒ· τὸ ἄρα DC plus the (square) on FC is equal to the (square) on
ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΒ ἴσον ἐστὶ τῷ ἀπὸ FD [Prop. 2.6]. And FC (is) equal to FB. Thus, the
τὴς ΖΔ. τῷ δὲ ἀπὸ τῆς ΖΔ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΖΒ, ΒΔ· (rectangle contained) by AD and DC plus the (square)
τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΒ ἴσον ἐστὶ on FB is equal to the (square) on FD. And the (square)
τοῖς ἀπὸ τῶν ΖΒ, ΒΔ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΖΒ· on FD is equal to the (sum of the squares) on FB and
λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ BD [Prop. 1.47]. Thus, the (rectangle contained) by AD

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STOIQEIWN gþ. ELEMENTS BOOK 3
ἐφαπτομένης. and DC plus the (square) on FB is equal to the (sum
Ἀλλὰ δὴ ἡ ΔΓΑ μὴ ἔστω διὰ τοῦ κέντρου τοῦ ΑΒΓ of the squares) on FB and BD. Let the (square) on

κύκλου, καὶ εἰλήφθω τὸ κέντρον τὸ Ε, καὶ ἀπὸ τοῦ Ε ἐπὶ FB have been subtracted from both. Thus, the remain-
τὴν ΑΓ κάθετος ἤχθω ἡ ΕΖ, καὶ ἐπεζεύχθωσαν αἱ ΕΒ, ΕΓ, ing (rectangle contained) by AD and DC is equal to the
ΕΔ· ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΕΒΔ. καὶ ἐπεὶ εὐθεῖά τις διὰ τοῦ (square) on the tangent DB.
κέντρου ἡ ΕΖ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ πρὸς And so let DCA not be through the center of cir-
ὀρθὰς τέμνει, καὶ δίχα αὐτὴν τέμνει· ἡ ΑΖ ἄρα τῇ ΖΓ ἐστιν cle ABC, and let the center E have been found, and
ἴση. καὶ ἐπεὶ εὐθεῖα ἡ ΑΓ τέτμηται δίχα κατὰ τὸ Ζ σημεῖον, let EF have been drawn from E, perpendicular to AC
πρόσκειται δὲ αὐτῇ ἡ ΓΔ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ [Prop. 1.12]. And let EB, EC, and ED have been joined.
ἀπὸ τῆς ΖΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. κοινὸν προσκείσθω (Angle) EBD (is) thus a right-angle [Prop. 3.18]. And
τὸ ἀπὸ τῆς ΖΕ· τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τῶν ἀπὸ τῶν since some straight-line, EF , through the center, cuts
ΓΖ, ΖΕ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΖΔ, ΖΕ. τοῖς δὲ ἀπὸ τῶν some (other) straight-line, AC, not through the center,
ΓΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΓ· ὀρθὴ γὰρ [ἐστιν] ἡ ὑπὸ at right-angles, it also cuts it in half [Prop. 3.3]. Thus,
ΕΖΓ [γωνία]· τοῖς δὲ ἀπὸ τῶν ΔΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς AF is equal to FC. And since the straight-line AC is cut
ΕΔ· τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΓ ἴσον in half at point F , let CD have been added to it. Thus, the
ἐστὶ τῷ ἀπὸ τῆς ΕΔ. ἴση δὲ ἡ ΕΓ τῂ ΕΒ· τὸ ἄρα ὑπὸ τῶν (rectangle contained) by AD and DC plus the (square)
ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΒ ἴσον ἐστὶ τῷ ἄπὸ τῆς ΕΔ. on FC is equal to the (square) on FD [Prop. 2.6]. Let
τῷ δὲ ἀπὸ τῆς ΕΔ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΕΒ, ΒΔ· ὀρθὴ γὰρ the (square) on FE have been added to both. Thus, the
ἡ ὑπὸ ΕΒΔ γωνία· τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ (rectangle contained) by AD and DC plus the (sum of
τῆς ΕΒ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΕΒ, ΒΔ. κοινὸν ἀφῃρήσθω the squares) on CF and FE is equal to the (sum of the
τὸ ἀπὸ τῆς ΕΒ· λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ squares) on FD and FE. But the (square) on EC is equal
τῷ ἀπὸ τῆς ΔΒ. to the (sum of the squares) on CF and FE. For [angle]
᾿Εὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, καὶ ἀπ᾿ αὐτοῦ EFC [is] a right-angle [Prop. 1.47]. And the (square)

πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν on ED is equal to the (sum of the squares) on DF and
τέμνῃ τὸν κύκλον, ἡ δὲ ἐφάπτηται, ἔσται τὸ ὑπὸ ὅλης τῆς FE [Prop. 1.47]. Thus, the (rectangle contained) by AD
τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε and DC plus the (square) on EC is equal to the (square)
σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς ἐφα- on ED. And EC (is) equal to EB. Thus, the (rectan-
πτομένης τετραγώνῳ· ὅπερ ἔδει δεῖξαι. gle contained) by AD and DC plus the (square) on EB

is equal to the (square) on ED. And the (sum of the
squares) on EB and BD is equal to the (square) on ED.

For EBD (is) a right-angle [Prop. 1.47]. Thus, the (rect-

angle contained) by AD and DC plus the (square) on
EB is equal to the (sum of the squares) on EB and BD.

Let the (square) on EB have been subtracted from both.

Thus, the remaining (rectangle contained) by AD and
DC is equal to the (square) on BD.

Thus, if some point is taken outside a circle, and two
straight-lines radiate from it towards the circle, and (one)

of them cuts the circle, and (the other) touches (it), then

the (rectangle contained) by the whole (straight-line)
cutting (the circle), and the (part of it) cut off outside

(the circle), between the point and the convex circumfer-

ence, will be equal to the square on the tangent (line).
(Which is) the very thing it was required to show.lzþ. Proposition 37

᾿Εὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ If some point is taken outside a circle, and two
σημείου πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ straight-lines radiate from the point towards the circle,
ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ προσπίπτῃ, ᾖ δὲ τὸ and one of them cuts the circle, and the (other) meets
ὑπὸ [τῆς] ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβα- (it), and the (rectangle contained) by the whole (straight-

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STOIQEIWN gþ. ELEMENTS BOOK 3
νομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας line) cutting (the circle), and the (part of it) cut off out-
ἴσον τῷ ἀπὸ τῆς προσπιπτούσης, ἡ προσπίπτουσα ἐφάψεται side (the circle), between the point and the convex cir-
τοῦ κύκλου. cumference, is equal to the (square) on the (straight-line)

meeting (the circle), then the (straight-line) meeting (the

circle) will touch the circle.

∆

Κύκλου γὰρ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον ἐκτὸς τὸ Δ, For let some point D have been taken outside circle
καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν δύο ABC, and let two straight-lines, DCA and DB, radiate
εὐθεῖαι αἱ ΔΓΑ, ΔΒ, καὶ ἡ μὲν ΔΓΑ τεμνέτω τὸν κύκλον, ἡ from D towards circle ABC, and let DCA cut the circle,
δὲ ΔΒ προσπιπτέτω, ἔστω δὲ τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον τῷ and let DB meet (the circle). And let the (rectangle con-
ἀπὸ τῆς ΔΒ. λέγω, ὅτι ἡ ΔΒ ἐφάπτεται τοῦ ΑΒΓ κύκλου. tained) by AD and DC be equal to the (square) on DB.
῎Ηχθω γὰρ τοῦ ΑΒΓ ἐφαπτομένη ἡ ΔΕ, καὶ εἰλήφθω τὸ I say that DB touches circle ABC.

κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ Ζ, καὶ ἐπεζεύχθωσαν For let DE have been drawn touching ABC [Prop.
αἱ ΖΕ, ΖΒ, ΖΔ. ἡ ἄρα ὑπὸ ΖΕΔ ὀρθή ἐστιν. καὶ ἐπεὶ ἡ ΔΕ 3.17], and let the center of the circle ABC have been
ἐφάπτεται τοῦ ΑΒΓ κύκλου, τέμνει δὲ ἡ ΔΓΑ, τὸ ἄρα ὑπὸ found, and let it be (at) F . And let FE, FB, and FD
τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΕ. ἦν δὲ καὶ τὸ ὑπὸ have been joined. (Angle) FED is thus a right-angle
τῶν ΑΔ, ΔΓ ἴσον τῷ ἀπὸ τῆς ΔΒ· τὸ ἄρα ἀπὸ τῆς ΔΕ [Prop. 3.18]. And since DE touches circle ABC, and
ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ· ἴση ἄρα ἡ ΔΕ τῇ ΔΒ. ἐστὶ δὲ DCA cuts (it), the (rectangle contained) by AD and DC
καὶ ἡ ΖΕ τῇ ΖΒ ἴση· δύο δὴ αἱ ΔΕ, ΕΖ δύο ταῖς ΔΒ, ΒΖ is thus equal to the (square) on DE [Prop. 3.36]. And the
ἴσαι εἰσίν· καὶ βάσις αὐτῶν κοινὴ ἡ ΖΔ· γωνία ἄρα ἡ ὑπὸ (rectangle contained) by AD and DC was also equal to
ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΒΖ ἐστιν ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΔΕΖ· the (square) on DB. Thus, the (square) on DE is equal
ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΔΒΖ. καί ἐστιν ἡ ΖΒ ἐκβαλλομένη to the (square) on DB. Thus, DE (is) equal to DB. And
διάμετρος· ἡ δὲ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾿ FE is also equal to FB. So the two (straight-lines) DE,
ἄκρας ἀγομένη ἐφάπτεται τοῦ κύκλου· ἡ ΔΒ ἄρα ἐφάπτεται EF are equal to the two (straight-lines) DB, BF (re-
τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δειχθήσεται, κἂν τὸ κέντρον spectively). And their base, FD, is common. Thus, angle
ἐπὶ τῆς ΑΓ τυγχάνῃ. DEF is equal to angle DBF [Prop. 1.8]. And DEF (is)
᾿Εὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ a right-angle. Thus, DBF (is) also a right-angle. And

σημείου πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ FB produced is a diameter, And a (straight-line) drawn
μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ προσπίπτῃ, ᾖ δὲ τὸ ὑπὸ at right-angles to a diameter of a circle, at its extremity,
ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ touches the circle [Prop. 3.16 corr.]. Thus, DB touches
τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ circle ABC. Similarly, (the same thing) can be shown,
τῆς προσπιπτούσης, ἡ προσπίπτουσα ἐφάψεται τοῦ κύκλου· even if the center happens to be on AC.
ὅπερ ἔδει δεῖξαι. Thus, if some point is taken outside a circle, and two

straight-lines radiate from the point towards the circle,
and one of them cuts the circle, and the (other) meets

(it), and the (rectangle contained) by the whole (straight-
line) cutting (the circle), and the (part of it) cut off out-

side (the circle), between the point and the convex cir-

cumference, is equal to the (square) on the (straight-line)
meeting (the circle), then the (straight-line) meeting (the

circle) will touch the circle. (Which is) the very thing it

107

STOIQEIWN gþ. ELEMENTS BOOK 3
was required to show.

108

ELEMENTS BOOK 4

Construction of Rectilinear Figures In and
Around Circles

109

STOIQEIWN dþ. ELEMENTS BOOK 4VOroi. Definitions
αʹ.Σχῆμα εὐθύγραμμον εἰς σχῆμα εὐθύγραμμον ἐγγράφ- 1. A rectilinear figure is said to be inscribed in

εσθαι λέγεται, ὅταν ἑκάστη τῶν τοῦ ἐγγραφομένου σχήματ- a(nother) rectilinear figure when the respective angles
ος γωνιῶν ἑκάστης πλευρᾶς τοῦ, εἰς ὃ ἐγγράφεται, ἅπτηται. of the inscribed figure touch the respective sides of the
βʹ. Σχῆμα δὲ ὁμοίως περὶ σχῆμα περιγράφεσθαι λέγεται, (figure) in which it is inscribed.

ὅταν ἑκάστη πλευρὰ τοῦ περιγραφομένου ἑκάστης γωνίας 2. And, similarly, a (rectilinear) figure is said to be cir-
τοῦ, περὶ ὃ περιγράφεται, ἅπτηται. cumscribed about a(nother rectilinear) figure when the
γʹ.Σχῆμα εὐθύγραμμον εἰς κύκλον ἐγγράφεσθαι λέγεται, respective sides of the circumscribed (figure) touch the

ὅταν ἑκάστη γωνία τοῦ ἐγγραφομένου ἅπτηται τῆς τοῦ respective angles of the (figure) about which it is circum-
κύκλου περιφερείας. scribed.
δʹ. Σχῆμα δὲ εὐθύγραμμον περὶ κύκλον περιγράφε- 3. A rectilinear figure is said to be inscribed in a cir-

σθαι λέγεται, ὅταν ἑκάστη πλευρὰ τοῦ περιγραφομένου cle when each angle of the inscribed (figure) touches the
ἐφάπτηται τῆς τοῦ κύκλου περιφερείας. circumference of the circle.
εʹ. Κύκλος δὲ εἰς σχῆμα ὁμοίως ἐγγράφεσθαι λέγεται, 4. And a rectilinear figure is said to be circumscribed

ὅταν ἡ τοῦ κύκλου περιφέρεια ἑκάστης πλευρᾶς τοῦ, εἰς ὃ about a circle when each side of the circumscribed (fig-
ἐγγράφεται, ἅπτηται. ure) touches the circumference of the circle.
ϛʹ. Κύκλος δὲ περὶ σχῆμα περιγράφεσθαι λέγεται, ὅταν 5. And, similarly, a circle is said to be inscribed in a

ἡ τοῦ κύκλου περιφέρεια ἑκάστης γωνίας τοῦ, περὶ ὃ πε- (rectilinear) figure when the circumference of the circle
ριγράφεται, ἅπτηται. touches each side of the (figure) in which it is inscribed.
ζʹ. Εὐθεῖα εἰς κύκλον ἐναρμόζεσθαι λέγεται, ὅταν τὰ 6. And a circle is said to be circumscribed about a

πέρατα αὐτῆς ἐπὶ τῆς περιφερείας ᾖ τοῦ κύκλου. rectilinear (figure) when the circumference of the circle
touches each angle of the (figure) about which it is cir-
cumscribed.

7. A straight-line is said to be inserted into a circle

when its extemities are on the circumference of the circle.aþ. Proposition 1
Εἰς τὸν δοθέντα κύκλον τῇ δοθείσῃ εὐθείᾳ μὴ μείζονι To insert a straight-line equal to a given straight-line

οὔσῃ τῆς τοῦ κύκλου διαμέτρου ἴσην εὐθεῖαν ἐναρμόσαι. into a circle, (the latter straight-line) not being greater
than the diameter of the circle.

∆

ΓΒ

B CE

῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, ἡ δὲ δοθεῖσα εὐθεῖα μὴ Let ABC be the given circle, and D the given straight-
μείζων τῆς τοῦ κύκλου διαμέτρου ἡ Δ. δεῖ δὴ εἰς τὸν ΑΒΓ line (which is) not greater than the diameter of the cir-
κύκλον τῇ Δ εὐθείᾳ ἴσην εὐθεῖαν ἐναρμόσαι. cle. So it is required to insert a straight-line, equal to the
῎Ηχθω τοῦ ΑΒΓ κύκλου διάμετρος ἡ ΒΓ. εἰ μὲν οὖν ἴση straight-line D, into the circle ABC.

ἐστὶν ἡ ΒΓ τῇ Δ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν· ἐνήρμοσται Let a diameter BC of circle ABC have been drawn.†

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STOIQEIWN dþ. ELEMENTS BOOK 4
γὰρ εἰς τὸν ΑΒΓ κύκλον τῇ Δ εὐθείᾳ ἴση ἡ ΒΓ. εἰ δὲ μείζων Therefore, if BC is equal to D then that (which) was
ἐστὶν ἡ ΒΓ τῆς Δ, κείσθω τῇ Δ ἴση ἡ ΓΕ, καὶ κέντρῳ prescribed has taken place. For the (straight-line) BC,
τῷ Γ διαστήματι δὲ τῷ ΓΕ κύκλος γεγράφθω ὁ ΕΑΖ, καὶ equal to the straight-line D, has been inserted into the
ἐπεζεύχθω ἡ ΓΑ. circle ABC. And if BC is greater than D then let CE be
᾿Επεὶ οὖν το Γ σημεῖον κέντρον ἐστὶ τοῦ ΕΑΖ κύκλου, made equal to D [Prop. 1.3], and let the circle EAF have

ἴση ἐστὶν ἡ ΓΑ τῇ ΓΕ. ἀλλὰ τῇ Δ ἡ ΓΕ ἐστιν ἴση· καὶ ἡ Δ been drawn with center C and radius CE. And let CA
ἄρα τῇ ΓΑ ἐστιν ἴση. have been joined.
Εἰς ἄρα τὸν δοθέντα κύκλον τὸν ΑΒΓ τῇ δοθείσῃ εὐθείᾳ Therefore, since the point C is the center of circle

τῇ Δ ἴση ἐνήρμοσται ἡ ΓΑ· ὅπερ ἔδει ποιῆσαι. EAF , CA is equal to CE. But, CE is equal to D. Thus,
D is also equal to CA.

Thus, CA, equal to the given straight-line D, has been
inserted into the given circle ABC. (Which is) the very

thing it was required to do.

† Presumably, by finding the center of the circle [Prop. 3.1], and then drawing a line through it.bþ. Proposition 2
Εἰς τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον To inscribe a triangle, equiangular with a given trian-

τρίγωνον ἐγγράψαι. gle, in a given circle.

Β Ε

∆ D

B E

῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, τὸ δὲ δοθὲν τριγωνον Let ABC be the given circle, and DEF the given tri-
τὸ ΔΕΖ· δεῖ δὴ εἰς τὸν ΑΒΓ κύκλον τῷ ΔΕΖ τριγώνῳ angle. So it is required to inscribe a triangle, equiangular
ἰσογώνιον τρίγωνον ἐγγράψαι. with triangle DEF , in circle ABC.

῎Ηχθω τοῦ ΑΒΓ κύκλου ἐφαπτομένη ἡ ΗΘ κατὰ τὸ Α, Let GH have been drawn touching circle ABC at A.†

καὶ συνεστάτω πρὸς τῇ ΑΘ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ And let (angle) HAC, equal to angle DEF , have been
τῷ Α τῇ ὑπὸ ΔΕΖ γωνίᾳ ἴση ἡ ὑπὸ ΘΑΓ, πρὸς δὲ τῇ ΑΗ constructed on the straight-line AH at the point A on it,
εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΔΖΕ [γωνίᾳ] and (angle) GAB, equal to [angle] DFE, on the straight-
ἴση ἡ ὑπὸ ΗΑΒ, καὶ ἐπεζεύχθω ἡ ΒΓ. line AG at the point A on it [Prop. 1.23]. And let BC
᾿Επεὶ οὖν κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΑΘ, have been joined.

καὶ ἀπὸ τῆς κατὰ τὸ Α ἐπαφῆς εἰς τὸν κύκλον διῆκται εὐθεῖα Therefore, since some straight-line AH touches the
ἡ ΑΓ, ἡ ἄρα ὑπὸ ΘΑΓ ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου circle ABC, and the straight-line AC has been drawn
τμήματι γωνίᾳ τῇ ὑπὸ ΑΒΓ. ἀλλ᾿ ἡ ὑπὸ ΘΑΓ τῇ ὑπὸ ΔΕΖ across (the circle) from the point of contact A, (angle)
ἐστιν ἴση· καὶ ἡ ὑπὸ ΑΒΓ ἄρα γωνία τῇ ὑπὸ ΔΕΖ ἐστιν HAC is thus equal to the angle ABC in the alternate
ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ἐστιν segment of the circle [Prop. 3.32]. But, HAC is equal to
ἴση· καὶ λοιπὴ ἄρα ἡ ὑπὸ ΒΑΓ λοιπῇ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση DEF . Thus, angle ABC is also equal to DEF . So, for the
[ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ, same (reasons), ACB is also equal to DFE. Thus, the re-
καὶ ἐγγέγραπται εἰς τὸν ΑΒΓ κύκλον]. maining (angle) BAC is equal to the remaining (angle)
Εἰς τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ EDF [Prop. 1.32]. [Thus, triangle ABC is equiangu-

ἰσογώνιον τρίγωνον ἐγγέγραπται· ὅπερ ἔδει ποιῆσαι. lar with triangle DEF , and has been inscribed in circle

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STOIQEIWN dþ. ELEMENTS BOOK 4
ABC].

Thus, a triangle, equiangular with the given triangle,

has been inscribed in the given circle. (Which is) the very
thing it was required to do.

† See the footnote to Prop. 3.34. gþ. Proposition 3
Περὶ τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον To circumscribe a triangle, equiangular with a given

τρίγωνον περιγράψαι. triangle, about a given circle.

∆

Λ Γ

L C N
G

D
M

E
K

῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, τὸ δὲ δοθὲν τρίγωνον Let ABC be the given circle, and DEF the given tri-
τὸ ΔΕΖ· δεῖ δὴ περὶ τὸν ΑΒΓ κύκλον τῷ ΔΕΖ τριγώνῳ angle. So it is required to circumscribe a triangle, equian-
ἰσογώνιον τρίγωνον περιγράψαι. gular with triangle DEF , about circle ABC.
᾿Εκβεβλήσθω ἡ ΕΖ ἐφ᾿ ἑκάτερα τὰ μέρη κατὰ τὰ Η, Θ Let EF have been produced in each direction to

σημεῖα, καὶ εἰλήφθω τοῦ ΑΒΓ κύκλου κέντρον τὸ Κ, καὶ points G and H . And let the center K of circle ABC
διήχθω, ὡς ἔτυχεν, εὐθεῖα ἡ ΚΒ, καὶ συνεστάτω πρὸς τῇ have been found [Prop. 3.1]. And let the straight-line
ΚΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Κ τῇ μὲν ὑπὸ ΔΕΗ KB have been drawn, at random, across (ABC). And
γωνίᾳ ἴση ἡ ὑπὸ ΒΚΑ, τῇ δὲ ὑπὸ ΔΖΘ ἴση ἡ ὑπὸ ΒΚΓ, καὶ let (angle) BKA, equal to angle DEG, have been con-
διὰ τῶν Α, Β, Γ σημείων ἤχθωσαν ἐφαπτόμεναι τοῦ ΑΒΓ structed on the straight-line KB at the point K on it,
κύκλου αἱ ΛΑΜ, ΜΒΝ, ΝΓΛ. and (angle) BKC, equal to DFH [Prop. 1.23]. And let
Καὶ ἐπεὶ ἐφάπτονται τοῦ ΑΒΓ κύκλου αἱ ΛΜ, ΜΝ, ΝΛ the (straight-lines) LAM , MBN , and NCL have been

κατὰ τὰ Α, Β, Γ σημεῖα, ἀπὸ δὲ τοῦ Κ κέντρου ἐπὶ τὰ Α, Β, drawn through the points A, B, and C (respectively),

Γ σημεῖα ἐπεζευγμέναι εἰσὶν αἱ ΚΑ, ΚΒ, ΚΓ, ὀρθαὶ ἄρα εἰσὶν touching the circle ABC.†

αἱ πρὸς τοῖς Α, Β, Γ σημείοις γωνίαι. καὶ ἐπεὶ τοῦ ΑΜΒΚ And since LM , MN , and NL touch circle ABC at
τετραπλεύρου αἱ τέσσαρες γωνίαι τέτρασιν ὀρθαῖς ἴσαι εἰσίν, points A, B, and C (respectively), and KA, KB, and
ἐπειδήπερ καὶ εἰς δύο τρίγωνα διαιρεῖται τὸ ΑΜΒΚ, καί εἰσιν KC are joined from the center K to points A, B, and
ὀρθαὶ αἱ ὑπὸ ΚΑΜ, ΚΒΜ γωνίαι, λοιπαὶ ἄρα αἱ ὑπὸ ΑΚΒ, C (respectively), the angles at points A, B, and C are
ΑΜΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν. εἰσὶ δὲ καὶ αἱ ὑπὸ ΔΕΗ, thus right-angles [Prop. 3.18]. And since the (sum of the)
ΔΕΖ δυσὶν ὀρθαῖς ἴσαι· αἱ ἄρα ὑπὸ ΑΚΒ, ΑΜΒ ταῖς ὑπὸ four angles of quadrilateral AMBK is equal to four right-
ΔΕΗ, ΔΕΖ ἴσαι εἰσίν, ὧν ἡ ὑπὸ ΑΚΒ τῇ ὑπὸ ΔΕΗ ἐστιν angles, inasmuch as AMBK (can) also (be) divided into
ἴση· λοιπὴ ἄρα ἡ ὑπὸ ΑΜΒ λοιπῇ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση. two triangles [Prop. 1.32], and angles KAM and KBM
ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἡ ὑπὸ ΛΝΒ τῇ ὑπὸ ΔΖΕ are (both) right-angles, the (sum of the) remaining (an-
ἐστιν ἴση· καὶ λοιπὴ ἄρα ἡ ὑπὸ ΜΛΝ [λοιπῇ] τῇ ὑπὸ ΕΔΖ gles), AKB and AMB, is thus equal to two right-angles.
ἐστιν ἴση. ἰσογώνιον ἄρα ἐστὶ τὸ ΛΜΝ τρίγωνον τῷ ΔΕΖ And DEG and DEF is also equal to two right-angles
τριγώνῳ· καὶ περιγέγραπται περὶ τὸν ΑΒΓ κύκλον. [Prop. 1.13]. Thus, AKB and AMB is equal to DEG
Περὶ τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ and DEF , of which AKB is equal to DEG. Thus, the re-

ἰσογώνιον τρίγωνον περιγέγραπται· ὅπερ ἔδει ποιῆσαι. mainder AMB is equal to the remainder DEF . So, sim-
ilarly, it can be shown that LNB is also equal to DFE.

Thus, the remaining (angle) MLN is also equal to the

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STOIQEIWN dþ. ELEMENTS BOOK 4
[remaining] (angle) EDF [Prop. 1.32]. Thus, triangle

LMN is equiangular with triangle DEF . And it has been

drawn around circle ABC.
Thus, a triangle, equiangular with the given triangle,

has been circumscribed about the given circle. (Which is)
the very thing it was required to do.

† See the footnote to Prop. 3.34. dþ. Proposition 4
Εἰς τὸ δοθὲν τρίγωνον κύκλον ἐγγράψαι. To inscribe a circle in a given triangle.

∆

Ε
G

B
F

῎Εστω τὸ δοθὲν τρίγωνον τὸ ΑΒΓ· δεῖ δὴ εἰς τὸ ΑΒΓ Let ABC be the given triangle. So it is required to
τρίγωνον κύκλον ἐγγράψαι. inscribe a circle in triangle ABC.
Τετμήσθωσαν αἱ ὑπὸ ΑΒΓ, ΑΓΒ γωνίαι δίχα ταῖς ΒΔ, Let the angles ABC and ACB have been cut in half by

ΓΔ εὐθείαις, καὶ συμβαλλέτωσαν ἀλλήλαις κατὰ τὸ Δ the straight-lines BD and CD (respectively) [Prop. 1.9],
σημεῖον, καὶ ἤχθωσαν ἀπὸ τοῦ Δ ἐπὶ τὰς ΑΒ, ΒΓ, ΓΑ and let them meet one another at point D, and let DE,
εὐθείας κάθετοι αἱ ΔΕ, ΔΖ, ΔΗ. DF , and DG have been drawn from point D, perpendic-
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΑΒΔ γωνία τῇ ὑπὸ ΓΒΔ, ular to the straight-lines AB, BC, and CA (respectively)

ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΒΕΔ ὀρθῇ τῇ ὑπὸ ΒΖΔ ἴση, δύο [Prop. 1.12].
δὴ τρίγωνά ἐστι τὰ ΕΒΔ, ΖΒΔ τὰς δύο γωνίας ταῖς δυσὶ And since angle ABD is equal to CBD, and the right-
γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν angle BED is also equal to the right-angle BFD, EBD
ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν κοινὴν αὐτῶν τὴν and FBD are thus two triangles having two angles equal
ΒΔ· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας to two angles, and one side equal to one side—the (one)
ἕξουσιν· ἴση ἄρα ἡ ΔΕ τῇ ΔΖ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΔΗ subtending one of the equal angles (which is) common to
τῇ ΔΖ ἐστιν ἴση. αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΔΕ, ΔΖ, ΔΗ the (triangles)—(namely), BD. Thus, they will also have
ἴσαι ἀλλήλαις εἰσίν· ὁ ἄρα κέντρῷ τῷ Δ καὶ διαστήματι ἑνὶ the remaining sides equal to the (corresponding) remain-
τῶν Ε, Ζ, Η κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν ing sides [Prop. 1.26]. Thus, DE (is) equal to DF . So,
σημείων καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΑ εὐθειῶν διὰ τὸ for the same (reasons), DG is also equal to DF . Thus,
ὀρθὰς εἶναι τὰς πρὸς τοῖς Ε, Ζ, Η σημείοις γωνίας. εἰ γὰρ the three straight-lines DE, DF , and DG are equal to
τεμεῖ αὐτάς, ἔσται ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς one another. Thus, the circle drawn with center D, and

ἀπ᾿ ἄκρας ἀγομένη ἐντὸς πίπτουσα τοῦ κύκλου· ὅπερ ἄτο- radius one of E, F , or G,† will also go through the re-
πον ἐδείχθη· οὐκ ἄρα ὁ κέντρῳ τῷ Δ διαστήματι δὲ ἑνὶ τῶν maining points, and will touch the straight-lines AB, BC,
Ε, Ζ, Η γραφόμενος κύκλος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΑ εὐθείας· and CA, on account of the angles at E, F , and G being
ἐφάψεται ἄρα αὐτῶν, καὶ ἔσται ὁ κύκλος ἐγγεγραμμένος εἰς right-angles. For if it cuts (one of) them then it will be
τὸ ΑΒΓ τρίγωνον. ἐγγεγράφθω ὡς ὁ ΖΗΕ. a (straight-line) drawn at right-angles to a diameter of
Εἰς ἄρα τὸ δοθὲν τρίγωνον τὸ ΑΒΓ κύκλος ἐγγέγραπται the circle, from its extremity, falling inside the circle. The

ὁ ΕΖΗ· ὅπερ ἔδει ποιῆσαι. very thing was shown (to be) absurd [Prop. 3.16]. Thus,
the circle drawn with center D, and radius one of E, F ,

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or G, does not cut the straight-lines AB, BC, and CA.

Thus, it will touch them and will be the circle inscribed

in triangle ABC. Let it have been (so) inscribed, like
FGE (in the figure).

Thus, the circle EFG has been inscribed in the given
triangle ABC. (Which is) the very thing it was required

to do.

† Here, and in the following propositions, it is understood that the radius is actually one of DE, DF , or DG.eþ. Proposition 5
Περὶ τὸ δοθὲν τρίγωνον κύκλον περιγράψαι. To circumscribe a circle about a given triangle.

∆

Α Α

ΓΓ

∆

Β
Ζ

Ε
Β

∆

ΖΖ

B
F

῎Εστω τὸ δοθὲν τρίγωνον τὸ ΑΒΓ· δεῖ δὲ περὶ τὸ δοθὲν Let ABC be the given triangle. So it is required to
τρίγωνον τὸ ΑΒΓ κύκλον περιγράψαι. circumscribe a circle about the given triangle ABC.
Τετμήσθωσαν αἱ ΑΒ, ΑΓ εὐθεῖαι δίχα κατὰ τὰ Δ, Ε Let the straight-lines AB and AC have been cut in

σημεῖα, καὶ ἀπὸ τῶν Δ, Ε σημείων ταῖς ΑΒ, ΑΓ πρὸς ὀρθὰς half at points D and E (respectively) [Prop. 1.10]. And
ἤχθωσαν αἱ ΔΖ, ΕΖ· συμπεσοῦνται δὴ ἤτοι ἐντὸς τοῦ ΑΒΓ let DF and EF have been drawn from points D and E,
τριγώνου ἢ ἐπὶ τῆς ΒΓ εὐθείας ἢ ἐκτὸς τῆς ΒΓ. at right-angles to AB and AC (respectively) [Prop. 1.11].
Συμπιπτέτωσαν πρότερον ἐντὸς κατὰ τὸ Ζ, καὶ ἐπεζεύχθ- So (DF and EF ) will surely either meet inside triangle

ωσαν αἱ ΖΒ, ΖΓ, ΖΑ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ ABC, on the straight-line BC, or beyond BC.
δὲ καὶ πρὸς ὀρθὰς ἡ ΔΖ, βάσις ἄρα ἡ ΑΖ βάσει τῇ ΖΒ ἐστιν Let them, first of all, meet inside (triangle ABC) at
ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΓΖ τῇ ΑΖ ἐστιν ἴση· (point) F , and let FB, FC, and FA have been joined.
ὥστε καὶ ἡ ΖΒ τῇ ΖΓ ἐστιν ἴση· αἱ τρεῖς ἄρα αἱ ΖΑ, ΖΒ, ΖΓ And since AD is equal to DB, and DF is common and
ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ at right-angles, the base AF is thus equal to the base FB
τῶν Α, Β, Γ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν [Prop. 1.4]. So, similarly, we can show that CF is also
σημείων, καὶ ἔσται περιγεγραμμένος ὁ κύκλος περὶ τὸ ΑΒΓ equal to AF . So that FB is also equal to FC. Thus, the
τρίγωνον. περιγεγράφθω ὡς ὁ ΑΒΓ. three (straight-lines) FA, FB, and FC are equal to one
Ἀλλὰ δὴ αἱ ΔΖ, ΕΖ συμπιπτέτωσαν ἐπὶ τῆς ΒΓ εὐθείας another. Thus, the circle drawn with center F , and radius

κατὰ τὸ Ζ, ὡς ἔχει ἐπὶ τῆς δευτέρας καταγραφῆς, καὶ one of A, B, or C, will also go through the remaining
ἐπεζεύχθω ἡ ΑΖ. ὁμοίως δὴ δείξομεν, ὅτι τὸ Ζ σημεῖον points. And the circle will have been circumscribed about
κέντρον ἐστὶ τοῦ περὶ τὸ ΑΒΓ τρίγωνον περιγραφομένου triangle ABC. Let it have been (so) circumscribed, like
κύκλου. ABC (in the first diagram from the left).
Ἀλλὰ δὴ αἱ ΔΖ, ΕΖ συμπιπτέτωσαν ἐκτὸς τοῦ ΑΒΓ And so, let DF and EF meet on the straight-line

τριγώνου κατὰ τὸ Ζ πάλιν, ὡς ἔχει ἐπὶ τῆς τρίτης κατα- BC at (point) F , like in the second diagram (from the
γραφῆς, καί ἐπεζεύχθωσαν αἱ ΑΖ, ΒΖ, ΓΖ. καὶ ἐπεὶ πάλιν left). And let AF have been joined. So, similarly, we can
ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΔΖ, βάσις show that point F is the center of the circle circumscribed
ἄρα ἡ ΑΖ βάσει τῇ ΒΖ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι about triangle ABC.
καὶ ἡ ΓΖ τῇ ΑΖ ἐστιν ἴση· ὥστε καὶ ἡ ΒΖ τῇ ΖΓ ἐστιν ἴση· And so, let DF and EF meet outside triangle ABC,
ὁ ἄρα [πάλιν] κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν ΖΑ, ΖΒ, again at (point) F , like in the third diagram (from the
ΖΓ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων, left). And let AF , BF , and CF have been joined. And,
καὶ ἔσται περιγεγραμμένος περὶ τὸ ΑΒΓ τρίγωνον. again, since AD is equal to DB, and DF is common and
Περὶ τὸ δοθὲν ἄρα τρίγωνον κύκλος περιγέγραπται· at right-angles, the base AF is thus equal to the base BF

ὅπερ ἔδει ποιῆσαι. [Prop. 1.4]. So, similarly, we can show that CF is also
equal to AF . So that BF is also equal to FC. Thus,

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STOIQEIWN dþ. ELEMENTS BOOK 4
[again] the circle drawn with center F , and radius one

of FA, FB, and FC, will also go through the remaining

points. And it will have been circumscribed about trian-
gle ABC.

Thus, a circle has been circumscribed about the given
triangle. (Which is) the very thing it was required to do.�þ. Proposition 6

Εἰς τὸν δοθέντα κύκλον τετράγωνον ἐγγράψαι. To inscribe a square in a given circle.

∆

E
D

῎Εστω ἡ δοθεὶς κύκλος ὁ ΑΒΓΔ· δεῖ δὴ εἰς τὸν ΑΒΓΔ Let ABCD be the given circle. So it is required to
κύκλον τετράγωνον ἐγγράψαι. inscribe a square in circle ABCD.
῎Ηχθωσαν τοῦ ΑΒΓΔ κύκλου δύο διάμετροι πρὸς ὀρθὰς Let two diameters of circle ABCD, AC and BD, have

ἀλλήλαις αἱ ΑΓ, ΒΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, been drawn at right-angles to one another.† And let AB,
ΔΑ. BC, CD, and DA have been joined.
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΔ· κέντρον γὰρ τὸ Ε· κοινὴ And since BE is equal to ED, for E (is) the center

δὲ καὶ πρὸς ὀρθὰς ἡ ΕΑ, βάσις ἄρα ἡ ΑΒ βάσει τῇ ΑΔ ἴση (of the circle), and EA is common and at right-angles,
ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ΒΓ, ΓΔ ἑκατέρᾳ the base AB is thus equal to the base AD [Prop. 1.4].
τῶν ΑΒ, ΑΔ ἴση ἐστίν· ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔ So, for the same (reasons), each of BC and CD is equal
τετράπλευρον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ ἡ to each of AB and AD. Thus, the quadrilateral ABCD
ΒΔ εὐθεῖα διάμετρός ἐστι τοῦ ΑΒΓΔ κύκλου, ἡμικύκλιον is equilateral. So I say that (it is) also right-angled. For
ἄρα ἐστὶ τὸ ΒΑΔ· ὀρθὴ ἄρα ἡ ὑπὸ ΒΑΔ γωνία. διὰ τὰ since the straight-line BD is a diameter of circle ABCD,
αὐτὰ δὴ καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΑ ὀρθή ἐστιν· BAD is thus a semi-circle. Thus, angle BAD (is) a right-
ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔ τετράπλευρον. ἐδείχθη δὲ angle [Prop. 3.31]. So, for the same (reasons), (angles)
καὶ ἰσόπλευρον· τετράγωνον ἄρα ἐστίν. καὶ ἐγγέγραπται εἰς ABC, BCD, and CDA are also each right-angles. Thus,
τὸν ΑΒΓΔ κύκλον. the quadrilateral ABCD is right-angled. And it was also
Εἰς ἄρα τὸν δοθέντα κύκλον τετράγωνον ἐγγέγραπται shown (to be) equilateral. Thus, it is a square [Def. 1.22].

τὸ ΑΒΓΔ· ὅπερ ἔδει ποιῆσαι. And it has been inscribed in circle ABCD.
Thus, the square ABCD has been inscribed in the

given circle. (Which is) the very thing it was required

to do.

† Presumably, by finding the center of the circle [Prop. 3.1], drawing a line through it, and then drawing a second line through it, at right-angles

to the first [Prop. 1.11]. zþ. Proposition 7
Περὶ τὸν δοθέντα κύκλον τετράγωνον περιγράψαι. To circumscribe a square about a given circle.

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STOIQEIWN dþ. ELEMENTS BOOK 4
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ· δεῖ δὴ περὶ τὸν ΑΒΓΔ Let ABCD be the given circle. So it is required to

κύκλον τετράγωνον περιγράψαι. circumscribe a square about circle ABCD.

Θ Γ Κ

Η Α Ζ

∆Β D

H C K

G A F

῎Ηχθωσαν τοῦ ΑΒΓΔ κύκλου δύο διάμετροι πρὸς ὀρθὰς Let two diameters of circle ABCD, AC and BD, have

ἀλλήλαις αἱ ΑΓ, ΒΔ, καὶ διὰ τῶν Α, Β, Γ, Δ σημείων ἤχθω- been drawn at right-angles to one another.† And let FG,
σαν ἐφαπτόμεναι τοῦ ΑΒΓΔ κύκλου αἱ ΖΗ, ΗΘ, ΘΚ, ΚΖ. GH , HK, and KF have been drawn through points A,

᾿Επεὶ οὖν ἐφάπτεται ἡ ΖΗ τοῦ ΑΒΓΔ κύκλου, ἀπὸ δὲ B, C, and D (respectively), touching circle ABCD.‡

τοῦ Ε κέντρου ἐπὶ τὴν κατὰ τὸ Α ἐπαφὴν ἐπέζευκται ἡ Therefore, since FG touches circle ABCD, and EA
ΕΑ, αἱ ἄρα πρὸς τῷ Α γωνίαι ὀρθαί εἰσιν. διὰ τὰ αὐτὰ has been joined from the center E to the point of contact
δὴ καὶ αἱ πρὸς τοῖς Β, Γ, Δ σημείοις γωνίαι ὀρθαί εἰσιν. A, the angles at A are thus right-angles [Prop. 3.18]. So,
καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΑΕΒ γωνία, ἐστὶ δὲ ὀρθὴ καὶ ἡ for the same (reasons), the angles at points B, C, and
ὑπὸ ΕΒΗ, παράλληλος ἄρα ἐστὶν ἡ ΗΘ τῇ ΑΓ. διὰ τὰ αὐτὰ D are also right-angles. And since angle AEB is a right-
δὴ καὶ ἡ ΑΓ τῇ ΖΚ ἐστι παράλληλος. ὥστε καὶ ἡ ΗΘ τῇ angle, and EBG is also a right-angle, GH is thus parallel
ΖΚ ἐστι παράλληλος. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα to AC [Prop. 1.29]. So, for the same (reasons), AC is
τῶν ΗΖ, ΘΚ τῇ ΒΕΔ ἐστι παράλληλος. παραλληλόγραμμα also parallel to FK. So that GH is also parallel to FK
ἄρα ἐστὶ τὰ ΗΚ, ΗΓ, ΑΚ, ΖΒ, ΒΚ· ἴση ἄρα ἐστὶν ἡ μὲν [Prop. 1.30]. So, similarly, we can show that GF and
ΗΖ τῇ ΘΚ, ἡ δὲ ΗΘ τῇ ΖΚ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ HK are each parallel to BED. Thus, GK, GC, AK, FB,
ΒΔ, ἀλλὰ καὶ ἡ μὲν ΑΓ ἑκατέρᾳ τῶν ΗΘ, ΖΚ, ἡ δὲ ΒΔ and BK are (all) parallelograms. Thus, GF is equal to
ἑκατέρᾳ τῶν ΗΖ, ΘΚ ἐστιν ἴση [καὶ ἑκατέρα ἄρα τῶν ΗΘ, HK, and GH to FK [Prop. 1.34]. And since AC is equal
ΖΚ ἑκατέρᾳ τῶν ΗΖ, ΘΚ ἐστιν ἴση], ἰσόπλευρον ἄρα ἐστὶ τὸ to BD, but AC (is) also (equal) to each of GH and FK,
ΖΗΘΚ τετράπλευρον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ and BD is equal to each of GF and HK [Prop. 1.34]
γὰρ παραλληλόγραμμόν ἐστι τὸ ΗΒΕΑ, καί ἐστιν ὀρθὴ ἡ [and each of GH and FK is thus equal to each of GF
ὑπὸ ΑΕΒ, ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΑΗΒ. ὁμοίως δὴ δείξομεν, and HK], the quadrilateral FGHK is thus equilateral.
ὅτι καὶ αἱ πρὸς τοῖς Θ, Κ, Ζ γωνίαι ὀρθαί εἰσιν. ὀρθογώνιον So I say that (it is) also right-angled. For since GBEA
ἄρα ἐστὶ τὸ ΖΗΘΚ. ἐδείχθη δὲ καὶ ἰσόπλευρον· τετράγωνον is a parallelogram, and AEB is a right-angle, AGB is
ἄρα ἐστίν. καὶ περιγέγραπται περὶ τὸν ΑΒΓΔ κύκλον. thus also a right-angle [Prop. 1.34]. So, similarly, we can
Περὶ τὸν δοθέντα ἄρα κύκλον τετράγωνον περιγέγραπται· show that the angles at H , K, and F are also right-angles.

ὅπερ ἔδει ποιῆσαι. Thus, FGHK is right-angled. And it was also shown (to
be) equilateral. Thus, it is a square [Def. 1.22]. And it

has been circumscribed about circle ABCD.

Thus, a square has been circumscribed about the
given circle. (Which is) the very thing it was required

to do.

† See the footnote to the previous proposition.
‡ See the footnote to Prop. 3.34.

116

STOIQEIWN dþ. ELEMENTS BOOK 4hþ. Proposition 8
Εἰς τὸ δοθὲν τετράγωνον κύκλον ἐγγράψαι. To inscribe a circle in a given square.
῎Εστω τὸ δοθὲν τετράγωνον τὸ ΑΒΓΔ. δεῖ δὴ εἰς τὸ Let the given square be ABCD. So it is required to

ΑΒΓΔ τετράγωνον κύκλον ἐγγράψαι. inscribe a circle in square ABCD.

Α Ε ∆

ΓΘΒ

Ζ
G

A E D

B H C
Τετμήσθω ἑκατέρα τῶν ΑΔ, ΑΒ δίχα κατὰ τὰ Ε, Ζ Let AD and AB each have been cut in half at points E

σημεῖα, καὶ διὰ μὲν τοῦ Ε ὁποτέρᾳ τῶν ΑΒ, ΓΔ παράλληλος and F (respectively) [Prop. 1.10]. And let EH have been
ἤχθω ὁ ΕΘ, διὰ δὲ τοῦ Ζ ὁποτέρᾳ τῶν ΑΔ, ΒΓ παράλληλος drawn through E, parallel to either of AB or CD, and let
ἤχθω ἡ ΖΚ· παραλληλόγραμμον ἄρα ἐστὶν ἕκαστον τῶν FK have been drawn through F , parallel to either of AD
ΑΚ, ΚΒ, ΑΘ, ΘΔ, ΑΗ, ΗΓ, ΒΗ, ΗΔ, καὶ αἱ ἀπεναντίον or BC [Prop. 1.31]. Thus, AK, KB, AH , HD, AG, GC,
αὐτῶν πλευραὶ δηλονότι ἴσαι [εἰσίν]. καὶ ἐπεὶ ἴση ἐστὶν ἡ BG, and GD are each parallelograms, and their opposite
ΑΔ τῇ ΑΒ, καί ἐστι τῆς μὲν ΑΔ ἡμίσεια ἡ ΑΕ, τῆς δὲ ΑΒ sides [are] manifestly equal [Prop. 1.34]. And since AD
ἡμίσεια ἡ ΑΖ, ἴση ἄρα καὶ ἡ ΑΕ τῇ ΑΖ· ὥστε καὶ αἱ ἀπε- is equal to AB, and AE is half of AD, and AF half of
ναντίον· ἴση ἄρα καὶ ἡ ΖΗ τῇ ΗΕ. ὁμοίως δὴ δείξομεν, ὅτι AB, AE (is) thus also equal to AF . So that the opposite
καὶ ἑκατέρα τῶν ΗΘ, ΗΚ ἑκατέρᾳ τῶν ΖΗ, ΗΕ ἐστιν ἴση· (sides are) also (equal). Thus, FG (is) also equal to GE.
αἱ τέσσαρες ἄρα αἱ ΗΕ, ΗΖ, ΗΘ, ΗΚ ἴσαι ἀλλήλαις [εἰσίν]. So, similarly, we can also show that each of GH and GK
ὁ ἄρα κέντρῳ μὲν τῷ Η διαστήματι δὲ ἑνὶ τῶν Ε, Ζ, Θ, Κ is equal to each of FG and GE. Thus, the four (straight-
κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων· καὶ lines) GE, GF , GH , and GK [are] equal to one another.
ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΔ, ΔΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι Thus, the circle drawn with center G, and radius one of
τὰς πρὸς τοῖς Ε, Ζ, Θ, Κ γωνίας· εἰ γὰρ τεμεῖ ὁ κύκλος τὰς E, F , H , or K, will also go through the remaining points.
ΑΒ, ΒΓ, ΓΔ, ΔΑ, ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς And it will touch the straight-lines AB, BC, CD, and
ἀπ᾿ ἄκρας ἀγομένη ἐντὸς πεσεῖται τοῦ κύκλου· ὅπερ ἄτοπον DA, on account of the angles at E, F , H , and K being
ἐδείχθη. οὐκ ἄρα ὁ κέντρῳ τῷ Η διαστήματι δὲ ἑνὶ τῶν Ε, right-angles. For if the circle cuts AB, BC, CD, or DA,
Ζ, Θ, Κ κύκλος γραφόμενος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΑ then a (straight-line) drawn at right-angles to a diameter
εὐθείας. ἐφάψεται ἄρα αὐτῶν καὶ ἔσται ἐγγεγραμμένος εἰς of the circle, from its extremity, will fall inside the circle.
τὸ ΑΒΓΔ τετράγωνον. The very thing was shown (to be) absurd [Prop. 3.16].
Εἰς ἄρα τὸ δοθὲν τετράγωνον κύκλος ἐγγέγραπται· Thus, the circle drawn with center G, and radius one of

ὅπερ ἔδει ποιῆσαι. E, F , H , or K, does not cut the straight-lines AB, BC,
CD, or DA. Thus, it will touch them, and will have been
inscribed in the square ABCD.

Thus, a circle has been inscribed in the given square.

(Which is) the very thing it was required to do.jþ. Proposition 9
Περὶ τὸ δοθὲν τετράγωνον κύκλον περιγράψαι. To circumscribe a circle about a given square.
῎Εστω τὸ δοθὲν τετράγωνον τὸ ΑΒΓΔ· δεῖ δὴ περὶ τὸ Let ABCD be the given square. So it is required to

ΑΒΓΔ τετράγωνον κύκλον περιγράψαι. circumscribe a circle about square ABCD.

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STOIQEIWN dþ. ELEMENTS BOOK 4
᾿Επιζευχθεῖσαι γὰρ αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας AC and BD being joined, let them cut one another at

κατὰ τὸ Ε. E.

∆

E
D

Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, κοινὴ δὲ ἡ ΑΓ, δύο And since DA is equal to AB, and AC (is) common,
δὴ αἱ ΔΑ, ΑΓ δυσὶ ταῖς ΒΑ, ΑΓ ἴσαι εἰσίν· καὶ βάσις ἡ the two (straight-lines) DA, AC are thus equal to the two
ΔΓ βάσει τῇ ΒΓ ἴση· γωνία ἄρα ἡ ὑπὸ ΔΑΓ γωνίᾳ τῇ ὑπὸ (straight-lines) BA, AC. And the base DC (is) equal to
ΒΑΓ ἴση ἐστίν· ἡ ἄρα ὑπὸ ΔΑΒ γωνία δίχα τέτμηται ὑπὸ the base BC. Thus, angle DAC is equal to angle BAC
τῆς ΑΓ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, [Prop. 1.8]. Thus, the angle DAB has been cut in half
ΒΓΔ, ΓΔΑ δίχα τέτμηται ὑπὸ τῶν ΑΓ, ΔΒ εὐθειῶν. καὶ by AC. So, similarly, we can show that ABC, BCD, and
ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΑΒ γωνία τῇ ὑπὸ ΑΒΓ, καί ἐστι CDA have each been cut in half by the straight-lines AC
τῆς μὲν ὑπὸ ΔΑΒ ἡμίσεια ἡ ὑπὸ ΕΑΒ, τῆς δὲ ὑπὸ ΑΒΓ and DB. And since angle DAB is equal to ABC, and
ἡμίσεια ἡ ὑπὸ ΕΒΑ, καὶ ἡ ὑπὸ ΕΑΒ ἄρα τῇ ὑπὸ ΕΒΑ ἐστιν EAB is half of DAB, and EBA half of ABC, EAB is
ἴση· ὥστε καὶ πλευρὰ ἡ ΕΑ τῇ ΕΒ ἐστιν ἴση. ὁμοίως δὴ thus also equal to EBA. So that side EA is also equal
δείξομεν, ὅτι καὶ ἑκατέρα τῶν ΕΑ, ΕΒ [εὐθειῶν] ἑκατέρᾳ to EB [Prop. 1.6]. So, similarly, we can show that each
τῶν ΕΓ, ΕΔ ἴση ἐστίν. αἱ τέσσαρες ἄρα αἱ ΕΑ, ΕΒ, ΕΓ, of the [straight-lines] EA and EB are also equal to each
ΕΔ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ε καὶ διαστήματι of EC and ED. Thus, the four (straight-lines) EA, EB,
ἑνὶ τῶν Α, Β, Γ, Δ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν EC, and ED are equal to one another. Thus, the circle
λοιπῶν σημείων καὶ ἔσται περιγεγραμμένος περὶ τὸ ΑΒΓΔ drawn with center E, and radius one of A, B, C, or D,
τετράγωνον. περιγεγράφθω ὡς ὁ ΑΒΓΔ. will also go through the remaining points, and will have
Περὶ τὸ δοθὲν ἄρα τετράγωνον κύκλος περιγέγραπται· been circumscribed about the square ABCD. Let it have

ὅπερ ἔδει ποιῆσαι. been (so) circumscribed, like ABCD (in the figure).
Thus, a circle has been circumscribed about the given

square. (Which is) the very thing it was required to do.iþ. Proposition 10
᾿Ισοσκελὲς τρίγωνον συστήσασθαι ἔχον ἑκατέραν τῶν To construct an isosceles triangle having each of the

πρὸς τῇ βάσει γωνιῶν διπλασίονα τῆς λοιπῆς. angles at the base double the remaining (angle).
᾿Εκκείσθω τις εὐθεῖα ἡ ΑΒ, καὶ τετμήσθω κατὰ τὸ Let some straight-line AB be taken, and let it have

Γ σημεῖον, ὥστε τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον been cut at point C so that the rectangle contained by
ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ· καὶ AB and BC is equal to the square on CA [Prop. 2.11].
κέντρῳ τῷ Α καὶ διαστήματι τῷ ΑΒ κύκλος γεγράφθω And let the circle BDE have been drawn with center A,
ὁ ΒΔΕ, καὶ ἐνηρμόσθω εἰς τὸν ΒΔΕ κύκλον τῇ ΑΓ εὐθείᾳ and radius AB. And let the straight-line BD, equal to
μὴ μείζονι οὔσῃ τῆς τοῦ ΒΔΕ κύκλου διαμέτρου ἴση εὐθεῖα the straight-line AC, being not greater than the diame-
ἡ ΒΔ· καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ, καὶ περιγεγράφθω ter of circle BDE, have been inserted into circle BDE
περὶ τὸ ΑΓΔ τρίγωνον κύκλος ὁ ΑΓΔ. [Prop. 4.1]. And let AD and DC have been joined. And

let the circle ACD have been circumscribed about trian-
gle ACD [Prop. 4.5].

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STOIQEIWN dþ. ELEMENTS BOOK 4
Α

∆

Γ C

Καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς And since the (rectangle contained) by AB and BC
ΑΓ, ἴση δὲ ἡ ΑΓ τῇ ΒΔ, τὸ ἄρα ὑπὸ τῶν ΑΒ, ΒΓ ἴσον is equal to the (square) on AC, and AC (is) equal to
ἐστὶ τῷ ἀπὸ τῆς ΒΔ. καὶ ἐπεὶ κύκλου τοῦ ΑΓΔ εἴληπταί τι BD, the (rectangle contained) by AB and BC is thus
σημεῖον ἐκτὸς τὸ Β, καὶ ἀπὸ τοῦ Β πρὸς τὸν ΑΓΔ κύκλον equal to the (square) on BD. And since some point B
προσπεπτώκασι δύο εὐθεῖαι αἱ ΒΑ, ΒΔ, καὶ ἡ μὲν αὐτῶν has been taken outside of circle ACD, and two straight-
τέμνει, ἡ δὲ προσπίπτει, καί ἐστι τὸ ὑπὸ τῶν ΑΒ, ΒΓ ἴσον lines BA and BD have radiated from B towards the cir-
τῷ ἀπὸ τῆς ΒΔ, ἡ ΒΔ ἄρα ἐφάπτεται τοῦ ΑΓΔ κύκλου. ἐπεὶ cle ACD, and (one) of them cuts (the circle), and (the
οὖν ἐφάπτεται μὲν ἡ ΒΔ, ἀπὸ δὲ τῆς κατὰ τὸ Δ ἐπαφῆς other) meets (the circle), and the (rectangle contained)
διῆκται ἡ ΔΓ, ἡ ἄρα ὑπὸ ΒΔΓ γωνιά ἴση ἐστὶ τῇ ἐν τῷ by AB and BC is equal to the (square) on BD, BD thus
ἐναλλὰξ τοῦ κύκλου τμήματι γωνίᾳ τῇ ὑπὸ ΔΑΓ. ἐπεὶ οὖν touches circle ACD [Prop. 3.37]. Therefore, since BD
ἴση ἐστὶν ἡ ὑπὸ ΒΔΓ τῇ ὑπὸ ΔΑΓ, κοινὴ προσκείσθω ἡ touches (the circle), and DC has been drawn across (the
ὑπὸ ΓΔΑ· ὅλη ἄρα ἡ ὑπὸ ΒΔΑ ἴση ἐστὶ δυσὶ ταῖς ὑπὸ ΓΔΑ, circle) from the point of contact D, the angle BDC is
ΔΑΓ. ἀλλὰ ταῖς ὑπὸ ΓΔΑ, ΔΑΓ ἴση ἐστὶν ἡ ἐκτὸς ἡ ὑπὸ thus equal to the angle DAC in the alternate segment of
ΒΓΔ· καὶ ἡ ὑπὸ ΒΔΑ ἄρα ἴση ἐστὶ τῇ ὑπὸ ΒΓΔ. ἀλλὰ ἡ the circle [Prop. 3.32]. Therefore, since BDC is equal
ὑπὸ ΒΔΑ τῇ ὑπὸ ΓΒΔ ἐστιν ἴση, ἐπεὶ καὶ πλευρὰ ἡ ΑΔ to DAC, let CDA have been added to both. Thus, the
τῇ ΑΒ ἐστιν ἴση· ὥστε καὶ ἡ ὑπὸ ΔΒΑ τῇ ὑπὸ ΒΓΔ ἐστιν whole of BDA is equal to the two (angles) CDA and
ἴση. αἱ τρεῖς ἄρα αἱ ὑπὸ ΒΔΑ, ΔΒΑ, ΒΓΑ ἴσαι ἀλλήλαις DAC. But, the external (angle) BCD is equal to CDA
εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΒΓΔ, and DAC [Prop. 1.32]. Thus, BDA is also equal to
ἴση ἐστὶ καὶ πλευρὰ ἡ ΒΔ πλευρᾷ τῇ ΔΓ. ἀλλὰ ἡ ΒΔ τῇ BCD. But, BDA is equal to CBD, since the side AD is
ΓΑ ὑπόκειται ἴση· καὶ ἡ ΓΑ ἄρα τῇ ΓΔ ἐστιν ἴση· ὥστε καὶ also equal to AB [Prop. 1.5]. So that DBA is also equal
γωνία ἡ ὑπὸ ΓΔΑ γωνίᾳ τῇ ὑπὸ ΔΑΓ ἐστιν ἴση· αἱ ἄρα to BCD. Thus, the three (angles) BDA, DBA, and BCD
ὑπὸ ΓΔΑ, ΔΑΓ τῆς ὑπὸ ΔΑΓ εἰσι διπλασίους. ἴση δὲ ἡ are equal to one another. And since angle DBC is equal
ὑπὸ ΒΓΔ ταῖς ὑπὸ ΓΔΑ, ΔΑΓ· καὶ ἡ ὑπὸ ΒΓΔ ἄρα τῆς ὑπὸ to BCD, side BD is also equal to side DC [Prop. 1.6].
ΓΑΔ ἐστι διπλῆ. ἴση δὲ ἡ ὑπὸ ΒΓΔ ἑκατέρᾳ τῶν ὑπὸ ΒΔΑ, But, BD was assumed (to be) equal to CA. Thus, CA
ΔΒΑ· καὶ ἑκατέρα ἄρα τῶν ὑπὸ ΒΔΑ, ΔΒΑ τῆς ὑπὸ ΔΑΒ is also equal to CD. So that angle CDA is also equal to
ἐστι διπλῆ. angle DAC [Prop. 1.5]. Thus, CDA and DAC is double
᾿Ισοσκελὲς ἄρα τρίγωνον συνέσταται τὸ ΑΒΔ ἔχον DAC. But BCD (is) equal to CDA and DAC. Thus,

ἑκατέραν τῶν πρὸς τῇ ΔΒ βάσει γωνιῶν διπλασίονα τῆς BCD is also double CAD. And BCD (is) equal to to
λοιπῆς· ὅπερ ἔδει ποιῆσαι. each of BDA and DBA. Thus, BDA and DBA are each

double DAB.

Thus, the isosceles triangle ABD has been con-
structed having each of the angles at the base BD double

the remaining (angle). (Which is) the very thing it was

required to do.iaþ. Proposition 11
Εἰς τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ To inscribe an equilateral and equiangular pentagon

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STOIQEIWN dþ. ELEMENTS BOOK 4
ἰσογώνιον ἐγγράψαι. in a given circle.

Θ∆

G H

῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ· δεῖ δὴ εἰς τὸν Let ABCDE be the given circle. So it is required to
ΑΒΓΔΕ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον inscribed an equilateral and equiangular pentagon in cir-
ἐγγράψαι. cle ABCDE.
᾿Εκκείσθω τρίγωνον ἰσοσκελὲς τὸ ΖΗΘ διπλασίονα Let the the isosceles triangle FGH be set up hav-

ἔχον ἑκατέραν τῶν πρὸς τοῖς Η, Θ γωνιῶν τῆς πρὸς τῷ Ζ, ing each of the angles at G and H double the (angle)
καὶ ἐγγεγράφθω εἰς τὸν ΑΒΓΔΕ κύκλον τῷ ΖΗΘ τριγώνῳ at F [Prop. 4.10]. And let triangle ACD, equiangular
ἰσογώνιον τρίγωνον τὸ ΑΓΔ, ὥστε τῇ μὲν πρὸς τῷ Ζ γωνίᾳ to FGH , have been inscribed in circle ABCDE, such
ἴσην εἶναι τὴν ὑπὸ ΓΑΔ, ἑκατέραν δὲ τῶν πρὸς τοῖς Η, Θ that CAD is equal to the angle at F , and the (angles)
ἴσην ἑκατέρᾳ τῶν ὑπὸ ΑΓΔ, ΓΔΑ· καὶ ἑκατέρα ἄρα τῶν at G and H (are) equal to ACD and CDA, respectively
ὑπὸ ΑΓΔ, ΓΔΑ τῆς ὑπὸ ΓΑΔ ἐστι διπλῆ. τετμήσθω δὴ [Prop. 4.2]. Thus, ACD and CDA are each double
ἑκατέρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ δίχα ὑπὸ ἑκατέρας τῶν ΓΕ, CAD. So let ACD and CDA have been cut in half by
ΔΒ εὐθειῶν, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΔΕ, ΕΑ. the straight-lines CE and DB, respectively [Prop. 1.9].
᾿Επεὶ οὖν ἑκατέρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ γωνιῶν δι- And let AB, BC, DE and EA have been joined.

πλασίων ἐστὶ τῆς ὑπὸ ΓΑΔ, καὶ τετμημέναι εἰσὶ δίχα ὑπὸ Therefore, since angles ACD and CDA are each dou-
τῶν ΓΕ, ΔΒ εὐθειῶν, αἱ πέντε ἄρα γωνίαι αἱ ὑπὸ ΔΑΓ, ble CAD, and are cut in half by the straight-lines CE and
ΑΓΕ, ΕΓΔ, ΓΔΒ, ΒΔΑ ἴσαι ἀλλήλαις εἰσίν. αἱ δὲ ἴσαι DB, the five angles DAC, ACE, ECD, CDB, and BDA
γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν· αἱ πέντε ἄρα πε- are thus equal to one another. And equal angles stand
ριφέρειαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ ἴσαι ἀλλήλαις εἰσίν. ὑπὸ upon equal circumferences [Prop. 3.26]. Thus, the five
δὲ τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν· αἱ πέντε circumferences AB, BC, CD, DE, and EA are equal to
ἄρα εὐθεῖαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ ἴσαι ἀλλήλαις εἰσίν· one another [Prop. 3.29]. Thus, the pentagon ABCDE
ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον. λέγω δή, is equilateral. So I say that (it is) also equiangular. For
ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἡ ΑΒ περιφέρεια τῇ ΔΕ πε- since the circumference AB is equal to the circumfer-
ριφερείᾳ ἐστὶν ἴση, κοινὴ προσκείσθω ἡ ΒΓΔ· ὅλη ἄρα ἡ ence DE, let BCD have been added to both. Thus, the
ΑΒΓΔ περιφέρια ὅλῃ τῇ ΕΔΓΒ περιφερείᾳ ἐστὶν ἴση. καὶ whole circumference ABCD is equal to the whole cir-
βέβηκεν ἐπὶ μὲν τῆς ΑΒΓΔ περιφερείας γωνία ἡ ὑπὸ ΑΕΔ, cumference EDCB. And the angle AED stands upon
ἐπὶ δὲ τῆς ΕΔΓΒ περιφερείας γωνία ἡ ὑπὸ ΒΑΕ· καὶ ἡ ὑπὸ circumference ABCD, and angle BAE upon circumfer-
ΒΑΕ ἄρα γωνία τῇ ὑπὸ ΑΕΔ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ ence EDCB. Thus, angle BAE is also equal to AED
καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΕ γωνιῶν ἑκατέρᾳ τῶν [Prop. 3.27]. So, for the same (reasons), each of the an-
ὑπὸ ΒΑΕ, ΑΕΔ ἐστιν ἴση· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕ gles ABC, BCD, and CDE is also equal to each of BAE
πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον. and AED. Thus, pentagon ABCDE is equiangular. And
Εἰς ἄρα τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε it was also shown (to be) equilateral.

καὶ ἰσογώνιον ἐγγέγραπται· ὅπερ ἔδει ποιῆσαι. Thus, an equilateral and equiangular pentagon has
been inscribed in the given circle. (Which is) the very

thing it was required to do.ibþ. Proposition 12
Περὶ τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε To circumscribe an equilateral and equiangular pen-

καὶ ἰσογώνιον περιγράψαι. tagon about a given circle.

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STOIQEIWN dþ. ELEMENTS BOOK 4
Θ

Κ Γ Λ

∆Β

ΕΑ

K C L

῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ· δεῖ δὲ περὶ τὸν Let ABCDE be the given circle. So it is required
ΑΒΓΔΕ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον to circumscribe an equilateral and equiangular pentagon
περιγράψαι. about circle ABCDE.
Νενοήσθω τοῦ ἐγγεγραμμένου πενταγώνου τῶν γωνιῶν Let A, B, C, D, and E have been conceived as the an-

σημεῖα τὰ Α, Β, Γ, Δ, Ε, ὥστε ἴσας εἶναι τὰς ΑΒ, ΒΓ, gular points of a pentagon having been inscribed (in cir-
ΓΔ, ΔΕ, ΕΑ περιφερείας· καὶ διὰ τῶν Α, Β, Γ, Δ, Ε cle ABCDE) [Prop. 3.11], such that the circumferences
ἤχθωσαν τοῦ κύκλου ἐφαπτόμεναι αἱ ΗΘ, ΘΚ, ΚΛ, ΛΜ, AB, BC, CD, DE, and EA are equal. And let GH , HK,
ΜΗ, καὶ εἰλήφθω τοῦ ΑΒΓΔΕ κύκλου κέντρον τὸ Ζ, καὶ KL, LM , and MG have been drawn through (points) A,

ἐπεζεύχθωσαν αἱ ΖΒ, ΖΚ, ΖΓ, ΖΛ, ΖΔ. B, C, D, and E (respectively), touching the circle.† And
Καὶ ἐπεὶ ἡ μὲν ΚΛ εὐθεῖα ἐφάπτεται τοῦ ΑΒΓΔΕ κατὰ let the center F of the circle ABCDE have been found

τὸ Γ, ἀπὸ δὲ τοῦ Ζ κέντρου ἐπὶ τὴν κατὰ τὸ Γ ἐπαφὴν [Prop. 3.1]. And let FB, FK, FC, FL, and FD have
ἐπέζευκται ἡ ΖΓ, ἡ ΖΓ ἄρα κάθετός ἐστιν ἐπὶ τὴν ΚΛ· ὀρθὴ been joined.
ἄρα ἐστὶν ἑκατέρα τῶν πρὸς τῷ Γ γωνιῶν. διὰ τὰ αὐτὰ δὴ And since the straight-line KL touches (circle) ABCDE
καὶ αἱ πρὸς τοῖς Β, Δ σημείοις γωνίαι ὀρθαί εἰσιν. καὶ ἐπεὶ at C, and FC has been joined from the center F to the
ὀρθή ἐστιν ἡ ὑπὸ ΖΓΚ γωνία, τὸ ἄρα ἀπὸ τῆς ΖΚ ἴσον ἐστὶ point of contact C, FC is thus perpendicular to KL
τοῖς ἀπὸ τῶν ΖΓ, ΓΚ. διὰ τὰ αὐτὰ δὴ καὶ τοῖς ἀπὸ τῶν ΖΒ, [Prop. 3.18]. Thus, each of the angles at C is a right-
ΒΚ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΚ· ὥστε τὰ ἀπὸ τῶν ΖΓ, ΓΚ angle. So, for the same (reasons), the angles at B and
τοῖς ἀπὸ τῶν ΖΒ, ΒΚ ἐστιν ἴσα, ὧν τὸ ἀπὸ τῆς ΖΓ τῷ ἀπὸ D are also right-angles. And since angle FCK is a right-
τῆς ΖΒ ἐστιν ἴσον· λοιπὸν ἄρα τὸ ἀπὸ τῆς ΓΚ τῷ ἀπὸ τῆς angle, the (square) on FK is thus equal to the (sum of
ΒΚ ἐστιν ἴσον. ἴση ἄρα ἡ ΒΚ τῇ ΓΚ. καὶ ἐπεὶ ἴση ἐστὶν the squares) on FC and CK [Prop. 1.47]. So, for the
ἡ ΖΒ τῇ ΖΓ, καὶ κοινὴ ἡ ΖΚ, δύο δὴ αἱ ΒΖ, ΖΚ δυσὶ ταῖς same (reasons), the (square) on FK is also equal to the
ΓΖ, ΖΚ ἴσαι εἰσίν· καὶ βάσις ἡ ΒΚ βάσει τῇ ΓΚ [ἐστιν] ἴση· (sum of the squares) on FB and BK. So that the (sum
γωνία ἄρα ἡ μὲν ὑπὸ ΒΖΚ [γωνίᾳ] τῇ ὑπὸ ΚΖΓ ἐστιν ἴση· of the squares) on FC and CK is equal to the (sum of
ἡ δὲ ὑπὸ ΒΚΖ τῇ ὑπὸ ΖΚΓ· διπλῆ ἄρα ἡ μὲν ὑπὸ ΒΖΓ τῆς the squares) on FB and BK, of which the (square) on
ὑπὸ ΚΖΓ, ἡ δὲ ὑπὸ ΒΚΓ τῆς ὑπὸ ΖΚΓ. διὰ τὰ αὐτὰ δὴ καὶ FC is equal to the (square) on FB. Thus, the remain-
ἡ μὲν ὑπὸ ΓΖΔ τῆς ὑπὸ ΓΖΛ ἐστι διπλῆ, ἡ δὲ ὑπὸ ΔΛΓ ing (square) on CK is equal to the remaining (square)
τῆς ὑπὸ ΖΛΓ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ περιφέρεια τῇ ΓΔ, on BK. Thus, BK (is) equal to CK. And since FB is
ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΒΖΓ τῇ ὑπὸ ΓΖΔ. καί ἐστιν ἡ equal to FC, and FK (is) common, the two (straight-
μὲν ὑπὸ ΒΖΓ τῆς ὑπὸ ΚΖΓ διπλῆ, ἡ δὲ ὑπὸ ΔΖΓ τῆς ὑπὸ lines) BF , FK are equal to the two (straight-lines) CF ,
ΛΖΓ· ἴση ἄρα καὶ ἡ ὑπὸ ΚΖΓ τῇ ὑπὸ ΛΖΓ· ἐστὶ δὲ καὶ ἡ FK. And the base BK [is] equal to the base CK. Thus,
ὑπὸ ΖΓΚ γωνία τῇ ὑπὸ ΖΓΛ ἴση. δύο δὴ τρίγωνά ἐστι τὰ angle BFK is equal to [angle] KFC [Prop. 1.8]. And
ΖΚΓ, ΖΛΓ τὰς δύο γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα BKF (is equal) to FKC [Prop. 1.8]. Thus, BFC (is)
καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΖΓ· double KFC, and BKC (is double) FKC. So, for the
καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει same (reasons), CFD is also double CFL, and DLC (is
καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ· ἴση ἄρα ἡ μὲν ΚΓ also double) FLC. And since circumference BC is equal
εὐθεῖα τῇ ΓΛ, ἡ δὲ ὑπὸ ΖΚΓ γωνία τῇ ὑπὸ ΖΛΓ. καὶ ἐπεὶ ἴση to CD, angle BFC is also equal to CFD [Prop. 3.27].
ἐστὶν ἡ ΚΓ τῇ ΓΛ, διπλῆ ἄρα ἡ ΚΛ τῆς ΚΓ. διὰ τὰ αὐτα δὴ And BFC is double KFC, and DFC (is double) LFC.
δειχθήσεται καὶ ἡ ΘΚ τῆς ΒΚ διπλῆ. καί ἐστιν ἡ ΒΚ τῇ ΚΓ Thus, KFC is also equal to LFC. And angle FCK is also
ἴση· καὶ ἡ ΘΚ ἄρα τῇ ΚΛ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται equal to FCL. So, FKC and FLC are two triangles hav-

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STOIQEIWN dþ. ELEMENTS BOOK 4
καὶ ἑκάστη τῶν ΘΗ, ΗΜ, ΜΛ ἑκατέρᾳ τῶν ΘΚ, ΚΛ ἴση· ing two angles equal to two angles, and one side equal
ἰσόπλευρον ἄρα ἐστὶ τὸ ΗΘΚΛΜ πεντάγωνον. λέγω δή, to one side, (namely) their common (side) FC. Thus,
ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΖΚΓ γωνία τῇ they will also have the remaining sides equal to the (cor-
ὑπὸ ΖΛΓ, καὶ ἐδείχθη τῆς μὲν ὑπὸ ΖΚΓ διπλῆ ἡ ὑπὸ ΘΚΛ, responding) remaining sides, and the remaining angle to
τῆς δὲ ὑπὸ ΖΛΓ διπλῆ ἡ ὑπὸ ΚΛΜ, καὶ ἡ ὑπὸ ΘΚΛ ἄρα the remaining angle [Prop. 1.26]. Thus, the straight-line
τῇ ὑπὸ ΚΛΜ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται καὶ ἑκάστη KC (is) equal to CL, and the angle FKC to FLC. And
τῶν ὑπὸ ΚΘΗ, ΘΗΜ, ΗΜΛ ἑκατέρᾳ τῶν ὑπὸ ΘΚΛ, ΚΛΜ since KC is equal to CL, KL (is) thus double KC. So,
ἴση· αἱ πέντε ἄρα γωνίαι αἱ ὑπὸ ΗΘΚ, ΘΚΛ, ΚΛΜ, ΛΜΗ, for the same (reasons), it can be shown that HK (is) also
ΜΗΘ ἴσαι ἀλλήλαις εἰσίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΗΘΚΛΜ double BK. And BK is equal to KC. Thus, HK is also
πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον, καὶ περιγέγραπται equal to KL. So, similarly, each of HG, GM , and ML
περὶ τὸν ΑΒΓΔΕ κύκλον. can also be shown (to be) equal to each of HK and KL.
[Περὶ τὸν δοθέντα ἄρα κύκλον πεντάγωνον ἰσόπλευρόν Thus, pentagon GHKLM is equilateral. So I say that

τε καὶ ἰσογώνιον περιγέγραπται]· ὅπερ ἔδει ποιῆσαι. (it is) also equiangular. For since angle FKC is equal to
FLC, and HKL was shown (to be) double FKC, and

KLM double FLC, HKL is thus also equal to KLM .

So, similarly, each of KHG, HGM , and GML can also
be shown (to be) equal to each of HKL and KLM . Thus,

the five angles GHK, HKL, KLM , LMG, and MGH

are equal to one another. Thus, the pentagon GHKLM
is equiangular. And it was also shown (to be) equilateral,

and has been circumscribed about circle ABCDE.
[Thus, an equilateral and equiangular pentagon has

been circumscribed about the given circle]. (Which is)

the very thing it was required to do.

† See the footnote to Prop. 3.34. igþ. Proposition 13
Εἰς τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ To inscribe a circle in a given pentagon, which is equi-

ἰσογώνιον, κύκλον ἐγγράψαι. lateral and equiangular.

Γ Κ ∆

Ζ
Β

Θ H

C K D

῎Εστω τὸ δοθὲν πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνι- Let ABCDE be the given equilateral and equiangular
ον τὸ ΑΒΓΔΕ· δεῖ δὴ εἰς τὸ ΑΒΓΔΕ πεντάγωνον κύκλον pentagon. So it is required to inscribe a circle in pentagon
ἐγγράψαι. ABCDE.
Τετμήσθω γὰρ ἑκατέρα τῶν ὑπὸ ΒΓΔ, ΓΔΕ γωνιῶν For let angles BCD and CDE have each been cut

δίχα ὑπὸ ἑκατέρας τῶν ΓΖ, ΔΖ εὐθειῶν· καὶ ἀπὸ τοῦ Ζ in half by each of the straight-lines CF and DF (re-
σημείου, καθ᾿ ὃ συμβάλλουσιν ἀλλήλαις αἱ ΓΖ, ΔΖ εὐθεῖαι, spectively) [Prop. 1.9]. And from the point F , at which
ἐπεζεύχθωσαν αἱ ΖΒ, ΖΑ, ΖΕ εὐθεῖαι. καὶ ἐπεὶ ἴση ἐστὶν the straight-lines CF and DF meet one another, let the

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STOIQEIWN dþ. ELEMENTS BOOK 4
ἡ ΒΓ τῇ ΓΔ, κοινὴ δὲ ἡ ΓΖ, δύο δὴ αἱ ΒΓ, ΓΖ δυσὶ ταῖς straight-lines FB, FA, and FE have been joined. And
ΔΓ, ΓΖ ἴσαι εἰσίν· καὶ γωνία ἡ ὑπὸ ΒΓΖ γωνίᾳ τῇ ὑπὸ since BC is equal to CD, and CF (is) common, the two
ΔΓΖ [ἐστιν] ἴση· βάσις ἄρα ἡ ΒΖ βάσει τῇ ΔΖ ἐστιν ἴση, (straight-lines) BC, CF are equal to the two (straight-
καὶ τὸ ΒΓΖ τρίγωνον τῷ ΔΓΖ τριγώνῳ ἐστιν ἴσον, καὶ αἱ lines) DC, CF . And angle BCF [is] equal to angle
λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾿ ἃς αἱ DCF . Thus, the base BF is equal to the base DF , and
ἴσαι πλευραὶ ὑποτείνουσιν· ἴση ἄρα ἡ ὑπὸ ΓΒΖ γωνία τῇ triangle BCF is equal to triangle DCF , and the remain-
ὑπὸ ΓΔΖ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ὑπὸ ΓΔΕ τῆς ὑπὸ ΓΔΖ, ing angles will be equal to the (corresponding) remain-
ἴση δὲ ἡ μὲν ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΒΓ, ἡ δὲ ὑπὸ ΓΔΖ τῇ ὑπὸ ing angles which the equal sides subtend [Prop. 1.4].
ΓΒΖ, καὶ ἡ ὑπὸ ΓΒΑ ἄρα τῆς ὑπὸ ΓΒΖ ἐστι διπλῆ· ἴση Thus, angle CBF (is) equal to CDF . And since CDE
ἄρα ἡ ὑπὸ ΑΒΖ γωνία τῇ ὑπὸ ΖΒΓ· ἡ ἄρα ὑπὸ ΑΒΓ γωνία is double CDF , and CDE (is) equal to ABC, and CDF
δίχα τέτμηται ὑπὸ τῆς ΒΖ εὐθείας. ὁμοίως δὴ δειχθήσεται, to CBF , CBA is thus also double CBF . Thus, angle
ὅτι καὶ ἑκατέρα τῶν ὑπὸ ΒΑΕ, ΑΕΔ δίχα τέτμηται ὑπὸ ABF is equal to FBC. Thus, angle ABC has been cut
ἑκατέρας τῶν ΖΑ, ΖΕ εὐθειῶν. ἤχθωσαν δὴ ἀπὸ τοῦ Ζ in half by the straight-line BF . So, similarly, it can be
σημείου ἐπὶ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθείας κάθετοι shown that BAE and AED have been cut in half by the
αἱ ΖΗ, ΖΘ, ΖΚ, ΖΛ, ΖΜ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΘΓΖ straight-lines FA and FE, respectively. So let FG, FH ,
γωνία τῇ ὑπὸ ΚΓΖ, ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΖΘΓ [ὀρθῇ] FK, FL, and FM have been drawn from point F , per-
τῇ ὑπὸ ΖΚΓ ἴση, δύο δὴ τρίγωνά ἐστι τὰ ΖΘΓ, ΖΚΓ τὰς pendicular to the straight-lines AB, BC, CD, DE, and
δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ EA (respectively) [Prop. 1.12]. And since angle HCF
πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΖΓ ὑποτείνουσαν ὑπὸ μίαν is equal to KCF , and the right-angle FHC is also equal
τῶν ἴσων γωνιῶν· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς to the [right-angle] FKC, FHC and FKC are two tri-
πλευραῖς ἴσας ἕξει· ἴση ἄρα ἡ ΖΘ κάθετος τῂ ΖΚ καθέτῳ. angles having two angles equal to two angles, and one
ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΖΛ, ΖΜ, ΖΗ side equal to one side, (namely) their common (side) FC,
ἑκατέρᾳ τῶν ΖΘ, ΖΚ ἴση ἐστίν· αἱ πέντε ἄρα εὐθεῖαι αἱ ΖΗ, subtending one of the equal angles. Thus, they will also
ΖΘ, ΖΚ, ΖΛ, ΖΜ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ζ have the remaining sides equal to the (corresponding)
διαστήματι δὲ ἑνὶ τῶν Η, Θ, Κ, Λ, Μ κύκλος γραφόμενος remaining sides [Prop. 1.26]. Thus, the perpendicular
ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, FH (is) equal to the perpendicular FK. So, similarly, it
ΓΔ, ΔΕ, ΕΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι τὰς πρὸς τοῖς Η, can be shown that FL, FM , and FG are each equal to
Θ, Κ, Λ, Μ σημείοις γωνίας. εἰ γὰρ οὐκ ἐφάψεται αὐτῶν, each of FH and FK. Thus, the five straight-lines FG,
ἀλλὰ τεμεῖ αὐτάς, συμβήσεται τὴν τῇ διαμέτρῳ τοῦ κύκλου FH , FK, FL, and FM are equal to one another. Thus,
πρὸς ὀρθὰς ἀπ᾿ ἄκρας ἀγομένην ἐντὸς πίπτειν τοῦ κύκλου· the circle drawn with center F , and radius one of G, H ,
ὅπερ ἄτοπον ἐδείχθη. οὐκ ἄρα ὁ κέντρῳ τῷ Ζ διαστήματι δὲ K, L, or M , will also go through the remaining points,
ἑνὶ τῶν Η, Θ, Κ, Λ, Μ σημείων γραφόμενος κύκλος τεμεῖ and will touch the straight-lines AB, BC, CD, DE, and
τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθείας· ἐφάψεται ἄρα αὐτῶν. EA, on account of the angles at points G, H , K, L, and
γεγράφθω ὡς ὁ ΗΘΚΛΜ. M being right-angles. For if it does not touch them, but
Εἰς ἄρα τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε cuts them, it follows that a (straight-line) drawn at right-

καὶ ἰσογώνιον, κύκλος ἐγγέγραπται· ὅπερ ἔδει ποιῆσαι. angles to the diameter of the circle, from its extremity,
falls inside the circle. The very thing was shown (to be)
absurd [Prop. 3.16]. Thus, the circle drawn with center

F , and radius one of G, H , K, L, or M , does not cut

the straight-lines AB, BC, CD, DE, or EA. Thus, it will
touch them. Let it have been drawn, like GHKLM (in

the figure).

Thus, a circle has been inscribed in the given pen-
tagon which is equilateral and equiangular. (Which is)

the very thing it was required to do.idþ. Proposition 14
Περὶ τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ To circumscribe a circle about a given pentagon which

ἰσογώνιον, κύκλον περιγράψαι. is equilateral and equiangular.
῎Εστω τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ Let ABCDE be the given pentagon which is equilat-

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STOIQEIWN dþ. ELEMENTS BOOK 4
ἰσογώνιον, τὸ ΑΒΓΔΕ· δεῖ δὴ περὶ τὸ ΑΒΓΔΕ πεντάγωνον eral and equiangular. So it is required to circumscribe a
κύκλον περιγράψαι. circle about the pentagon ABCDE.

∆

C D

Τετμήσθω δὴ ἑκατέρα τῶν ὑπὸ ΒΓΔ, ΓΔΕ γωνιῶν So let angles BCD and CDE have been cut in half by
δίχα ὑπὸ ἑκατέρας τῶν ΓΖ, ΔΖ, καὶ ἀπὸ τοῦ Ζ σημείου, the (straight-lines) CF and DF , respectively [Prop. 1.9].
καθ᾿ ὃ συμβάλλουσιν αἱ εὐθεῖαι, ἐπὶ τὰ Β, Α, Ε σημεῖα And let the straight-lines FB, FA, and FE have been
ἐπεζεύχθωσαν εὐθεῖαι αἱ ΖΒ, ΖΑ, ΖΕ. ὁμοίως δὴ τῷ πρὸ joined from point F , at which the straight-lines meet,
τούτου δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ὑπὸ ΓΒΑ, ΒΑΕ, to the points B, A, and E (respectively). So, similarly,
ΑΕΔ γωνιῶν δίχα τέτμηται ὑπὸ ἑκάστης τῶν ΖΒ, ΖΑ, ΖΕ to the (proposition) before this (one), it can be shown
εὐθειῶν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΒΓΔ γωνία τῇ ὑπὸ ΓΔΕ, that angles CBA, BAE, and AED have also been cut
καί ἐστι τῆς μὲν ὑπὸ ΒΓΔ ἡμίσεια ἡ ὑπὸ ΖΓΔ, τῆς δὲ ὑπὸ in half by the straight-lines FB, FA, and FE, respec-
ΓΔΕ ἡμίσεια ἡ ὑπὸ ΓΔΖ, καὶ ἡ ὑπὸ ΖΓΔ ἄρα τῇ ὑπὸ ΖΔΓ tively. And since angle BCD is equal to CDE, and FCD
ἐστιν ἴση· ὥστε καὶ πλευρὰ ἡ ΖΓ πλευρᾷ τῇ ΖΔ ἐστιν ἴση. is half of BCD, and CDF half of CDE, FCD is thus
ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΖΒ, ΖΑ, ΖΕ also equal to FDC. So that side FC is also equal to side
ἑκατέρᾳ τῶν ΖΓ, ΖΔ ἐστιν ἴση· αἱ πέντε ἄρα εὐθεῖαι αἱ FD [Prop. 1.6]. So, similarly, it can be shown that FB,
ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ ἴσαι ἀλλήλαις εἰσίν. ὀ ἄρα κέντρῳ FA, and FE are also each equal to each of FC and FD.
τῷ Ζ καὶ διαστήματι ἑνὶ τῶν ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ κύκλος Thus, the five straight-lines FA, FB, FC, FD, and FE
γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται πε- are equal to one another. Thus, the circle drawn with
ριγεγραμμένος. περιγεγράφθω καὶ ἔστω ὁ ΑΒΓΔΕ. center F , and radius one of FA, FB, FC, FD, or FE,
Περὶ ἄρα τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε will also go through the remaining points, and will have

καὶ ἰσογώνιον, κύκλος περιγέγραπται· ὅπερ ἔδει ποιῆσαι. been circumscribed. Let it have been (so) circumscribed,
and let it be ABCDE.

Thus, a circle has been circumscribed about the given
pentagon, which is equilateral and equiangular. (Which

is) the very thing it was required to do.ieþ. Proposition 15
Εἰς τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ To inscribe an equilateral and equiangular hexagon in

ἰσογώνιον ἐγγράψαι. a given circle.
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕΖ· δεῖ δὴ εἰς τὸν Let ABCDEF be the given circle. So it is required to

ΑΒΓΔΕΖ κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον inscribe an equilateral and equiangular hexagon in circle
ἐγγράψαι. ABCDEF .
῎Ηχθω τοῦ ΑΒΓΔΕΖ κύκλου διάμετρος ἡ ΑΔ, καὶ Let the diameter AD of circle ABCDEF have been

εἰλήφθω τὸ κέντρον τοῦ κύκλου τὸ Η, καὶ κέντρῳ μὲν drawn,† and let the center G of the circle have been
τῷ Δ διαστήματι δὲ τῷ ΔΗ κύκλος γεγράφθω ὁ ΕΗΓΘ, found [Prop. 3.1]. And let the circle EGCH have been
καὶ ἐπιζευχθεῖσαι αἱ ΕΗ, ΓΗ διήχθωσαν ἐπὶ τὰ Β, Ζ σημεῖα, drawn, with center D, and radius DG. And EG and CG
καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΖ, ΖΑ· λέγω, ὅτι being joined, let them have been drawn across (the cir-

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STOIQEIWN dþ. ELEMENTS BOOK 4
τὸ ΑΒΓΔΕΖ ἑξάγωνον ἰσόπλευρόν τέ ἐστι καὶ ἰσογώνιον. cle) to points B and F (respectively). And let AB, BC,

CD, DE, EF , and FA have been joined. I say that the

hexagon ABCDEF is equilateral and equiangular.

∆

Γ C

᾿Επεὶ γὰρ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓΔΕΖ For since point G is the center of circle ABCDEF ,
κύκλου, ἴση ἐστὶν ἡ ΗΕ τῇ ΗΔ. πάλιν, ἐπεὶ τὸ Δ σημεῖον GE is equal to GD. Again, since point D is the cen-
κέντρον ἐστὶ τοῦ ΗΓΘ κύκλου, ἴση ἐστὶν ἡ ΔΕ τῇ ΔΗ. ter of circle GCH , DE is equal to DG. But, GE was
ἀλλ᾿ ἡ ΗΕ τῇ ΗΔ ἐδείχθη ἴση· καὶ ἡ ΗΕ ἄρα τῇ ΕΔ ἴση shown (to be) equal to GD. Thus, GE is also equal to
ἐστίν· ἰσόπλευρον ἄρα ἐστὶ τὸ ΕΗΔ τρίγωνον· καὶ αἱ τρεῖς ED. Thus, triangle EGD is equilateral. Thus, its three
ἄρα αὐτοῦ γωνίαι αἱ ὑπὸ ΕΗΔ, ΗΔΕ, ΔΕΗ ἴσαι ἀλλήλαις angles EGD, GDE, and DEG are also equal to one an-
εἰσίν, ἐπειδήπερ τῶν ἰσοσκελῶν τριγώνων αἱ πρὸς τῇ βάσει other, inasmuch as the angles at the base of isosceles tri-
γωνίαι ἴσαι ἀλλήλαις εἰσίν· καί εἰσιν αἱ τρεῖς τοῦ τριγώνου angles are equal to one another [Prop. 1.5]. And the
γωνίαι δυσὶν ὀρθαῖς ἴσαι· ἡ ἄρα ὑπὸ ΕΗΔ γωνία τρίτον ἐστὶ three angles of the triangle are equal to two right-angles
δύο ὀρθῶν. ὁμοίως δὴ δειχθήσεται καὶ ἡ ὑπὸ ΔΗΓ τρίτον [Prop. 1.32]. Thus, angle EGD is one third of two right-
δύο ὀρθῶν. καὶ ἐπεὶ ἡ ΓΗ εὐθεῖα ἐπὶ τὴν ΕΒ σταθεῖσα τὰς angles. So, similarly, DGC can also be shown (to be)
ἐφεξῆς γωνίας τὰς ὑπὸ ΕΗΓ, ΓΗΒ δυσὶν ὀρθαῖς ἴσας ποιεῖ, one third of two right-angles. And since the straight-line
καὶ λοιπὴ ἄρα ἡ ὑπὸ ΓΗΒ τρίτον ἐστὶ δύο ὀρθῶν· αἱ ἄρα CG, standing on EB, makes adjacent angles EGC and
ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ γωνίαι ἴσαι ἀλλήλαις εἰσίν· ὥστε καὶ CGB equal to two right-angles [Prop. 1.13], the remain-
αἱ κατὰ κορυφὴν αὐταῖς αἱ ὑπὸ ΒΗΑ, ΑΗΖ, ΖΗΕ ἴσαι εἰσὶν ing angle CGB is thus also one third of two right-angles.
[ταῖς ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ]. αἱ ἓξ ἄρα γωνίαι αἱ ὑπὸ ΕΗΔ, Thus, angles EGD, DGC, and CGB are equal to one an-
ΔΗΓ, ΓΗΒ, ΒΗΑ, ΑΗΖ, ΖΗΕ ἴσαι ἀλλήλαις εἰσίν. αἱ δὲ other. And hence the (angles) opposite to them BGA,
ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν· αἱ ἓξ ἄρα πε- AGF , and FGE are also equal [to EGD, DGC, and
ριφέρειαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΖ, ΖΑ ἴσαι ἀλλήλαις εἰσίν. CGB (respectively)] [Prop. 1.15]. Thus, the six angles
ὑπὸ δὲ τὰς ἴσας περιφερείας αἱ ἴσαι εὐθεῖαι ὑποτείνουσιν· EGD, DGC, CGB, BGA, AGF , and FGE are equal
αἱ ἓξ ἄρα εὐθεῖαι ἴσαι ἀλλήλαις εἰσίν· ἰσόπλευρον ἄρα ἐστὶ to one another. And equal angles stand on equal cir-
το ΑΒΓΔΕΖ ἑξάγωνον. λέγω δή, ὅτι καὶ ἰσογώνιον. ἐπεὶ cumferences [Prop. 3.26]. Thus, the six circumferences
γὰρ ἴση ἐστὶν ἡ ΖΑ περιφέρεια τῇ ΕΔ περιφερείᾳ, κοινὴ AB, BC, CD, DE, EF , and FA are equal to one an-
προσκείσθω ἡ ΑΒΓΔ περιφέρεια· ὅλη ἄρα ἡ ΖΑΒΓΔ ὅλῃ other. And equal circumferences are subtended by equal
τῇ ΕΔΓΒΑ ἐστιν ἴση· καὶ βέβηκεν ἐπὶ μὲν τῆς ΖΑΒΓΔ straight-lines [Prop. 3.29]. Thus, the six straight-lines
περιφερείας ἡ ὑπὸ ΖΕΔ γωνία, ἐπὶ δὲ τῆς ΕΔΓΒΑ περι- (AB, BC, CD, DE, EF , and FA) are equal to one
φερείας ἡ ὑπὸ ΑΖΕ γωνία· ἴση ἄρα ἡ ὑπὸ ΑΖΕ γωνία τῇ another. Thus, hexagon ABCDEF is equilateral. So,
ὑπὸ ΔΕΖ. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ αἱ λοιπαὶ γωνίαι I say that (it is) also equiangular. For since circumfer-
τοῦ ΑΒΓΔΕΖ ἑξαγώνου κατὰ μίαν ἴσαι εἰσὶν ἑκατέρᾳ τῶν ence FA is equal to circumference ED, let circumference
ὑπὸ ΑΖΕ, ΖΕΔ γωνιῶν· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕΖ ABCD have been added to both. Thus, the whole of
ἑξάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον· καὶ ἐγγέγραπται εἰς FABCD is equal to the whole of EDCBA. And angle
τὸν ΑΒΓΔΕΖ κύκλον. FED stands on circumference FABCD, and angle AFE
Εἰς ἄρα τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε on circumference EDCBA. Thus, angle AFE is equal

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STOIQEIWN dþ. ELEMENTS BOOK 4
καὶ ἰσογώνιον ἐγγέγραπται· ὅπερ ἔδει ποιῆσαι. to DEF [Prop. 3.27]. Similarly, it can also be shown

that the remaining angles of hexagon ABCDEF are in-

dividually equal to each of the angles AFE and FED.
Thus, hexagon ABCDEF is equiangular. And it was also

shown (to be) equilateral. And it has been inscribed in
circle ABCDE.

Thus, an equilateral and equiangular hexagon has

been inscribed in the given circle. (Which is) the very
thing it was required to do.Pìrisma. Corollary

᾿Εκ δὴ τούτου φανερόν, ὅτι ἡ τοῦ ἑξαγώνου πλευρὰ ἴση So, from this, (it is) manifest that a side of the
ἐστὶ τῇ ἐκ τοῦ κέντρου τοῦ κύκλου. hexagon is equal to the radius of the circle.
῾Ομοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν κατὰ And similarly to a pentagon, if we draw tangents

τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν, to the circle through the (sixfold) divisions of the (cir-
περιγραφήσεται περὶ τὸν κύκλον ἑξάγωνον ἰσόπλευρόν cumference of the) circle, an equilateral and equiangu-
τε καὶ ἰσογώνιον ἀκολούθως τοῖς ἐπὶ τοῦ πενταγώνου lar hexagon can be circumscribed about the circle, analo-
εἰρημένοις. καὶ ἔτι διὰ τῶν ὁμοίων τοῖς ἐπὶ τοῦ πενταγώνου gously to the aforementioned pentagon. And, further, by
εἰρημένοις εἰς τὸ δοθὲν ἑξάγωνον κύκλον ἐγγράψομέν τε (means) similar to the aforementioned pentagon, we can
καὶ περιγράψομεν· ὅπερ ἔδει ποιῆσαι. inscribe and circumscribe a circle in (and about) a given

hexagon. (Which is) the very thing it was required to do.

† See the footnote to Prop. 4.6. i�þ. Proposition 16
Εἰς τὸν δοθέντα κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν To inscribe an equilateral and equiangular fifteen-

τε καὶ ἰσογώνιον ἐγγράψαι. sided figure in a given circle.

Γ ∆

C D

῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ· δεῖ δὴ εἰς τὸν ΑΒΓΔ Let ABCD be the given circle. So it is required to in-
κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον scribe an equilateral and equiangular fifteen-sided figure
ἐγγράψαι. in circle ABCD.
᾿Εγγεγράφθω εἰς τὸν ΑΒΓΔ κύκλον τριγώνου μὲν ἰσο- Let the side AC of an equilateral triangle inscribed

πλεύρου τοῦ εἰς αὐτὸν ἐγγραφομένου πλευρὰ ἡ ΑΓ, πεν- in (the circle) [Prop. 4.2], and (the side) AB of an (in-
ταγώνου δὲ ἰσοπλεύρου ἡ ΑΒ· οἵων ἄρα ἐστὶν ὁ ΑΒΓΔ scribed) equilateral pentagon [Prop. 4.11], have been in-
κύκλος ἴσων τμήματων δεκαπέντε, τοιούτων ἡ μὲν ΑΒΓ scribed in circle ABCD. Thus, just as the circle ABCD
περιφέρεια τρίτον οὖσα τοῦ κύκλου ἔσται πέντε, ἡ δὲ ΑΒ is (made up) of fifteen equal pieces, the circumference
περιφέρεια πέμτον οὖσα τοῦ κύκλου ἔσται τριῶν· λοιπὴ ἄρα ABC, being a third of the circle, will be (made up) of five

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STOIQEIWN dþ. ELEMENTS BOOK 4
ἡ ΒΓ τῶν ἴσων δύο. τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε· such (pieces), and the circumference AB, being a fifth of
ἑκατέρα ἄρα τῶν ΒΕ, ΕΓ περιφερειῶν πεντεκαιδέκατόν ἐστι the circle, will be (made up) of three. Thus, the remain-
τοῦ ΑΒΓΔ κύκλου. der BC (will be made up) of two equal (pieces). Let (cir-
᾿Εὰν ἄρα ἐπιζεύξαντες τὰς ΒΕ, ΕΓ ἴσας αὐταῖς κατὰ τὸ cumference) BC have been cut in half at E [Prop. 3.30].

συνεχὲς εὐθείας ἐναρμόσωμεν εἰς τὸν ΑΒΓΔ[Ε] κύκλον, Thus, each of the circumferences BE and EC is one fif-
ἔσται εἰς αὐτὸν ἐγγεγραμμένον πεντεκαιδεκάγωνον ἰσόπλευ- teenth of the circle ABCDE.
ρόν τε καὶ ἰσογώνιον· ὅπερ ἔδει ποιῆσαι. Thus, if, joining BE and EC, we continuously in-
῾Ομοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν sert straight-lines equal to them into circle ABCD[E]

κατὰ τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου [Prop. 4.1], then an equilateral and equiangular fifteen-
ἀγάγωμεν, περιγραφήσεται περὶ τὸν κύκλον πεντεκαι- sided figure will have been inserted into (the circle).
δεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον. ἔτι δὲ διὰ (Which is) the very thing it was required to do.
τῶν ὁμοίων τοῖς ἐπὶ τοῦ πενταγώνου δείξεων καὶ εἰς τὸ And similarly to the pentagon, if we draw tangents to
δοθὲν πεντεκαιδεκάγωνον κύκλον ἐγγράψομέν τε καὶ πε- the circle through the (fifteenfold) divisions of the (cir-
ριγράψομεν· ὅπερ ἔδει ποιῆσαι. cumference of the) circle, we can circumscribe an equilat-

eral and equiangular fifteen-sided figure about the circle.

And, further, through similar proofs to the pentagon, we
can also inscribe and circumscribe a circle in (and about)

a given fifteen-sided figure. (Which is) the very thing it

was required to do.

127

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ELEMENTS BOOK 5

Proportion†

†The theory of proportion set out in this book is generally attributed to Eudoxus of Cnidus. The novel feature of this theory is its ability to deal
with irrational magnitudes, which had hitherto been a major stumbling block for Greek mathematicians. Throughout the footnotes in this book,
α, β, γ, etc., denote general (possibly irrational) magnitudes, whereas m, n, l, etc., denote positive integers.

129

STOIQEIWN eþ. ELEMENTS BOOK 5VOroi. Definitions
αʹ. Μέρος ἐστὶ μέγεθος μεγέθους τὸ ἔλασσον τοῦ 1. A magnitude is a part of a(nother) magnitude, the

μείζονος, ὅταν καταμετρῇ τὸ μεῖζον. lesser of the greater, when it measures the greater.†

βʹ. Πολλαπλάσιον δὲ τὸ μεῖζον τοῦ ἐλάττονος, ὅταν κα- 2. And the greater (magnitude is) a multiple of the
ταμετρῆται ὑπὸ τοῦ ἐλάττονος. lesser when it is measured by the lesser.
γʹ. Λόγος ἐστὶ δύο μεγεθῶν ὁμογενῶν ἡ κατὰ πη- 3. A ratio is a certain type of condition with respect to

λικότητά ποια σχέσις. size of two magnitudes of the same kind.‡

δʹ. Λόγον ἔχειν πρὸς ἄλληλα μεγέθη λέγεται, ἃ δύναται 4. (Those) magnitudes are said to have a ratio with re-
πολλαπλασιαζόμενα ἀλλήλων ὑπερέχειν. spect to one another which, being multiplied, are capable

εʹ. ᾿Εν τῷ αὐτῷ λόγῳ μεγέθη λέγεται εἶναι πρῶτον of exceeding one another.§

πρὸς δεύτερον καὶ τρίτον πρὸς τέταρτον, ὅταν τὰ τοῦ 5. Magnitudes are said to be in the same ratio, the first
πρώτου καί τρίτου ἰσάκις πολλαπλάσια τῶν τοῦ δευτέρου to the second, and the third to the fourth, when equal
καὶ τετάρτου ἰσάκις πολλαπλασίων καθ᾿ ὁποιονοῦν πολλα- multiples of the first and the third either both exceed, are
πλασιασμὸν ἑκάτερον ἑκατέρου ἢ ἅμα ὑπερέχῃ ἢ ἅμα ἴσα ᾖ both equal to, or are both less than, equal multiples of the
ἢ ἅμα ἐλλείπῇ ληφθέντα κατάλληλα. second and the fourth, respectively, being taken in corre-
ϛʹ. Τὰ δὲ τὸν αὐτὸν ἔχοντα λόγον μεγέθη ἀνάλογον sponding order, according to any kind of multiplication

καλείσθω. whatever.¶

ζʹ. ῞Οταν δὲ τῶν ἰσάκις πολλαπλασίων τὸ μὲν τοῦ 6. And let magnitudes having the same ratio be called
πρώτου πολλαπλάσιον ὑπερέχῃ τοῦ τοῦ δευτέρου πολ- proportional.∗

λαπλασίου, τὸ δὲ τοῦ τρίτου πολλαπλάσιον μὴ ὑπερέχῃ 7. And when for equal multiples (as in Def. 5), the
τοῦ τοῦ τετάρτου πολλαπλασίου, τότε τὸ πρῶτον πρὸς τὸ multiple of the first (magnitude) exceeds the multiple of
δεύτερον μείζονα λόγον ἔχειν λέγεται, ἤπερ τὸ τρίτον πρὸς the second, and the multiple of the third (magnitude)
τὸ τέταρτον. does not exceed the multiple of the fourth, then the first
ηʹ. Ἀναλογία δὲ ἐν τρισὶν ὅροις ἐλαχίστη ἐστίν. (magnitude) is said to have a greater ratio to the second
θʹ. ῞Οταν δὲ τρία μεγέθη ἀνάλογον ᾖ, τὸ πρῶτον πρὸς than the third (magnitude has) to the fourth.

τὸ τρίτον διπλασίονα λόγον ἔχειν λέγεται ἤπερ πρὸς τὸ 8. And a proportion in three terms is the smallest

δεύτερον. (possible).$

ιʹ. ῞Οταν δὲ τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ πρῶτον 9. And when three magnitudes are proportional, the

πρὸς τὸ τέταρτον τριπλασίονα λόγον ἔχειν λέγεται ἤπερ first is said to have to the third the squared‖ ratio of that

πρὸς τὸ δεύτερον, καὶ ἀεὶ ἑξῆς ὁμοίως, ὡς ἂν ἡ ἀναλογία (it has) to the second.††

ὑπάρχῃ. 10. And when four magnitudes are (continuously)
ιαʹ. ῾Ομόλογα μεγέθη λέγεται τὰ μὲν ἡγούμενα τοῖς proportional, the first is said to have to the fourth the

ἡγουμένοις τὰ δὲ ἑπόμενα τοῖς ἑπομένοις. cubed‡‡ ratio of that (it has) to the second.§§ And so on,
ιβʹ. ᾿Εναλλὰξ λόγος ἐστὶ λῆψις τοῦ ἡγουμένου πρὸς τὸ similarly, in successive order, whatever the (continuous)

ἡγούμενον καὶ τοῦ ἑπομένου πρὸς τὸ ἑπόμενον. proportion might be.
ιγʹ. Ἀνάπαλιν λόγος ἐστὶ λῆψις τοῦ ἑπομένου ὡς 11. These magnitudes are said to be corresponding

ἡγουμένου πρὸς τὸ ἡγούμενον ὡς ἑπόμενον. (magnitudes): the leading to the leading (of two ratios),
ιδʹ. Σύνθεσις λόγου ἐστὶ λῆψις τοῦ ἡγουμένου μετὰ τοῦ and the following to the following.

ἑπομένου ὡς ἑνὸς πρὸς αὐτὸ τὸ ἑπόμενον. 12. An alternate ratio is a taking of the (ratio of the)
ιεʹ.Διαίρεσις λόγου ἐστὶ λῆψις τῆς ὑπεροχῆς, ᾗ ὑπερέχει leading (magnitude) to the leading (of two equal ratios),

τὸ ἡγούμενον τοῦ ἑπομένου, πρὸς αὐτὸ τὸ ἑπόμενον. and (setting it equal to) the (ratio of the) following (mag-

ιϛʹ. Ἀναστροφὴ λόγου ἐστὶ λῆψις τοῦ ἡγουμένου πρὸς nitude) to the following.¶¶

τὴν ὑπεροχήν, ᾗ ὑπερέχει τὸ ἡγούμενον τοῦ ἑπομένου. 13. An inverse ratio is a taking of the (ratio of the) fol-
ιζʹ. Δι᾿ ἴσου λόγος ἐστὶ πλειόνων ὄντων μεγεθῶν καὶ lowing (magnitude) as the leading and the leading (mag-

ἄλλων αὐτοῖς ἴσων τὸ πλῆθος σύνδυο λαμβανομένων καὶ nitude) as the following.∗∗

ἐν τῷ αὐτῷ λόγῳ, ὅταν ᾖ ὡς ἐν τοῖς πρώτοις μεγέθεσι τὸ 14. A composition of a ratio is a taking of the (ratio of
πρῶτον πρὸς τὸ ἔσχατον, οὕτως ἐν τοῖς δευτέροις μεγέθεσι the) leading plus the following (magnitudes), as one, to

τὸ πρῶτον πρὸς τὸ ἔσχατον· ἢ ἄλλως· λῆψις τῶν ἄκρων the following (magnitude) by itself.$$

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STOIQEIWN eþ. ELEMENTS BOOK 5
καθ᾿ ὑπεξαίρεσιν τῶν μέσων. 15. A separation of a ratio is a taking of the (ratio
ιηʹ. Τεταραγμένη δὲ ἀναλογία ἐστίν, ὅταν τριῶν ὄντων of the) excess by which the leading (magnitude) exceeds

μεγεθῶν καὶ ἄλλων αὐτοῖς ἴσων τὸ πλῆθος γίνηται ὡς μὲν the following to the following (magnitude) by itself.‖‖

ἐν τοῖς πρώτοις μεγέθεσιν ἡγούμενον πρὸς ἐπόμενον, οὕτως 16. A conversion of a ratio is a taking of the (ratio
ἐν τοῖς δευτέροις μεγέθεσιν ἡγούμενον πρὸς ἑπόμενον, ὡς of the) leading (magnitude) to the excess by which the

δὲ ἐν τοῖς πρώτοις μεγέθεσιν ἑπόμενον πρὸς ἄλλο τι, οὕτως leading (magnitude) exceeds the following.†††

ἐν τοῖς δευτέροις ἄλλο τι πρὸς ἡγούμενον. 17. There being several magnitudes, and other (mag-
nitudes) of equal number to them, (which are) also in the

same ratio taken two by two, a ratio via equality (or ex
aequali) occurs when as the first is to the last in the first

(set of) magnitudes, so the first (is) to the last in the sec-

ond (set of) magnitudes. Or alternately, (it is) a taking of
the (ratio of the) outer (magnitudes) by the removal of

the inner (magnitudes).‡‡‡

18. There being three magnitudes, and other (magni-

tudes) of equal number to them, a perturbed proportion

occurs when as the leading is to the following in the first
(set of) magnitudes, so the leading (is) to the following

in the second (set of) magnitudes, and as the following

(is) to some other (i.e., the remaining magnitude) in the
first (set of) magnitudes, so some other (is) to the leading

in the second (set of) magnitudes.§§§

† In other words, α is said to be a part of β if β = m α.

‡ In modern notation, the ratio of two magnitudes, α and β, is denoted α : β.

§ In other words, α has a ratio with respect to β if m α > β and n β > α, for some m and n.

¶ In other words, α : β :: γ : δ if and only if m α > n β whenever m γ > n δ, and m α = n β whenever m γ = n δ, and m α < n β whenever m γ < n δ, for all m and n. This definition is the kernel of Eudoxus’ theory of proportion, and is valid even if α, β, etc., are irrational. ∗ Thus if α and β have the same ratio as γ and δ then they are proportional. In modern notation, α : β :: γ : δ. $ In modern notation, a proportion in three terms—α, β, and γ—is written: α : β :: β : γ. ‖ Literally, “double”. †† In other words, if α : β :: β : γ then α : γ :: α 2 : β 2. ‡‡ Literally, “triple”. §§ In other words, if α : β :: β : γ :: γ : δ then α : δ :: α 3 : β 3. ¶¶ In other words, if α : β :: γ : δ then the alternate ratio corresponds to α : γ :: β : δ. ∗∗ In other words, if α : β then the inverse ratio corresponds to β : α. $$ In other words, if α : β then the composed ratio corresponds to α + β : β. ‖‖ In other words, if α : β then the separated ratio corresponds to α − β : β. ††† In other words, if α : β then the converted ratio corresponds to α : α − β. ‡‡‡ In other words, if α, β, γ are the first set of magnitudes, and δ, ǫ, ζ the second set, and α : β : γ :: δ : ǫ : ζ, then the ratio via equality (or ex aequali) corresponds to α : γ :: δ : ζ. §§§ In other words, if α, β, γ are the first set of magnitudes, and δ, ǫ, ζ the second set, and α : β :: δ : ǫ as well as β : γ :: ζ : δ, then the proportion is said to be perturbed. aþ. Proposition 1† ᾿Εὰν ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ If there are any number of magnitudes whatsoever πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν (which are) equal multiples, respectively, of some (other) ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ magnitudes, of equal number (to them), then as many 131 STOIQEIWN eþ. ELEMENTS BOOK 5 πάντα τῶν πάντων. times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second). Ζ Η Β Γ Θ ∆Α Ε F A G B C H D E ῎Εστω ὁποσαοῦν μεγέθη τὰ ΑΒ, ΓΔ ὁποσωνοῦν με- Let there be any number of magnitudes whatsoever, γεθῶν τῶν Ε, Ζ ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις AB, CD, (which are) equal multiples, respectively, of πολλαπλάσιον· λέγω, ὅτι ὁσαπλάσιόν ἐστι τὸ ΑΒ τοῦ Ε, some (other) magnitudes, E, F , of equal number (to τοσαυταπλάσια ἔσται καὶ τὰ ΑΒ, ΓΔ τῶν Ε, Ζ. them). I say that as many times as AB is (divisible) by E, ᾿Επεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Ε καὶ so many times will AB, CD also be (divisible) by E, F . τὸ ΓΔ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μεγέθη ἴσα τῷ Ε, For since AB, CD are equal multiples of E, F , thus τοσαῦτα καὶ ἐν τῷ ΓΔ ἴσα τῷ Ζ. διῃρήσθω τὸ μὲν ΑΒ εἰς τὰ as many magnitudes as (there) are in AB equal to E, so τῷ Ε μεγέθη ἴσα τὰ ΑΗ, ΗΒ, τὸ δὲ ΓΔ εἰς τὰ τῷ Ζ ἴσα τὰ many (are there) also in CD equal to F . Let AB have ΓΘ, ΘΔ· ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΒ τῷ πλήθει been divided into magnitudes AG, GB, equal to E, and τῶν ΓΘ, ΘΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ μὲν ΑΗ τῷ Ε, τὸ δὲ ΓΘ CD into (magnitudes) CH , HD, equal to F . So, the τῷ Ζ, ἴσον ἄρα τὸ ΑΗ τῷ Ε, καὶ τὰ ΑΗ, ΓΘ τοῖς Ε, Ζ. διὰ number of (divisions) AG, GB will be equal to the num- τὰ αὐτὰ δὴ ἴσον ἐστὶ τὸ ΗΒ τῷ Ε, καὶ τὰ ΗΒ, ΘΔ τοῖς Ε, ber of (divisions) CH , HD. And since AG is equal to E, Ζ· ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ ἴσα τῷ Ε, τοσαῦτα καὶ ἐν τοῖς and CH to F , AG (is) thus equal to E, and AG, CH to E, ΑΒ, ΓΔ ἴσα τοῖς Ε, Ζ· ὁσαπλάσιον ἄρα ἐστὶ τὸ ΑΒ τοῦ Ε, F . So, for the same (reasons), GB is equal to E, and GB, τοσαυταπλάσια ἔσται καὶ τὰ ΑΒ, ΓΔ τῶν Ε, Ζ. HD to E, F . Thus, as many (magnitudes) as (there) are ᾿Εὰν ἄρα ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν in AB equal to E, so many (are there) also in AB, CD ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, equal to E, F . Thus, as many times as AB is (divisible) ὁσαπλάσιόν ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια by E, so many times will AB, CD also be (divisible) by ἔσται καὶ τὰ πάντα τῶν πάντων· ὅπερ ἔδει δεῖξαι. E, F . Thus, if there are any number of magnitudes what- soever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisi- ble) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second). (Which is) the very thing it was required to show. † In modern notation, this proposition reads m α + m β + · · · = m (α + β + · · · ).bþ. Proposition 2† ᾿Εὰν πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον If a first (magnitude) and a third are equal multiples τετάρτου, ᾖ δὲ καὶ πέμπτον δευτέρου ἰσάκις πολλαπλάσιον of a second and a fourth (respectively), and a fifth (mag- καὶ ἕκτον τετάρτου, καὶ συντεθὲν πρῶτον καὶ πέμπτον nitude) and a sixth (are) also equal multiples of the sec- δευτέρου ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον ond and fourth (respectively), then the first (magnitude) τετάρτου. and the fifth, being added together, and the third and the Πρῶτον γὰρ τὸ ΑΒ δευτέρου τοῦ Γ ἰσάκις ἔστω πολ- sixth, (being added together), will also be equal multiples λαπλάσιον καὶ τρίτον τὸ ΔΕ τετάρτου τοῦ Ζ, ἔστω δὲ καὶ of the second (magnitude) and the fourth (respectively). πέμπτον τὸ ΒΗ δευτέρου τοῦ Γ ἰσάκις πολλαπλάσιον καὶ For let a first (magnitude) AB and a third DE be ἕκτον τὸ ΕΘ τετάρτου τοῦ Ζ· λέγω, ὅτι καὶ συντεθὲν equal multiples of a second C and a fourth F (respec- πρῶτον καὶ πέμπτον τὸ ΑΗ δευτέρου τοῦ Γ ἰσάκις ἔσται tively). And let a fifth (magnitude) BG and a sixth EH πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΔΘ τετάρτου τοῦ Ζ. also be (other) equal multiples of the second C and the fourth F (respectively). I say that the first (magnitude) and the fifth, being added together, (to give) AG, and the third (magnitude) and the sixth, (being added together, 132 STOIQEIWN eþ. ELEMENTS BOOK 5 to give) DH , will also be equal multiples of the second (magnitude) C and the fourth F (respectively). Ζ Α Β Η ∆ Ε Θ Γ H A B G C F D E ᾿Επεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ τὸ For since AB and DE are equal multiples of C and F ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ ἴσα τῷ Γ, τοσαῦτα καὶ (respectively), thus as many (magnitudes) as (there) are ἐν τῷ ΔΕ ἴσα τῷ Ζ. διὰ τὰ αὐτὰ δὴ καὶ ὅσα ἐστὶν ἐν τῷ ΒΗ in AB equal to C, so many (are there) also in DE equal to ἴσα τῷ Γ, τοσαῦτα καὶ ἐν τῷ ΕΘ ἴσα τῷ Ζ· ὅσα ἄρα ἐστὶν ἐν F . And so, for the same (reasons), as many (magnitudes) ὅλῳ τῷ ΑΗ ἴσα τῷ Γ, τοσαῦτα καὶ ἐν ὅλῳ τῷ ΔΘ ἴσα τῷ Ζ· as (there) are in BG equal to C, so many (are there) ὁσαπλάσιον ἄρα ἐστὶ τὸ ΑΗ τοῦ Γ, τοσαυταπλάσιον ἔσται also in EH equal to F . Thus, as many (magnitudes) as καὶ τὸ ΔΘ τοῦ Ζ. καὶ συντεθὲν ἄρα πρῶτον καὶ πέμπτον τὸ (there) are in the whole of AG equal to C, so many (are ΑΗ δευτέρου τοῦ Γ ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον there) also in the whole of DH equal to F . Thus, as many καὶ ἕκτον τὸ ΔΘ τετάρτου τοῦ Ζ. times as AG is (divisible) by C, so many times will DH ᾿Εὰν ἄρα πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ also be divisible by F . Thus, the first (magnitude) and τρίτον τετάρτου, ᾖ δὲ καὶ πέμπτον δευτέρου ἰσάκις πολ- the fifth, being added together, (to give) AG, and the λαπλάσιον καὶ ἕκτον τετάρτου, καὶ συντεθὲν πρῶτον καὶ third (magnitude) and the sixth, (being added together, πέμπτον δευτέρου ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ to give) DH , will also be equal multiples of the second ἕκτον τετάρτου· ὅπερ ἔδει δεῖξαι. (magnitude) C and the fourth F (respectively). Thus, if a first (magnitude) and a third are equal mul- tiples of a second and a fourth (respectively), and a fifth (magnitude) and a sixth (are) also equal multiples of the second and fourth (respectively), then the first (magni- tude) and the fifth, being added together, and the third and sixth, (being added together), will also be equal mul- tiples of the second (magnitude) and the fourth (respec- tively). (Which is) the very thing it was required to show. † In modern notation, this propostion reads m α + n α = (m + n) α.gþ. Proposition 3† ᾿Εὰν πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον If a first (magnitude) and a third are equal multiples τετάρτου, ληφθῇ δὲ ἰσάκις πολλαπλάσια τοῦ τε πρώτου of a second and a fourth (respectively), and equal multi- καὶ τρίτου, καὶ δι᾿ ἴσου τῶν ληφθέντων ἑκάτερον ἑκατέρου ples are taken of the first and the third, then, via equality, ἰσάκις ἔσται πολλαπλάσιον τὸ μὲν τοῦ δευτέρου τὸ δὲ τοῦ the (magnitudes) taken will also be equal multiples of the τετάρτου. second (magnitude) and the fourth, respectively. Πρῶτον γὰρ τὸ Α δευτέρου τοῦ Β ἰσάκις ἔστω πολ- For let a first (magnitude) A and a third C be equal λαπλάσιον καὶ τρίτον τὸ Γ τετάρτου τοῦ Δ, καὶ εἰλήφθω multiples of a second B and a fourth D (respectively), τῶν Α, Γ ἰσάκις πολλαπλάσια τὰ ΕΖ, ΗΘ· λέγω, ὅτι ἰσάκις and let the equal multiples EF and GH have been taken ἐστὶ πολλαπλάσιον τὸ ΕΖ τοῦ Β καὶ τὸ ΗΘ τοῦ Δ. of A and C (respectively). I say that EF and GH are ᾿Επεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΕΖ τοῦ Α καὶ equal multiples of B and D (respectively). τὸ ΗΘ τοῦ Γ, ὅσα ἄρα ἐστὶν ἐν τῷ ΕΖ ἴσα τῷ Α, τοσαῦτα For since EF and GH are equal multiples of A and καὶ ἐν τῷ ΗΘ ἴσα τῷ Γ. διῃρήσθω τὸ μὲν ΕΖ εἰς τὰ τῷ Α C (respectively), thus as many (magnitudes) as (there) μεγέθη ἴσα τὰ ΕΚ, ΚΖ, τὸ δὲ ΗΘ εἰς τὰ τῷ Γ ἴσα τὰ ΗΛ, are in EF equal to A, so many (are there) also in GH 133 STOIQEIWN eþ. ELEMENTS BOOK 5 ΛΘ· ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΕΚ, ΚΖ τῷ πλήθει τῶν equal to C. Let EF have been divided into magnitudes ΗΛ, ΛΘ. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Α τοῦ Β καὶ EK, KF equal to A, and GH into (magnitudes) GL, LH τὸ Γ τοῦ Δ, ἴσον δὲ τὸ μὲν ΕΚ τῷ Α, τὸ δὲ ΗΛ τῷ Γ, equal to C. So, the number of (magnitudes) EK, KF ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΕΚ τοῦ Β καὶ τὸ ΗΛ τοῦ will be equal to the number of (magnitudes) GL, LH . Δ. διὰ τὰ αὐτὰ δὴ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΚΖ τοῦ Β And since A and C are equal multiples of B and D (re- καὶ τὸ ΛΘ τοῦ Δ. ἐπεὶ οὖν πρῶτον τὸ ΕΚ δευτέρου τοῦ Β spectively), and EK (is) equal to A, and GL to C, EK ἴσάκις ἐστὶ πολλαπλάσιον καὶ τρίτον τὸ ΗΛ τετάρτου τοῦ and GL are thus equal multiples of B and D (respec- Δ, ἔστι δὲ καὶ πέμπτον τὸ ΚΖ δευτέρου τοῦ Β ἰσάκις πολ- tively). So, for the same (reasons), KF and LH are equal λαπλάσιον καὶ ἕκτον τὸ ΛΘ τετάρτου τοῦ Δ, καὶ συντεθὲν multiples of B and D (respectively). Therefore, since the ἄρα πρῶτον καὶ πέμπτον τὸ ΕΖ δευτέρου τοῦ Β ἰσάκις ἐστὶ first (magnitude) EK and the third GL are equal mul- πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΗΘ τετάρτου τοῦ Δ. tiples of the second B and the fourth D (respectively), and the fifth (magnitude) KF and the sixth LH are also equal multiples of the second B and the fourth D (re- spectively), then the first (magnitude) and fifth, being added together, (to give) EF , and the third (magnitude) and sixth, (being added together, to give) GH , are thus also equal multiples of the second (magnitude) B and the fourth D (respectively) [Prop. 5.2]. Α Γ Κ ΖΕ ∆ Η Λ Θ Β A E K F HG C D B L ᾿Εὰν ἄρα πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον Thus, if a first (magnitude) and a third are equal mul- καὶ τρίτον τετάρτου, ληφθῇ δὲ τοῦ πρώτου καὶ τρίτου tiples of a second and a fourth (respectively), and equal ἰσάκις πολλαπλάσια, καὶ δι᾿ ἴσου τῶν ληφθέντων ἑκάτερον multiples are taken of the first and the third, then, via ἑκατέρου ἰσάκις ἔσται πολλαπλάσιον τὸ μὲν τοῦ δευτέρου equality, the (magnitudes) taken will also be equal mul- τὸ δὲ τοῦ τετάρτου· ὅπερ ἔδει δεῖξαι. tiples of the second (magnitude) and the fourth, respec- tively. (Which is) the very thing it was required to show. † In modern notation, this proposition reads m(n α) = (m n) α.dþ. Proposition 4† ᾿Εὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ If a first (magnitude) has the same ratio to a second τρίτον πρὸς τέταρτον, καὶ τὰ ἰσάκις πολλαπλάσια τοῦ τε that a third (has) to a fourth then equal multiples of the πρώτου καὶ τρίτου πρὸς τὰ ἰσάκις πολλαπλάσια τοῦ δευτέρου first (magnitude) and the third will also have the same καὶ τετάρτου καθ᾿ ὁποιονοῦν πολλαπλασιασμὸν τὸν αὐτὸν ratio to equal multiples of the second and the fourth, be- ἕξει λόγον ληφθέντα κατάλληλα. ing taken in corresponding order, according to any kind Πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω of multiplication whatsoever. λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, καὶ εἰλήφθω For let a first (magnitude) A have the same ratio to τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Ε, Ζ, τῶν δὲ Β, Δ a second B that a third C (has) to a fourth D. And let ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Η, Θ· λέγω, ὅτι equal multiples E and F have been taken of A and C ἐστὶν ὡς τὸ Ε πρὸς τὸ Η, οὕτως τὸ Ζ πρὸς τὸ Θ. (respectively), and other random equal multiples G and 134 STOIQEIWN eþ. ELEMENTS BOOK 5 H of B and D (respectively). I say that as E (is) to G, so F (is) to H . Ν Μ Γ ∆ Α Κ Η Ε Β Ζ Θ Λ N A B E G K M C D F H L Εἰλήφθω γὰρ τῶν μὲν Ε, Ζ ἰσάκις πολλαπλάσια τὰ Κ, For let equal multiples K and L have been taken of E Λ, τῶν δὲ Η, Θ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, and F (respectively), and other random equal multiples Ν. M and N of G and H (respectively). [Καὶ] ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ μὲν Ε τοῦ Α, τὸ [And] since E and F are equal multiples of A and δὲ Ζ τοῦ Γ, καὶ εἴληπται τῶν Ε, Ζ ἴσάκις πολλαπλάσια τὰ Κ, C (respectively), and the equal multiples K and L have Λ, ἴσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ Κ τοῦ Α καὶ τὸ Λ τοῦ been taken of E and F (respectively), K and L are thus Γ. διὰ τὰ αὐτὰ δὴ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Μ τοῦ Β καὶ equal multiples of A and C (respectively) [Prop. 5.3]. So, τὸ Ν τοῦ Δ. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ for the same (reasons), M and N are equal multiples of πρὸς τὸ Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια B and D (respectively). And since as A is to B, so C (is) τὰ Κ, Λ, τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια to D, and the equal multiples K and L have been taken τὰ Μ, Ν, εἰ ἄρα ὑπερέχει τὸ Κ τοῦ Μ, ὑπερέχει καὶ τὸ Λ of A and C (respectively), and the other random equal τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι multiples M and N of B and D (respectively), then if K τὰ μὲν Κ, Λ τῶν Ε, Ζ ἰσάκις πολλαπλάσια, τὰ δὲ Μ, Ν τῶν exceeds M then L also exceeds N , and if (K is) equal (to Η, Θ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· ἔστιν ἄρα ὡς τὸ M then L is also) equal (to N), and if (K is) less (than M Ε πρὸς τὸ Η, οὕτως τὸ Ζ πρὸς τὸ Θ. then L is also) less (than N) [Def. 5.5]. And K and L are ᾿Εὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον equal multiples of E and F (respectively), and M and N καὶ τρίτον πρὸς τέταρτον, καὶ τὰ ἰσάκις πολλαπλάσια τοῦ τε other random equal multiples of G and H (respectively). πρώτου καὶ τρίτου πρὸς τὰ ἰσάκις πολλαπλάσια τοῦ δευτέρου Thus, as E (is) to G, so F (is) to H [Def. 5.5]. καὶ τετάρτου τὸν αὐτὸν ἕξει λόγον καθ᾿ ὁποιονοῦν πολλα- Thus, if a first (magnitude) has the same ratio to a πλασιασμὸν ληφθέντα κατάλληλα· ὅπερ ἔδει δεῖξαι. second that a third (has) to a fourth then equal multi- ples of the first (magnitude) and the third will also have the same ratio to equal multiples of the second and the fourth, being taken in corresponding order, according to any kind of multiplication whatsoever. (Which is) the very thing it was required to show. † In modern notation, this proposition reads that if α : β :: γ : δ then m α : n β :: m γ : n δ, for all m and n. 135 STOIQEIWN eþ. ELEMENTS BOOK 5eþ. Proposition 5† ᾿Εὰν μέγεθος μεγέθους ἰσάκις ᾖ πολλαπλάσιον, ὅπερ If a magnitude is the same multiple of a magnitude ἀφαιρεθὲν ἀφαιρεθέντος, καὶ τὸ λοιπὸν τοῦ λοιποῦ ἰσάκις that a (part) taken away (is) of a (part) taken away (re- ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστι τὸ ὅλον τοῦ ὅλου. spectively) then the remainder will also be the same mul- tiple of the remainder as that which the whole (is) of the whole (respectively). ΒΑ Ε Η Γ Ζ ∆ D A G C E F B Μέγεθος γὰρ τὸ ΑΒ μεγέθους τοῦ ΓΔ ἰσάκις ἔστω πολ- For let the magnitude AB be the same multiple of the λαπλάσιον, ὅπερ ἀφαιρεθὲν τὸ ΑΕ ἀφαιρεθέντος τοῦ ΓΖ· magnitude CD that the (part) taken away AE (is) of the λέγω, ὅτι καὶ λοιπὸν τὸ ΕΒ λοιποῦ τοῦ ΖΔ ἰσάκις ἔσται (part) taken away CF (respectively). I say that the re- πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ὅλον τὸ ΑΒ ὅλου τοῦ ΓΔ. mainder EB will also be the same multiple of the remain- ῾Οσαπλάσιον γάρ ἐστι τὸ ΑΕ τοῦ ΓΖ, τοσαυταπλάσιον der FD as that which the whole AB (is) of the whole CD γεγονέτω καὶ τὸ ΕΒ τοῦ ΓΗ. (respectively). Καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ For as many times as AE is (divisible) by CF , so many ΕΒ τοῦ ΗΓ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ times let EB also have been made (divisible) by CG. καὶ τὸ ΑΒ τοῦ ΗΖ. κεῖται δὲ ἰσάκις πολλαπλάσιον τὸ ΑΕ And since AE and EB are equal multiples of CF and τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΓΔ. ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ GC (respectively), AE and AB are thus equal multiples ΑΒ ἑκατέρου τῶν ΗΖ, ΓΔ· ἴσον ἄρα τὸ ΗΖ τῷ ΓΔ. κοινὸν of CF and GF (respectively) [Prop. 5.1]. And AE and ἀφῃρήσθω τὸ ΓΖ· λοιπὸν ἄρα τὸ ΗΓ λοιπῷ τῷ ΖΔ ἴσον AB are assumed (to be) equal multiples of CF and CD ἐστίν. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ (respectively). Thus, AB is an equal multiple of each καὶ τὸ ΕΒ τοῦ ΗΓ, ἴσον δὲ τὸ ΗΓ τῷ ΔΖ, ἰσάκις ἄρα ἐστὶ of GF and CD. Thus, GF (is) equal to CD. Let CF πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΕΒ τοῦ ΖΔ. ἰσάκις δὲ have been subtracted from both. Thus, the remainder ὑπόκειται πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΓΔ· GC is equal to the remainder FD. And since AE and ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΕΒ τοῦ ΖΔ καὶ τὸ ΑΒ EB are equal multiples of CF and GC (respectively), τοῦ ΓΔ. καὶ λοιπὸν ἄρα τὸ ΕΒ λοιποῦ τοῦ ΖΔ ἰσάκις ἔσται and GC (is) equal to DF , AE and EB are thus equal πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ὅλον τὸ ΑΒ ὅλου τοῦ ΓΔ. multiples of CF and FD (respectively). And AE and ᾿Εὰν ἄρα μέγεθος μεγέθους ἰσάκις ᾖ πολλαπλάσιον, AB are assumed (to be) equal multiples of CF and CD ὅπερ ἀφαιρεθὲν ἀφαιρεθέντος, καὶ τὸ λοιπὸν τοῦ λοιποῦ (respectively). Thus, EB and AB are equal multiples of ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστι καὶ τὸ ὅλον FD and CD (respectively). Thus, the remainder EB will τοῦ ὅλου· ὅπερ ἔδει δεῖξαι. also be the same multiple of the remainder FD as that which the whole AB (is) of the whole CD (respectively). Thus, if a magnitude is the same multiple of a magni- tude that a (part) taken away (is) of a (part) taken away (respectively) then the remainder will also be the same multiple of the remainder as that which the whole (is) of the whole (respectively). (Which is) the very thing it was required to show. † In modern notation, this proposition reads m α − m β = m (α − β).�þ. Proposition 6† ᾿Εὰν δύο μεγέθη δύο μεγεθῶν ἰσάκις ᾖ πολλαπλάσια, If two magnitudes are equal multiples of two (other) καὶ ἀφαιρεθέντα τινὰ τῶν αὐτῶν ἰσάκις ᾖ πολλαπλάσια, καὶ magnitudes, and some (parts) taken away (from the for- τὰ λοιπὰ τοῖς αὐτοῖς ἤτοι ἴσα ἐστὶν ἢ ἰσάκις αὐτῶν πολ- mer magnitudes) are equal multiples of the latter (mag- λαπλάσια. nitudes, respectively), then the remainders are also either Δύο γὰρ μεγέθη τὰ ΑΒ, ΓΔ δύο μεγεθῶν τῶν Ε, Ζ equal to the latter (magnitudes), or (are) equal multiples 136 STOIQEIWN eþ. ELEMENTS BOOK 5 ἰσάκις ἔστω πολλαπλάσια, καὶ ἀφαιρεθέντα τὰ ΑΗ, ΓΘ τῶν of them (respectively). αὐτῶν τῶν Ε, Ζ ἰσάκις ἔστω πολλαπλάσια· λέγω, ὅτι καὶ For let two magnitudes AB and CD be equal multi- λοιπὰ τὰ ΗΒ, ΘΔ τοῖς Ε, Ζ ἤτοι ἴσα ἐστὶν ἢ ἰσάκις αὐτῶν ples of two magnitudes E and F (respectively). And let πολλαπλάσια. the (parts) taken away (from the former) AG and CH be equal multiples of E and F (respectively). I say that the remainders GB and HD are also either equal to E and F (respectively), or (are) equal multiples of them. Ζ Η Β Γ Θ ∆ Α Κ Ε F G B C H D A K E ῎Εστω γὰρ πρότερον τὸ ΗΒ τῷ Ε ἴσον· λέγω, ὅτι καὶ For let GB be, first of all, equal to E. I say that HD is τὸ ΘΔ τῷ Ζ ἴσον ἐστίν. also equal to F . Κείσθω γὰρ τῷ Ζ ἴσον τὸ ΓΚ. ἐπεὶ ἰσάκις ἐστὶ πολ- For let CK be made equal to F . Since AG and CH λαπλάσιον τὸ ΑΗ τοῦ Ε καὶ τὸ ΓΘ τοῦ Ζ, ἴσον δὲ τὸ μὲν ΗΒ are equal multiples of E and F (respectively), and GB τῷ Ε, τὸ δὲ ΚΓ τῷ Ζ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΒ (is) equal to E, and KC to F , AB and KH are thus equal τοῦ Ε καὶ τὸ ΚΘ τοῦ Ζ. ἰσάκις δὲ ὑπόκειται πολλαπλάσιον multiples of E and F (respectively) [Prop. 5.2]. And AB τὸ ΑΒ τοῦ Ε καὶ τὸ ΓΔ τοῦ Ζ· ἴσάκις ἄρα ἐστὶ πολλαπλάσιον and CD are assumed (to be) equal multiples of E and F τὸ ΚΘ τοῦ Ζ καὶ τὸ ΓΔ τοῦ Ζ. ἐπεὶ οὖν ἑκάτερον τῶν ΚΘ, (respectively). Thus, KH and CD are equal multiples of ΓΔ τοῦ Ζ ἰσάκις ἐστὶ πολλαπλάσιον, ἴσον ἄρα ἐστὶ τὸ ΚΘ F and F (respectively). Therefore, KH and CD are each τῷ ΓΔ. κοινὸν ἀφῃρήσθω τὸ ΓΘ· λοιπὸν ἄρα τὸ ΚΓ λοιπῷ equal multiples of F . Thus, KH is equal to CD. Let CH τῷ ΘΔ ἴσον ἐστίν. ἀλλὰ τὸ Ζ τῷ ΚΓ ἐστιν ἴσον· καὶ τὸ have be taken away from both. Thus, the remainder KC ΘΔ ἄρα τῷ Ζ ἴσον ἐστίν. ὥστε εἰ τὸ ΗΒ τῷ Ε ἴσον ἐστίν, is equal to the remainder HD. But, F is equal to KC. καὶ τὸ ΘΔ ἴσον ἔσται τῷ Ζ. Thus, HD is also equal to F . Hence, if GB is equal to E ῾Ομοίως δὴ δείξομεν, ὅτι, κᾂν πολλαπλάσιον ᾖ τὸ ΗΒ then HD will also be equal to F . τοῦ Ε, τοσαυταπλάσιον ἔσται καὶ τὸ ΘΔ τοῦ Ζ. So, similarly, we can show that even if GB is a multi- ᾿Εὰν ἄρα δύο μεγέθη δύο μεγεθῶν ἰσάκις ᾖ πολ- ple of E then HD will also be the same multiple of F . λαπλάσια, καὶ ἀφαιρεθέντα τινὰ τῶν αὐτῶν ἰσάκις ᾖ πολ- Thus, if two magnitudes are equal multiples of two λαπλάσια, καὶ τὰ λοιπὰ τοῖς αὐτοῖς ἤτοι ἴσα ἐστὶν ἢ ἰσάκις (other) magnitudes, and some (parts) taken away (from αὐτῶν πολλαπλάσια· ὅπερ ἔδει δεῖξαι. the former magnitudes) are equal multiples of the latter (magnitudes, respectively), then the remainders are also either equal to the latter (magnitudes), or (are) equal multiples of them (respectively). (Which is) the very thing it was required to show. † In modern notation, this proposition reads m α − n α = (m − n) α.zþ. Proposition 7 Τὰ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον καὶ τὸ αὐτὸ Equal (magnitudes) have the same ratio to the same πρὸς τὰ ἴσα. (magnitude), and the latter (magnitude has the same ra- ῎Εστω ἴσα μεγέθη τὰ Α, Β, ἄλλο δέ τι, ὃ ἔτυχεν, tio) to the equal (magnitudes). μέγεθος τὸ Γ· λέγω, ὅτι ἑκάτερον τῶν Α, Β πρὸς τὸ Γ Let A and B be equal magnitudes, and C some other τὸν αὐτὸν ἔχει λόγον, καὶ τὸ Γ πρὸς ἑκάτερον τῶν Α, Β. random magnitude. I say that A and B each have the 137 STOIQEIWN eþ. ELEMENTS BOOK 5 same ratio to C, and (that) C (has the same ratio) to each of A and B. Ζ Α ∆ Β Ε Γ D B A C F E Εἰλήφθω γὰρ τῶν μὲν Α, Β ἰσάκις πολλαπλάσια τὰ Δ, For let the equal multiples D and E have been taken Ε, τοῦ δὲ Γ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον τὸ Ζ. of A and B (respectively), and the other random multiple ᾿Επεὶ οὖν ἰσάκις ἐστὶ πολλαπλάσιον τὸ Δ τοῦ Α καὶ τὸ F of C. Ε τοῦ Β, ἴσον δὲ τὸ Α τῷ Β, ἴσον ἄρα καὶ τὸ Δ τῷ Ε. ἄλλο Therefore, since D and E are equal multiples of A δέ, ὅ ἔτυχεν, τὸ Ζ. Εἰ ἄρα ὑπερέχει τὸ Δ τοῦ Ζ, ὑπερέχει and B (respectively), and A (is) equal to B, D (is) thus καὶ τὸ Ε τοῦ Ζ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. also equal to E. And F (is) different, at random. Thus, if καί ἐστι τὰ μὲν Δ, Ε τῶν Α, Β ἰσάκις πολλαπλάσια, τὸ δὲ D exceeds F then E also exceeds F , and if (D is) equal Ζ τοῦ Γ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον· ἔστιν ἄρα ὡς τὸ Α (to F then E is also) equal (to F ), and if (D is) less πρὸς τὸ Γ, οὕτως τὸ Β πρὸς τὸ Γ. (than F then E is also) less (than F ). And D and E are Λέγω [δή], ὅτι καὶ τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν equal multiples of A and B (respectively), and F another αὐτὸν ἔχει λόγον. random multiple of C. Thus, as A (is) to C, so B (is) to Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, C [Def. 5.5]. ὅτι ἴσον ἐστὶ τὸ Δ τῷ Ε· ἄλλο δέ τι τὸ Ζ· εἰ ἄρα ὑπερέχει [So] I say that C† also has the same ratio to each of A τὸ Ζ τοῦ Δ, ὑπερέχει καὶ τοῦ Ε, καὶ εἰ ἴσον, ἴσον, καὶ εἰ and B. ἔλαττον, ἔλαττον. καί ἐστι τὸ μὲν Ζ τοῦ Γ πολλαπλάσιον, For, similarly, we can show, by the same construction, τὰ δὲ Δ, Ε τῶν Α, Β ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· that D is equal to E. And F (has) some other (value). ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως τὸ Γ πρὸς τὸ Β. Thus, if F exceeds D then it also exceeds E, and if (F is) Τὰ ἴσα ἄρα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον καὶ τὸ equal (to D then it is also) equal (to E), and if (F is) less αὐτὸ πρὸς τὰ ἴσα. (than D then it is also) less (than E). And F is a multiple of C, and D and E other random equal multiples of A and B. Thus, as C (is) to A, so C (is) to B [Def. 5.5]. Thus, equal (magnitudes) have the same ratio to the same (magnitude), and the latter (magnitude has the same ratio) to the equal (magnitudes).Pìrisma. Corollary‡ ᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν μεγέθη τινὰ ἀνάλογον So (it is) clear, from this, that if some magnitudes are ᾖ, καὶ ἀνάπαλιν ἀνάλογον ἔσται. ὅπερ ἔδει δεῖξαι. proportional then they will also be proportional inversely. (Which is) the very thing it was required to show. † The Greek text has “E”, which is obviously a mistake. ‡ In modern notation, this corollary reads that if α : β :: γ : δ then β : α :: δ : γ.hþ. Proposition 8 Τῶν ἀνίσων μεγεθῶν τὸ μεῖζον πρὸς τὸ αὐτὸ μείζονα For unequal magnitudes, the greater (magnitude) has λόγον ἔχει ἤπερ τὸ ἔλαττον. καὶ τὸ αὐτὸ πρὸς τὸ ἔλαττον a greater ratio than the lesser to the same (magnitude). μείζονα λόγον ἔχει ἤπερ πρὸς τὸ μεῖζον. And the latter (magnitude) has a greater ratio to the ῎Εστω ἄνισα μεγέθη τὰ ΑΒ, Γ, καὶ ἔστω μεῖζον τὸ ΑΒ, lesser (magnitude) than to the greater. ἄλλο δέ, ὃ ἔτυχεν, τὸ Δ· λέγω, ὅτι τὸ ΑΒ πρὸς τὸ Δ Let AB and C be unequal magnitudes, and let AB be μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Δ, καὶ τὸ Δ πρὸς the greater (of the two), and D another random magni- τὸ Γ μείζονα λόγον ἔχει ἤπερ πρὸς τὸ ΑΒ. tude. I say that AB has a greater ratio to D than C (has) to D, and (that) D has a greater ratio to C than (it has) to AB. 138 STOIQEIWN eþ. ELEMENTS BOOK 5 GEA B A E BH JZ Z H JG KDLMN NM LD K C EA B A E B K D L M N N M L D K F G H F G H C ᾿Επεὶ γὰρ μεῖζόν ἐστι τὸ ΑΒ τοῦ Γ, κείσθω τῷ Γ ἴσον For since AB is greater than C, let BE be made equal τὸ ΒΕ· τὸ δὴ ἔλασσον τῶν ΑΕ, ΕΒ πολλαπλασιαζόμενον to C. So, the lesser of AE and EB, being multiplied, will ἔσται ποτὲ τοῦ Δ μεῖζον. ἔστω πρότερον τὸ ΑΕ ἔλαττον sometimes be greater than D [Def. 5.4]. First of all, let τοῦ ΕΒ, καὶ πεπολλαπλασιάσθω τὸ ΑΕ, καὶ ἔστω αὐτοῦ AE be less than EB, and let AE have been multiplied, πολλαπλάσιον τὸ ΖΗ μεῖζον ὂν τοῦ Δ, καὶ ὁσαπλάσιόν ἐστι and let FG be a multiple of it which (is) greater than τὸ ΖΗ τοῦ ΑΕ, τοσαυταπλάσιον γεγονέτω καὶ τὸ μὲν ΗΘ D. And as many times as FG is (divisible) by AE, so τοῦ ΕΒ τὸ δὲ Κ τοῦ Γ· καὶ εἰλήφθω τοῦ Δ διπλάσιον μὲν many times let GH also have become (divisible) by EB, τὸ Λ, τριπλάσιον δὲ τὸ Μ, καὶ ἑξῆς ἑνὶ πλεῖον, ἕως ἂν τὸ and K by C. And let the double multiple L of D have λαμβανόμενον πολλαπλάσιον μὲν γένηται τοῦ Δ, πρώτως δὲ been taken, and the triple multiple M , and several more, μεῖζον τοῦ Κ. εἰλήφθω, καὶ ἔστω τὸ Ν τετραπλάσιον μὲν (each increasing) in order by one, until the (multiple) τοῦ Δ, πρώτως δὲ μεῖζον τοῦ Κ. taken becomes the first multiple of D (which is) greater ᾿Επεὶ οὖν τὸ Κ τοῦ Ν πρώτως ἐστὶν ἔλαττον, τὸ Κ ἄρα than K. Let it have been taken, and let it also be the τοῦ Μ οὔκ ἐστιν ἔλαττον. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον quadruple multiple N of D—the first (multiple) greater τὸ ΖΗ τοῦ ΑΕ καὶ τὸ ΗΘ τοῦ ΕΒ, ἰσάκις ἄρα ἐστὶ πολ- than K. λαπλάσιον τὸ ΖΗ τοῦ ΑΕ καὶ τὸ ΖΘ τοῦ ΑΒ. ἰσάκις δέ Therefore, since K is less than N first, K is thus not ἐστι πολλαπλάσιον τὸ ΖΗ τοῦ ΑΕ καὶ τὸ Κ τοῦ Γ· ἰσάκις less than M . And since FG and GH are equal multi- ἄρα ἐστὶ πολλαπλάσιον τὸ ΖΘ τοῦ ΑΒ καὶ τὸ Κ τοῦ Γ. τὰ ples of AE and EB (respectively), FG and FH are thus ΖΘ, Κ ἄρα τῶν ΑΒ, Γ ἰσάκις ἐστὶ πολλαπλάσια. πάλιν, ἐπεὶ equal multiples of AE and AB (respectively) [Prop. 5.1]. ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΕΒ καὶ τὸ Κ τοῦ Γ, And FG and K are equal multiples of AE and C (re- ἴσον δὲ τὸ ΕΒ τῷ Γ, ἴσον ἄρα καὶ τὸ ΗΘ τῷ Κ. τὸ δὲ Κ spectively). Thus, FH and K are equal multiples of AB τοῦ Μ οὔκ ἐστιν ἔλαττον· οὐδ᾿ ἄρα τὸ ΗΘ τοῦ Μ ἔλαττόν and C (respectively). Thus, FH , K are equal multiples ἐστιν. μεῖζον δὲ τὸ ΖΗ τοῦ Δ· ὅλον ἄρα τὸ ΖΘ συναμ- of AB, C. Again, since GH and K are equal multiples φοτέρων τῶν Δ, Μ μεῖζόν ἐστιν. ἀλλὰ συναμφότερα τὰ Δ, of EB and C, and EB (is) equal to C, GH (is) thus also Μ τῷ Ν ἐστιν ἴσα, ἐπειδήπερ τὸ Μ τοῦ Δ τριπλάσιόν ἐστιν, equal to K. And K is not less than M . Thus, GH not less συναμφότερα δὲ τὰ Μ, Δ τοῦ Δ ἐστι τετραπλάσια, ἔστι δὲ than M either. And FG (is) greater than D. Thus, the καὶ τὸ Ν τοῦ Δ τετραπλάσιον· συναμφότερα ἄρα τὰ Μ, Δ whole of FH is greater than D and M (added) together. τῷ Ν ἴσα ἐστίν. ἀλλὰ τὸ ΖΘ τῶν Μ, Δ μεῖζόν ἐστιν· τὸ But, D and M (added) together is equal to N , inasmuch ΖΘ ἄρα τοῦ Ν ὑπερέχει· τὸ δὲ Κ τοῦ Ν οὐχ ὑπερέχει. καί as M is three times D, and M and D (added) together is ἐστι τὰ μὲν ΖΘ, Κ τῶν ΑΒ, Γ ἰσάκις πολλαπλάσια, τὸ δὲ Ν four times D, and N is also four times D. Thus, M and D τοῦ Δ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον· τὸ ΑΒ ἄρα πρὸς τὸ (added) together is equal to N . But, FH is greater than Δ μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Δ. M and D. Thus, FH exceeds N . And K does not exceed Λέγω δή, ὅτι καὶ τὸ Δ πρὸς τὸ Γ μείζονα λόγον ἔχει N . And FH , K are equal multiples of AB, C, and N ἤπερ τὸ Δ πρὸς τὸ ΑΒ. another random multiple of D. Thus, AB has a greater Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ratio to D than C (has) to D [Def. 5.7]. ὅτι τὸ μὲν Ν τοῦ Κ ὑπερέχει, τὸ δὲ Ν τοῦ ΖΘ οὐχ ὑπερέχει. So, I say that D also has a greater ratio to C than D καί ἐστι τὸ μὲν Ν τοῦ Δ πολλαπλάσιον, τὰ δὲ ΖΘ, Κ τῶν (has) to AB. ΑΒ, Γ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· τὸ Δ ἄρα πρὸς For, similarly, by the same construction, we can show τὸ Γ μείζονα λόγον ἔχει ἤπερ τὸ Δ πρὸς τὸ ΑΒ. that N exceeds K, and N does not exceed FH . And Ἀλλὰ δὴ τὸ ΑΕ τοῦ ΕΒ μεῖζον ἔστω. τὸ δὴ ἔλαττον N is a multiple of D, and FH , K other random equal τὸ ΕΒ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ Δ μεῖζον. πε- multiples of AB, C (respectively). Thus, D has a greater 139 STOIQEIWN eþ. ELEMENTS BOOK 5 πολλαπλασιάσθω, καὶ ἔστω τὸ ΗΘ πολλαπλάσιον μὲν τοῦ ratio to C than D (has) to AB [Def. 5.5]. ΕΒ, μεῖζον δὲ τοῦ Δ· καὶ ὁσαπλάσιόν ἐστι τὸ ΗΘ τοῦ ΕΒ, And so let AE be greater than EB. So, the lesser, τοσαυταπλάσιον γεγονέτω καὶ τὸ μὲν ΖΗ τοῦ ΑΕ, τὸ δὲ Κ EB, being multiplied, will sometimes be greater than D. τοῦ Γ. ὁμοίως δὴ δείξομεν, ὅτι τὰ ΖΘ, Κ τῶν ΑΒ, Γ ἰσάκις Let it have been multiplied, and let GH be a multiple of ἐστὶ πολλαπλάσια· καὶ εἰλήφθω ὁμοίως τὸ Ν πολλαπλάσιον EB (which is) greater than D. And as many times as μὲν τοῦ Δ, πρώτως δὲ μεῖζον τοῦ ΖΗ· ὥστε πάλιν τὸ ΖΗ GH is (divisible) by EB, so many times let FG also have τοῦ Μ οὔκ ἐστιν ἔλασσον. μεῖζον δὲ τὸ ΗΘ τοῦ Δ· ὅλον become (divisible) by AE, and K by C. So, similarly ἄρα τὸ ΖΘ τῶν Δ, Μ, τουτέστι τοῦ Ν, ὑπερέχει. τὸ δὲ Κ (to the above), we can show that FH and K are equal τοῦ Ν οὐχ ὑπερέχει, ἐπειδήπερ καὶ τὸ ΖΗ μεῖζον ὂν τοῦ multiples of AB and C (respectively). And, similarly (to ΗΘ, τουτέστι τοῦ Κ, τοῦ Ν οὐχ ὑπερέχει. καὶ ὡσαύτως the above), let the multiple N of D, (which is) the first κατακολουθοῦντες τοῖς ἐπάνω περαίνομεν τὴν ἀπόδειξιν. (multiple) greater than FG, have been taken. So, FG Τῶν ἄρα ἀνίσων μεγεθῶν τὸ μεῖζον πρὸς τὸ αὐτὸ is again not less than M . And GH (is) greater than D. μείζονα λόγον ἔχει ἤπερ τὸ ἔλαττον· καὶ τὸ αὐτὸ πρὸς τὸ Thus, the whole of FH exceeds D and M , that is to say ἔλαττον μείζονα λόγον ἔχει ἤπερ πρὸς τὸ μεῖζον· ὅπερ ἔδει N . And K does not exceed N , inasmuch as FG, which δεῖξαι. (is) greater than GH—that is to say, K—also does not exceed N . And, following the above (arguments), we (can) complete the proof in the same manner. Thus, for unequal magnitudes, the greater (magni- tude) has a greater ratio than the lesser to the same (mag- nitude). And the latter (magnitude) has a greater ratio to the lesser (magnitude) than to the greater. (Which is) the very thing it was required to show.jþ. Proposition 9 Τὰ πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχοντα λὸγον ἴσα ἀλλήλοις (Magnitudes) having the same ratio to the same ἐστίν· καὶ πρὸς ἃ τὸ αὐτὸ τὸν αὐτὸν ἕχει λόγον, ἐκεῖνα ἴσα (magnitude) are equal to one another. And those (mag- ἐστίν. nitudes) to which the same (magnitude) has the same ratio are equal. Β Γ Α B C A ᾿Εχέτω γὰρ ἑκάτερον τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν For let A and B each have the same ratio to C. I say λόγον· λέγω, ὅτι ἴσον ἐστὶ τὸ Α τῷ Β. that A is equal to B. Εἰ γὰρ μή, οὐκ ἂν ἑκάτερον τῶν Α, Β πρὸς τὸ Γ τὸν For if not, A and B would not each have the same αὐτὸν εἶχε λόγον· ἔχει δέ· ἴσον ἄρα ἐστὶ τὸ Α τῷ Β. ratio to C [Prop. 5.8]. But they do. Thus, A is equal to ᾿Εχέτω δὴ πάλιν τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν B. λόγον· λέγω, ὅτι ἴσον ἐστὶ τὸ Α τῷ Β. So, again, let C have the same ratio to each of A and Εἰ γὰρ μή, οὐκ ἂν τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν B. I say that A is equal to B. αὐτὸν εἶχε λόγον· ἔχει δέ· ἴσον ἄρα ἐστὶ τὸ Α τῷ Β. For if not, C would not have the same ratio to each of Τὰ ἄρα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχοντα λόγον ἴσα A and B [Prop. 5.8]. But it does. Thus, A is equal to B. ἀλλήλοις ἐστίν· καὶ πρὸς ἃ τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον, Thus, (magnitudes) having the same ratio to the same ἐκεῖνα ἴσα ἐστίν· ὅπερ ἔδει δεῖξαι. (magnitude) are equal to one another. And those (magni- tudes) to which the same (magnitude) has the same ratio are equal. (Which is) the very thing it was required to show.iþ. Proposition 10 Τῶν πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον For (magnitudes) having a ratio to the same (mag- ἔχον ἐκεῖνο μεῖζόν ἐστιν· πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον nitude), that (magnitude which) has the greater ratio is 140 STOIQEIWN eþ. ELEMENTS BOOK 5 ἔχει, ἐκεῖνο ἔλαττόν ἐστιν. (the) greater. And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser. Β Γ Α C A B ᾿Εχέτω γὰρ τὸ Α πρὸς τὸ Γ μείζονα λόγον ἤπερ τὸ Β For let A have a greater ratio to C than B (has) to C. πρὸς τὸ Γ· λέγω, ὅτι μεῖζόν ἐστι τὸ Α τοῦ Β. I say that A is greater than B. Εἰ γὰρ μή, ἤτοι ἴσον ἐστὶ τὸ Α τῷ Β ἢ ἔλασσον. ἴσον For if not, A is surely either equal to or less than B. μὲν οὖν οὔκ ἐστὶ τὸ Α τῷ Β· ἑκάτερον γὰρ ἂν τῶν Α, Β In fact, A is not equal to B. For (then) A and B would πρὸς τὸ Γ τὸν αὐτὸν εἶχε λόγον. οὐκ ἔχει δέ· οὐκ ἄρα ἴσον each have the same ratio to C [Prop. 5.7]. But they do ἐστὶ τὸ Α τῷ Β. οὐδὲ μὴν ἔλασσόν ἐστι τὸ Α τοῦ Β· τὸ Α not. Thus, A is not equal to B. Neither, indeed, is A less γὰρ ἂν πρὸς τὸ Γ ἐλάσσονα λόγον εἶχεν ἤπερ τὸ Β πρὸς τὸ than B. For (then) A would have a lesser ratio to C than Γ. οὐκ ἔχει δέ· οὐκ ἄρα ἔλασσόν ἐστι τὸ Α τοῦ Β. ἐδείχθη B (has) to C [Prop. 5.8]. But it does not. Thus, A is not δὲ οὐδὲ ἴσον· μεῖζον ἄρα ἐστὶ τὸ Α τοῦ Β. less than B. And it was shown not (to be) equal either. ᾿Εχέτω δὴ πάλιν τὸ Γ πρὸς τὸ Β μείζονα λόγον ἤπερ τὸ Thus, A is greater than B. Γ πρὸς τὸ Α· λέγω, ὅτι ἔλασσόν ἐστι τὸ Β τοῦ Α. So, again, let C have a greater ratio to B than C (has) Εἰ γὰρ μή, ἤτοι ἴσον ἐστὶν ἢ μεῖζον. ἴσον μὲν οὖν οὔκ to A. I say that B is less than A. ἐστι τὸ Β τῷ Α· τὸ Γ γὰρ ἂν πρὸς ἑκάτερον τῶν Α, Β τὸν For if not, (it is) surely either equal or greater. In fact, αὐτὸν εἶχε λόγον. οὐκ ἔχει δέ· οὐκ ἄρα ἴσον ἐστὶ τὸ Α B is not equal to A. For (then) C would have the same τῷ Β. οὐδὲ μὴν μεῖζόν ἐστι τὸ Β τοῦ Α· τὸ Γ γὰρ ἂν πρὸς ratio to each of A and B [Prop. 5.7]. But it does not. τὸ Β ἐλάσσονα λόγον εἶχεν ἤπερ πρὸς τὸ Α. οὐκ ἔχει δέ· Thus, A is not equal to B. Neither, indeed, is B greater οὐκ ἄρα μεῖζόν ἐστι τὸ Β τοῦ Α. ἐδείχθη δέ, ὅτι οὐδὲ ἴσον· than A. For (then) C would have a lesser ratio to B than ἔλαττον ἄρα ἐστὶ τὸ Β τοῦ Α. (it has) to A [Prop. 5.8]. But it does not. Thus, B is not Τῶν ἄρα πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον greater than A. And it was shown that (it is) not equal ἔχον μεῖζόν ἐστιν· καὶ πρὸς ὃ τὸ αὐτὸ μείζονα λόγον ἔχει, (to A) either. Thus, B is less than A. ἐκεῖνο ἔλαττόν ἐστιν· ὅπερ ἔδει δεῖξαι. Thus, for (magnitudes) having a ratio to the same (magnitude), that (magnitude which) has the greater ratio is (the) greater. And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser. (Which is) the very thing it was required to show.iaþ. Proposition 11† Οἱ τῷ αὐτῷ λόγῳ οἱ αὐτοὶ καὶ ἀλλήλοις εἰσὶν οἱ αὐτοί. (Ratios which are) the same with the same ratio are also the same with one another. ΕΑ Β Η Λ Γ ∆ Θ Μ Ν Κ Ζ N G A B L H M D C E F K ῎Εστωσαν γὰρ ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς For let it be that as A (is) to B, so C (is) to D, and as τὸ Δ, ὡς δὲ τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ· λέγω, C (is) to D, so E (is) to F . I say that as A is to B, so E ὅτι ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ. (is) to F . Εἰλήφθω γὰρ τῶν Α, Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, Θ, For let the equal multiples G, H , K have been taken Κ, τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ of A, C, E (respectively), and the other random equal Λ, Μ, Ν. multiples L, M , N of B, D, F (respectively). Καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ And since as A is to B, so C (is) to D, and the equal Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Η, Θ, multiples G and H have been taken of A and C (respec- τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, tively), and the other random equal multiples L and M εἰ ἄρα ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Θ τοῦ Μ, καὶ εἰ of B and D (respectively), thus if G exceeds L then H ἴσον ἐστίν, ἴσον, καὶ εἰ ἐλλείπει, ἐλλείπει. πάλιν, ἐπεί ἐστιν also exceeds M , and if (G is) equal (to L then H is also) 141 STOIQEIWN eþ. ELEMENTS BOOK 5 ὡς τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ, καὶ εἴληπται equal (to M), and if (G is) less (than L then H is also) τῶν Γ, Ε ἰσάκις πολλαπλάσια τὰ Θ, Κ, τῶν δὲ Δ, Ζ ἄλλα, less (than M) [Def. 5.5]. Again, since as C is to D, so ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, Ν, εἰ ἄρα ὑπερέχει τὸ E (is) to F , and the equal multiples H and K have been Θ τοῦ Μ, ὑπερέχει καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ taken of C and E (respectively), and the other random ἔλλατον, ἔλαττον. ἀλλὰ εἰ ὑπερεῖχε τὸ Θ τοῦ Μ, ὑπερεῖχε equal multiples M and N of D and F (respectively), thus καὶ τὸ Η τοῦ Λ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον· if H exceeds M then K also exceeds N , and if (H is) ὥστε καὶ εἰ ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Κ τοῦ equal (to M then K is also) equal (to N), and if (H is) Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ less (than M then K is also) less (than N) [Def. 5.5]. But μὲν Η, Κ τῶν Α, Ε ἰσάκις πολλαπλάσια, τὰ δὲ Λ, Ν τῶν Β, (we saw that) if H was exceeding M then G was also ex- Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· ἔστιν ἄρα ὡς τὸ Α ceeding L, and if (H was) equal (to M then G was also) πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ. equal (to L), and if (H was) less (than M then G was Οἱ ἄρα τῷ αὐτῷ λόγῳ οἱ αὐτοὶ καὶ ἀλλήλοις εἰσὶν οἱ also) less (than L). And, hence, if G exceeds L then K αὐτοί· ὅπερ ἔδει δεῖξαι. also exceeds N , and if (G is) equal (to L then K is also) equal (to N), and if (G is) less (than L then K is also) less (than N). And G and K are equal multiples of A and E (respectively), and L and N other random equal multiples of B and F (respectively). Thus, as A is to B, so E (is) to F [Def. 5.5]. Thus, (ratios which are) the same with the same ratio are also the same with one another. (Which is) the very thing it was required to show. † In modern notation, this proposition reads that if α : β :: γ : δ and γ : δ :: ǫ : ζ then α : β :: ǫ : ζ.ibþ. Proposition 12† ᾿Εὰν ᾖ ὁποσαοῦν μεγέθη ἀνάλογον, ἔσται ὡς ἓν τῶν If there are any number of magnitudes whatsoever ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ (which are) proportional then as one of the leading (mag- ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα. nitudes is) to one of the following, so will all of the lead- ing (magnitudes) be to all of the following. Ε Β ∆ Ζ Λ Μ ΝΚ Θ Η Α Γ N A C E B D F G H K L M ῎Εστωσαν ὁποσαοῦν μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, Let there be any number of magnitudes whatsoever, Ε, Ζ, ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ τὸ Ε A, B, C, D, E, F , (which are) proportional, (so that) as πρὸς το Ζ· λέγω, ὅτι ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὰ A (is) to B, so C (is) to D, and E to F . I say that as A is Α, Γ, Ε πρὸς τὰ Β, Δ, Ζ. to B, so A, C, E (are) to B, D, F . Εἰλήφθω γὰρ τῶν μὲν Α, Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, For let the equal multiples G, H , K have been taken Θ, Κ, τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια of A, C, E (respectively), and the other random equal τὰ Λ, Μ, Ν. multiples L, M , N of B, D, F (respectively). Καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ And since as A is to B, so C (is) to D, and E to F , and Δ, καὶ τὸ Ε πρὸς τὸ Ζ, καὶ εἴληπται τῶν μὲν Α, Γ, Ε ἰσάκις the equal multiples G, H , K have been taken of A, C, E πολλαπλάσια τὰ Η, Θ, Κ τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, (respectively), and the other random equal multiples L, ἰσάκις πολλαπλάσια τὰ Λ, Μ, Ν, εἰ ἄρα ὑπερέχει τὸ Η τοῦ Λ, M , N of B, D, F (respectively), thus if G exceeds L then ὑπερέχει καὶ τὸ Θ τοῦ Μ, καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, H also exceeds M , and K (exceeds) N , and if (G is) καὶ εἰ ἔλαττον, ἔλαττον. ὥστε καὶ εἰ ὑπερέχει τὸ Η τοῦ Λ, equal (to L then H is also) equal (to M , and K to N), 142 STOIQEIWN eþ. ELEMENTS BOOK 5 ὑπερέχει καὶ τὰ Η, Θ, Κ τῶν Λ, Μ, Ν, καὶ εἰ ἴσον, ἴσα, καὶ and if (G is) less (than L then H is also) less (than M , εἰ ἔλαττον, ἔλαττονα. καί ἐστι τὸ μὲν Η καὶ τὰ Η, Θ, Κ and K than N) [Def. 5.5]. And, hence, if G exceeds L τοῦ Α καὶ τῶν Α, Γ, Ε ἰσάκις πολλαπλάσια, ἐπειδήπερ ἐὰν then G, H , K also exceed L, M , N , and if (G is) equal ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ πλῆθος (to L then G, H , K are also) equal (to L, M , N) and ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν ἐστιν if (G is) less (than L then G, H , K are also) less (than ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ πάντα L, M , N). And G and G, H , K are equal multiples of τῶν πάντων. διὰ τὰ αὐτὰ δὴ καὶ τὸ Λ καὶ τὰ Λ, Μ, Ν τοῦ A and A, C, E (respectively), inasmuch as if there are Β καὶ τῶν Β, Δ, Ζ ἰσάκις ἐστὶ πολλαπλάσια· ἔστιν ἄρα ὡς any number of magnitudes whatsoever (which are) equal τὸ Α πρὸς τὸ Β, οὕτως τὰ Α, Γ, Ε πρὸς τὰ Β, Δ, Ζ. multiples, respectively, of some (other) magnitudes, of ᾿Εὰν ἄρα ᾖ ὁποσαοῦν μεγέθη ἀνάλογον, ἔσται ὡς ἓν equal number (to them), then as many times as one of the τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ (first) magnitudes is (divisible) by one (of the second), ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα· ὅπερ ἔδει δεῖξαι. so many times will all (of the first magnitudes) also (be divisible) by all (of the second) [Prop. 5.1]. So, for the same (reasons), L and L, M , N are also equal multiples of B and B, D, F (respectively). Thus, as A is to B, so A, C, E (are) to B, D, F (respectively). Thus, if there are any number of magnitudes whatso- ever (which are) proportional then as one of the leading (magnitudes is) to one of the following, so will all of the leading (magnitudes) be to all of the following. (Which is) the very thing it was required to show. † In modern notation, this proposition reads that if α : α′ :: β : β′ :: γ : γ′ etc. then α : α′ :: (α + β + γ + · · · ) : (α′ + β′ + γ′ + · · · ).igþ. Proposition 13† ᾿Εὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἒχῃ λόγον καὶ If a first (magnitude) has the same ratio to a second τρίτον πρὸς τέταρτον, τρίτον δὲ πρὸς τέταρτον μείζονα that a third (has) to a fourth, and the third (magnitude) λόγον ἔχῃ ἢ πέμπτον πρὸς ἕκτον, καὶ πρῶτον πρὸς δεύτερον has a greater ratio to the fourth than a fifth (has) to a μείζονα λόγον ἕξει ἢ πέμπτον πρὸς ἕκτον. sixth, then the first (magnitude) will also have a greater ratio to the second than the fifth (has) to the sixth. Λ Α Β Μ Ν Γ ∆ Η Κ Ε Ζ Θ L M N A B C D G K E F H Πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω For let a first (magnitude) A have the same ratio to a λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, τρίτον δὲ τὸ Γ second B that a third C (has) to a fourth D, and let the πρὸς τέταρτον τὸ Δ μείζονα λόγον ἐχέτω ἢ πέμπτον τὸ Ε third (magnitude) C have a greater ratio to the fourth πρὸς ἕκτον τὸ Ζ. λέγω, ὅτι καὶ πρῶτον τὸ Α πρὸς δεύτερον D than a fifth E (has) to a sixth F . I say that the first τὸ Β μείζονα λόγον ἕξει ἤπερ πέμπτον τὸ Ε πρὸς ἕκτον τὸ (magnitude) A will also have a greater ratio to the second Ζ. B than the fifth E (has) to the sixth F . ᾿Επεὶ γὰρ ἔστι τινὰ τῶν μὲν Γ, Ε ἰσάκις πολλαπλάσια, For since there are some equal multiples of C and τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια, καὶ τὸ μὲν E, and other random equal multiples of D and F , (for τοῦ Γ πολλαπλάσιον τοῦ τοῦ Δ πολλαπλασίου ὑπερέχει, which) the multiple of C exceeds the (multiple) of D, τὸ δὲ τοῦ Ε πολλαπλάσιον τοῦ τοῦ Ζ πολλαπλασίου οὐχ and the multiple of E does not exceed the multiple of F ὑπερέχει, εἰλήφθω, καὶ ἔστω τῶν μὲν Γ, Ε ἰσάκις πολ- [Def. 5.7], let them have been taken. And let G and H be λαπλάσια τὰ Η, Θ, τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις equal multiples of C and E (respectively), and K and L πολλαπλάσια τὰ Κ, Λ, ὥστε τὸ μὲν Η τοῦ Κ ὑπερέχειν, τὸ other random equal multiples of D and F (respectively), δὲ Θ τοῦ Λ μὴ ὑπερέχειν· καὶ ὁσαπλάσιον μέν ἐστι τὸ Η such that G exceeds K, but H does not exceed L. And as τοῦ Γ, τοσαυταπλάσιον ἔστω καὶ τὸ Μ τοῦ Α, ὁσαπλάσιον many times as G is (divisible) by C, so many times let M δὲ τὸ Κ τοῦ Δ, τοσαυταπλάσιον ἔστω καὶ τὸ Ν τοῦ Β. be (divisible) by A. And as many times as K (is divisible) 143 STOIQEIWN eþ. ELEMENTS BOOK 5 Καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς by D, so many times let N be (divisible) by B. τὸ Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ And since as A is to B, so C (is) to D, and the equal Μ, Η, τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ multiples M and G have been taken of A and C (respec- Ν, Κ, εἰ ἄρα ὑπερέχει τὸ Μ τοῦ Ν, ὑπερέχει καὶ τὸ Η τοῦ tively), and the other random equal multiples N and K Κ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλλατον. ὑπερέχει δὲ of B and D (respectively), thus if M exceeds N then G τὸ Η τοῦ Κ· ὑπερέχει ἄρα καὶ τὸ Μ τοῦ Ν. τὸ δὲ Θ τοῦ exceeds K, and if (M is) equal (to N then G is also) Λ οὐχ ὑπερέχει· καί ἐστι τὰ μὲν Μ, Θ τῶν Α, Ε ἰσάκις equal (to K), and if (M is) less (than N then G is also) πολλαπλάσια, τὰ δὲ Ν, Λ τῶν Β, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις less (than K) [Def. 5.5]. And G exceeds K. Thus, M πολλαπλάσια· τὸ ἄρα Α πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ also exceeds N . And H does not exceeds L. And M and τὸ Ε πρὸς τὸ Ζ. H are equal multiples of A and E (respectively), and N ᾿Εὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἒχῃ λόγον and L other random equal multiples of B and F (respec- καὶ τρίτον πρὸς τέταρτον, τρίτον δὲ πρὸς τέταρτον μείζονα tively). Thus, A has a greater ratio to B than E (has) to λόγον ἔχῃ ἢ πέμπτον πρὸς ἕκτον, καὶ πρῶτον πρὸς δεύτερον F [Def. 5.7]. μείζονα λόγον ἕξει ἢ πέμπτον πρὸς ἕκτον· ὅπερ ἔδει δεῖξαι. Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and a third (magni- tude) has a greater ratio to a fourth than a fifth (has) to a sixth, then the first (magnitude) will also have a greater ratio to the second than the fifth (has) to the sixth. (Which is) the very thing it was required to show. † In modern notation, this proposition reads that if α : β :: γ : δ and γ : δ > ǫ : ζ then α : β > ǫ : ζ.idþ. Proposition 14†
᾿Εὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ If a first (magnitude) has the same ratio to a second

τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, that a third (has) to a fourth, and the first (magnitude)
καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, is greater than the third, then the second will also be
κἂν ἔλαττον, ἔλαττον. greater than the fourth. And if (the first magnitude is)

equal (to the third then the second will also be) equal (to
the fourth). And if (the first magnitude is) less (than the

third then the second will also be) less (than the fourth).

∆

Γ CA

B D
Πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β αὐτὸν ἐχέτω For let a first (magnitude) A have the same ratio to a

λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, μεῖζον δὲ ἔστω second B that a third C (has) to a fourth D. And let A be
τὸ Α τοῦ Γ· λέγω, ὅτι καὶ τὸ Β τοῦ Δ μεῖζόν ἐστιν. greater than C. I say that B is also greater than D.
᾿Επεὶ γὰρ τὸ Α τοῦ Γ μεῖζόν ἐστιν, ἄλλο δέ, ὃ ἔτυχεν, For since A is greater than C, and B (is) another ran-

[μέγεθος] τὸ Β, τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει dom [magnitude], A thus has a greater ratio to B than C
ἤπερ τὸ Γ πρὸς τὸ Β. ὡς δὲ τὸ Α πρὸς τὸ Β, οὕτως τὸ (has) to B [Prop. 5.8]. And as A (is) to B, so C (is) to
Γ πρὸς τὸ Δ· καὶ τὸ Γ ἄρα πρὸς τὸ Δ μείζονα λόγον ἔχει D. Thus, C also has a greater ratio to D than C (has) to
ἤπερ τὸ Γ πρὸς τὸ Β. πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον B. And that (magnitude) to which the same (magnitude)
ἔχει, ἐκεῖνο ἔλασσόν ἐστιν· ἔλασσον ἄρα τὸ Δ τοῦ Β· ὥστε has a greater ratio is the lesser [Prop. 5.10]. Thus, D (is)
μεῖζόν ἐστι τὸ Β τοῦ Δ. less than B. Hence, B is greater than D.
῾Ομοίως δὴ δεῖξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον So, similarly, we can show that even if A is equal to C

ἔσται καὶ τὸ Β τῷ Δ, κἄν ἔλασσον ᾖ τὸ Α τοῦ Γ, ἔλασσον then B will also be equal to D, and even if A is less than
ἔσται καὶ τὸ Β τοῦ Δ. C then B will also be less than D.
᾿Εὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον Thus, if a first (magnitude) has the same ratio to a

καὶ τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, second that a third (has) to a fourth, and the first (mag-
καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, nitude) is greater than the third, then the second will also
κἂν ἔλαττον, ἔλαττον· ὅπερ ἔδει δεῖξαι. be greater than the fourth. And if (the first magnitude is)

144

STOIQEIWN eþ. ELEMENTS BOOK 5
equal (to the third then the second will also be) equal (to

the fourth). And if (the first magnitude is) less (than the

third then the second will also be) less (than the fourth).
(Which is) the very thing it was required to show.

† In modern notation, this proposition reads that if α : β :: γ : δ then α T γ as β T δ.ieþ. Proposition 15†
Τὰ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει Parts have the same ratio as similar multiples, taken

λόγον ληφθέντα κατάλληλα. in corresponding order.

ΘΗΑ
Γ

Ζ
∆ Κ Λ

A G H

D K L
F

C
B

῎Εστω γὰρ ἰσάκις πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ το ΔΕ For let AB and DE be equal multiples of C and F
τοῦ Ζ· λέγω, ὅτι ἐστὶν ὡς τὸ Γ πρὸς τὸ Ζ, οὕτως τὸ ΑΒ (respectively). I say that as C is to F , so AB (is) to DE.
πρὸς τὸ ΔΕ. For since AB and DE are equal multiples of C and
᾿Επεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ F (respectively), thus as many magnitudes as there are

τὸ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μεγέθη ἴσα τῷ in AB equal to C, so many (are there) also in DE equal
Γ, τοσαῦτα καὶ ἐν τῷ ΔΕ ἴσα τῷ Ζ. διῃρήσθω τὸ μὲν ΑΒ to F . Let AB have been divided into (magnitudes) AG,
εἰς τὰ τῷ Γ ἴσα τὰ ΑΗ, ΗΘ, ΘΒ, τὸ δὲ ΔΕ εἰς τὰ τῷ Ζ GH , HB, equal to C, and DE into (magnitudes) DK,
ἴσα τὰ ΔΚ, ΚΛ, ΛΕ· ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, KL, LE, equal to F . So, the number of (magnitudes)
ΗΘ, ΘΒ τῷ πλήθει τῶν ΔΚ, ΚΛ, ΛΕ. καὶ ἐπεὶ ἴσα ἐστὶ τὰ AG, GH , HB will equal the number of (magnitudes)
ΑΗ, ΗΘ, ΘΒ ἀλλήλοις, ἔστι δὲ καὶ τὰ ΔΚ, ΚΛ, ΛΕ ἴσα DK, KL, LE. And since AG, GH , HB are equal to one
ἀλλήλοις, ἔστιν ἄρα ὡς τὸ ΑΗ πρὸς τὸ ΔΚ, οὕτως τὸ ΗΘ another, and DK, KL, LE are also equal to one another,
πρὸς τὸ ΚΛ, καὶ τὸ ΘΒ πρὸς τὸ ΛΕ. ἔσται ἄρα καὶ ὡς ἓν thus as AG is to DK, so GH (is) to KL, and HB to LE
τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ [Prop. 5.7]. And, thus (for proportional magnitudes), as
ἡγουμένα πρὸς ἅπαντα τὰ ἑπόμενα· ἔστιν ἄρα ὡς τὸ ΑΗ one of the leading (magnitudes) will be to one of the fol-
πρὸς τὸ ΔΚ, οὕτως τὸ ΑΒ πρὸς τὸ ΔΕ. ἴσον δὲ τὸ μὲν ΑΗ lowing, so all of the leading (magnitudes will be) to all of
τῷ Γ, τὸ δὲ ΔΚ τῷ Ζ· ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ζ οὕτως the following [Prop. 5.12]. Thus, as AG is to DK, so AB
τὸ ΑΒ πρὸς τὸ ΔΕ. (is) to DE. And AG is equal to C, and DK to F . Thus,
Τὰ ἄρα μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν as C is to F , so AB (is) to DE.

ἔχει λόγον ληφθέντα κατάλληλα· ὅπερ ἔδει δεῖξαι. Thus, parts have the same ratio as similar multiples,
taken in corresponding order. (Which is) the very thing
it was required to show.

† In modern notation, this proposition reads that α : β :: m α : m β.i�þ. Proposition 16†
᾿Εὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, καὶ ἐναλλὰξ ἀνάλογον If four magnitudes are proportional then they will also

ἔσται. be proportional alternately.
῎Εστω τέσσαρα μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, ὡς τὸ Let A, B, C and D be four proportional magnitudes,

Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ· λέγω, ὅτι καὶ ἐναλλὰξ (such that) as A (is) to B, so C (is) to D. I say that they
[ἀνάλογον] ἔσται, ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Β πρὸς τὸ will also be [proportional] alternately, (so that) as A (is)
Δ. to C, so B (is) to D.
Εἰλήφθω γὰρ τῶν μὲν Α, Β ἰσάκις πολλαπλάσια τὰ Ε, For let the equal multiples E and F have been taken

Ζ, τῶν δὲ Γ, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Η, of A and B (respectively), and the other random equal
Θ. multiples G and H of C and D (respectively).

145

STOIQEIWN eþ. ELEMENTS BOOK 5
Θ

∆

Καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Ε τοῦ Α καὶ τὸ Ζ And since E and F are equal multiples of A and B
τοῦ Β, τὰ δὲ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν (respectively), and parts have the same ratio as similar
ἔχει λόγον, ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς multiples [Prop. 5.15], thus as A is to B, so E (is) to F .
τὸ Ζ. ὡς δὲ τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ· καὶ ὡς But as A (is) to B, so C (is) to D. And, thus, as C (is)
ἄρα τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ. πάλιν, ἐπεὶ τὰ to D, so E (is) to F [Prop. 5.11]. Again, since G and H
Η, Θ τῶν Γ, Δ ἰσάκις ἐστὶ πολλαπλάσια, ἔστιν ἄρα ὡς τὸ Γ are equal multiples of C and D (respectively), thus as C
πρὸς τὸ Δ, οὕτως τὸ Η πρὸς τὸ Θ. ὡς δὲ τὸ Γ πρὸς τὸ Δ, is to D, so G (is) to H [Prop. 5.15]. But as C (is) to D,
[οὕτως] τὸ Ε πρὸς τὸ Ζ· καὶ ὡς ἄρα τὸ Ε πρὸς τὸ Ζ, οὕτως [so] E (is) to F . And, thus, as E (is) to F , so G (is) to
τὸ Η πρὸς τὸ Θ. ἐὰν δὲ τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ δὲ H [Prop. 5.11]. And if four magnitudes are proportional,
πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου and the first is greater than the third then the second will
μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἄν ἔλαττον, ἔλαττον. εἰ ἄρα also be greater than the fourth, and if (the first is) equal
ὑπερέχει τὸ Ε τοῦ Η, ὑπερέχει καὶ τὸ Ζ τοῦ Θ, καὶ εἰ ἴσον, (to the third then the second will also be) equal (to the
ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Ε, Ζ τῶν fourth), and if (the first is) less (than the third then the
Α, Β ἰσάκις πολλαπλάσια, τὰ δὲ Η, Θ τῶν Γ, Δ ἄλλα, ἃ second will also be) less (than the fourth) [Prop. 5.14].
ἔτυχεν, ἰσάκις πολλαπλάσια· ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, Thus, if E exceeds G then F also exceeds H , and if (E is)
οὕτως τὸ Β πρὸς τὸ Δ. equal (to G then F is also) equal (to H), and if (E is) less
᾿Εὰν ἄρα τέσσαρα μεγέθη ἀνάλογον ᾖ, καὶ ἐναλλὰξ (than G then F is also) less (than H). And E and F are

ἀνάλογον ἔσται· ὅπερ ἔδει δεῖξαι. equal multiples of A and B (respectively), and G and H
other random equal multiples of C and D (respectively).

Thus, as A is to C, so B (is) to D [Def. 5.5].
Thus, if four magnitudes are proportional then they

will also be proportional alternately. (Which is) the very

thing it was required to show.

† In modern notation, this proposition reads that if α : β :: γ : δ then α : γ :: β : δ.izþ. Proposition 17†
᾿Εὰν συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα If composed magnitudes are proportional then they

ἀνάλογον ἔσται. will also be proportional (when) separarted.

ΓΑ Β ∆

Η Θ Κ Ξ

ΖΕ

Μ Ν ΠΛ

A E B F D

PNM

῎Εστω συγκείμενα μεγέθη ἀνάλογον τὰ ΑΒ, ΒΕ, ΓΔ, Let AB, BE, CD, and DF be composed magnitudes
ΔΖ, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ· (which are) proportional, (so that) as AB (is) to BE, so
λέγω, ὅτι καὶ διαιρεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΕ πρὸς CD (is) to DF . I say that they will also be proportional
τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΔΖ. (when) separated, (so that) as AE (is) to EB, so CF (is)
Εἰλήφθω γὰρ τῶν μὲν ΑΕ, ΕΒ, ΓΖ, ΖΔ ἰσάκις πολ- to DF .

λαπλάσια τὰ ΗΘ, ΘΚ, ΛΜ, ΜΝ, τῶν δὲ ΕΒ, ΖΔ ἄλλα, ἃ For let the equal multiples GH , HK, LM , and MN
ἔτυχεν, ἰσάκις πολλαπλάσια τὰ ΚΞ, ΝΠ. have been taken of AE, EB, CF , and FD (respectively),
Καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ and the other random equal multiples KO and NP of

τὸ ΘΚ τοῦ ΕΒ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ EB and FD (respectively).

146

STOIQEIWN eþ. ELEMENTS BOOK 5
ΑΕ καὶ τὸ ΗΚ τοῦ ΑΒ. ἰσάκις δέ ἐστι πολλαπλάσιον τὸ ΗΘ And since GH and HK are equal multiples of AE and
τοῦ ΑΕ καὶ τὸ ΛΜ τοῦ ΓΖ· ἰσάκις ἄρα ἐστὶ πολλαπλάσιον EB (respectively), GH and GK are thus equal multiples
τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΜ τοῦ ΓΖ. πάλιν, ἐπεὶ ἰσάκις ἐστὶ of AE and AB (respectively) [Prop. 5.1]. But GH and
πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΜΝ τοῦ ΖΔ, ἰσάκις ἄρα LM are equal multiples of AE and CF (respectively).
ἐστὶ πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΛΝ τοῦ ΓΔ. ἰσάκις Thus, GK and LM are equal multiples of AB and CF
δὲ ἦν πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΗΚ τοῦ ΑΒ· (respectively). Again, since LM and MN are equal mul-
ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΝ τοῦ tiples of CF and FD (respectively), LM and LN are thus
ΓΔ. τὰ ΗΚ, ΛΝ ἄρα τῶν ΑΒ, ΓΔ ἰσάκις ἐστὶ πολλαπλάσια. equal multiples of CF and CD (respectively) [Prop. 5.1].
πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλασίον τὸ ΘΚ τοῦ ΕΒ καὶ τὸ And LM and GK were equal multiples of CF and AB
ΜΝ τοῦ ΖΔ, ἔστι δὲ καὶ τὸ ΚΞ τοῦ ΕΒ ἰσάκις πολλαπλάσιον (respectively). Thus, GK and LN are equal multiples
καὶ τὸ ΝΠ τοῦ ΖΔ, καὶ συντεθὲν τὸ ΘΞ τοῦ ΕΒ ἰσάκις ἐστὶ of AB and CD (respectively). Thus, GK, LN are equal
πολλαπλάσιον καὶ τὸ ΜΠ τοῦ ΖΔ. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒ multiples of AB, CD. Again, since HK and MN are
πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ, καὶ εἴληπται τῶν equal multiples of EB and FD (respectively), and KO
μὲν ΑΒ, ΓΔ ἰσάκις πολλαπλάσια τὰ ΗΚ, ΛΝ, τῶν δὲ ΕΒ, and NP are also equal multiples of EB and FD (respec-
ΖΔ ἰσάκις πολλαπλάσια τὰ ΘΞ, ΜΠ, εἰ ἄρα ὑπερέχει τὸ tively), then, added together, HO and MP are also equal
ΗΚ τοῦ ΘΞ, ὑπερέχει καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ εἰ ἴσον, ἴσον, multiples of EB and FD (respectively) [Prop. 5.2]. And
καὶ εἰ ἔλαττον, ἔλαττον. ὑπερεχέτω δὴ τὸ ΗΚ τοῦ ΘΞ, since as AB (is) to BE, so CD (is) to DF , and the equal
καὶ κοινοῦ ἀφαιρεθέντος τοῦ ΘΚ ὑπερέχει ἄρα καὶ τὸ ΗΘ multiples GK, LN have been taken of AB, CD, and the
τοῦ ΚΞ. ἀλλα εἰ ὑπερεῖχε τὸ ΗΚ τοῦ ΘΞ ὑπερεῖχε καὶ τὸ equal multiples HO, MP of EB, FD, thus if GK exceeds
ΛΝ τοῦ ΜΠ· ὑπερέχει ἄρα καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ κοινοῦ HO then LN also exceeds MP , and if (GK is) equal (to
ἀφαιρεθέντος τοῦ ΜΝ ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ· ὥστε HO then LN is also) equal (to MP ), and if (GK is) less
εἰ ὑπερέχει τὸ ΗΘ τοῦ ΚΞ, ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ. (than HO then LN is also) less (than MP ) [Def. 5.5].
ὁμοίως δὴ δεῖξομεν, ὅτι κἂν ἴσον ᾖ τὸ ΗΘ τῷ ΚΞ, ἴσον So let GK exceed HO, and thus, HK being taken away
ἔσται καὶ τὸ ΛΜ τῷ ΝΠ, κἂν ἔλαττον, ἔλαττον. καί ἐστι τὰ from both, GH exceeds KO. But (we saw that) if GK
μὲν ΗΘ, ΛΜ τῶν ΑΕ, ΓΖ ἰσάκις πολλαπλάσια, τὰ δὲ ΚΞ, was exceeding HO then LN was also exceeding MP .
ΝΠ τῶν ΕΒ, ΖΔ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· ἔστιν Thus, LN also exceeds MP , and, MN being taken away
ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. from both, LM also exceeds NP . Hence, if GH exceeds
᾿Εὰν ἄρα συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαι- KO then LM also exceeds NP . So, similarly, we can

ρεθέντα ἀνάλογον ἔσται· ὅπερ ἔδει δεῖξαι. show that even if GH is equal to KO then LM will also
be equal to NP , and even if (GH is) less (than KO then

LM will also be) less (than NP ). And GH , LM are equal

multiples of AE, CF , and KO, NP other random equal
multiples of EB, FD. Thus, as AE is to EB, so CF (is)

to FD [Def. 5.5].

Thus, if composed magnitudes are proportional then
they will also be proportional (when) separarted. (Which

is) the very thing it was required to show.

† In modern notation, this proposition reads that if α + β : β :: γ + δ : δ then α : β :: γ : δ.ihþ. Proposition 18†
᾿Εὰν διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα If separated magnitudes are proportional then they

ἀνάλογον ἔσται. will also be proportional (when) composed.

Α Β

∆

Ζ Η C

A B

DGF

῎Εστω διῃρημένα μεγέθη ἀνάλογον τὰ ΑΕ, ΕΒ, ΓΖ, ΖΔ, Let AE, EB, CF , and FD be separated magnitudes
ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ· λέγω, (which are) proportional, (so that) as AE (is) to EB, so
ὅτι καὶ συντεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, CF (is) to FD. I say that they will also be proportional

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οὕτως τὸ ΓΔ πρὸς τὸ ΖΔ. (when) composed, (so that) as AB (is) to BE, so CD (is)
Εἰ γὰρ μή ἐστὶν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ to FD.

πρὸς τὸ ΔΖ, ἔσται ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ For if (it is) not (the case that) as AB is to BE, so
ἤτοι πρὸς ἔλασσόν τι τοῦ ΔΖ ἢ πρὸς μεῖζον. CD (is) to FD, then it will surely be (the case that) as
῎Εστω πρότερον πρὸς ἔλασσον τὸ ΔΗ. καὶ ἐπεί ἐστιν ὡς AB (is) to BE, so CD is either to some (magnitude) less

τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΗ, συγκείμενα than DF , or (some magnitude) greater (than DF ).‡

μεγέθη ἀνάλογόν ἐστιν· ὥστε καὶ διαιρεθέντα ἀνάλογον Let it, first of all, be to (some magnitude) less (than
ἔσται. ἔστιν ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΗ πρὸς DF ), (namely) DG. And since composed magnitudes
τὸ ΗΔ. ὑπόκειται δὲ καὶ ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ are proportional, (so that) as AB is to BE, so CD (is) to
ΓΖ πρὸς τὸ ΖΔ. καὶ ὡς ἄρα τὸ ΓΗ πρὸς τὸ ΗΔ, οὕτως τὸ DG, they will thus also be proportional (when) separated
ΓΖ πρὸς τὸ ΖΔ. μεῖζον δὲ τὸ πρῶτον τὸ ΓΗ τοῦ τρίτου τοῦ [Prop. 5.17]. Thus, as AE is to EB, so CG (is) to GD.
ΓΖ· μεῖζον ἄρα καὶ τὸ δεύτερον τὸ ΗΔ τοῦ τετάρτου τοῦ But it was also assumed that as AE (is) to EB, so CF
ΖΔ. ἀλλὰ καὶ ἔλαττον· ὅπερ ἐστὶν ἀδύνατον· οὐκ ἄρα ἐστὶν (is) to FD. Thus, (it is) also (the case that) as CG (is)
ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς ἔλασσον τοῦ to GD, so CF (is) to FD [Prop. 5.11]. And the first
ΖΔ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ πρὸς μεῖζον· πρὸς αὐτὸ (magnitude) CG (is) greater than the third CF . Thus,
ἄρα. the second (magnitude) GD (is) also greater than the
᾿Εὰν ἄρα διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα fourth FD [Prop. 5.14]. But (it is) also less. The very

ἀνάλογον ἔσται· ὅπερ ἔδει δεῖξαι. thing is impossible. Thus, (it is) not (the case that) as AB
is to BE, so CD (is) to less than FD. Similarly, we can
show that neither (is it the case) to greater (than FD).

Thus, (it is the case) to the same (as FD).

Thus, if separated magnitudes are proportional then
they will also be proportional (when) composed. (Which

is) the very thing it was required to show.

† In modern notation, this proposition reads that if α : β :: γ : δ then α + β : β :: γ + δ : δ.
‡ Here, Euclid assumes, without proof, that a fourth magnitude proportional to three given magnitudes can always be found.ijþ. Proposition 19†
᾿Εὰν ᾖ ὡς ὅλον πρὸς ὅλον, οὕτως ἀφαιρεθὲν πρὸς ἀφαι- If as the whole is to the whole so the (part) taken

ρεθέν, καὶ τὸ λοιπὸν πρὸς τὸ λοιπὸν ἔσται ὡς ὅλον πρὸς away is to the (part) taken away then the remainder to
ὅλον. the remainder will also be as the whole (is) to the whole.

ΒΑ

Ζ ∆ D

῎Εστω γὰρ ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ, οὕτως For let the whole AB be to the whole CD as the (part)
ἀφαιρεθὲν τὸ ΑΕ πρὸς ἀφειρεθὲν τὸ ΓΖ· λέγω, ὅτι καὶ taken away AE (is) to the (part) taken away CF . I say
λοιπὸν τὸ ΕΒ πρὸς λοιπὸν τὸ ΖΔ ἔσται ὡς ὅλον τὸ ΑΒ that the remainder EB to the remainder FD will also be
πρὸς ὅλον τὸ ΓΔ. as the whole AB (is) to the whole CD.
᾿Επεὶ γάρ ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ ΑΕ For since as AB is to CD, so AE (is) to CF , (it is)

πρὸς τὸ ΓΖ, καὶ ἐναλλὰξ ὡς τὸ ΒΑ πρὸς τὸ ΑΕ, οὕτως also (the case), alternately, (that) as BA (is) to AE, so
τὸ ΔΓ πρὸς τὸ ΓΖ. καὶ ἐπεὶ συγκείμενα μεγέθη ἀνάλογόν DC (is) to CF [Prop. 5.16]. And since composed magni-
ἐστιν, καὶ διαιρεθέντα ἀνάλογον ἔσται, ὡς τὸ ΒΕ πρὸς τὸ tudes are proportional then they will also be proportional
ΕΑ, οὕτως τὸ ΔΖ πρὸς τὸ ΓΖ· καὶ ἐναλλάξ, ὡς τὸ ΒΕ πρὸς (when) separated, (so that) as BE (is) to EA, so DF (is)
τὸ ΔΖ, οὕτως τὸ ΕΑ πρὸς τὸ ΖΓ. ὡς δὲ τὸ ΑΕ πρὸς τὸ ΓΖ, to CF [Prop. 5.17]. Also, alternately, as BE (is) to DF ,
οὕτως ὑπόκειται ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ. καὶ λοιπὸν so EA (is) to FC [Prop. 5.16]. And it was assumed that
ἄρα τὸ ΕΒ πρὸς λοιπὸν τὸ ΖΔ ἔσται ὡς ὅλον τὸ ΑΒ πρὸς as AE (is) to CF , so the whole AB (is) to the whole CD.
ὅλον τὸ ΓΔ. And, thus, as the remainder EB (is) to the remainder
᾿Εὰν ἄρα ᾖ ὡς ὅλον πρὸς ὅλον, οὕτως ἀφαιρεθὲν πρὸς FD, so the whole AB will be to the whole CD.

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ἀφαιρεθέν, καὶ τὸ λοιπὸν πρὸς τὸ λοιπὸν ἔσται ὡς ὅλον Thus, if as the whole is to the whole so the (part)
πρὸς ὅλον [ὅπερ ἔδει δεῖξαι]. taken away is to the (part) taken away then the remain-
[Καὶ ἐπεὶ ἐδείχθη ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ ΕΒ der to the remainder will also be as the whole (is) to

πρὸς τὸ ΖΔ, καὶ ἐναλλὰξ ὡς τὸ ΑΒ πρὸς τὸ ΒΕ οὕτως τὸ the whole. [(Which is) the very thing it was required to
ΓΔ πρὸς τὸ ΖΔ, συγκείμενα ἄρα μεγέθη ἀνάλογόν ἐστιν· show.]
ἐδείχθη δὲ ὡς τὸ ΒΑ πρὸς τὸ ΑΕ, οὕτως τὸ ΔΓ πρὸς τὸ [And since it was shown (that) as AB (is) to CD, so
ΓΖ· καί ἐστιν ἀναστρέψαντι]. EB (is) to FD, (it is) also (the case), alternately, (that)

as AB (is) to BE, so CD (is) to FD. Thus, composed
magnitudes are proportional. And it was shown (that)

as BA (is) to AE, so DC (is) to CF . And (the latter) is

converted (from the former).]Pìrisma. Corollary‡
᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν συγκείμενα μεγέθη So (it is) clear, from this, that if composed magni-

ἀνάλογον ᾖ, καὶ ἀναστρέψαντι ἀνάλογον ἔσται· ὅπερ ἔδει tudes are proportional then they will also be proportional
δεῖξαι. (when) converted. (Which is) the very thing it was re-

quired to show.

† In modern notation, this proposition reads that if α : β :: γ : δ then α : β :: α − γ : β − δ.
‡ In modern notation, this corollary reads that if α : β :: γ : δ then α : α − β :: γ : γ − δ.kþ. Proposition 20†
᾿Εὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, If there are three magnitudes, and others of equal

σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγω, δι᾿ ἴσου δὲ τὸ number to them, (being) also in the same ratio taken two
πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου by two, and (if), via equality, the first is greater than the
μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. third then the fourth will also be greater than the sixth.

And if (the first is) equal (to the third then the fourth

will also be) equal (to the sixth). And if (the first is) less

(than the third then the fourth will also be) less (than the
sixth).

∆

A D

῎Εστω τρία μεγέθη τὰ Α, Β, Γ, καὶ ἄλλα αὐτοῖς ἴσα τὸ Let A, B, and C be three magnitudes, and D, E, F
πλῆθος τὰ Δ, Ε, Ζ, σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, other (magnitudes) of equal number to them, (being) in
ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ Ε, ὡς δὲ τὸ Β the same ratio taken two by two, (so that) as A (is) to B,
πρὸς τὸ Γ, οὕτως τὸ Ε πρὸς τὸ Ζ, δι᾿ ἴσου δὲ μεῖζον ἔστω so D (is) to E, and as B (is) to C, so E (is) to F . And let
τὸ Α τοῦ Γ· λέγω, ὅτι καὶ τὸ Δ τοῦ Ζ μεῖζον ἔσται, κἂν A be greater than C, via equality. I say that D will also
ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. be greater than F . And if (A is) equal (to C then D will
᾿Επεὶ γὰρ μεῖζόν ἐστι τὸ Α τοῦ Γ, ἄλλο δέ τι τὸ Β, τὸ δὲ also be) equal (to F ). And if (A is) less (than C then D

μεῖζον πρὸς τὸ αὐτὸ μείζονα λόγον ἔχει ἤπερ τὸ ἔλαττον, will also be) less (than F ).
τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ For since A is greater than C, and B some other (mag-
Β. ἀλλ᾿ ὡς μὲν τὸ Α πρὸς τὸ Β [οὕτως] τὸ Δ πρὸς τὸ Ε, ὡς nitude), and the greater (magnitude) has a greater ratio
δὲ τὸ Γ πρὸς τὸ Β, ἀνάπαλιν οὕτως τὸ Ζ πρὸς τὸ Ε· καὶ τὸ than the lesser to the same (magnitude) [Prop. 5.8], A
Δ ἄρα πρὸς τὸ Ε μείζονα λόγον ἔχει ἤπερ τὸ Ζ πρὸς τὸ Ε. thus has a greater ratio to B than C (has) to B. But as A
τῶν δὲ πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον ἔχον (is) to B, [so] D (is) to E. And, inversely, as C (is) to B,
μεῖζόν ἐστιν. μεῖζον ἄρα τὸ Δ τοῦ Ζ. ὁμοίως δὴ δείξομεν, so F (is) to E [Prop. 5.7 corr.]. Thus, D also has a greater
ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον ἔσται καὶ τὸ Δ τῷ Ζ, κἂν ratio to E than F (has) to E [Prop. 5.13]. And for (mag-

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ἔλαττον, ἔλαττον. nitudes) having a ratio to the same (magnitude), that
᾿Εὰν ἄρα ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, having the greater ratio is greater [Prop. 5.10]. Thus,

σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγω, δι᾿ ἴσου δὲ τὸ D (is) greater than F . Similarly, we can show that even if
πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου A is equal to C then D will also be equal to F , and even
μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον· ὅπερ if (A is) less (than C then D will also be) less (than F ).
ἔδει δεῖξαι. Thus, if there are three magnitudes, and others of

equal number to them, (being) also in the same ratio

taken two by two, and (if), via equality, the first is greater
than the third, then the fourth will also be greater than

the sixth. And if (the first is) equal (to the third then the

fourth will also be) equal (to the sixth). And (if the first
is) less (than the third then the fourth will also be) less

(than the sixth). (Which is) the very thing it was required
to show.

† In modern notation, this proposition reads that if α : β :: δ : ǫ and β : γ :: ǫ : ζ then α T γ as δ T ζ.kaþ. Proposition 21†
᾿Εὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος If there are three magnitudes, and others of equal

σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τετα- number to them, (being) also in the same ratio taken two
ραγμένη αὐτῶν ἡ ἀναλογία, δι᾿ ἴσου δὲ τὸ πρῶτον τοῦ by two, and (if) their proportion (is) perturbed, and (if),
τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, via equality, the first is greater than the third then the
κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. fourth will also be greater than the sixth. And if (the first

is) equal (to the third then the fourth will also be) equal
(to the sixth). And if (the first is) less (than the third then

the fourth will also be) less (than the sixth).

∆ D

῎Εστω τρία μεγέθη τὰ Α, Β, Γ καὶ ἄλλα αὐτοῖς ἴσα τὸ Let A, B, and C be three magnitudes, and D, E, F
πλῆθος τὰ Δ, Ε, Ζ, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ other (magnitudes) of equal number to them, (being) in
λόγῳ, ἔστω δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, ὡς μὲν τὸ the same ratio taken two by two. And let their proportion
Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ, ὡς δὲ τὸ Β πρὸς τὸ Γ, be perturbed, (so that) as A (is) to B, so E (is) to F , and
οὕτως τὸ Δ πρὸς τὸ Ε, δι᾿ ἴσου δὲ τὸ Α τοῦ Γ μεῖζον ἔστω· as B (is) to C, so D (is) to E. And let A be greater than
λέγω, ὅτι καὶ τὸ Δ τοῦ Ζ μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν C, via equality. I say that D will also be greater than F .
ἒλαττον, ἒλαττον. And if (A is) equal (to C then D will also be) equal (to
᾿Επεὶ γὰρ μεῖζόν ἐστι τὸ Α τοῦ Γ, ἄλλο δέ τι τὸ Β, τὸ F ). And if (A is) less (than C then D will also be) less

Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. (than F ).
ἀλλ᾿ ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ, ὡς For since A is greater than C, and B some other (mag-
δὲ τὸ Γ πρὸς τὸ Β, ἀνάπαλιν οὕτως τὸ Ε πρὸς τὸ Δ. καὶ nitude), A thus has a greater ratio to B than C (has) to
τὸ Ε ἄρα πρὸς τὸ Ζ μείζονα λόγον ἔχει ἤπερ τὸ Ε πρὸς τὸ B [Prop. 5.8]. But as A (is) to B, so E (is) to F . And,
Δ. πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλασσόν inversely, as C (is) to B, so E (is) to D [Prop. 5.7 corr.].
ἐστιν· ἔλασσον ἄρα ἐστὶ τὸ Ζ τοῦ Δ· μεῖζον ἄρα ἐστὶ τὸ Δ Thus, E also has a greater ratio to F than E (has) to D
τοῦ Ζ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον [Prop. 5.13]. And that (magnitude) to which the same
ἔσται καὶ τὸ Δ τῷ Ζ, κἂν ἔλαττον, ἔλαττον. (magnitude) has a greater ratio is (the) lesser (magni-
᾿Εὰν ἄρα ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, tude) [Prop. 5.10]. Thus, F is less than D. Thus, D is

σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τετα- greater than F . Similarly, we can show that even if A is
ραγμένη αὐτῶν ἡ ἀναλογία, δι᾿ ἴσου δὲ τὸ πρῶτον τοῦ τρίτου equal to C then D will also be equal to F , and even if (A
μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, κἂν ἴσον, is) less (than C then D will also be) less (than F ).

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ἴσον, κἂν ἔλαττον, ἔλαττον· ὅπερ ἔδει δεῖξαι. Thus, if there are three magnitudes, and others of

equal number to them, (being) also in the same ratio

taken two by two, and (if) their proportion (is) per-
turbed, and (if), via equality, the first is greater than the

third then the fourth will also be greater than the sixth.
And if (the first is) equal (to the third then the fourth

will also be) equal (to the sixth). And if (the first is) less

(than the third then the fourth will also be) less (than the
sixth). (Which is) the very thing it was required to show.

† In modern notation, this proposition reads that if α : β :: ǫ : ζ and β : γ :: δ : ǫ then α T γ as δ T ζ.kbþ. Proposition 22†
᾿Εὰν ᾖ ὁποσαοῦν μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, If there are any number of magnitudes whatsoever,

σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ δι᾿ ἴσου ἐν and (some) other (magnitudes) of equal number to them,
τῷ αὐτῷ λόγῳ ἔσται. (which are) also in the same ratio taken two by two, then

they will also be in the same ratio via equality.

ΓΑ

∆

Μ M

῎Εστω ὁποσαοῦν μεγέθη τὰ Α, Β, Γ καὶ ἄλλα αὐτοῖς ἴσα Let there be any number of magnitudes whatsoever,
τὸ πλῆθος τὰ Δ, Ε, Ζ, σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ A, B, C, and (some) other (magnitudes), D, E, F , of
λόγῳ, ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ Ε, ὡς equal number to them, (which are) in the same ratio
δὲ τὸ Β πρὸς τὸ Γ, οὕτως τὸ Ε πρὸς τὸ Ζ· λέγω, ὅτι καὶ taken two by two, (so that) as A (is) to B, so D (is) to
δι᾿ ἴσου ἐν τῷ αὐτῳ λόγῳ ἔσται. E, and as B (is) to C, so E (is) to F . I say that they will
Εἰλήφθω γὰρ τῶν μὲν Α, Δ ἰσάκις πολλαπλάσια τὰ Η, also be in the same ratio via equality. (That is, as A is to

Θ, τῶν δὲ Β, Ε ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Κ, C, so D is to F .)
Λ, καὶ ἔτι τῶν Γ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ For let the equal multiples G and H have been taken
Μ, Ν. of A and D (respectively), and the other random equal
Καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ multiples K and L of B and E (respectively), and the

Ε, καὶ εἴληπται τῶν μὲν Α, Δ ἰσάκις πολλαπλάσια τὰ Η, Θ, yet other random equal multiples M and N of C and F
τῶν δὲ Β, Ε ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Κ, Λ, (respectively).
ἔστιν ἄρα ὡς τὸ Η πρὸς τὸ Κ, οὕτως τὸ Θ πρὸς τὸ Λ. δὶα And since as A is to B, so D (is) to E, and the equal
τὰ αὐτὰ δὴ καὶ ὡς τὸ Κ πρὸς τὸ Μ, οὕτως τὸ Λ πρὸς τὸ multiples G and H have been taken of A and D (respec-
Ν. ἐπεὶ οὖν τρία μεγέθη ἐστὶ τὰ Η, Κ, Μ, καὶ ἄλλα αὐτοῖς tively), and the other random equal multiples K and L
ἴσα τὸ πλῆθος τὰ Θ, Λ, Ν, σύνδυο λαμβανόμενα καὶ ἐν τῷ of B and E (respectively), thus as G is to K, so H (is) to
αὐτῷ λόγῳ, δι᾿ ἴσου ἄρα, εἰ ὑπερέχει τὸ Η τοῦ Μ, ὑπερέχει L [Prop. 5.4]. And, so, for the same (reasons), as K (is)
καὶ τὸ Θ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. to M , so L (is) to N . Therefore, since G, K, and M are
καί ἐστι τὰ μὲν Η, Θ τῶν Α, Δ ἰσάκις πολλαπλάσια, τὰ δὲ three magnitudes, and H , L, and N other (magnitudes)
Μ, Ν τῶν Γ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια. ἔστιν of equal number to them, (which are) also in the same
ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ζ. ratio taken two by two, thus, via equality, if G exceeds M
᾿Εὰν ἄρα ᾖ ὁποσαοῦν μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ then H also exceeds N , and if (G is) equal (to M then H

πλῆθος, σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, καὶ δι᾿ ἴσου is also) equal (to N), and if (G is) less (than M then H is
ἐν τῷ αὐτῷ λόγῳ ἔσται· ὅπερ ἔδει δεῖξαι. also) less (than N) [Prop. 5.20]. And G and H are equal

multiples of A and D (respectively), and M and N other

random equal multiples of C and F (respectively). Thus,
as A is to C, so D (is) to F [Def. 5.5].

Thus, if there are any number of magnitudes what-

soever, and (some) other (magnitudes) of equal number
to them, (which are) also in the same ratio taken two by

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two, then they will also be in the same ratio via equality.

(Which is) the very thing it was required to show.

† In modern notation, this proposition reads that if α : β :: ǫ : ζ and β : γ :: ζ : η and γ : δ :: η : θ then α : δ :: ǫ : θ.kgþ. Proposition 23†
᾿Εὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος If there are three magnitudes, and others of equal

σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τεταραγμένη number to them, (being) in the same ratio taken two by
αὐτῶν ἡ ἀναλογία, καὶ δι᾿ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται. two, and (if) their proportion is perturbed, then they will

also be in the same ratio via equality.

∆

῎Εστω τρία μεγέθη τὰ Α, Β, Γ καὶ ἄλλα αὐτοῖς ἴσα τὸ Let A, B, and C be three magnitudes, and D, E and F
πλῆθος σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ τὰ Δ, Ε, Ζ, other (magnitudes) of equal number to them, (being) in
ἔστω δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, ὡς μὲν τὸ Α πρὸς the same ratio taken two by two. And let their proportion
τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ, ὡς δὲ τὸ Β πρὸς τὸ Γ, οὕτως be perturbed, (so that) as A (is) to B, so E (is) to F , and
τὸ Δ πρὸς τὸ Ε· λέγω, ὅτι ἐστὶν ὡς τὸ Α πρὸς τὸ Γ, οὕτως as B (is) to C, so D (is) to E. I say that as A is to C, so
τὸ Δ πρὸς τὸ Ζ. D (is) to F .
Εἰλήφθω τῶν μὲν Α, Β, Δ ἰσάκις πολλαπλάσια τὰ Η, Θ, Let the equal multiples G, H , and K have been taken

Κ, τῶν δὲ Γ, Ε, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ of A, B, and D (respectively), and the other random
Λ, Μ, Ν. equal multiples L, M , and N of C, E, and F (respec-
Καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσια τὰ Η, Θ τῶν Α, Β, τὰ tively).

δὲ μέρη τοὶς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον, And since G and H are equal multiples of A and B
ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Η πρὸς τὸ Θ. διὰ (respectively), and parts have the same ratio as similar
τὰ αὐτὰ δὴ καὶ ὡς τὸ Ε πρὸς τὸ Ζ, οὕτως τὸ Μ πρὸς τὸ Ν· multiples [Prop. 5.15], thus as A (is) to B, so G (is) to
καί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ· καὶ ὡς H . And, so, for the same (reasons), as E (is) to F , so M
ἄρα τὸ Η πρὸς τὸ Θ, οὕτως τὸ Μ πρὸς τὸ Ν. καὶ ἐπεί ἐστιν (is) to N . And as A is to B, so E (is) to F . And, thus, as
ὡς τὸ Β πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ε, καὶ ἐναλλὰξ G (is) to H , so M (is) to N [Prop. 5.11]. And since as B
ὡς τὸ Β πρὸς τὸ Δ, οὕτως τὸ Γ πρὸς τὸ Ε. καὶ ἐπεὶ τὰ is to C, so D (is) to E, also, alternately, as B (is) to D, so
Θ, Κ τῶν Β, Δ ἰσάκις ἐστὶ πολλαπλάσια, τὰ δὲ μέρη τοῖς C (is) to E [Prop. 5.16]. And since H and K are equal
ἰσάκις πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς multiples of B and D (respectively), and parts have the
τὸ Β πρὸς τὸ Δ, οὕτως τὸ Θ πρὸς τὸ Κ. ἀλλ᾿ ὡς τὸ Β πρὸς same ratio as similar multiples [Prop. 5.15], thus as B is
τὸ Δ, οὕτως τὸ Γ πρὸς τὸ Ε· καὶ ὡς ἄρα τὸ Θ πρὸς τὸ Κ, to D, so H (is) to K. But, as B (is) to D, so C (is) to
οὕτως τὸ Γ πρὸς τὸ Ε. πάλιν, ἐπεὶ τὰ Λ, Μ τῶν Γ, Ε ἰσάκις E. And, thus, as H (is) to K, so C (is) to E [Prop. 5.11].
ἐστι πολλαπλάσια, ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ε, οὕτως τὸ Again, since L and M are equal multiples of C and E (re-
Λ πρὸς τὸ Μ. ἀλλ᾿ ὡς τὸ Γ πρὸς τὸ Ε, οὕτως τὸ Θ πρὸς spectively), thus as C is to E, so L (is) to M [Prop. 5.15].
τὸ Κ· καὶ ὡς ἄρα τὸ Θ πρὸς τὸ Κ, οὕτως τὸ Λ πρὸς τὸ Μ, But, as C (is) to E, so H (is) to K. And, thus, as H (is)
καὶ ἐναλλὰξ ὡς τὸ Θ πρὸς τὸ Λ, τὸ Κ πρὸς τὸ Μ. ἐδείχθη to K, so L (is) to M [Prop. 5.11]. Also, alternately, as H
δὲ καὶ ὡς τὸ Η πρὸς τὸ Θ, οὕτως τὸ Μ πρὸς τὸ Ν. ἐπεὶ (is) to L, so K (is) to M [Prop. 5.16]. And it was also
οὖν τρία μεγέθη ἐστὶ τὰ Η, Θ, Λ, καὶ ἄλλα αὐτοις ἴσα τὸ shown (that) as G (is) to H , so M (is) to N . Therefore,
πλῆθος τὰ Κ, Μ, Ν σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, since G, H , and L are three magnitudes, and K, M , and
καί ἐστιν αὐτῶν τεταραγμένη ἡ ἀναλογία, δι᾿ ἴσου ἄρα, εἰ N other (magnitudes) of equal number to them, (being)
ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, in the same ratio taken two by two, and their proportion
ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Η, Κ τῶν Α, is perturbed, thus, via equality, if G exceeds L then K
Δ ἰσάκις πολλαπλάσια, τὰ δὲ Λ, Ν τῶν Γ, Ζ. ἔστιν ἄρα ὡς also exceeds N , and if (G is) equal (to L then K is also)
τὸ Α πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ζ. equal (to N), and if (G is) less (than L then K is also)
᾿Εὰν ἄρα ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος less (than N) [Prop. 5.21]. And G and K are equal mul-

σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τεταραγμένη tiples of A and D (respectively), and L and N of C and

152

STOIQEIWN eþ. ELEMENTS BOOK 5
αὐτῶν ἡ ἀναλογία, καὶ δι᾿ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται· ὅπερ F (respectively). Thus, as A (is) to C, so D (is) to F
ἔδει δεῖξαι. [Def. 5.5].

Thus, if there are three magnitudes, and others of
equal number to them, (being) in the same ratio taken

two by two, and (if) their proportion is perturbed, then
they will also be in the same ratio via equality. (Which is)

the very thing it was required to show.

† In modern notation, this proposition reads that if α : β :: ǫ : ζ and β : γ :: δ : ǫ then α : γ :: δ : ζ.kdþ. Proposition 24†
᾿Εὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ If a first (magnitude) has to a second the same ratio

τρίτον πρὸς τέταρτον, ἔχῃ δὲ καὶ πέμπτον πρὸς δεύτερον that third (has) to a fourth, and a fifth (magnitude) also
τὸν αὐτὸν λόγον καὶ ἕκτον πρὸς τέταρτον, καὶ συντεθὲν has to the second the same ratio that a sixth (has) to the
πρῶτον καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν ἕξει λόγον fourth, then the first (magnitude) and the fifth, added
καὶ τρίτον καὶ ἕκτον πρὸς τέταρτον. together, will also have the same ratio to the second that

the third (magnitude) and sixth (added together, have)

to the fourth.

Θ∆

Α Η

Ε E

D H

A G
B

Πρῶτον γὰρ τὸ ΑΒ πρὸς δεύρερον τὸ Γ τὸν αὐτὸν ἐχέτω For let a first (magnitude) AB have the same ratio to
λόγον καὶ τρίτον τὸ ΔΕ πρὸς τέταρτον τὸ Ζ, ἐχέτω δὲ καὶ a second C that a third DE (has) to a fourth F . And let
πέμπτον τὸ ΒΗ πρὸς δεύτερον τὸ Γ τὸν αὐτὸν λόγον καὶ a fifth (magnitude) BG also have the same ratio to the
ἕκτον τὸ ΕΘ πρὸς τέταρτον τὸ Ζ· λέγω, ὅτι καὶ συντεθὲν second C that a sixth EH (has) to the fourth F . I say
πρῶτον καὶ πέμπτον τὸ ΑΗ πρὸς δεύτερον τὸ Γ τὸν αὐτὸν that the first (magnitude) and the fifth, added together,
ἕξει λόγον, καὶ τρίτον καὶ ἕκτον τὸ ΔΘ πρὸς τέταρτον τὸ AG, will also have the same ratio to the second C that the
Ζ. third (magnitude) and the sixth, (added together), DH ,
᾿Επεὶ γάρ ἐστιν ὡς τὸ ΒΗ πρὸς τὸ Γ, οὕτως τὸ ΕΘ πρὸς (has) to the fourth F .

τὸ Ζ, ἀνάπαλιν ἄρα ὡς τὸ Γ πρὸς τὸ ΒΗ, οὕτως τὸ Ζ πρὸς For since as BG is to C, so EH (is) to F , thus, in-
τὸ ΕΘ. ἐπεὶ οὖν ἐστιν ὡς τὸ ΑΒ πρὸς τὸ Γ, οὕτως τὸ ΔΕ versely, as C (is) to BG, so F (is) to EH [Prop. 5.7 corr.].
πρὸς τὸ Ζ, ὡς δὲ τὸ Γ πρὸς τὸ ΒΗ, οὕτως τὸ Ζ πρὸς τὸ Therefore, since as AB is to C, so DE (is) to F , and as C
ΕΘ, δι᾿ ἴσου ἄρα ἐστὶν ὡς τὸ ΑΒ πρὸς τὸ ΒΗ, οὕτως τὸ ΔΕ (is) to BG, so F (is) to EH , thus, via equality, as AB is to
πρὸς τὸ ΕΘ. καὶ ἐπεὶ διῃρημένα μεγέθη ἀνάλογόν ἐστιν, καὶ BG, so DE (is) to EH [Prop. 5.22]. And since separated
συντεθέντα ἀνάλογον ἔσται· ἔστιν ἄρα ὡς τὸ ΑΗ πρὸς τὸ magnitudes are proportional then they will also be pro-
ΗΒ, οὕτως τὸ ΔΘ πρὸς τὸ ΘΕ. ἔστι δὲ καὶ ὡς τὸ ΒΗ πρὸς portional (when) composed [Prop. 5.18]. Thus, as AG is
τὸ Γ, οὕτως τὸ ΕΘ πρὸς τὸ Ζ· δι᾿ ἴσου ἄρα ἐστὶν ὡς τὸ ΑΗ to GB, so DH (is) to HE. And, also, as BG is to C, so
πρὸς τὸ Γ, οὕτως τὸ ΔΘ πρὸς τὸ Ζ. EH (is) to F . Thus, via equality, as AG is to C, so DH
᾿Εὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον (is) to F [Prop. 5.22].

καὶ τρίτον πρὸς τέταρτον, ἔχῃ δὲ καὶ πέμπτον πρὸς δεύτερον Thus, if a first (magnitude) has to a second the same
τὸν αὐτὸν λόγον καὶ ἕκτον πρὸς τέταρτον, καὶ συντεθὲν ratio that a third (has) to a fourth, and a fifth (magni-
πρῶτον καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν ἕξει λόγον tude) also has to the second the same ratio that a sixth
καὶ τρίτον καὶ ἕκτον πρὸς τέταρτον· ὅπερ ἔδει δεὶξαι. (has) to the fourth, then the first (magnitude) and the

fifth, added together, will also have the same ratio to the
second that the third (magnitude) and the sixth (added

153

STOIQEIWN eþ. ELEMENTS BOOK 5
together, have) to the fourth. (Which is) the very thing it

was required to show.

† In modern notation, this proposition reads that if α : β :: γ : δ and ǫ : β :: ζ : δ then α + ǫ : β :: γ + ζ : δ.keþ. Proposition 25†
᾿Εὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ μέγιστον [αὐτῶν] If four magnitudes are proportional then the (sum of

καὶ τὸ ἐλάχιστον δύο τῶν λοιπῶν μείζονά ἐστιν. the) largest and the smallest [of them] is greater than the
(sum of the) remaining two (magnitudes).

Θ
Γ

Α Β

∆

A B

῎Εστω τέσσαρα μεγέθη ἀνάλογον τὰ ΑΒ, ΓΔ, Ε, Ζ, ὡς Let AB, CD, E, and F be four proportional magni-
τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ Ε πρὸς τὸ Ζ, ἔστω δὲ μέγιστον tudes, (such that) as AB (is) to CD, so E (is) to F . And
μὲν αὐτῶν τὸ ΑΒ, ἐλάχιστον δὲ τὸ Ζ· λέγω, ὅτι τὰ ΑΒ, Ζ let AB be the greatest of them, and F the least. I say that
τῶν ΓΔ, Ε μείζονά ἐστιν. AB and F is greater than CD and E.
Κείσθω γὰρ τῷ μὲν Ε ἴσον τὸ ΑΗ, τῷ δὲ Ζ ἴσον τὸ ΓΘ. For let AG be made equal to E, and CH equal to F .
᾿Επεὶ [οὖν] ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ Ε [In fact,] since as AB is to CD, so E (is) to F , and

πρὸς τὸ Ζ, ἴσον δὲ τὸ μὲν Ε τῷ ΑΗ, τὸ δὲ Ζ τῷ ΓΘ, ἔστιν E (is) equal to AG, and F to CH , thus as AB is to CD,
ἄρα ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ ΑΗ πρὸς τὸ ΓΘ. so AG (is) to CH . And since the whole AB is to the
καὶ ἐπεί ἐστιν ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ, οὕτως whole CD as the (part) taken away AG (is) to the (part)
ἀφαιρεθὲν τὸ ΑΗ πρὸς ἀφαιρεθὲν τὸ ΓΘ, καὶ λοιπὸν ἄρα taken away CH , thus the remainder GB will also be to
τὸ ΗΒ πρὸς λοιπὸν τὸ ΘΔ ἔσται ὡς ὅλον τὸ ΑΒ πρὸς ὅλον the remainder HD as the whole AB (is) to the whole CD
τὸ ΓΔ. μεῖζον δὲ τὸ ΑΒ τοῦ ΓΔ· μεῖζον ἄρα καὶ τὸ ΗΒ τοῦ [Prop. 5.19]. And AB (is) greater than CD. Thus, GB
ΘΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ μὲν ΑΗ τῷ Ε, τὸ δὲ ΓΘ τῷ Ζ, (is) also greater than HD. And since AG is equal to E,
τὰ ἄρα ΑΗ, Ζ ἴσα ἐστὶ τοῖς ΓΘ, Ε. καὶ [ἐπεὶ] ἐὰν [ἀνίσοις and CH to F , thus AG and F is equal to CH and E.
ἴσα προστεθῇ, τὰ ὅλα ἄνισά ἐστιν, ἐὰν ἄρα] τῶν ΗΒ, ΘΔ And [since] if [equal (magnitudes) are added to unequal
ἀνίσων ὄντων καὶ μείζονος τοῦ ΗΒ τῷ μὲν ΗΒ προστεθῇ (magnitudes) then the wholes are unequal, thus if] AG
τὰ ΑΗ, Ζ, τῷ δὲ ΘΔ προστεθῇ τὰ ΓΘ, Ε, συνάγεται τὰ and F are added to GB, and CH and E to HD—GB
ΑΒ, Ζ μείζονα τῶν ΓΔ, Ε. and HD being unequal, and GB greater—it is inferred
᾿Εὰν ἄρα τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ μέγιστον that AB and F (is) greater than CD and E.

αὐτῶν καὶ τὸ ἐλάχιστον δύο τῶν λοιπῶν μείζονά ἐστιν. ὅπερ Thus, if four magnitudes are proportional then the
ἔδει δεῖξαι. (sum of the) largest and the smallest of them is greater

than the (sum of the) remaining two (magnitudes).
(Which is) the very thing it was required to show.

† In modern notation, this proposition reads that if α : β :: γ : δ, and α is the greatest and δ the least, then α + δ > β + γ.

154

ELEMENTS BOOK 6

Similar Figures

155

STOIQEIWN �þ. ELEMENTS BOOK 6VOroi. Definitions
αʹ. ῞Ομοια σχήματα εὐθύγραμμά ἐστιν, ὅσα τάς τε 1. Similar rectilinear figures are those (which) have

γωνίας ἴσας ἔχει κατὰ μίαν καὶ τὰς περὶ τὰς ἴσας γωνίας (their) angles separately equal and the (corresponding)
πλευρὰς ἀνάλογον. sides about the equal angles proportional.
βʹ. ῎Ακρον καὶ μέσον λόγον εὐθεῖα τετμῆσθαι λέγεται, 2. A straight-line is said to have been cut in extreme

ὅταν ᾖ ὡς ἡ ὅλη πρὸς τὸ μεῖζον τμῆμα, οὕτως τὸ μεῖζον and mean ratio when as the whole is to the greater seg-
πρὸς τὸ ἔλαττὸν. ment so the greater (segment is) to the lesser.
γʹ. ῞Υψος ἐστὶ πάντος σχήματος ἡ ἀπὸ τῆς κορυφῆς ἐπὶ 3. The height of any figure is the (straight-line) drawn

τὴν βάσιν κάθετος ἀγομένη. from the vertex perpendicular to the base.aþ. Proposition 1†
Τὰ τρίγωνα καὶ τὰ παραλληλόγραμμα τὰ ὑπὸ τὸ αὐτὸ Triangles and parallelograms which are of the same

ὕψος ὄντα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις. height are to one another as their bases.

Θ ΛΚ

Η Β Γ ∆

Α F

H K L

G B C D

῎Εστω τρίγωνα μὲν τὰ ΑΒΓ, ΑΓΔ, παραλληλόγραμμα Let ABC and ACD be triangles, and EC and CF par-
δὲ τὰ ΕΓ, ΓΖ ὑπὸ τὸ αὐτὸ ὕψος τὸ ΑΓ· λέγω, ὅτι ἐστὶν ὡς allelograms, of the same height AC. I say that as base BC
ἡ ΒΓ βάσις πρὸς τὴν ΓΔ βάσις, οὕτως τὸ ΑΒΓ τρίγωνον is to base CD, so triangle ABC (is) to triangle ACD, and
πρὸς τὸ ΑΓΔ τρίγωνον, καὶ τὸ ΕΓ παραλληλόγραμμον πρὸς parallelogram EC to parallelogram CF .
τὸ ΓΖ παραλληλόγραμμον. For let the (straight-line) BD have been produced in
᾿Εκβεβλήσθω γὰρ ἡ ΒΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Θ, Λ each direction to points H and L, and let [any number]

σημεῖα, καὶ κείσθωσαν τῇ μὲν ΒΓ βάσει ἴσαι [ὁσαιδηποτοῦν] (of straight-lines) BG and GH be made equal to base
αἱ ΒΗ, ΗΘ, τῇ δὲ ΓΔ βάσει ἴσαι ὁσαιδηποτοῦν αἱ ΔΚ, ΚΛ, BC, and any number (of straight-lines) DK and KL
καὶ ἐπεζεύχθωσαν αἱ ΑΗ, ΑΘ, ΑΚ, ΑΛ. equal to base CD. And let AG, AH , AK, and AL have
Καὶ ἐπεὶ ἴσαι εἰσὶν αἱ ΓΒ, ΒΗ, ΗΘ ἀλλήλαις, ἴσα ἐστὶ καὶ been joined.

τὰ ΑΘΗ, ΑΗΒ, ΑΒΓ τρίγωνα ἀλλήλοις. ὁσαπλασίων ἄρα And since CB, BG, and GH are equal to one another,
ἐστὶν ἡ ΘΓ βάσις τῆς ΒΓ βάσεως, τοσαυταπλάσιόν ἐστι triangles AHG, AGB, and ABC are also equal to one
καὶ τὸ ΑΘΓ τρίγωνον τοῦ ΑΒΓ τριγώνου. διὰ τὰ αὐτὰ another [Prop. 1.38]. Thus, as many times as base HC
δὴ ὁσαπλασίων ἐστὶν ἡ ΛΓ βάσις τῆς ΓΔ βάσεως, τοσαυ- is (divisible by) base BC, so many times is triangle AHC
ταπλάσιόν ἐστι καὶ τὸ ΑΛΓ τρίγωνον τοῦ ΑΓΔ τριγώνου· also (divisible) by triangle ABC. So, for the same (rea-
καὶ εἰ ἴση ἐστὶν ἡ ΘΓ βάσις τῇ ΓΛ βάσει, ἴσον ἐστὶ καὶ τὸ sons), as many times as base LC is (divisible) by base
ΑΘΓ τρίγωνον τῳ ΑΓΛ τριγώνῳ, καὶ εἰ ὑπερέχει ἡ ΘΓ βάσις CD, so many times is triangle ALC also (divisible) by
τῆς ΓΛ βάσεως, ὑπερέχει καὶ τὸ ΑΘΓ τρίγωνον τοῦ ΑΓΛ triangle ACD. And if base HC is equal to base CL then
τριγώνου, καὶ εἰ ἐλάσσων, ἔλασσον. τεσσάρων δὴ ὄντων triangle AHC is also equal to triangle ACL [Prop. 1.38].
μεγεθῶν δύο μὲν βάσεων τῶν ΒΓ, ΓΔ, δύο δὲ τριγώνων And if base HC exceeds base CL then triangle AHC

τῶν ΑΒΓ, ΑΓΔ εἴληπται ἰσάκις πολλαπλάσια τῆς μὲν ΒΓ also exceeds triangle ACL.‡ And if (HC is) less (than
βάσεως καὶ τοῦ ΑΒΓ τριγώνου ἥ τε ΘΓ βάσις καὶ τὸ ΑΘΓ CL then AHC is also) less (than ACL). So, their being
τρίγωνον, τῆς δὲ ΓΔ βάσεως καὶ τοῦ ΑΔΓ τριγώνου ἄλλα, four magnitudes, two bases, BC and CD, and two trian-

156

STOIQEIWN �þ. ELEMENTS BOOK 6
ἃ ἔτυχεν, ἰσάκις πολλαπλάσια ἥ τε ΛΓ βάσις καὶ τὸ ΑΛΓ gles, ABC and ACD, equal multiples have been taken of
τρίγωνον· καὶ δέδεικται, ὅτι, εἰ ὑπερέχει ἡ ΘΓ βάσις τῆς ΓΛ base BC and triangle ABC—(namely), base HC and tri-
βάσεως, ὑπερέχει καὶ τὸ ΑΘΓ τρίγωνον τοῦ ΑΛΓ τριγώνου, angle AHC—and other random equal multiples of base
καί εἰ ἴση, ἴσον, καὶ εἰ ἔλασσων, ἔλασσον· ἔστιν ἄρα ὡς ἡ CD and triangle ADC—(namely), base LC and triangle
ΒΓ βάσις πρὸς τὴν ΓΔ βάσιν, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς ALC. And it has been shown that if base HC exceeds
τὸ ΑΓΔ τρίγωνον. base CL then triangle AHC also exceeds triangle ALC,
Καὶ ἐπεὶ τοῦ μὲν ΑΒΓ τριγώνου διπλάσιόν ἐστι τὸ ΕΓ and if (HC is) equal (to CL then AHC is also) equal (to

παραλληλόγραμμον, τοῦ δὲ ΑΓΔ τριγώνου διπλάσιόν ἐστι ALC), and if (HC is) less (than CL then AHC is also)
τὸ ΖΓ παραλληλόγραμμον, τὰ δὲ μέρη τοῖς ὡσαύτως πολ- less (than ALC). Thus, as base BC is to base CD, so
λαπλασίοις τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς τὸ ΑΒΓ triangle ABC (is) to triangle ACD [Def. 5.5]. And since
τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, οὕτως τὸ ΕΓ παραλ- parallelogram EC is double triangle ABC, and parallelo-
ληλόγραμμον πρὸς τὸ ΖΓ παραλληλόγραμμον. ἐπεὶ οὖν gram FC is double triangle ACD [Prop. 1.34], and parts
ἐδείχθη, ὡς μὲν ἡ ΒΓ βάσις πρὸς τὴν ΓΔ, οὕτως τὸ ΑΒΓ have the same ratio as similar multiples [Prop. 5.15], thus
τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, ὡς δὲ τὸ ΑΒΓ τρίγωνον as triangle ABC is to triangle ACD, so parallelogram EC
πρὸς τὸ ΑΓΔ τρίγωνον, οὕτως τὸ ΕΓ παραλληλόγραμμον (is) to parallelogram FC. In fact, since it was shown that
πρὸς τὸ ΓΖ παραλληλόγραμμον, καὶ ὡς ἄρα ἡ ΒΓ βάσις πρὸς as base BC (is) to CD, so triangle ABC (is) to triangle
τὴν ΓΔ βάσιν, οὕτως τὸ ΕΓ παραλληλόγραμμον πρὸς τὸ ΖΓ ACD, and as triangle ABC (is) to triangle ACD, so par-
παραλληλόγραμμον. allelogram EC (is) to parallelogram CF , thus, also, as
Τὰ ἄρα τρίγωνα καὶ τὰ παραλληλόγραμμα τὰ ὑπὸ τὸ base BC (is) to base CD, so parallelogram EC (is) to

αὐτὸ ὕψος ὄντα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις· ὅπερ ἔδει parallelogram FC [Prop. 5.11].
δεῖξαι. Thus, triangles and parallelograms which are of the

same height are to one another as their bases. (Which is)

the very thing it was required to show.

† As is easily demonstrated, this proposition holds even when the triangles, or parallelograms, do not share a common side, and/or are not

right-angled.

‡ This is a straight-forward generalization of Prop. 1.38.bþ. Proposition 2
᾿Εὰν τριγώνου παρὰ μίαν τῶν πλευρῶν ἀχθῇ τις εὐθεῖα, If some straight-line is drawn parallel to one of the

ἀνάλογον τεμεῖ τὰς τοῦ τριγώνου πλευράς· καὶ ἐὰν αἱ τοῦ sides of a triangle then it will cut the (other) sides of the
τριγώνου πλευραὶ ἀνάλογον τμηθῶσιν, ἡ ἐπὶ τὰς τομὰς ἐπι- triangle proportionally. And if (two of) the sides of a tri-
ζευγνυμένη εὐθεῖα παρὰ τὴν λοιπὴν ἔσται τοῦ τριγώνου angle are cut proportionally then the straight-line joining
πλευράν. the cutting (points) will be parallel to the remaining side

of the triangle.

∆

Α A

D E

Τριγώνου γὰρ τοῦ ΑΒΓ παράλληλος μιᾷ τῶν πλευρῶν For let DE have been drawn parallel to one of the
τῇ ΒΓ ἤχθω ἡ ΔΕ· λέγω, ὅτι ἐστὶν ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, sides BC of triangle ABC. I say that as BD is to DA, so
οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ. CE (is) to EA.

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STOIQEIWN �þ. ELEMENTS BOOK 6
᾿Επεζεύχθωσαν γὰρ αἱ ΒΕ, ΓΔ. For let BE and CD have been joined.
῎Ισον ἄρα ἐστὶ τὸ ΒΔΕ τρίγωνον τῷ ΓΔΕ τριγώνῳ· Thus, triangle BDE is equal to triangle CDE. For

ἐπὶ γὰρ τῆς αὐτῆς βάσεώς ἐστι τῆς ΔΕ καὶ ἐν ταῖς αὐταῖς they are on the same base DE and between the same
παραλλήλοις ταῖς ΔΕ, ΒΓ· ἄλλο δέ τι τὸ ΑΔΕ τρίγωνον. τὰ parallels DE and BC [Prop. 1.38]. And ADE is some
δὲ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον· ἔστιν ἄρα ὡς τὸ other triangle. And equal (magnitudes) have the same ra-
ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ [τρίγωνον], οὕτως τὸ ΓΔΕ tio to the same (magnitude) [Prop. 5.7]. Thus, as triangle
τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον. αλλ᾿ ὡς μὲν τὸ ΒΔΕ BDE is to [triangle] ADE, so triangle CDE (is) to trian-
τρίγωνον πρὸς τὸ ΑΔΕ, οὕτως ἡ ΒΔ πρὸς τὴν ΔΑ· ὑπὸ gle ADE. But, as triangle BDE (is) to triangle ADE, so
γὰρ τὸ αὐτὸ ὕψος ὄντα τὴν ἀπὸ τοῦ Ε ἐπὶ τὴν ΑΒ κάθετον (is) BD to DA. For, having the same height—(namely),
ἀγομένην πρὸς ἄλληλά εἰσιν ὡς αἱ βάσεις. διὰ τὰ αὐτὰ δὴ the (straight-line) drawn from E perpendicular to AB—
ὡς τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ, οὕτως ἡ ΓΕ πρὸς τὴν they are to one another as their bases [Prop. 6.1]. So, for
ΕΑ· καὶ ὡς ἄρα ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν the same (reasons), as triangle CDE (is) to ADE, so CE
ΕΑ. (is) to EA. And, thus, as BD (is) to DA, so CE (is) to
Ἀλλὰ δὴ αἱ τοῦ ΑΒΓ τριγώνου πλευραὶ αἱ ΑΒ, ΑΓ EA [Prop. 5.11].

ἀνάλογον τετμήσθωσαν, ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως And so, let the sides AB and AC of triangle ABC
ἡ ΓΕ πρὸς τὴν ΕΑ, καὶ ἐπεζεύχθω ἡ ΔΕ· λέγω, ὅτι have been cut proportionally (such that) as BD (is) to
παράλληλός ἐστιν ἡ ΔΕ τῇ ΒΓ. DA, so CE (is) to EA. And let DE have been joined. I
Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ say that DE is parallel to BC.

ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ, ἀλλ᾿ ὡς For, by the same construction, since as BD is to DA,
μὲν ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως τὸ ΒΔΕ τρίγωνον πρὸς so CE (is) to EA, but as BD (is) to DA, so triangle BDE
τὸ ΑΔΕ τρίγωνον, ὡς δὲ ἡ ΓΕ πρὸς τὴν ΕΑ, οὕτως τὸ (is) to triangle ADE, and as CE (is) to EA, so triangle
ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, καὶ ὡς ἄρα τὸ CDE (is) to triangle ADE [Prop. 6.1], thus, also, as tri-
ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, οὕτως τὸ ΓΔΕ angle BDE (is) to triangle ADE, so triangle CDE (is)
τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον. ἑκάτερον ἄρα τῶν ΒΔΕ, to triangle ADE [Prop. 5.11]. Thus, triangles BDE and
ΓΔΕ τριγώνων πρὸς τὸ ΑΔΕ τὸν αὐτὸν ἔχει λόγον. ἴσον CDE each have the same ratio to ADE. Thus, triangle
ἄρα ἐστὶ τὸ ΒΔΕ τρίγωνον τῷ ΓΔΕ τριγώνῳ· καί εἰσιν ἐπὶ BDE is equal to triangle CDE [Prop. 5.9]. And they are
τῆς αὐτῆς βάσεως τῆς ΔΕ. τὰ δὲ ἴσα τρίγωνα καὶ ἐπὶ τῆς on the same base DE. And equal triangles, which are
αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν. also on the same base, are also between the same paral-
παράλληλος ἄρα ἐστὶν ἡ ΔΕ τῇ ΒΓ. lels [Prop. 1.39]. Thus, DE is parallel to BC.
᾿Εὰν ἄρα τριγώνου παρὰ μίαν τῶν πλευρῶν ἀχθῇ τις Thus, if some straight-line is drawn parallel to one of

εὐθεῖα, ἀνάλογον τεμεῖ τὰς τοῦ τριγώνου πλευράς· καὶ ἐὰν the sides of a triangle, then it will cut the (other) sides
αἱ τοῦ τριγώνου πλευραὶ ἀνάλογον τμηθῶσιν, ἡ ἐπὶ τὰς of the triangle proportionally. And if (two of) the sides
τομὰς ἐπιζευγνυμένη εὐθεῖα παρὰ τὴν λοιπὴν ἔσται τοῦ of a triangle are cut proportionally, then the straight-line
τριγώνου πλευράν· ὅπερ ἔδει δεῖξαι. joining the cutting (points) will be parallel to the remain-

ing side of the triangle. (Which is) the very thing it was
required to show.gþ. Proposition 3

᾿Εὰν τριγώνου ἡ γωνία δίχα τμηθῇ, ἡ δὲ τέμνουσα τὴν If an angle of a triangle is cut in half, and the straight-
γωνίαν εὐθεῖα τέμνῃ καὶ τὴν βάσιν, τὰ τῆς βάσεως τμήματα line cutting the angle also cuts the base, then the seg-
τὸν αὐτὸν ἕξει λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς· ments of the base will have the same ratio as the remain-
καὶ ἐὰν τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἔχῃ λόγον ταῖς ing sides of the triangle. And if the segments of the base
λοιπαῖς τοῦ τριγώνου πλευραῖς, ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν have the same ratio as the remaining sides of the trian-
τομὴν ἐπιζευγνυμένη εὐθεῖα δίχα τεμεῖ τὴν τοῦ τριγώνου gle, then the straight-line joining the vertex to the cutting
γωνίαν. (point) will cut the angle of the triangle in half.
῎Εστω τρίγωνον τὸ ΑΒΓ, καὶ τετμήσθω ἡ ὑπὸ ΒΑΓ Let ABC be a triangle. And let the angle BAC have

γωνία δίχα ὑπὸ τῆς ΑΔ εὐθείας· λέγω, ὅτι ἐστὶν ὡς ἡ ΒΔ been cut in half by the straight-line AD. I say that as BD
πρὸς τὴν ΓΔ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ. is to CD, so BA (is) to AC.
῎Ηχθω γὰρ διὰ τοῦ Γ τῇ ΔΑ παράλληλος ἡ ΓΕ, καὶ For let CE have been drawn through (point) C par-

διαχθεῖσα ἡ ΒΑ συμπιπτέτω αὐτῇ κατὰ τὸ Ε. allel to DA. And, BA being drawn through, let it meet

158

STOIQEIWN �þ. ELEMENTS BOOK 6
(CE) at (point) E.†

Β ∆ Γ

B D C

Καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΔ, ΕΓ εὐθεῖα ἐνέπεσεν And since the straight-line AC falls across the parallel
ἡ ΑΓ, ἡ ἄρα ὑπὸ ΑΓΕ γωνία ἴση ἐστὶ τῇ ὑπὸ ΓΑΔ. ἀλλ᾿ (straight-lines) AD and EC, angle ACE is thus equal to
ἡ ὑπὸ ΓΑΔ τῇ ὑπὸ ΒΑΔ ὑπόκειται ἴση· καὶ ἡ ὑπὸ ΒΑΔ CAD [Prop. 1.29]. But, (angle) CAD is assumed (to
ἄρα τῇ ὑπὸ ΑΓΕ ἐστιν ἴση. πάλιν, ἐπεὶ εἰς παραλλήλους be) equal to BAD. Thus, (angle) BAD is also equal to
τὰς ΑΔ, ΕΓ εὐθεῖα ἐνέπεσεν ἡ ΒΑΕ, ἡ ἐκτὸς γωνία ἡ ὑπὸ ACE. Again, since the straight-line BAE falls across the
ΒΑΔ ἴση ἐστὶ τῇ ἐντὸς τῇ ὑπὸ ΑΕΓ. ἐδείχθη δὲ καὶ ἡ ὑπὸ parallel (straight-lines) AD and EC, the external angle
ΑΓΕ τῇ ὑπὸ ΒΑΔ ἴση· καὶ ἡ ὑπὸ ΑΓΕ ἄρα γωνία τῇ ὑπὸ BAD is equal to the internal (angle) AEC [Prop. 1.29].
ΑΕΓ ἐστιν ἴση· ὥστε καὶ πλευρὰ ἡ ΑΕ πλευρᾷ τῇ ΑΓ ἐστιν And (angle) ACE was also shown (to be) equal to BAD.
ἴση. καὶ ἐπεὶ τριγώνου τοῦ ΒΓΕ παρὰ μίαν τῶν πλευρῶν Thus, angle ACE is also equal to AEC. And, hence, side
τὴν ΕΓ ἦκται ἡ ΑΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΔ πρὸς τὴν AE is equal to side AC [Prop. 1.6]. And since AD has
ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΕ. ἴση δὲ ἡ ΑΕ τῇ ΑΓ· ὡς ἄρα been drawn parallel to one of the sides EC of triangle
ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ. BCE, thus, proportionally, as BD is to DC, so BA (is)
Ἀλλὰ δὴ ἔστω ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς to AE [Prop. 6.2]. And AE (is) equal to AC. Thus, as

τὴν ΑΓ, καὶ ἐπεζεύχθω ἡ ΑΔ· λέγω, ὅτι δίχα τέτμηται ἡ BD (is) to DC, so BA (is) to AC.
ὑπὸ ΒΑΓ γωνία ὑπὸ τῆς ΑΔ εὐθείας. And so, let BD be to DC, as BA (is) to AC. And let
Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ ΒΔ AD have been joined. I say that angle BAC has been cut

πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ, ἀλλὰ καὶ ὡς ἡ ΒΔ in half by the straight-line AD.
πρὸς τὴν ΔΓ, οὕτως ἐστὶν ἡ ΒΑ πρὸς τὴν ΑΕ· τριγώνου For, by the same construction, since as BD is to DC,
γὰρ τοῦ ΒΓΕ παρὰ μίαν τὴν ΕΓ ἦκται ἡ ΑΔ· καὶ ὡς ἄρα ἡ so BA (is) to AC, then also as BD (is) to DC, so BA is
ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΕ. ἴση ἄρα ἡ ΑΓ to AE. For AD has been drawn parallel to one (of the
τῇ ΑΕ· ὥστε καὶ γωνία ἡ ὑπὸ ΑΕΓ τῇ ὑπὸ ΑΓΕ ἐστιν ἴση. sides) EC of triangle BCE [Prop. 6.2]. Thus, also, as
ἀλλ᾿ ἡ μὲν ὑπὸ ΑΕΓ τῇ ἐκτὸς τῇ ὑπὸ ΒΑΔ [ἐστιν] ἴση, ἡ BA (is) to AC, so BA (is) to AE [Prop. 5.11]. Thus, AC
δὲ ὑπὸ ΑΓΕ τῇ ἐναλλὰξ τῇ ὑπὸ ΓΑΔ ἐστιν ἴση· καὶ ἡ ὑπὸ (is) equal to AE [Prop. 5.9]. And, hence, angle AEC
ΒΑΔ ἄρα τῇ ὑπὸ ΓΑΔ ἐστιν ἴση. ἡ ἄρα ὑπὸ ΒΑΓ γωνία is equal to ACE [Prop. 1.5]. But, AEC [is] equal to the
δίχα τέτμηται ὑπὸ τῆς ΑΔ εὐθείας. external (angle) BAD, and ACE is equal to the alternate
᾿Εὰν ἄρα τριγώνου ἡ γωνία δίχα τμηθῇ, ἡ δὲ τέμνουσα (angle) CAD [Prop. 1.29]. Thus, (angle) BAD is also

τὴν γωνίαν εὐθεῖα τέμνῃ καὶ τὴν βάσιν, τὰ τῆς βάσεως equal to CAD. Thus, angle BAC has been cut in half by
τμήματα τὸν αὐτὸν ἕξει λόγον ταῖς λοιπαῖς τοῦ τριγώνου the straight-line AD.
πλευραῖς· καὶ ἐὰν τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἔχῃ Thus, if an angle of a triangle is cut in half, and the
λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς, ἡ ἀπὸ τῆς κο- straight-line cutting the angle also cuts the base, then the
ρυφῆς ἐπὶ τὴν τομὴν ἐπιζευγνυμένη εὐθεῖα δίχα τέμνει τὴν segments of the base will have the same ratio as the re-
τοῦ τριγώνου γωνίαν· ὅπερ ἔδει δεῖξαι. maining sides of the triangle. And if the segments of the

base have the same ratio as the remaining sides of the

triangle, then the straight-line joining the vertex to the
cutting (point) will cut the angle of the triangle in half.

(Which is) the very thing it was required to show.

† The fact that the two straight-lines meet follows because the sum of ACE and CAE is less than two right-angles, as can easily be demonstrated.

159

STOIQEIWN �þ. ELEMENTS BOOK 6
See Post. 5. dþ. Proposition 4
Τῶν ἰσογωνίων τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ In equiangular triangles the sides about the equal an-

περὶ τὰς ἴσας γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας gles are proportional, and those (sides) subtending equal
ὑποτείνουσαι. angles correspond.

∆

Β Γ Ε

CB E

῎Εστω ἰσογώνια τρίγωνα τὰ ΑΒΓ, ΔΓΕ ἴσην ἔχοντα τὴν Let ABC and DCE be equiangular triangles, having
μὲν ὑπὸ ΑΒΓ γωνίαν τῇ ὑπὸ ΔΓΕ, τὴν δὲ ὑπὸ ΒΑΓ τῇ ὑπὸ angle ABC equal to DCE, and (angle) BAC to CDE,
ΓΔΕ καὶ ἔτι τὴν ὑπὸ ΑΓΒ τῇ ὑπὸ ΓΕΔ· λέγω, ὅτι τῶν ΑΒΓ, and, further, (angle) ACB to CED. I say that in trian-
ΔΓΕ τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας gles ABC and DCE the sides about the equal angles are
γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι. proportional, and those (sides) subtending equal angles
Κείσθω γὰρ ἐπ᾿ εὐθείας ἡ ΒΓ τῇ ΓΕ. καὶ ἐπεὶ αἱ ὑπὸ correspond.

ΑΒΓ, ΑΓΒ γωνίαι δύο ὀρθῶν ἐλάττονές εἰσιν, ἴση δὲ Let BC be placed straight-on to CE. And since
ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΕΓ, αἱ ἄρα ὑπὸ ΑΒΓ, ΔΕΓ δύο angles ABC and ACB are less than two right-angles
ὀρθῶν ἐλάττονές εἰσιν· αἱ ΒΑ, ΕΔ ἄρα ἐκβαλλόμεναι συμ- [Prop 1.17], and ACB (is) equal to DEC, thus ABC
πεσοῦνται. ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν κατὰ τὸ Ζ. and DEC are less than two right-angles. Thus, BA and
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΓΕ γωνία τῇ ὑπὸ ΑΒΓ, ED, being produced, will meet [C.N. 5]. Let them have

παράλληλός ἐστιν ἡ ΒΖ τῇ ΓΔ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ been produced, and let them meet at (point) F .
ΑΓΒ τῇ ὑπὸ ΔΕΓ, παράλληλός ἐστιν ἡ ΑΓ τῇ ΖΕ. παραλ- And since angle DCE is equal to ABC, BF is parallel
ληλόγραμμον ἄρα ἐστὶ τὸ ΖΑΓΔ· ἴση ἄρα ἡ μὲν ΖΑ τῇ ΔΓ, to CD [Prop. 1.28]. Again, since (angle) ACB is equal to
ἡ δὲ ΑΓ τῇ ΖΔ. καὶ ἐπεὶ τριγώνου τοῦ ΖΒΕ παρὰ μίαν τὴν DEC, AC is parallel to FE [Prop. 1.28]. Thus, FACD
ΖΕ ἦκται ἡ ΑΓ, ἐστιν ἄρα ὡς ἡ ΒΑ πρὸς τὴν ΑΖ, οὕτως ἡ is a parallelogram. Thus, FA is equal to DC, and AC to
ΒΓ πρὸς τὴν ΓΕ. ἴση δὲ ἡ ΑΖ τῇ ΓΔ· ὡς ἄρα ἡ ΒΑ πρὸς τὴν FD [Prop. 1.34]. And since AC has been drawn parallel
ΓΔ, οὕτως ἡ ΒΓ πρὸς τὴν ΓΕ, καὶ ἐναλλὰξ ὡς ἡ ΑΒ πρὸς to one (of the sides) FE of triangle FBE, thus as BA
τὴν ΒΓ, οὕτως ἡ ΔΓ πρὸς τὴν ΓΕ. πάλιν, ἐπεὶ παράλληλός is to AF , so BC (is) to CE [Prop. 6.2]. And AF (is)
ἐστιν ἡ ΓΔ τῇ ΒΖ, ἔστιν ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΕ, οὕτως equal to CD. Thus, as BA (is) to CD, so BC (is) to CE,
ἡ ΖΔ πρὸς τὴν ΔΕ. ἴση δὲ ἡ ΖΔ τῇ ΑΓ· ὡς ἄρα ἡ ΒΓ πρὸς and, alternately, as AB (is) to BC, so DC (is) to CE
τὴν ΓΕ, οὕτως ἡ ΑΓ πρὸς τὴν ΔΕ, καὶ ἐναλλὰξ ὡς ἡ ΒΓ [Prop. 5.16]. Again, since CD is parallel to BF , thus as
πρὸς τὴν ΓΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΔ. ἐπεὶ οὖν ἐδείχθη BC (is) to CE, so FD (is) to DE [Prop. 6.2]. And FD
ὡς μὲν ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΔΓ πρὸς τὴν ΓΕ, ὡς (is) equal to AC. Thus, as BC is to CE, so AC (is) to
δὲ ἡ ΒΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΔ, δι᾿ ἴσου DE, and, alternately, as BC (is) to CA, so CE (is) to
ἄρα ὡς ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΓΔ πρὸς τὴν ΔΕ. ED [Prop. 6.2]. Therefore, since it was shown that as
Τῶν ἄρα ἰσογωνίων τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ AB (is) to BC, so DC (is) to CE, and as BC (is) to CA,

αἱ περὶ τὰς ἴσας γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας so CE (is) to ED, thus, via equality, as BA (is) to AC, so
ὑποτείνουσαι· ὅπερ ἔδει δεῖξαι. CD (is) to DE [Prop. 5.22].

Thus, in equiangular triangles the sides about the

equal angles are proportional, and those (sides) subtend-

160

STOIQEIWN �þ. ELEMENTS BOOK 6
ing equal angles correspond. (Which is) the very thing it

was required to show.eþ. Proposition 5
᾿Εὰν δύο τρίγωνα τὰς πλευρὰς ἀνάλογον ἔχῃ, ἰσογώνια If two triangles have proportional sides then the trian-

ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾿ ἃς αἱ gles will be equiangular, and will have the angles which
ὁμόλογοι πλευραὶ ὑποτείνουσιν. corresponding sides subtend equal.

Β
Γ

∆

Ε E

B
C

A D

῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς πλευρὰς ἀνάλογον Let ABC and DEF be two triangles having propor-
ἔχοντα, ὡς μὲν τὴν ΑΒ πρὸς τὴν ΒΓ, οὕτως τὴν ΔΕ πρὸς tional sides, (so that) as AB (is) to BC, so DE (is) to
τὴν ΕΖ, ὡς δὲ τὴν ΒΓ πρὸς τὴν ΓΑ, οὕτως τὴν ΕΖ πρὸς EF , and as BC (is) to CA, so EF (is) to FD, and, fur-
τὴν ΖΔ, καὶ ἔτι ὡς τὴν ΒΑ πρὸς τὴν ΑΓ, οὕτως τὴν ΕΔ ther, as BA (is) to AC, so ED (is) to DF . I say that
πρὸς τὴν ΔΖ. λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον triangle ABC is equiangular to triangle DEF , and (that
τῷ ΔΕΖ τριγώνῳ καὶ ἴσας ἕξουσι τὰς γωνίας, ὑφ᾿ ἃς αἱ the triangles) will have the angles which corresponding
ὁμόλογοι πλευραὶ ὑποτείνουσιν, τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ sides subtend equal. (That is), (angle) ABC (equal) to
ΔΕΖ, τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ καὶ ἔτι τὴν ὑπὸ ΒΑΓ DEF , BCA to EFD, and, further, BAC to EDF .
τῇ ὑπὸ ΕΔΖ. For let (angle) FEG, equal to angle ABC, and (an-
Συνεστάτω γὰρ πρὸς τῇ ΕΖ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ gle) EFG, equal to ACB, have been constructed on the

σημείοις τοῖς Ε, Ζ τῇ μὲν ὑπὸ ΑΒΓ γωνίᾳ ἴση ἡ ὑπὸ ΖΕΗ, straight-line EF at the points E and F on it (respectively)
τῇ δὲ ὑπὸ ΑΓΒ ἴση ἡ ὑπὸ ΕΖΗ· λοιπὴ ἄρα ἡ πρὸς τῷ Α [Prop. 1.23]. Thus, the remaining (angle) at A is equal
λοιπῇ τῇ πρὸς τῷ Η ἐστιν ἴση. to the remaining (angle) at G [Prop. 1.32].
῎Ισογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΕΗΖ [τριγών- Thus, triangle ABC is equiangular to [triangle] EGF .

ῳ]. τῶν ἄρα ΑΒΓ, ΕΗΖ τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ Thus, for triangles ABC and EGF , the sides about the
αἱ περὶ τὰς ἴσας γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας equal angles are proportional, and (those) sides subtend-
ὑποτείνουσαι· ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, [οὕτως] ing equal angles correspond [Prop. 6.4]. Thus, as AB
ἡ ΗΕ πρὸς τὴν ΕΖ. ἀλλ᾿ ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως is to BC, [so] GE (is) to EF . But, as AB (is) to BC,
ὑπόκειται ἡ ΔΕ πρὸς τὴν ΕΖ· ὡς ἄρα ἡ ΔΕ πρὸς τὴν ΕΖ, so, it was assumed, (is) DE to EF . Thus, as DE (is) to
οὕτως ἡ ΗΕ πρὸς τὴν ΕΖ. ἑκατέρα ἄρα τῶν ΔΕ, ΗΕ πρὸς EF , so GE (is) to EF [Prop. 5.11]. Thus, DE and GE
τὴν ΕΖ τὸν αὐτὸν ἔχει λόγον· ἴση ἄρα ἐστὶν ἡ ΔΕ τῇ ΗΕ. each have the same ratio to EF . Thus, DE is equal to
διὰ τὰ αὐτὰ δὴ καὶ ἡ ΔΖ τῇ ΗΖ ἐστιν ἴση. ἐπεὶ οὖν ἴση ἐστὶν GE [Prop. 5.9]. So, for the same (reasons), DF is also
ἡ ΔΕ τῇ ΕΗ, κοινὴ δὲ ἡ ΕΖ, δύο δὴ αἱ ΔΕ, ΕΖ δυσὶ ταῖς ΗΕ, equal to GF . Therefore, since DE is equal to EG, and
ΕΖ ἴσαι εἰσίν· καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΗ [ἐστιν] ἴση· γωνία EF (is) common, the two (sides) DE, EF are equal to
ἄρα ἡ ὑπὸ ΔΕΖ γωνίᾳ τῇ ὑπὸ ΗΕΖ ἐστιν ἴση, καὶ τὸ ΔΕΖ the two (sides) GE, EF (respectively). And base DF
τρίγωνον τῷ ΗΕΖ τριγώνῳ ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς [is] equal to base FG. Thus, angle DEF is equal to
λοιπαῖς γωνίαις ἴσαι, ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν. angle GEF [Prop. 1.8], and triangle DEF (is) equal
ἴση ἄρα ἐστὶ καὶ ἡ μὲν ὑπὸ ΔΖΕ γωνία τῇ ὑπὸ ΗΖΕ, ἡ δὲ to triangle GEF , and the remaining angles (are) equal
ὑπὸ ΕΔΖ τῇ ὑπὸ ΕΗΖ. καὶ ἐπεὶ ἡ μὲν ὑπὸ ΖΕΔ τῇ ὑπὸ to the remaining angles which the equal sides subtend
ΗΕΖ ἐστιν ἴση, ἀλλ᾿ ἡ ὑπὸ ΗΕΖ τῇ ὑπὸ ΑΒΓ, καὶ ἡ ὑπὸ [Prop. 1.4]. Thus, angle DFE is also equal to GFE, and

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ΑΒΓ ἄρα γωνία τῇ ὑπὸ ΔΕΖ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ (angle) EDF to EGF . And since (angle) FED is equal
ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ἐστιν ἴση, καὶ ἔτι ἡ πρὸς τῷ Α τῇ to GEF , and (angle) GEF to ABC, angle ABC is thus
πρὸς τῷ Δ· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ also equal to DEF . So, for the same (reasons), (angle)
τριγώνῳ. ACB is also equal to DFE, and, further, the (angle) at A
᾿Εὰν ἄρα δύο τρίγωνα τὰς πλευρὰς ἀνάλογον ἔχῃ, to the (angle) at D. Thus, triangle ABC is equiangular

ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾿ to triangle DEF .
ἃς αἱ ὁμόλογοι πλευραὶ ὑποτείνουσιν· ὅπερ ἔδει δεῖξαι. Thus, if two triangles have proportional sides then the

triangles will be equiangular, and will have the angles
which corresponding sides subtend equal. (Which is) the

very thing it was required to show.�þ. Proposition 6
᾿Εὰν δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, περὶ If two triangles have one angle equal to one angle,

δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ἰσογώνια ἔσται and the sides about the equal angles proportional, then
τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾿ ἃς αἱ ὁμόλογοι the triangles will be equiangular, and will have the angles
πλευραὶ ὑποτείνουσιν. which corresponding sides subtend equal.

Β Γ

∆

B C

E F

῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ μίαν γωνίαν τὴν ὑπὸ Let ABC and DEF be two triangles having one angle,
ΒΑΓ μιᾷ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴσην ἔχοντα, περὶ δὲ τὰς ἴσας BAC, equal to one angle, EDF (respectively), and the
γωνίας τὰς πλευρὰς ἀνάλογον, ὡς τὴν ΒΑ πρὸς τὴν ΑΓ, sides about the equal angles proportional, (so that) as BA
οὕτως τὴν ΕΔ πρὸς τὴν ΔΖ· λέγω, ὅτι ἰσογώνιόν ἐστι τὸ (is) to AC, so ED (is) to DF . I say that triangle ABC is
ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ καὶ ἴσην ἕξει τὴν ὑπὸ ΑΒΓ equiangular to triangle DEF , and will have angle ABC
γωνίαν τῇ ὑπὸ ΔΕΖ, τὴν δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ. equal to DEF , and (angle) ACB to DFE.
Συνεστάτω γὰρ πρὸς τῇ ΔΖ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ For let (angle) FDG, equal to each of BAC and

σημείοις τοῖς Δ, Ζ ὁποτέρᾳ μὲν τῶν ὑπὸ ΒΑΓ, ΕΔΖ ἴση EDF , and (angle) DFG, equal to ACB, have been con-
ἡ ὑπὸ ΖΔΗ, τῇ δὲ ὑπὸ ΑΓΒ ἴση ἡ ὑπὸ ΔΖΗ· λοιπὴ ἄρα ἡ structed on the straight-line AF at the points D and F on
πρὸς τῷ Β γωνία λοιπῇ τῇ πρὸς τῷ Η ἴση ἐστίν. it (respectively) [Prop. 1.23]. Thus, the remaining angle
᾿Ισογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΗΖ at B is equal to the remaining angle at G [Prop. 1.32].

τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως Thus, triangle ABC is equiangular to triangle DGF .
ἡ ΗΔ πρὸς τὴν ΔΖ. ὑπόκειται δὲ καὶ ὡς ἡ ΒΑ πρὸς τὴν ΑΓ, Thus, proportionally, as BA (is) to AC, so GD (is) to
οὕτως ἡ ΕΔ πρὸς τὴν ΔΖ· καὶ ὡς ἄρα ἡ ΕΔ πρὸς τὴν ΔΖ, DF [Prop. 6.4]. And it was also assumed that as BA
οὕτως ἡ ΗΔ πρὸς τὴν ΔΖ. ἴση ἄρα ἡ ΕΔ τῇ ΔΗ· καὶ κοινὴ is) to AC, so ED (is) to DF . And, thus, as ED (is)
ἡ ΔΖ· δύο δὴ αἱ ΕΔ, ΔΖ δυσὶ ταῖς ΗΔ, ΔΖ ἴσας εἰσίν· καὶ to DF , so GD (is) to DF [Prop. 5.11]. Thus, ED (is)
γωνία ἡ ὑπὸ ΕΔΖ γωνίᾳ τῇ ὑπὸ ΗΔΖ [ἐστιν] ἴση· βάσις equal to DG [Prop. 5.9]. And DF (is) common. So, the
ἄρα ἡ ΕΖ βάσει τῇ ΗΖ ἐστιν ἴση, καὶ τὸ ΔΕΖ τρίγωνον τῷ two (sides) ED, DF are equal to the two (sides) GD,
ΗΔΖ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς DF (respectively). And angle EDF [is] equal to angle
γωνίαις ἴσας ἔσονται, ὐφ᾿ ἃς ἴσας πλευραὶ ὑποτείνουσιν. ἴση GDF . Thus, base EF is equal to base GF , and triangle
ἄρα ἐστὶν ἡ μὲν ὑπὸ ΔΖΗ τῇ ὑπο ΔΖΕ, ἡ δὲ ὑπὸ ΔΗΖ DEF is equal to triangle GDF , and the remaining angles

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STOIQEIWN �þ. ELEMENTS BOOK 6
τῇ ὑπὸ ΔΕΖ. ἀλλ᾿ ἡ ὑπὸ ΔΖΗ τῇ ὑπὸ ΑΓΒ ἐστιν ἴση· καὶ will be equal to the remaining angles which the equal
ἡ ὑπὸ ΑΓΒ ἄρα τῇ ὑπὸ ΔΖΕ ἐστιν ἴση. ὑπόκειται δὲ καὶ sides subtend [Prop. 1.4]. Thus, (angle) DFG is equal
ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ἴση· καὶ λοιπὴ ἄρα ἡ πρὸς τῷ Β to DFE, and (angle) DGF to DEF . But, (angle) DFG
λοιπῇ τῇ πρὸς τῷ Ε ἴση ἐστίν· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ is equal to ACB. Thus, (angle) ACB is also equal to
τρίγωνον τῷ ΔΕΖ τριγώνῳ. DFE. And (angle) BAC was also assumed (to be) equal
᾿Εὰν ἄρα δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, to EDF . Thus, the remaining (angle) at B is equal to the

περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ἰσογώνια remaining (angle) at E [Prop. 1.32]. Thus, triangle ABC
ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾿ ἃς αἱ is equiangular to triangle DEF .
ὁμόλογοι πλευραὶ ὑποτείνουσιν· ὅπερ ἔδει δεῖξαι. Thus, if two triangles have one angle equal to one

angle, and the sides about the equal angles proportional,

then the triangles will be equiangular, and will have the
angles which corresponding sides subtend equal. (Which

is) the very thing it was required to show.zþ. Proposition 7
᾿Εὰν δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, If two triangles have one angle equal to one angle,

περὶ δὲ ἄλλας γωνίας τὰς πλευρὰς ἀνάλογον, τῶν δὲ and the sides about other angles proportional, and the
λοιπῶν ἑκατέραν ἅμα ἤτοι ἐλάσσονα ἢ μὴ ἐλάσσονα ὀρθῆς, remaining angles either both less than, or both not less
ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, περὶ than, right-angles, then the triangles will be equiangular,
ἃς ἀνάλογόν εἰσιν αἱ πλευραί. and will have the angles about which the sides are pro-

portional equal.

∆

Β
Ε

῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ μίαν γωνίαν μιᾷ Let ABC and DEF be two triangles having one an-
γωνίᾳ ἴσην ἔχοντα τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ, περὶ δὲ gle, BAC, equal to one angle, EDF (respectively), and
ἄλλας γωνίας τὰς ὑπὸ ΑΒΓ, ΔΕΖ τὰς πλευρὰς ἀνάλογον, the sides about (some) other angles, ABC and DEF (re-
ὡς τὴν ΑΒ πρὸς τὴν ΒΓ, οὕτως τὴν ΔΕ πρὸς τὴν ΕΖ, spectively), proportional, (so that) as AB (is) to BC, so
τῶν δὲ λοιπῶν τῶν πρὸς τοῖς Γ, Ζ πρότερον ἑκατέραν DE (is) to EF , and the remaining (angles) at C and F ,
ἅμα ἐλάσσονα ὀρθῆς· λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓ first of all, both less than right-angles. I say that triangle
τρίγωνον τῷ ΔΕΖ τριγώνῳ, καὶ ἴση ἔσται ἡ ὑπὸ ΑΒΓ γωνία ABC is equiangular to triangle DEF , and (that) angle
τῇ ὑπὸ ΔΕΖ, καὶ λοιπὴ δηλονότι ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς ABC will be equal to DEF , and (that) the remaining
τῷ Ζ ἴση. (angle) at C (will be) manifestly equal to the remaining
Εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ, (angle) at F .

μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ὑπὸ ΑΒΓ. καὶ συ- For if angle ABC is not equal to (angle) DEF then
νεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ one of them is greater. Let ABC be greater. And let (an-
Β τῇ ὑπὸ ΔΕΖ γωνίᾳ ἴση ἡ ὑπὸ ΑΒΗ. gle) ABG, equal to (angle) DEF , have been constructed
Καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν Α γωνία τῇ Δ, ἡ δὲ ὑπὸ ΑΒΗ on the straight-line AB at the point B on it [Prop. 1.23].

τῇ ὑπὸ ΔΕΖ, λοιπὴ ἄρα ἡ ὑπὸ ΑΗΒ λοιπῇ τῇ ὑπὸ ΔΖΕ And since angle A is equal to (angle) D, and (angle)
ἐστιν ἴση. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ ABG to DEF , the remaining (angle) AGB is thus equal

163

STOIQEIWN �þ. ELEMENTS BOOK 6
τριγώνῳ. ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΒΗ, οὕτως ἡ ΔΕ to the remaining (angle) DFE [Prop. 1.32]. Thus, trian-
πρὸς τὴν ΕΖ. ὡς δὲ ἡ ΔΕ πρὸς τὴν ΕΖ, [οὕτως] ὑπόκειται ἡ gle ABG is equiangular to triangle DEF . Thus, as AB is
ΑΒ πρὸς τὴν ΒΓ· ἡ ΑΒ ἄρα πρὸς ἑκατέραν τῶν ΒΓ, ΒΗ τὸν to BG, so DE (is) to EF [Prop. 6.4]. And as DE (is) to
αὐτὸν ἔχει λόγον· ἴση ἄρα ἡ ΒΓ τῇ ΒΗ. ὥστε καὶ γωνία ἡ EF , [so] it was assumed (is) AB to BC. Thus, AB has
πρὸς τῷ Γ γωνίᾳ τῇ ὑπὸ ΒΗΓ ἐστιν ἴση. ἐλάττων δὲ ὀρθῆς the same ratio to each of BC and BG [Prop. 5.11]. Thus,
ὑπόκειται ἡ πρὸς τῷ Γ· ἐλάττων ἄρα ἐστὶν ὀρθῆς καὶ ὑπὸ BC (is) equal to BG [Prop. 5.9]. And, hence, the angle
ΒΗΓ· ὥστε ἡ ἐφεξῆς αὐτῇ γωνία ἡ ὑπὸ ΑΗΒ μείζων ἐστὶν at C is equal to angle BGC [Prop. 1.5]. And the angle
ὀρθῆς. καὶ ἐδείχθη ἴση οὖσα τῇ πρὸς τῷ Ζ· καὶ ἡ πρὸς τῷ Ζ at C was assumed (to be) less than a right-angle. Thus,
ἄρα μείζων ἐστὶν ὀρθῆς. ὑπόκειται δὲ ἐλάσσων ὀρθῆς· ὅπερ (angle) BGC is also less than a right-angle. Hence, the
ἐστὶν ἄτοπον. οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ γωνία τῇ adjacent angle to it, AGB, is greater than a right-angle
ὑπὸ ΔΕΖ· ἴση ἄρα. ἔστι δὲ καὶ ἡ πρὸς τῷ Α ἴση τῇ πρὸς τῷ [Prop. 1.13]. And (AGB) was shown to be equal to the
Δ· καὶ λοιπὴ ἄρα ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς τῷ Ζ ἴση ἐστίν. (angle) at F . Thus, the (angle) at F is also greater than a
ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. right-angle. But it was assumed (to be) less than a right-
Ἀλλὰ δὴ πάλιν ὑποκείσθω ἑκατέρα τῶν πρὸς τοῖς Γ, Ζ μὴ angle. The very thing is absurd. Thus, angle ABC is not

ἐλάσσων ὀρθῆς· λέγω πάλιν, ὅτι καὶ οὕτως ἐστὶν ἰσογώνιον unequal to (angle) DEF . Thus, (it is) equal. And the
τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. (angle) at A is also equal to the (angle) at D. And thus
Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, the remaining (angle) at C is equal to the remaining (an-

ὅτι ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ· ὥστε καὶ γωνία ἡ πρὸς τῷ Γ τῇ gle) at F [Prop. 1.32]. Thus, triangle ABC is equiangular
ὑπὸ ΒΗΓ ἴση ἐστίν. οὐκ ἐλάττων δὲ ὀρθῆς ἡ πρὸς τῷ Γ· to triangle DEF .
οὐκ ἐλάττων ἄρα ὀρθῆς οὐδὲ ἡ ὑπὸ ΒΗΓ. τριγώνου δὴ τοῦ But, again, let each of the (angles) at C and F be
ΒΗΓ αἱ δύο γωνίαι δύο ὀρθῶν οὔκ εἰσιν ἐλάττονες· ὅπερ assumed (to be) not less than a right-angle. I say, again,
ἐστὶν ἀδύνατον. οὐκ ἄρα πάλιν ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ that triangle ABC is equiangular to triangle DEF in this
γωνία τῇ ὑπὸ ΔΕΖ· ἴση ἄρα. ἔστι δὲ καὶ ἡ πρὸς τῷ Α τῇ case also.
πρὸς τῷ Δ ἴση· λοιπὴ ἄρα ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς τῷ Ζ For, with the same construction, we can similarly
ἴση ἐστίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ show that BC is equal to BG. Hence, also, the angle
τριγώνῳ. at C is equal to (angle) BGC. And the (angle) at C (is)
᾿Εὰν ἄρα δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, not less than a right-angle. Thus, BGC (is) not less than

περὶ δὲ ἄλλας γωνίας τὰς πλευρὰς ἀνάλογον, τῶν δὲ λοιπῶν a right-angle either. So, in triangle BGC the (sum of)
ἑκατέραν ἅμα ἐλάττονα ἢ μὴ ἐλάττονα ὀρθῆς, ἰσογώνια two angles is not less than two right-angles. The very
ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, περὶ ἃς ἀνάλογόν thing is impossible [Prop. 1.17]. Thus, again, angle ABC
εἰσιν αἱ πλευραί· ὅπερ ἔδει δεῖξαι. is not unequal to DEF . Thus, (it is) equal. And the (an-

gle) at A is also equal to the (angle) at D. Thus, the
remaining (angle) at C is equal to the remaining (angle)

at F [Prop. 1.32]. Thus, triangle ABC is equiangular to

triangle DEF .
Thus, if two triangles have one angle equal to one

angle, and the sides about other angles proportional, and
the remaining angles both less than, or both not less than,

right-angles, then the triangles will be equiangular, and

will have the angles about which the sides (are) propor-
tional equal. (Which is) the very thing it was required to

show.hþ. Proposition 8
᾿Εὰν ἐν ὀρθογωνίῳ τριγώνῳ ἀπό τῆς ὀρθῆς γωνίας ἐπὶ If, in a right-angled triangle, a (straight-line) is drawn

τὴν βάσιν κάθετος ἀχθῇ, τὰ πρὸς τῇ καθέτῳ τρίγωνα ὅμοιά from the right-angle perpendicular to the base then the
ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις. triangles around the perpendicular are similar to the
῎Εστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν whole (triangle), and to one another.

ὑπὸ ΒΑΓ γωνίαν, καὶ ἤχθω ἀπὸ τοῦ Α ἐπὶ τὴν ΒΓ κάθετος Let ABC be a right-angled triangle having the angle
ἡ ΑΔ· λέγω, ὅτι ὅμοιόν ἐστιν ἑκάτερον τῶν ΑΒΔ, ΑΔΓ BAC a right-angle, and let AD have been drawn from

164

STOIQEIWN �þ. ELEMENTS BOOK 6
τριγώνων ὅλῳ τῷ ΑΒΓ καὶ ἔτι ἀλλήλοις. A, perpendicular to BC [Prop. 1.12]. I say that triangles

ABD and ADC are each similar to the whole (triangle)

ABC and, further, to one another.

Β Γ∆

B CD

᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΑΔΒ· ὀρθὴ For since (angle) BAC is equal to ADB—for each
γὰρ ἑκατέρα· καὶ κοινὴ τῶν δύο τριγώνων τοῦ τε ΑΒΓ καὶ (are) right-angles—and the (angle) at B (is) common
τοῦ ΑΒΔ ἡ πρὸς τῷ Β, λοιπὴ ἄρα ἡ ὑπὸ ΑΓΒ λοιπῇ τῇ to the two triangles ABC and ABD, the remaining (an-
ὑπο ΒΑΔ ἐστιν ἴση· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον gle) ACB is thus equal to the remaining (angle) BAD
τῷ ΑΒΔ τριγώνῳ. ἔστιν ἄρα ὡς ἡ ΒΓ ὑποτείνουσα τὴν [Prop. 1.32]. Thus, triangle ABC is equiangular to tri-
ὀρθὴν τοῦ ΑΒΓ τριγώνου πρὸς τὴν ΒΑ ὑποτείνουσαν τὴν angle ABD. Thus, as BC, subtending the right-angle in
ὀρθὴν τοῦ ΑΒΔ τριγώνου, οὕτως αὐτὴ ἡ ΑΒ ὑποτείνουσα triangle ABC, is to BA, subtending the right-angle in tri-
τὴν πρὸς τῷ Γ γωνίαν τοῦ ΑΒΓ τριγώνου πρὸς τὴν ΒΔ angle ABD, so the same AB, subtending the angle at C
ὑποτείνουσαν τὴν ἴσην τὴν ὑπὸ ΒΑΔ τοῦ ΑΒΔ τριγώνου, in triangle ABC, (is) to BD, subtending the equal (an-
καὶ ἔτι ἡ ΑΓ πρὸς τὴν ΑΔ ὑποτείνουσαν τὴν πρὸς τῷ Β gle) BAD in triangle ABD, and, further, (so is) AC to
γωνίαν κοινὴν τῶν δύο τριγώνων. τὸ ΑΒΓ ἄρα τρίγωνον AD, (both) subtending the angle at B common to the
τῷ ΑΒΔ τριγώνῳ ἰσογώνιόν τέ ἐστι καὶ τὰς περὶ τὰς ἴσας two triangles [Prop. 6.4]. Thus, triangle ABC is equian-
γωνίας πλευρὰς ἀνάλογον ἔχει. ὅμοιον ἄμα [ἐστὶ] τὸ ΑΒΓ gular to triangle ABD, and has the sides about the equal
τρίγωνον τῷ ΑΒΔ τριγώνῳ. ὁμοίως δὴ δείξομεν, ὅτι καὶ angles proportional. Thus, triangle ABC [is] similar to
τῷ ΑΔΓ τριγώνῳ ὅμοιόν ἐστι τὸ ΑΒΓ τρίγωνον· ἑκάτερον triangle ABD [Def. 6.1]. So, similarly, we can show that
ἄρα τῶν ΑΒΔ, ΑΔΓ [τριγώνων] ὅμοιόν ἐστιν ὅλῳ τῷ ΑΒΓ. triangle ABC is also similar to triangle ADC. Thus, [tri-
Λέγω δή, ὅτι καὶ ἀλλήλοις ἐστὶν ὅμοια τὰ ΑΒΔ, ΑΔΓ angles] ABD and ADC are each similar to the whole

τρίγωνα. (triangle) ABC.
᾿Επεὶ γὰρ ὀρθὴ ἡ ὑπὸ ΒΔΑ ὀρθῇ τῇ ὑπὸ ΑΔΓ ἐστιν So I say that triangles ABD and ADC are also similar

ἴση, ἀλλὰ μὴν καὶ ἡ ὑπὸ ΒΑΔ τῇ πρὸς τῷ Γ ἐδείχθη ἴση, to one another.
καὶ λοιπὴ ἄρα ἡ πρὸς τῷ Β λοιπῇ τῇ ὑπὸ ΔΑΓ ἐστιν ἴση· For since the right-angle BDA is equal to the right-
ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΔ τρίγωνον τῷ ΑΔΓ τριγώνῳ. angle ADC, and, indeed, (angle) BAD was also shown
ἔστιν ἄρα ὡς ἡ ΒΔ τοῦ ΑΒΔ τριγώνου ὑποτείνουσα τὴν (to be) equal to the (angle) at C, thus the remaining (an-
ὑπὸ ΒΑΔ πρὸς τὴν ΔΑ τοῦ ΑΔΓ τριγώνου ὑποτείνουσαν gle) at B is also equal to the remaining (angle) DAC
τὴν πρὸς τῷ Γ ἴσην τῇ ὑπὸ ΒΑΔ, οὕτως αὐτὴ ἡ ΑΔ τοῦ [Prop. 1.32]. Thus, triangle ABD is equiangular to trian-
ΑΒΔ τριγώνου ὑποτείνουσα τὴν πρὸς τῷ Β γωνίαν πρὸς gle ADC. Thus, as BD, subtending (angle) BAD in tri-
τὴν ΔΓ ὑποτείνουσαν τὴν ὑπὸ ΔΑΓ τοῦ ΑΔΓ τριγώνου angle ABD, is to DA, subtending the (angle) at C in tri-
ἴσην τῇ πρὸς τῷ Β, καὶ ἔτι ἡ ΒΑ πρὸς τὴν ΑΓ ὑποτείνουσαι angle ADC, (which is) equal to (angle) BAD, so (is) the
τὰς ὀρθάς· ὅμοιον ἄρα ἐστὶ τὸ ΑΒΔ τρίγωνον τῷ ΑΔΓ same AD, subtending the angle at B in triangle ABD, to
τριγώνῳ. DC, subtending (angle) DAC in triangle ADC, (which
᾿Εὰν ἄρα ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας is) equal to the (angle) at B, and, further, (so is) BA to

ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, τὰ πρὸς τῇ καθέτῳ τρίγωνα AC, (each) subtending right-angles [Prop. 6.4]. Thus,
ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις [ὅπερ ἔδει δεῖξαι]. triangle ABD is similar to triangle ADC [Def. 6.1].

Thus, if, in a right-angled triangle, a (straight-line)

is drawn from the right-angle perpendicular to the base

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then the triangles around the perpendicular are similar

to the whole (triangle), and to one another. [(Which is)

the very thing it was required to show.]Pìrisma. Corollary
᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν ἐν ὀρθογωνίῳ τριγώνῳ So (it is) clear, from this, that if, in a right-angled tri-

ἀπὸ τῆς ὀρθῆς γωνάις ἐπὶ τὴν βάσις κάθετος ἀχθῇ, ἡ angle, a (straight-line) is drawn from the right-angle per-
ἀχθεῖσα τῶν τῆς βάσεως τμημάτων μέση ἀνάλογόν ἐστιν· pendicular to the base then the (straight-line so) drawn

ὅπερ ἔδει δεῖξαι. is in mean proportion to the pieces of the base.† (Which
is) the very thing it was required to show.

† In other words, the perpendicular is the geometric mean of the pieces.jþ. Proposition 9
Τῆς δοθείσης εὐθείας τὸ προσταχθὲν μέρος ἀφελεῖν. To cut off a prescribed part from a given straight-line.

ΒΑ

∆

A F

῎Εστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ· δεῖ δὴ τῆς ΑΒ τὸ προ- Let AB be the given straight-line. So it is required to
σταχθὲν μέρος ἀφελεῖν. cut off a prescribed part from AB.
᾿Επιτετάχθω δὴ τὸ τρίτον. [καὶ] διήθχω τις ἀπὸ τοῦ Α So let a third (part) have been prescribed. [And] let

εὐθεῖα ἡ ΑΓ γωνίαν περιέχουσα μετὰ τῆς ΑΒ τυχοῦσαν· καὶ some straight-line AC have been drawn from (point) A,
εἰλήφθω τυχὸν σημεῖον ἐπὶ τῆς ΑΓ τὸ Δ, καὶ κείσθωσαν encompassing a random angle with AB. And let a ran-
τῇ ΑΔ ἴσαι αἱ ΔΕ, ΕΓ. καὶ ἐπεζεύχθω ἡ ΒΓ, καὶ διὰ τοῦ Δ dom point D have been taken on AC. And let DE and
παράλληλος αὐτῇ ἤχθω ἡ ΔΖ. EC be made equal to AD [Prop. 1.3]. And let BC have
᾿Επεὶ οὖν τριγώνου τοῦ ΑΒΓ παρὰ μίαν τῶν πλευρῶν been joined. And let DF have been drawn through D

τὴν ΒΓ ἦκται ἡ ΖΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΓΔ πρὸς τὴν parallel to it [Prop. 1.31].
ΔΑ, οὕτως ἡ ΒΖ πρὸς τὴν ΖΑ. διπλῆ δὲ ἡ ΓΔ τῆς ΔΑ· Therefore, since FD has been drawn parallel to one
διπλῆ ἄρα καὶ ἡ ΒΖ τῆς ΖΑ· τριπλῆ ἄρα ἡ ΒΑ τῆς ΑΖ. of the sides, BC, of triangle ABC, then, proportionally,
Τῆς ἄρα δοθείσης εὐθείας τῆς ΑΒ τὸ ἐπιταχθὲν τρίτον as CD is to DA, so BF (is) to FA [Prop. 6.2]. And CD

μέρος ἀφῄρηται τὸ ΑΖ· ὅπερ ἔδει ποιῆσαι. (is) double DA. Thus, BF (is) also double FA. Thus,
BA (is) triple AF .

Thus, the prescribed third part, AF , has been cut off
from the given straight-line, AB. (Which is) the very

thing it was required to do.iþ. Proposition 10
Τὴν δοθεῖσαν εὐθεῖαν ἄτμητον τῇ δοθείσῃ τετμημένῃ To cut a given uncut straight-line similarly to a given

ὁμοίως τεμεῖν. cut (straight-line).

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∆

Κ
Θ

Ζ Η Β

GF B

῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἅτμητος ἡ ΑΒ, ἡ δὲ τε- Let AB be the given uncut straight-line, and AC a
τμημένη ἡ ΑΓ κατὰ τὰ Δ, Ε σημεῖα, καὶ κείσθωσαν ὥστε (straight-line) cut at points D and E, and let (AC) be
γωνίαν τυχοῦσαν περιέχειν, καὶ ἐπεζεύχθω ἡ ΓΒ, καὶ διὰ laid down so as to encompass a random angle (with AB).
τῶν Δ, Ε τῇ ΒΓ παράλληλοι ἤχθωσαν αἱ ΔΖ, ΕΗ, διὰ δὲ And let CB have been joined. And let DF and EG have
τοῦ Δ τῇ ΑΒ παράλληλος ἤχθω ἡ ΔΘΚ. been drawn through (points) D and E (respectively),
Παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΖΘ, ΘΒ· parallel to BC, and let DHK have been drawn through

ἴση ἄρα ἡ μὲν ΔΘ τῇ ΖΗ, ἡ δὲ ΘΚ τῇ ΗΒ. καὶ ἐπεὶ τριγώνου (point) D, parallel to AB [Prop. 1.31].
τοῦ ΔΚΓ παρὰ μίαν τῶν πλευρῶν τὴν ΚΓ εὐθεῖα ἦκται ἡ Thus, FH and HB are each parallelograms. Thus,
ΘΕ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ DH (is) equal to FG, and HK to GB [Prop. 1.34]. And
ΚΘ πρὸς τὴν ΘΔ. ἴση δὲ ἡ μὲν ΚΘ τῇ ΒΗ, ἡ δὲ ΘΔ τῇ since the straight-line HE has been drawn parallel to one
ΗΖ. ἔστιν ἄρα ὡς ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν of the sides, KC, of triangle DKC, thus, proportionally,
ΗΖ. πάλιν, ἐπεὶ τριγώνου τοῦ ΑΗΕ παρὰ μίαν τῶν πλευρῶν as CE is to ED, so KH (is) to HD [Prop. 6.2]. And
τὴν ΗΕ ἦκται ἡ ΖΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΕΔ πρὸς τὴν KH (is) equal to BG, and HD to GF . Thus, as CE is
ΔΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ. ἐδείχθη δὲ καὶ ὡς ἡ ΓΕ to ED, so BG (is) to GF . Again, since FD has been
πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΖ· ἔστιν ἄρα ὡς μὲν drawn parallel to one of the sides, GE, of triangle AGE,
ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΖ, ὡς δὲ ἡ ΕΔ thus, proportionally, as ED is to DA, so GF (is) to FA
πρὸς τὴν ΔΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ. [Prop. 6.2]. And it was also shown that as CE (is) to
῾Η ἄρα δοθεῖσα εὐθεῖα ἄτμητος ἡ ΑΒ τῇ δοθείσῃ εὐθείᾳ ED, so BG (is) to GF . Thus, as CE is to ED, so BG (is)

τετμημένῃ τῇ ΑΓ ὁμοίως τέτμηται· ὅπερ ἔδει ποιῆσαι· to GF , and as ED (is) to DA, so GF (is) to FA.
Thus, the given uncut straight-line, AB, has been cut

similarly to the given cut straight-line, AC. (Which is)

the very thing it was required to do.iaþ. Proposition 11
Δύο δοθεισῶν εὐθειῶν τρίτην ἀνάλογον προσευρεῖν. To find a third (straight-line) proportional to two
῎Εστωσαν αἱ δοθεῖσαι [δύο εὐθεῖαι] αἱ ΒΑ, ΑΓ καὶ given straight-lines.

κείσθωσαν γωνίαν περιέχουσαι τυχοῦσαν. δεῖ δὴ τῶν ΒΑ, Let BA and AC be the [two] given [straight-lines],
ΑΓ τρίτην ἀνάλογον προσευρεῖν. ἐκβεβλήσθωσαν γὰρ ἐπὶ and let them be laid down encompassing a random angle.
τὰ Δ, Ε σημεῖα, καὶ κείσθω τῇ ΑΓ ἴση ἡ ΒΔ, καὶ ἐπεζεύχθω So it is required to find a third (straight-line) proportional
ἡ ΒΓ, καὶ διὰ τοῦ Δ παράλληλος αὐτῇ ἤχθω ἡ ΔΕ. to BA and AC. For let (BA and AC) have been produced
᾿Επεὶ οὖν τριγώνου τοῦ ΑΔΕ παρὰ μίαν τῶν πλευρῶν to points D and E (respectively), and let BD be made

τὴν ΔΕ ἦκται ἡ ΒΓ, ἀνάλογόν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν equal to AC [Prop. 1.3]. And let BC have been joined.
ΒΔ, οὕτως ἡ ΑΓ πρὸς τὴν ΓΕ. ἴση δὲ ἡ ΒΔ τῇ ΑΓ. ἔστιν And let DE have been drawn through (point) D parallel
ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΑΓ, οὕτως ἡ ΑΓ πρὸς τὴν ΓΕ. to it [Prop. 1.31].

Therefore, since BC has been drawn parallel to one

of the sides DE of triangle ADE, proportionally, as AB is
to BD, so AC (is) to CE [Prop. 6.2]. And BD (is) equal

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to AC. Thus, as AB is to AC, so AC (is) to CE.

∆

Δύο ἄρα δοθεισῶν εὐθειῶν τῶν ΑΒ, ΑΓ τρίτη ἀνάλογον Thus, a third (straight-line), CE, has been found
αὐταῖς προσεύρηται ἡ ΓΕ· ὅπερ ἔδει ποιῆσαι. (which is) proportional to the two given straight-lines,

AB and AC. (Which is) the very thing it was required to

do.ibþ. Proposition 12
Τριῶν δοθεισῶν εὐθειῶν τετάρτην ἀνάλογον προ- To find a fourth (straight-line) proportional to three

σευρεῖν. given straight-lines.

∆ Θ

H F

῎Εστωσαν αἱ δοθεῖσαι τρεῖς εὐθεῖαι αἱ Α, Β, Γ· δεῖ δὴ Let A, B, and C be the three given straight-lines. So
τῶν Α, Β, Γ τετράτην ἀνάλογον προσευρεῖν. it is required to find a fourth (straight-line) proportional
᾿Εκκείσθωσαν δύο εὐθεῖαι αἱ ΔΕ, ΔΖ γωνίαν περιέχους- to A, B, and C.

αι [τυχοῦσαν] τὴν ὑπὸ ΕΔΖ· καὶ κείσθω τῇ μὲν Α ἴση ἡ ΔΗ, Let the two straight-lines DE and DF be set out en-
τῇ δὲ Β ἴση ἡ ΗΕ, καὶ ἔτι τῇ Γ ἴση ἡ ΔΘ· καὶ ἐπιζευχθείσης compassing the [random] angle EDF . And let DG be
τῆς ΗΘ παράλληλος αὐτῇ ἤχθω διὰ τοῦ Ε ἡ ΕΖ. made equal to A, and GE to B, and, further, DH to C
᾿Επεὶ οὖν τριγώνου τοῦ ΔΕΖ παρὰ μίαν τὴν ΕΖ ἦκται ἡ [Prop. 1.3]. And GH being joined, let EF have been

ΗΘ, ἔστιν ἄρα ὡς ἡ ΔΗ πρὸς τὴν ΗΕ, οὕτως ἡ ΔΘ πρὸς drawn through (point) E parallel to it [Prop. 1.31].
τὴν ΘΖ. ἴση δὲ ἡ μὲν ΔΗ τῇ Α, ἡ δὲ ΗΕ τῇ Β, ἡ δὲ ΔΘ τῇ Therefore, since GH has been drawn parallel to one
Γ· ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν ΘΖ. of the sides EF of triangle DEF , thus as DG is to GE,
Τριῶν ἄρα δοθεισῶν εὐθειῶν τῶν Α, Β, Γ τετάρτη so DH (is) to HF [Prop. 6.2]. And DG (is) equal to A,

ἀνάλογον προσεύρηται ἡ ΘΖ· ὅπερ ἔδει ποιῆσαι. and GE to B, and DH to C. Thus, as A is to B, so C (is)

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to HF .

Thus, a fourth (straight-line), HF , has been found

(which is) proportional to the three given straight-lines,
A, B, and C. (Which is) the very thing it was required to

do.igþ. Proposition 13
Δύο δοθεισῶν εὐθειῶν μέσην ἀνάλογον προσευρεῖν. To find the (straight-line) in mean proportion to two

given straight-lines.†

ΓΑ Β

∆

CA B

῎Εστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΑΒ, ΒΓ· δεῖ δὴ τῶν Let AB and BC be the two given straight-lines. So it
ΑΒ, ΒΓ μέσην ἀνάλογον προσευρεῖν. is required to find the (straight-line) in mean proportion
Κείσθωσαν ἐπ᾿ εὐθείας, καὶ γεγράφθω ἐπὶ τῆς ΑΓ to AB and BC.

ἡμικύκλιον τὸ ΑΔΓ, καὶ ἤχθω ἀπὸ τοῦ Β σημείου τῇ ΑΓ Let (AB and BC) be laid down straight-on (with re-
εὐθείᾳ πρὸς ὀρθὰς ἡ ΒΑ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ. spect to one another), and let the semi-circle ADC have
᾿Επεὶ ἐν ἡμικυκλίῳ γωνία ἐστὶν ἡ ὑπὸ ΑΔΓ, ὀρθή ἐστιν. been drawn on AC [Prop. 1.10]. And let BD have been

καὶ ἐπεὶ ἐν ὀρθογωνίῳ τριγώνῳ τῷ ΑΔΓ ἀπὸ τῆς ὀρθῆς drawn from (point) B, at right-angles to AC [Prop. 1.11].
γωνίας ἐπὶ τὴν βάσιν κάθετος ἦκται ἡ ΔΒ, ἡ ΔΒ ἄρα τῶν And let AD and DC have been joined.
τῆς βάσεως τμημάτων τῶν ΑΒ, ΒΓ μέση ἀνάλογόν ἐστιν. And since ADC is an angle in a semi-circle, it is a
Δύο ἄρα δοθεισῶν εὐθειῶν τῶν ΑΒ, ΒΓ μέση ἀνάλογον right-angle [Prop. 3.31]. And since, in the right-angled

προσεύρηται ἡ ΔΒ· ὅπερ ἔδει ποιῆσαι. triangle ADC, the (straight-line) DB has been drawn
from the right-angle perpendicular to the base, DB is

thus the mean proportional to the pieces of the base, AB
and BC [Prop. 6.8 corr.].

Thus, DB has been found (which is) in mean propor-

tion to the two given straight-lines, AB and BC. (Which
is) the very thing it was required to do.

† In other words, to find the geometric mean of two given straight-lines.idþ. Proposition 14
Τῶν ἴσων τε καὶ ἴσογωνίων παραλληλογράμμων ἀντι- In equal and equiangular parallelograms the sides

πεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας· καὶ ὧν ἰσο- about the equal angles are reciprocally proportional.
γωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ And those equiangular parallelograms in which the sides
περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα. about the equal angles are reciprocally proportional are
῎Εστω ἴσα τε καὶ ἰσογώνια παραλληλόγραμμα τὰ ΑΒ, equal.

ΒΓ ἴσας ἔχοντα τὰς πρὸς τῷ Β γωνίας, καὶ κείσθωσαν ἐπ᾿ Let AB and BC be equal and equiangular parallelo-
εὐθείας αἱ ΔΒ, ΒΕ· ἐπ᾿ εὐθείας ἄρα εἰσὶ καὶ αἱ ΖΒ, ΒΗ. grams having the angles at B equal. And let DB and BE
λέγω, ὅτι τῶν ΑΒ, ΒΓ ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ be laid down straight-on (with respect to one another).
τὰς ἴσας γωνίας, τουτέστιν, ὅτι ἐστὶν ὡς ἡ ΔΒ πρὸς τὴν Thus, FB and BG are also straight-on (with respect to
ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ. one another) [Prop. 1.14]. I say that the sides of AB and

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STOIQEIWN �þ. ELEMENTS BOOK 6
BC about the equal angles are reciprocally proportional,

that is to say, that as DB is to BE, so GB (is) to BF .

Β
ΗΖ

∆

F
B

E C

Συμπεπληρώσθω γὰρ τὸ ΖΕ παραλληλόγραμμον. ἐπεὶ For let the parallelogram FE have been completed.
οὖν ἴσον ἐστὶ τὸ ΑΒ παραλληλόγραμμον τῷ ΒΓ παραλλη- Therefore, since parallelogram AB is equal to parallelo-
λογράμμῳ, ἄλλο δέ τι τὸ ΖΕ, ἔστιν ἄρα ὡς τὸ ΑΒ πρὸς τὸ gram BC, and FE (is) some other (parallelogram), thus
ΖΕ, οὕτως τὸ ΒΓ πρὸς τὸ ΖΕ. ἀλλ᾿ ὡς μὲν τὸ ΑΒ πρὸς as (parallelogram) AB is to FE, so (parallelogram) BC
τὸ ΖΕ, οὕτως ἡ ΔΒ πρὸς τὴν ΒΕ, ὡς δὲ τὸ ΒΓ πρὸς τὸ (is) to FE [Prop. 5.7]. But, as (parallelogram) AB (is) to
ΖΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ· καὶ ὡς ἄρα ἡ ΔΒ πρὸς τὴν FE, so DB (is) to BE, and as (parallelogram) BC (is) to
ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ. τῶν ἄρα ΑΒ, ΒΓ παραλ- FE, so GB (is) to BF [Prop. 6.1]. Thus, also, as DB (is)
ληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας to BE, so GB (is) to BF . Thus, in parallelograms AB
γωνίας. and BC the sides about the equal angles are reciprocally
Ἀλλὰ δὴ ἔστω ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς proportional.

τὴν ΒΖ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒ παραλληλόγραμμον τῷ And so, let DB be to BE, as GB (is) to BF . I say that
ΒΓ παραλληλογράμμῳ. parallelogram AB is equal to parallelogram BC.
᾿Επεὶ γάρ ἐστιν ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ For since as DB is to BE, so GB (is) to BF , but as

πρὸς τὴν ΒΖ, ἀλλ᾿ ὡς μὲν ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως τὸ DB (is) to BE, so parallelogram AB (is) to parallelo-
ΑΒ παραλληλόγραμμον πρὸς τὸ ΖΕ παραλληλόγραμμον, ὡς gram FE, and as GB (is) to BF , so parallelogram BC
δὲ ἡ ΗΒ πρὸς τὴν ΒΖ, οὕτως τὸ ΒΓ παραλληλόγραμμον (is) to parallelogram FE [Prop. 6.1], thus, also, as (par-
πρὸς τὸ ΖΕ παραλληλόγραμμον, καὶ ὡς ἄρα τὸ ΑΒ πρὸς allelogram) AB (is) to FE, so (parallelogram) BC (is)
τὸ ΖΕ, οὕτως τὸ ΒΓ πρὸς τὸ ΖΕ· ἴσον ἄρα ἐστὶ τὸ ΑΒ to FE [Prop. 5.11]. Thus, parallelogram AB is equal to
παραλληλόγραμμον τῷ ΒΓ παραλληλογράμμῳ. parallelogram BC [Prop. 5.9].
Τῶν ἄρα ἴσων τε καὶ ἰσογωνίων παραλληλογράμμων Thus, in equal and equiangular parallelograms the

ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας· καὶ ὧν sides about the equal angles are reciprocally propor-
ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ tional. And those equiangular parallelograms in which
αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα· ὅπερ ἔδει δεῖξαι. the sides about the equal angles are reciprocally propor-

tional are equal. (Which is) the very thing it was required

to show.ieþ. Proposition 15
Τῶν ἴσων καὶ μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων In equal triangles also having one angle equal to one

ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας· καὶ ὧν (angle) the sides about the equal angles are reciprocally
μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ proportional. And those triangles having one angle equal
πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα. to one angle for which the sides about the equal angles
῎Εστω ἴσα τρίγωνα τὰ ΑΒΓ, ΑΔΕ μίαν μιᾷ ἴσην ἔχοντα (are) reciprocally proportional are equal.

γωνίαν τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΔΑΕ· λέγω, ὅτι τῶν ΑΒΓ, Let ABC and ADE be equal triangles having one an-
ΑΔΕ τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας gle equal to one (angle), (namely) BAC (equal) to DAE.
γωνίας, τουτέστιν, ὅτι ἐστὶν ὡς ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως I say that, in triangles ABC and ADE, the sides about the

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STOIQEIWN �þ. ELEMENTS BOOK 6
ἡ ΕΑ πρὸς τὴν ΑΒ. equal angles are reciprocally proportional, that is to say,

that as CA is to AD, so EA (is) to AB.

∆

Κείσθω γὰρ ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΓΑ τῇ ΑΔ· ἐπ᾿ For let CA be laid down so as to be straight-on (with
εὐθείας ἄρα ἐστὶ καὶ ἡ ΕΑ τῇ ΑΒ. καὶ ἐπεζεύχθω ἡ ΒΔ. respect) to AD. Thus, EA is also straight-on (with re-
᾿Επεὶ οὖν ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΔΕ τριγώνῳ, spect) to AB [Prop. 1.14]. And let BD have been joined.

ἄλλο δέ τι τὸ ΒΑΔ, ἔστιν ἄρα ὡς τὸ ΓΑΒ τρίγωνον πρὸς Therefore, since triangle ABC is equal to triangle
τὸ ΒΑΔ τρίγωνον, οὕτως τὸ ΕΑΔ τρίγωνον πρὸς τὸ ΒΑΔ ADE, and BAD (is) some other (triangle), thus as tri-
τρίγωνον. ἀλλ᾿ ὡς μὲν τὸ ΓΑΒ πρὸς τὸ ΒΑΔ, οὕτως ἡ angle CAB is to triangle BAD, so triangle EAD (is) to
ΓΑ πρὸς τὴν ΑΔ, ὡς δὲ τὸ ΕΑΔ πρὸς τὸ ΒΑΔ, οὕτως ἡ triangle BAD [Prop. 5.7]. But, as (triangle) CAB (is)
ΕΑ πρὸς τὴν ΑΒ. καὶ ὡς ἄρα ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως to BAD, so CA (is) to AD, and as (triangle) EAD (is)
ἡ ΕΑ πρὸς τὴν ΑΒ. τῶν ΑΒΓ, ΑΔΕ ἄρα τριγώνων ἀντι- to BAD, so EA (is) to AB [Prop. 6.1]. And thus, as CA
πεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. (is) to AD, so EA (is) to AB. Thus, in triangles ABC and
Ἀλλὰ δὴ ἀντιπεπονθέτωσαν αἱ πλευραὶ τῶν ΑΒΓ, ΑΔΕ ADE the sides about the equal angles (are) reciprocally

τριγώνων, καὶ ἔστω ὡς ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ proportional.
πρὸς τὴν ΑΒ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ And so, let the sides of triangles ABC and ADE be
ΑΔΕ τριγώνῳ. reciprocally proportional, and (thus) let CA be to AD,
᾿Επιζευχθείσης γὰρ πάλιν τῆς ΒΔ, ἐπεί ἐστιν ὡς ἡ ΓΑ as EA (is) to AB. I say that triangle ABC is equal to

πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ, ἀλλ᾿ ὡς μὲν triangle ADE.
ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ For, BD again being joined, since as CA is to AD, so
ΒΑΔ τρίγωνον, ὡς δὲ ἡ ΕΑ πρὸς τὴν ΑΒ, οὕτως τὸ ΕΑΔ EA (is) to AB, but as CA (is) to AD, so triangle ABC
τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, ὡς ἄρα τὸ ΑΒΓ τρίγωνον (is) to triangle BAD, and as EA (is) to AB, so triangle
πρὸς τὸ ΒΑΔ τρίγωνον, οὕτως τὸ ΕΑΔ τρίγωνον πρὸς EAD (is) to triangle BAD [Prop. 6.1], thus as triangle
τὸ ΒΑΔ τρίγωνον. ἑκάτερον ἄρα τῶν ΑΒΓ, ΕΑΔ πρὸς ABC (is) to triangle BAD, so triangle EAD (is) to tri-
τὸ ΒΑΔ τὸν αὐτὸν ἔχει λόγον. ἴσων ἄρα ἐστὶ τὸ ΑΒΓ angle BAD. Thus, (triangles) ABC and EAD each have
[τρίγωνον] τῷ ΕΑΔ τριγώνῳ. the same ratio to BAD. Thus, [triangle] ABC is equal to
Τῶν ἄρα ἴσων καὶ μίαν μιᾷ ἴσην ἐχόντων γωνίαν triangle EAD [Prop. 5.9].

τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας Thus, in equal triangles also having one angle equal to
γωνίας· καὶ ὧς μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων one (angle) the sides about the equal angles (are) recip-
ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἐκεῖνα rocally proportional. And those triangles having one an-
ἴσα ἐστὶν· ὅπερ ἔδει δεῖξαι. gle equal to one angle for which the sides about the equal

angles (are) reciprocally proportional are equal. (Which

is) the very thing it was required to show.i�þ. Proposition 16
᾿Εὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν If four straight-lines are proportional then the rect-

ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν angle contained by the (two) outermost is equal to the
μέσων περιεχομένῳ ὀρθογωνίῳ· κἂν τὸ ὑπὸ τῶν ἄκρων rectangle contained by the middle (two). And if the rect-

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STOIQEIWN �þ. ELEMENTS BOOK 6
περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ὑπὸ τῶν μέσων περιε- angle contained by the (two) outermost is equal to the
χομένῳ ὀρθογωνίῳ, αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται. rectangle contained by the middle (two) then the four

straight-lines will be proportional.

Β ∆

ΖΕ

Α Γ

E F

῎Εστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ ΑΒ, ΓΔ, Ε, Ζ, Let AB, CD, E, and F be four proportional straight-
ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ· λέγω, ὅτι lines, (such that) as AB (is) to CD, so E (is) to F . I say
τὸ ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ that the rectangle contained by AB and F is equal to the
ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ. rectangle contained by CD and E.
῎Ηχθωσαν [γὰρ] ἀπὸ τῶν Α, Γ σημείων ταῖς ΑΒ, ΓΔ [For] let AG and CH have been drawn from points

εὐθείαις πρὸς ὀρθὰς αἱ ΑΗ, ΓΘ, καὶ κείσθω τῇ μὲν Ζ ἴση A and C at right-angles to the straight-lines AB and CD
ἡ ΑΗ, τῇ δὲ Ε ἴση ἡ ΓΘ. καὶ συμπεπληρώσθω τὰ ΒΗ, ΔΘ (respectively) [Prop. 1.11]. And let AG be made equal to
παραλληλόγραμμα. F , and CH to E [Prop. 1.3]. And let the parallelograms
Καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς BG and DH have been completed.

τὴν Ζ, ἴση δὲ ἡ μὲν Ε τῇ ΓΘ, ἡ δὲ Ζ τῇ ΑΗ, ἔστιν ἄρα ὡς And since as AB is to CD, so E (is) to F , and E
ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΓΘ πρὸς τὴν ΑΗ. τῶν ΒΗ, (is) equal CH , and F to AG, thus as AB is to CD, so
ΔΘ ἄρα παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ CH (is) to AG. Thus, in the parallelograms BG and DH
περὶ τὰς ἴσας γωνίας. ὧν δὲ ἰσογωνίων παραλληλογράμμων the sides about the equal angles are reciprocally propor-
ἀντιπεπόνθασιν αἱ πλευραί αἱ περὶ τὰς ἴσας γωνάις, ἴσα ἐστὶν tional. And those equiangular parallelograms in which
ἐκεῖνα· ἴσον ἄρα ἐστὶ τὸ ΒΗ παραλληλόγραμμον τῷ ΔΘ the sides about the equal angles are reciprocally propor-
παραλληλογράμμῳ. καί ἐστι τὸ μὲν ΒΗ τὸ ὑπὸ τῶν ΑΒ, Ζ· tional are equal [Prop. 6.14]. Thus, parallelogram BG
ἴση γὰρ ἡ ΑΗ τῇ Ζ· τὸ δὲ ΔΘ τὸ ὑπὸ τῶν ΓΔ, Ε· ἴση γὰρ ἡ is equal to parallelogram DH . And BG is the (rectangle
Ε τῇ ΓΘ· τὸ ἄρα ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον contained) by AB and F . For AG (is) equal to F . And
ἴσον ἐστὶ τῷ ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογώνιῳ. DH (is) the (rectangle contained) by CD and E. For E
Ἀλλὰ δὴ τὸ ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον (is) equal to CH . Thus, the rectangle contained by AB

ἴσον ἔστω τῷ ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ. and F is equal to the rectangle contained by CD and E.
λέγω, ὅτι αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται, ὡς ἡ ΑΒ And so, let the rectangle contained by AB and F be
πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ. equal to the rectangle contained by CD and E. I say that
Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ τὸ ὑπὸ τῶν the four straight-lines will be proportional, (so that) as

ΑΒ, Ζ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΓΔ, Ε, καί ἐστι τὸ μὲν ὑπὸ AB (is) to CD, so E (is) to F .
τῶν ΑΒ, Ζ τὸ ΒΗ· ἴση γάρ ἐστιν ἡ ΑΗ τῇ Ζ· τὸ δὲ ὑπὸ τῶν For, with the same construction, since the (rectangle
ΓΔ, Ε τὸ ΔΘ· ἴση γὰρ ἡ ΓΘ τῇ Ε· τὸ ἄρα ΒΗ ἴσον ἐστὶ contained) by AB and F is equal to the (rectangle con-
τῷ ΔΘ. καί ἐστιν ἰσογώνια. τῶν δὲ ἴσων καὶ ἰσογωνίων tained) by CD and E. And BG is the (rectangle con-
παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς tained) by AB and F . For AG is equal to F . And DH
ἴσας γωνίας. ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΓΘ (is) the (rectangle contained) by CD and E. For CH
πρὸς τὴν ΑΗ. ἴση δὲ ἡ μὲν ΓΘ τῇ Ε, ἡ δὲ ΑΗ τῇ Ζ· ἔστιν (is) equal to E. BG is thus equal to DH . And they
ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ. are equiangular. And in equal and equiangular parallel-
᾿Εὰν ἄρα τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ograms the sides about the equal angles are reciprocally

ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν proportional [Prop. 6.14]. Thus, as AB is to CD, so CH
μέσων περιεχομένῳ ὀρθογωνίῳ· κἂν τὸ ὑπὸ τῶν ἄκρων (is) to AG. And CH (is) equal to E, and AG to F . Thus,
περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ὑπὸ τῶν μέσων περιε- as AB is to CD, so E (is) to F .
χομένῳ ὀρθογωνίῳ, αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται· Thus, if four straight-lines are proportional then the
ὅπερ ἔδει δεῖξαι. rectangle contained by the (two) outermost is equal to

the rectangle contained by the middle (two). And if the

rectangle contained by the (two) outermost is equal to

172

STOIQEIWN �þ. ELEMENTS BOOK 6
the rectangle contained by the middle (two) then the four

straight-lines will be proportional. (Which is) the very

thing it was required to show.izþ. Proposition 17
᾿Εὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων If three straight-lines are proportional then the rect-

περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης τε- angle contained by the (two) outermost is equal to the
τραγώνῳ· κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον square on the middle (one). And if the rectangle con-
ἴσον ᾖ τῷ ἀπὸ τῆς μέσης τετραγώνῳ, αἱ τρεῖς εὐθεῖαι tained by the (two) outermost is equal to the square on
ἀνάλογον ἔσονται. the middle (one) then the three straight-lines will be pro-

portional.

∆

C
῎Εστωσαν τρεῖς εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, ὡς ἡ Α Let A, B and C be three proportional straight-lines,

πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ· λέγω, ὅτι τὸ ὑπὸ τῶν (such that) as A (is) to B, so B (is) to C. I say that the
Α, Γ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς Β rectangle contained by A and C is equal to the square on
τετραγώνῳ. B.
Κείσθω τῇ Β ἴση ἡ Δ. Let D be made equal to B [Prop. 1.3].
Καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν And since as A is to B, so B (is) to C, and B (is)

Γ, ἴση δὲ ἡ Β τῇ Δ, ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, ἡ Δ πρὸς equal to D, thus as A is to B, (so) D (is) to C. And if
τὴν Γ. ἐὰν δὲ τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν four straight-lines are proportional then the [rectangle]
ἄκρων περιεχόμενον [ὀρθογώνιον] ἴσον ἐστὶ τῷ ὑπὸ τῶν contained by the (two) outermost is equal to the rectan-
μέσων περιεχομένῳ ὀρθογωνίῳ. τὸ ἄρα ὑπὸ τῶν Α, Γ ἴσον gle contained by the middle (two) [Prop. 6.16]. Thus,
ἐστὶ τῷ ὑπὸ τῶν Β, Δ. ἀλλὰ τὸ ὑπὸ τῶν Β, Δ τὸ ἀπὸ τῆς Β the (rectangle contained) by A and C is equal to the
ἐστιν· ἴση γὰρ ἡ Β τῇ Δ· τὸ ἄρα ὑπὸ τῶν Α, Γ περιεχόμενον (rectangle contained) by B and D. But, the (rectangle
ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς Β τετραγώνῳ. contained) by B and D is the (square) on B. For B (is)
Ἀλλὰ δὴ τὸ ὑπὸ τῶν Α, Γ ἴσον ἔστω τῷ ἀπὸ τῆς Β· equal to D. Thus, the rectangle contained by A and C is

λέγω, ὅτι ἐστὶν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ. equal to the square on B.
Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ τὸ ὑπὸ τῶν Α, And so, let the (rectangle contained) by A and C be

Γ ἴσον ἐστὶ τῷ ἀπὸ τῆς Β, ἀλλὰ τὸ ἀπὸ τῆς Β τὸ ὑπὸ τῶν equal to the (square) on B. I say that as A is to B, so B
Β, Δ ἐστιν· ἴση γὰρ ἡ Β τῇ Δ· τὸ ἄρα ὑπὸ τῶν Α, Γ ἴσον (is) to C.
ἐστὶ τῷ ὑπὸ τῶν Β, Δ. ἐὰν δὲ τὸ ὑπὸ τῶν ἄκρων ἴσον ᾖ τῷ For, with the same construction, since the (rectangle
ὑπὸ τῶν μέσων, αἱ τέσσαρες εὐθεῖαι ἀνάλογόν εἰσιν. ἔστιν contained) by A and C is equal to the (square) on B.
ἄρα ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Δ πρὸς τὴν Γ. ἴση δὲ ἡ Β But, the (square) on B is the (rectangle contained) by B
τῇ Δ· ὡς ἄρα ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ. and D. For B (is) equal to D. The (rectangle contained)
᾿Εὰν ἄρα τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων by A and C is thus equal to the (rectangle contained) by

περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης τε- B and D. And if the (rectangle contained) by the (two)
τραγώνῳ· κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον outermost is equal to the (rectangle contained) by the
ἴσον ᾖ τῷ ἀπὸ τῆς μέσης τετραγώνῳ, αἱ τρεῖς εὐθεῖαι middle (two) then the four straight-lines are proportional
ἀνάλογον ἔσονται· ὅπερ ἔδει δεῖξαι. [Prop. 6.16]. Thus, as A is to B, so D (is) to C. And B

(is) equal to D. Thus, as A (is) to B, so B (is) to C.

Thus, if three straight-lines are proportional then the

rectangle contained by the (two) outermost is equal to
the square on the middle (one). And if the rectangle con-

tained by the (two) outermost is equal to the square on
the middle (one) then the three straight-lines will be pro-

portional. (Which is) the very thing it was required to

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STOIQEIWN �þ. ELEMENTS BOOK 6
show.ihþ. Proposition 18

Ἀπὸ τῆς δοθείσης εὐθείας τῷ δοθέντι εὐθυγράμμῳ To describe a rectilinear figure similar, and simi-
ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον ἀναγράψαι. larly laid down, to a given rectilinear figure on a given

straight-line.

∆

Γ Α

A B

῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν Let AB be the given straight-line, and CE the given
εὐθύγραμμον τὸ ΓΕ· δεῖ δὴ ἀπὸ τῆς ΑΒ εὐθείας τῷ ΓΕ rectilinear figure. So it is required to describe a rectilinear
εὐθυγράμμῳ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον figure similar, and similarly laid down, to the rectilinear
ἀναγράψαι. figure CE on the straight-line AB.
᾿Επεζεύχθω ἡ ΔΖ, καὶ συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ Let DF have been joined, and let GAB, equal to the

καὶ τοῖς πρὸς αὐτῇ σημείοις τοῖς Α, Β τῇ μὲν πρὸς τῷ Γ angle at C, and ABG, equal to (angle) CDF , have been
γωνίᾳ ἴση ἡ ὑπὸ ΗΑΒ, τῇ δὲ ὑπὸ ΓΔΖ ἴση ἡ ὑπὸ ΑΒΗ. λοιπὴ constructed on the straight-line AB at the points A and
ἄρα ἡ ὑπὸ ΓΖΔ τῇ ὑπὸ ΑΗΒ ἐστιν ἴση· ἰσογώνιον ἄρα ἐστὶ B on it (respectively) [Prop. 1.23]. Thus, the remain-
τὸ ΖΓΔ τρίγωνον τῷ ΗΑΒ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ing (angle) CFD is equal to AGB [Prop. 1.32]. Thus,
ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, οὕτως ἡ ΖΓ πρὸς τὴν ΗΑ, καὶ ἡ triangle FCD is equiangular to triangle GAB. Thus,
ΓΔ πρὸς τὴν ΑΒ. πάλιν συνεστάτω πρὸς τῇ ΒΗ εὐθείᾳ καὶ proportionally, as FD is to GB, so FC (is) to GA, and
τοῖς πρὸς αὐτῇ σημείοις τοῖς Β, Η τῇ μὲν ὑπὸ ΔΖΕ γωνίᾳ CD to AB [Prop. 6.4]. Again, let BGH , equal to an-
ἴση ἡ ὑπὸ ΒΗΘ, τῇ δὲ ὑπὸ ΖΔΕ ἴση ἡ ὑπὸ ΗΒΘ. λοιπὴ ἄρα gle DFE, and GBH equal to (angle) FDE, have been
ἡ πρὸς τῷ Ε λοιπῇ τῇ πρὸς τῷ Θ ἐστιν ἴση· ἰσογώνιον ἄρα constructed on the straight-line BG at the points G and
ἐστὶ τὸ ΖΔΕ τρίγωνον τῷ ΗΘΒ τριγώνῳ· ἀνάλογον ἄρα B on it (respectively) [Prop. 1.23]. Thus, the remain-
ἐστὶν ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, οὕτως ἡ ΖΕ πρὸς τὴν ΗΘ καὶ ing (angle) at E is equal to the remaining (angle) at H
ἡ ΕΔ πρὸς τὴν ΘΒ. ἐδείχθη δὲ καὶ ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, [Prop. 1.32]. Thus, triangle FDE is equiangular to tri-
οὕτως ἡ ΖΓ πρὸς τὴν ΗΑ καὶ ἡ ΓΔ πρὸς τὴν ΑΒ· καὶ ὡς angle GHB. Thus, proportionally, as FD is to GB, so
ἄρα ἡ ΖΓ πρὸς τὴν ΑΗ, οὕτως ἥ τε ΓΔ πρὸς τὴν ΑΒ καὶ ἡ FE (is) to GH , and ED to HB [Prop. 6.4]. And it was
ΖΕ πρὸς τὴν ΗΘ καὶ ἔτι ἡ ΕΑ πρὸς τὴν ΘΒ. καὶ ἐπεὶ ἴση also shown (that) as FD (is) to GB, so FC (is) to GA,
ἐστὶν ἡ μὲν ὑπὸ ΓΖΔ γωνία τῇ ὑπὸ ΑΗΒ, ἡ δὲ ὑπὸ ΔΖΕ τῇ and CD to AB. Thus, also, as FC (is) to AG, so CD (is)
ὑπὸ ΒΗΘ, ὅλη ἄρα ἡ ὑπὸ ΓΖΕ ὅλῃ τῇ ὑπὸ ΑΗΘ ἐστιν ἴση. to AB, and FE to GH , and, further, ED to HB. And
διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΒΘ ἐστιν ἴση. ἔστι since angle CFD is equal to AGB, and DFE to BGH ,
δὲ καὶ ἡ μὲν πρὸς τῷ Γ τῇ πρὸς τῷ Α ἴση, ἡ δὲ πρὸς τῷ Ε thus the whole (angle) CFE is equal to the whole (an-
τῇ πρὸς τῷ Θ. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΘ τῷ ΓΕ· καὶ τὰς gle) AGH . So, for the same (reasons), (angle) CDE is
περὶ τὰς ἴσας γωνίας αὐτῶν πλευρὰς ἀνάλογον ἔχει· ὅμοιον also equal to ABH . And the (angle) at C is also equal
ἄρα ἐστὶ τὸ ΑΘ εὐθύγραμμον τῷ ΓΕ εὐθυγράμμῳ. to the (angle) at A, and the (angle) at E to the (angle)
Ἀπὸ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τῷ δοθέντι at H . Thus, (figure) AH is equiangular to CE. And (the

εὐθυγράμμῳ τῷ ΓΕ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγρα- two figures) have the sides about their equal angles pro-
μμον ἀναγέγραπται τὸ ΑΘ· ὅπερ ἔδει ποιῆσαι. portional. Thus, the rectilinear figure AH is similar to the

rectilinear figure CE [Def. 6.1].

Thus, the rectilinear figure AH , similar, and similarly
laid down, to the given rectilinear figure CE has been

constructed on the given straight-line AB. (Which is) the

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STOIQEIWN �þ. ELEMENTS BOOK 6
very thing it was required to do.ijþ. Proposition 19

Τὰ ὅμοια τρίγωνα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ ἐστὶ Similar triangles are to one another in the squared†

τῶν ὁμολόγων πλευρῶν. ratio of (their) corresponding sides.

Β Η Γ

∆

Ε EG C

῎Εστω ὅμοια τρίγωνα τὰ ΑΒΓ, ΔΕΖ ἴσην ἔχοντα τὴν Let ABC and DEF be similar triangles having the
πρὸς τῷ Β γωνίαν τῇ πρὸς τῷ Ε, ὡς δὲ τὴν ΑΒ πρὸς τὴν ΒΓ, angle at B equal to the (angle) at E, and AB to BC, as
οὕτως τὴν ΔΕ πρὸς τὴν ΕΖ, ὥστε ὁμόλογον εἶναι τὴν ΒΓ DE (is) to EF , such that BC corresponds to EF . I say
τῇ ΕΖ· λέγω, ὅτι τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον that triangle ABC has a squared ratio to triangle DEF
διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. with respect to (that side) BC (has) to EF .
Εἰλήφθω γὰρ τῶν ΒΓ, ΕΖ τρίτη ἀνάλογον ἡ ΒΗ, ὥστε For let a third (straight-line), BG, have been taken

εἶναι ὡς τὴν ΒΓ πρὸς τὴν ΕΖ, οὕτως τὴν ΕΖ πρὸς τὴν ΒΗ· (which is) proportional to BC and EF , so that as BC
καὶ ἐπεζεύχθω ἡ ΑΗ. (is) to EF , so EF (is) to BG [Prop. 6.11]. And let AG
᾿Επεὶ οὖν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΔΕ πρὸς have been joined.

τὴν ΕΖ, ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν ΔΕ, οὕτως ἡ Therefore, since as AB is to BC, so DE (is) to EF ,
ΒΓ πρὸς τὴν ΕΖ. ἀλλ᾿ ὡς ἡ ΒΓ πρὸς ΕΖ, οὕτως ἐστιν ἡ ΕΖ thus, alternately, as AB is to DE, so BC (is) to EF
πρὸς ΒΗ. καὶ ὡς ἄρα ἡ ΑΒ πρὸς ΔΕ, οὕτως ἡ ΕΖ πρὸς ΒΗ· [Prop. 5.16]. But, as BC (is) to EF , so EF is to BG.
τῶν ΑΒΗ, ΔΕΖ ἄρα τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ And, thus, as AB (is) to DE, so EF (is) to BG. Thus,
περὶ τὰς ἴσας γωνάις. ὧν δὲ μίαν μιᾷ ἴσην ἐχόντων γωνίαν for triangles ABG and DEF , the sides about the equal
τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνάις, angles are reciprocally proportional. And those triangles
ἴσα ἐστὶν ἐκεῖνα. ἴσον ἄρα ἐστὶ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ having one (angle) equal to one (angle) for which the
τριγώνῳ. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς τὴν ΕΖ, οὕτως ἡ sides about the equal angles are reciprocally proportional
ΕΖ πρὸς τὴν ΒΗ, ἐὰν δὲ τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἡ are equal [Prop. 6.15]. Thus, triangle ABG is equal to
πρώτη πρὸς τὴν τρίτην διπλασίονα λόγον ἔχει ἤπερ πρὸς triangle DEF . And since as BC (is) to EF , so EF (is)
τὴν δευτέραν, ἡ ΒΓ ἄρα πρὸς τὴν ΒΗ διπλασίονα λόγον to BG, and if three straight-lines are proportional then
ἔχει ἤπερ ἡ ΓΒ πρὸς τὴν ΕΖ. ὡς δὲ ἡ ΓΒ πρὸς τὴν ΒΗ, the first has a squared ratio to the third with respect to
οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΒΗ τρίγωνον· καὶ τὸ the second [Def. 5.9], BC thus has a squared ratio to BG
ΑΒΓ ἄρα τρίγωνον πρὸς τὸ ΑΒΗ διπλασίονα λόγον ἔχει with respect to (that) CB (has) to EF . And as CB (is)
ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ἴσον δὲ τὸ ΑΒΗ τρίγωνον τῷ to BG, so triangle ABC (is) to triangle ABG [Prop. 6.1].
ΔΕΖ τριγώνῳ. καὶ τὸ ΑΒΓ ἄρα τρίγωνον πρὸς τὸ ΔΕΖ Thus, triangle ABC also has a squared ratio to (triangle)
τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ABG with respect to (that side) BC (has) to EF . And
Τὰ ἄρα ὅμοια τρίγωνα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ triangle ABG (is) equal to triangle DEF . Thus, trian-

ἐστὶ τῶν ὁμολόγων πλευρῶν. [ὅπερ ἔδει δεῖξαι.] gle ABC also has a squared ratio to triangle DEF with
respect to (that side) BC (has) to EF .

Thus, similar triangles are to one another in the
squared ratio of (their) corresponding sides. [(Which

is) the very thing it was required to show].Pìrisma. Corollary
᾿Εκ δὴ τούτου φανερόν, ὅτι, ἐὰν τρεῖς εὐθεῖαι ἀνάλογον So it is clear, from this, that if three straight-lines are

ὦσιν, ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ proportional, then as the first is to the third, so the figure

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STOIQEIWN �þ. ELEMENTS BOOK 6
τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ (described) on the first (is) to the similar, and similarly
ὁμοίως ἀναγραφόμενον. ὅπερ ἔδει δεῖξαι. described, (figure) on the second. (Which is) the very

thing it was required to show.

† Literally, “double”. kþ. Proposition 20
Τὰ ὅμοια πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ Similar polygons can be divided into equal numbers

εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ πολύγωνον of similar triangles corresponding (in proportion) to the
πρὸς τὸ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος wholes, and one polygon has to the (other) polygon a
πλευρὰ πρὸς τὴν ὁμόλογον πλευράν. squared ratio with respect to (that) a corresponding side

(has) to a corresponding side.

Β Η
Λ

∆

L
B

῎Εστω ὅμοια πολύγωνα τὰ ΑΒΓΔΕ, ΖΗΘΚΛ, ὁμόλογος Let ABCDE and FGHKL be similar polygons, and
δὲ ἔστω ἡ ΑΒ τῇ ΖΗ· λέγω, ὅτι τὰ ΑΒΓΔΕ, ΖΗΘΚΛ let AB correspond to FG. I say that polygons ABCDE
πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ εἰς ἴσα τὸ and FGHKL can be divided into equal numbers of simi-
πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ ΑΒΓΔΕ πολύγωνον lar triangles corresponding (in proportion) to the wholes,
πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ and (that) polygon ABCDE has a squared ratio to poly-
ΑΒ πρὸς τὴν ΖΗ. gon FGHKL with respect to that AB (has) to FG.
᾿Επεζεύχθωσαν αἱ ΒΕ, ΕΓ, ΗΛ, ΛΘ. Let BE, EC, GL, and LH have been joined.
Καὶ ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓΔΕ πολύγωνον τῷ And since polygon ABCDE is similar to polygon

ΖΗΘΚΛ πολυγώνῳ, ἴση ἐστὶν ἡ ὑπὸ ΒΑΕ γωνία τῇ ὑπὸ FGHKL, angle BAE is equal to angle GFL, and as BA
ΗΖΛ. καί ἐστιν ὡς ἡ ΒΑ πρὸς ΑΕ, οὕτως ἡ ΗΖ πρὸς ΖΛ. is to AE, so GF (is) to FL [Def. 6.1]. Therefore, since
ἐπεὶ οὖν δύο τρίγωνά ἐστι τὰ ΑΒΕ, ΖΗΛ μίαν γωνίαν μιᾷ ABE and FGL are two triangles having one angle equal
γωνίᾳ ἴσην ἔχοντα, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς to one angle and the sides about the equal angles propor-
ἀνάλογον, ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΕ τρίγωνον τῷ ΖΗΛ tional, triangle ABE is thus equiangular to triangle FGL
τριγώνῳ· ὥστε καὶ ὅμοιον· ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΒΕ γωνία [Prop. 6.6]. Hence, (they are) also similar [Prop. 6.4,
τῇ ὑπὸ ΖΗΛ. ἔστι δὲ καὶ ὅλη ἡ ὑπὸ ΑΒΓ ὅλῃ τῇ ὑπὸ Def. 6.1]. Thus, angle ABE is equal to (angle) FGL.
ΖΗΘ ἴση διὰ τὴν ὁμοιότητα τῶν πολυγώνων· λοιπὴ ἄρα ἡ And the whole (angle) ABC is equal to the whole (angle)
ὑπὸ ΕΒΓ γωνία τῇ ὑπὸ ΛΗΘ ἐστιν ἴση. καὶ ἐπεὶ διὰ τὴν FGH , on account of the similarity of the polygons. Thus,
ὁμοιότητα τῶν ΑΒΕ, ΖΗΛ τριγώνων ἐστὶν ὡς ἡ ΕΒ πρὸς the remaining angle EBC is equal to LGH . And since,
ΒΑ, οὕτως ἡ ΛΗ πρὸς ΗΖ, ἀλλὰ μὴν καὶ διὰ τὴν ὁμοιότητα on account of the similarity of triangles ABE and FGL,
τῶν πολυγώνων ἐστὶν ὡς ἡ ΑΒ πρὸς ΒΓ, οὕτως ἡ ΖΗ πρὸς as EB is to BA, so LG (is) to GF , but also, on account of
ΗΘ, δι᾿ ἴσου ἄρα ἐστὶν ὡς ἡ ΕΒ πρὸς ΒΓ, οὕτως ἡ ΛΗ πρὸς the similarity of the polygons, as AB is to BC, so FG (is)
ΗΘ, καὶ περὶ τὰς ἴσας γωνάις τὰς ὑπὸ ΕΒΓ, ΛΗΘ αἱ πλευραὶ to GH , thus, via equality, as EB is to BC, so LG (is) to
ἀνάλογόν εἰσιν· ἰσογώνιον ἄρα ἐστὶ τὸ ΕΒΓ τρίγωνον τῷ GH [Prop. 5.22], and the sides about the equal angles,
ΛΗΘ τριγώνῳ· ὥστε καὶ ὅμοιόν ἐστι τὸ ΕΒΓ τρίγωνον EBC and LGH , are proportional. Thus, triangle EBC is
τῷ ΛΗΘ τριγώνω. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΕΓΔ τρίγωνον equiangular to triangle LGH [Prop. 6.6]. Hence, triangle
ὅμοιόν ἐστι τῷ ΛΘΚ τριγώνῳ. τὰ ἄρα ὅμοια πολύγωνα τὰ EBC is also similar to triangle LGH [Prop. 6.4, Def. 6.1].
ΑΒΓΔΕ, ΖΗΘΚΛ εἴς τε ὅμοια τρίγωνα διῄρηται καὶ εἰς ἴσα So, for the same (reasons), triangle ECD is also similar

176

STOIQEIWN �þ. ELEMENTS BOOK 6
τὸ πλῆθος. to triangle LHK. Thus, the similar polygons ABCDE
Λέγω, ὅτι καὶ ὁμόλογα τοῖς ὅλοις, τουτέστιν ὥστε and FGHKL have been divided into equal numbers of

ἀνάλογον εἶναι τὰ τρίγωνα, καὶ ἡγούμενα μὲν εἶναι τὰ ΑΒΕ, similar triangles.
ΕΒΓ, ΕΓΔ, ἑπόμενα δὲ αὐτῶν τὰ ΖΗΛ, ΛΗΘ, ΛΘΚ, καὶ I also say that (the triangles) correspond (in propor-
ὅτι τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον tion) to the wholes. That is to say, the triangles are
διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν proportional: ABE, EBC, and ECD are the leading
ὁμόλογον πλευράν, τουτέστιν ἡ ΑΒ πρὸς τὴν ΖΗ. (magnitudes), and their (associated) following (magni-
᾿Επεζεύχθωσαν γὰρ αἱ ΑΓ, ΖΘ. καὶ ἐπεὶ διὰ τὴν tudes are) FGL, LGH , and LHK (respectively). (I) also

ὁμοιότητα τῶν πολυγώνων ἴση ἐστὶν ἡ ὑπὸ ΑΒΓ γωνία τῇ (say) that polygon ABCDE has a squared ratio to poly-
ὑπὸ ΖΗΘ, καί ἐστιν ὡς ἡ ΑΒ πρὸς ΒΓ, οὕτως ἡ ΖΗ πρὸς gon FGHKL with respect to (that) a corresponding side
ΗΘ, ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΖΗΘ τριγώνῳ· (has) to a corresponding side—that is to say, (side) AB
ἴση ἄρα ἐστὶν ἡ μὲν ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΗΖΘ, ἡ δὲ ὑπὸ to FG.
ΒΓΑ τῇ ὑπὸ ΗΘΖ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΒΑΜ γωνία For let AC and FH have been joined. And since angle
τῇ ὑπὸ ΗΖΝ, ἔστι δὲ καὶ ἡ ὑπὸ ΑΒΜ τῇ ὑπὸ ΖΗΝ ἴση, ABC is equal to FGH , and as AB is to BC, so FG (is) to
καὶ λοιπὴ ἄρα ἡ ὑπὸ ΑΜΒ λοιπῇ τῇ ὑπὸ ΖΝΗ ἴση ἐστίν· GH , on account of the similarity of the polygons, triangle
ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΜ τρίγωνον τῷ ΖΗΝ τριγώνῳ. ABC is equiangular to triangle FGH [Prop. 6.6]. Thus,
ὁμοίως δὴ δεῖξομεν, ὅτι καὶ τὸ ΒΜΓ τρίγωνον ἰσογώνιόν angle BAC is equal to GFH , and (angle) BCA to GHF .
ἐστι τῷ ΗΝΘ τριγώνῳ. ἀνάλογον ἄρα ἐστίν, ὡς μὲν ἡ ΑΜ And since angle BAM is equal to GFN , and (angle)
πρὸς ΜΒ, οὕτως ἡ ΖΝ πρὸς ΝΗ, ὡς δὲ ἡ ΒΜ πρὸς ΜΓ, ABM is also equal to FGN (see earlier), the remaining
οὕτως ἡ ΗΝ πρὸς ΝΘ· ὥστε καὶ δι᾿ ἴσου, ὡς ἡ ΑΜ πρὸς (angle) AMB is thus also equal to the remaining (angle)
ΜΓ, οὕτως ἡ ΖΝ πρὸς ΝΘ. ἀλλ᾿ ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως FNG [Prop. 1.32]. Thus, triangle ABM is equiangular
τὸ ΑΒΜ [τρίγωνον] πρὸς τὸ ΜΒΓ, καὶ τὸ ΑΜΕ πρὸς τὸ to triangle FGN . So, similarly, we can show that triangle
ΕΜΓ· πρὸς ἄλληλα γάρ εἰσιν ὡς αἱ βάσεις. καὶ ὡς ἄρα BMC is also equiangular to triangle GNH . Thus, pro-
ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπόμενων, οὕτως ἅπαντα portionally, as AM is to MB, so FN (is) to NG, and as
τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα· ὡς ἄρα τὸ ΑΜΒ BM (is) to MC, so GN (is) to NH [Prop. 6.4]. Hence,
τρίγωνον πρὸς τὸ ΒΜΓ, οὕτως τὸ ΑΒΕ πρὸς τὸ ΓΒΕ. αλλ᾿ also, via equality, as AM (is) to MC, so FN (is) to NH
ὡς τὸ ΑΜΒ πρὸς τὸ ΒΜΓ, οὕτως ἡ ΑΜ πρὸς ΜΓ· καὶ [Prop. 5.22]. But, as AM (is) to MC, so [triangle] ABM
ὡς ἄρα ἡ ΑΜ πρὸς ΜΓ, οὕτως τὸ ΑΒΕ τρίγωνον πρὸς τὸ is to MBC, and AME to EMC. For they are to one an-
ΕΒΓ τρίγωνον. διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ ΖΝ πρὸς ΝΘ, other as their bases [Prop. 6.1]. And as one of the leading
οὕτως τὸ ΖΗΛ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον. καί ἐστιν (magnitudes) is to one of the following (magnitudes), so
ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως ἡ ΖΝ πρὸς ΝΘ· καὶ ὡς ἄρα (the sum of) all the leading (magnitudes) is to (the sum
τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΒΕΓ τρίγωνον, οὕτως τὸ ΖΗΛ of) all the following (magnitudes) [Prop. 5.12]. Thus, as
τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον, καὶ ἐναλλὰξ ὡς τὸ ΑΒΕ triangle AMB (is) to BMC, so (triangle) ABE (is) to
τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΒΕΓ τρίγωνον CBE. But, as (triangle) AMB (is) to BMC, so AM (is)
πρὸς τὸ ΗΛΘ τρίγωνον. ὁμοίως δὴ δείξομεν ἐπιζευχθεισῶν to MC. Thus, also, as AM (is) to MC, so triangle ABE
τῶν ΒΔ, ΗΚ, ὅτι καὶ ὡς τὸ ΒΕΓ τρίγωνον πρὸς τὸ ΛΗΘ (is) to triangle EBC. And so, for the same (reasons), as
τρίγωνον, οὕτως τὸ ΕΓΔ τρίγωνον πρὸς τὸ ΛΘΚ τρίγωνον. FN (is) to NH , so triangle FGL (is) to triangle GLH .
καὶ ἐπεί ἐστιν ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, And as AM is to MC, so FN (is) to NH . Thus, also, as
οὕτως τὸ ΕΒΓ πρὸς τὸ ΛΗΘ, καὶ ἔτι τὸ ΕΓΔ πρὸς τὸ ΛΘΚ, triangle ABE (is) to triangle BEC, so triangle FGL (is)
καὶ ὡς ἄρα ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως to triangle GLH , and, alternately, as triangle ABE (is)
ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα· ἔστιν ἄρα to triangle FGL, so triangle BEC (is) to triangle GLH
ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ [Prop. 5.16]. So, similarly, we can also show, by joining
ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον. ἀλλὰ τὸ BD and GK, that as triangle BEC (is) to triangle LGH ,
ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον διπλασίονα λόγον so triangle ECD (is) to triangle LHK. And since as tri-
ἔχει ἤπερ ἡ ΑΒ ὁμόλογος πλευρὰ πρὸς τὴν ΖΗ ὁμόλογον angle ABE is to triangle FGL, so (triangle) EBC (is)
πλευράν· τὰ γὰρ ὅμοια τρίγωνα ἐν διπλασίονι λόγῳ ἐστὶ to LGH , and, further, (triangle) ECD to LHK, and also
τῶν ὁμολόγων πλευρῶν. καὶ τὸ ΑΒΓΔΕ ἄρα πολύγωνον as one of the leading (magnitudes is) to one of the fol-
πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ lowing, so (the sum of) all the leading (magnitudes is) to
ΑΒ ὁμόλογος πλευρὰ πρὸς τὴν ΖΗ ὁμόλογον πλευράν. (the sum of) all the following [Prop. 5.12], thus as trian-
Τὰ ἄρα ὅμοια πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται gle ABE is to triangle FGL, so polygon ABCDE (is) to

καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ polygon FGHKL. But, triangle ABE has a squared ratio

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STOIQEIWN �þ. ELEMENTS BOOK 6
πολύγωνον πρὸς τὸ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ to triangle FGL with respect to (that) the corresponding
ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν [ὅπερ ἔδει side AB (has) to the corresponding side FG. For, similar
δεῖξαι]. triangles are in the squared ratio of corresponding sides

[Prop. 6.14]. Thus, polygon ABCDE also has a squared

ratio to polygon FGHKL with respect to (that) the cor-
responding side AB (has) to the corresponding side FG.

Thus, similar polygons can be divided into equal num-

bers of similar triangles corresponding (in proportion) to
the wholes, and one polygon has to the (other) polygon a

squared ratio with respect to (that) a corresponding side

(has) to a corresponding side. [(Which is) the very thing
it was required to show].Pìrisma. Corollary

῾Ωσαύτως δὲ καὶ ἐπὶ τῶν [ὁμοίων] τετραπλεύρων δειχθή- And, in the same manner, it can also be shown for
σεται, ὅτι ἐν διπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. [similar] quadrilaterals that they are in the squared ratio
ἐδείχθη δὲ καὶ ἐπὶ τῶν τριγώνων· ὥστε καὶ καθόλου τὰ of (their) corresponding sides. And it was also shown for
ὅμοια εὐθύγραμμα σχήματα πρὸς ἄλληλα ἐν διπλασίονι triangles. Hence, in general, similar rectilinear figures are
λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. ὅπερ ἔδει δεῖξαι. also to one another in the squared ratio of (their) corre-

sponding sides. (Which is) the very thing it was required
to show.kaþ. Proposition 21

Τὰ τῷ αὐτῷ εὐθυγράμμῳ ὅμοια καὶ ἀλλήλοις ἐστὶν (Rectilinear figures) similar to the same rectilinear fig-
ὅμοια. ure are also similar to one another.

Α
Β

A
B

῎Εστω γὰρ ἑκάτερον τῶν Α, Β εὐθυγράμμων τῷ Γ Let each of the rectilinear figures A and B be similar
ὅμοιον· λέγω, ὅτι καὶ τὸ Α τῷ Β ἐστιν ὅμοιον. to (the rectilinear figure) C. I say that A is also similar to
᾿Επεὶ γὰρ ὅμοιόν ἐστι τὸ Α τῷ Γ, ἰσογώνιόν τέ ἐστιν B.

αὐτῷ καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. For since A is similar to C, (A) is equiangular to (C),
πάλιν, ἐπεὶ ὅμοιόν ἐστι τὸ Β τῷ Γ, ἰσογώνιόν τέ ἐστιν and has the sides about the equal angles proportional
αὐτῷ καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. [Def. 6.1]. Again, since B is similar to C, (B) is equian-
ἑκάτερον ἄρα τῶν Α, Β τῷ Γ ἰσογώνιόν τέ ἐστι καὶ τὰς gular to (C), and has the sides about the equal angles
περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει [ὥστε καὶ τὸ proportional [Def. 6.1]. Thus, A and B are each equian-
Α τῷ Β ἰσογώνιόν τέ ἐστι καὶ τὰς περὶ τὰς ἴσας γωνίας gular to C, and have the sides about the equal angles

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STOIQEIWN �þ. ELEMENTS BOOK 6
πλευρὰς ἀνάλογον ἔχει]. ὅμοιον ἄρα ἐστὶ τὸ Α τῷ Β· ὅπερ proportional [hence, A is also equiangular to B, and has
ἔδει δεῖξαι. the sides about the equal angles proportional]. Thus, A

is similar to B [Def. 6.1]. (Which is) the very thing it was
required to show.kbþ. Proposition 22

᾿Εὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, καὶ τὰ ἀπ᾿ αὐτῶν If four straight-lines are proportional then similar, and
εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον similarly described, rectilinear figures (drawn) on them
ἔσται· κἂν τὰ ἀπ᾿ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως will also be proportional. And if similar, and similarly
ἀναγεγραμμένα ἀνάλογον ᾖ, καὶ αὐτὰι αἱ εὐθεῖαι ἀνάλογον described, rectilinear figures (drawn) on them are pro-
ἔσονται. portional then the straight-lines themselves will also be

proportional.

Κ
Λ

Ξ Ο

Π Ρ

Η Θ

Α Β ∆

Μ Ν

A B C D

E
G

K
L

M N

Q R

O P

S
F

῎Εστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ ΑΒ, ΓΔ, ΕΖ, Let AB, CD, EF , and GH be four proportional
ΗΘ, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, καὶ straight-lines, (such that) as AB (is) to CD, so EF (is)
ἀναγεγράφθωσαν ἀπὸ μὲν τῶν ΑΒ, ΓΔ ὅμοιά τε καὶ ὁμοίως to GH . And let the similar, and similarly laid out, rec-
κείμενα εὐθύγραμμα τὰ ΚΑΒ, ΛΓΔ, ἀπὸ δὲ τῶν ΕΖ, ΗΘ tilinear figures KAB and LCD have been described on
ὅμοιά τε καὶ ὁμοίως κείμενα εὐθύγραμμα τὰ ΜΖ, ΝΘ· λέγω, AB and CD (respectively), and the similar, and similarly
ὅτι ἐστὶν ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ laid out, rectilinear figures MF and NH on EF and GH
ΝΘ. (respectively). I say that as KAB is to LCD, so MF (is)
Εἰλήφθω γὰρ τῶν μὲν ΑΒ, ΓΔ τρίτη ἀνάλογον ἡ Ξ, τῶν to NH .

δὲ ΕΖ, ΗΘ τρίτη ἀνάλογον ἡ Ο. καὶ ἐπεί ἐστιν ὡς μὲν ἡ ΑΒ For let a third (straight-line) O have been taken
πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, ὡς δὲ ἡ ΓΔ πρὸς (which is) proportional to AB and CD, and a third
τὴν Ξ, οὕτως ἡ ΗΘ πρὸς τὴν Ο, δι᾿ ἴσου ἄρα ἐστὶν ὡς ἡ (straight-line) P proportional to EF and GH [Prop. 6.11].
ΑΒ πρὸς τὴν Ξ, οὕτως ἡ ΕΖ πρὸς τὴν Ο. ἀλλ᾿ ὡς μὲν ἡ And since as AB is to CD, so EF (is) to GH , and as CD
ΑΒ πρὸς τὴν Ξ, οὕτως [καὶ] τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, ὡς δὲ (is) to O, so GH (is) to P , thus, via equality, as AB is to
ἡ ΕΖ πρὸς τὴν Ο, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ· καὶ ὡς ἄρα O, so EF (is) to P [Prop. 5.22]. But, as AB (is) to O, so
τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ. [also] KAB (is) to LCD, and as EF (is) to P , so MF
Ἀλλὰ δὴ ἔστω ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ (is) to NH [Prop. 5.19 corr.]. And, thus, as KAB (is) to

πρὸς τὸ ΝΘ· λέγω, ὅτι ἐστὶ καὶ ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, LCD, so MF (is) to NH .
οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. εἰ γὰρ μή ἐστιν, ὡς ἡ ΑΒ πρὸς And so let KAB be to LCD, as MF (is) to NH . I say
τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, ἔστω ὡς ἡ ΑΒ πρὸς τὴν also that as AB is to CD, so EF (is) to GH . For if as AB
ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΠΡ, καὶ ἀναγεγράφθω ἀπὸ τῆς is to CD, so EF (is) not to GH , let AB be to CD, as EF

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STOIQEIWN �þ. ELEMENTS BOOK 6
ΠΡ ὁποτέρῳ τῶν ΜΖ, ΝΘ ὅμοιόν τε καὶ ὁμοίως κείμενον (is) to QR [Prop. 6.12]. And let the rectilinear figure SR,
εὐθύγραμμον τὸ ΣΡ. similar, and similarly laid down, to either of MF or NH ,
᾿Επεὶ οὖν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς have been described on QR [Props. 6.18, 6.21].

τὴν ΠΡ, καὶ ἀναγέγραπται ἀπὸ μὲν τῶν ΑΒ, ΓΔ ὅμοιά τε Therefore, since as AB is to CD, so EF (is) to QR,
καὶ ὁμοίως κείμενα τὰ ΚΑΒ, ΛΓΔ, ἀπὸ δὲ τῶν ΕΖ, ΠΡ and the similar, and similarly laid out, (rectilinear fig-
ὅμοιά τε καὶ ὁμοίως κείμενα τὰ ΜΖ, ΣΡ, ἔστιν ἄρα ὡς τὸ ures) KAB and LCD have been described on AB and
ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΣΡ. ὑπόκειται CD (respectively), and the similar, and similarly laid out,
δὲ καὶ ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ (rectilinear figures) MF and SR on EF and QR (re-
ΝΘ· καὶ ὡς ἄρα τὸ ΜΖ πρὸς τὸ ΣΡ, οὕτως τὸ ΜΖ πρὸς sespectively), thus as KAB is to LCD, so MF (is) to
τὸ ΝΘ. τὸ ΜΖ ἄρα πρὸς ἑκάτερον τῶν ΝΘ, ΣΡ τὸν αὐτὸν SR (see above). And it was also assumed that as KAB
ἔχει λόγον· ἴσον ἄρα ἐστὶ τὸ ΝΘ τῷ ΣΡ. ἔστι δὲ αὐτῷ καὶ (is) to LCD, so MF (is) to NH . Thus, also, as MF (is)
ὅμοιον καὶ ὁμοίως κείμενον· ἴση ἄρα ἡ ΗΘ τῇ ΠΡ. καὶ ἐπεί to SR, so MF (is) to NH [Prop. 5.11]. Thus, MF has
ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΠΡ, ἴση the same ratio to each of NH and SR. Thus, NH is equal
δὲ ἡ ΠΡ τῇ ΗΘ, ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ to SR [Prop. 5.9]. And it is also similar, and similarly laid

ΕΖ πρὸς τὴν ΗΘ. out, to it. Thus, GH (is) equal to QR.† And since AB is
᾿Εὰν ἄρα τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, καὶ τὰ ἀπ᾿ to CD, as EF (is) to QR, and QR (is) equal to GH , thus

αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα as AB is to CD, so EF (is) to GH .
ἀνάλογον ἔσται· κἂν τὰ ἀπ᾿ αὐτῶν εὐθύγραμμα ὅμοιά τε Thus, if four straight-lines are proportional, then sim-
καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ᾖ, καὶ αὐτὰι αἱ εὐθεῖαι ilar, and similarly described, rectilinear figures (drawn)
ἀνάλογον ἔσονται· ὅπερ ἔδει δεῖξαι. on them will also be proportional. And if similar, and

similarly described, rectilinear figures (drawn) on them

are proportional then the straight-lines themselves will
also be proportional. (Which is) the very thing it was

required to show.

† Here, Euclid assumes, without proof, that if two similar figures are equal then any pair of corresponding sides is also equal.kgþ. Proposition 23
Τὰ ἰσογώνια παραλληλόγραμμα πρὸς ἄλληλα λόγον ἔχει Equiangular parallelograms have to one another the

τὸν συγκείμενον ἐκ τῶν πλευρῶν. ratio compounded† out of (the ratios of) their sides.
῎Εστω ἰσογώνια παραλληλόγραμμα τὰ ΑΓ, ΓΖ ἴσην Let AC and CF be equiangular parallelograms having

ἔχοντα τὴν ὑπὸ ΒΓΔ γωνίαν τῇ ὑπὸ ΕΓΗ· λέγω, ὅτι τὸ ΑΓ angle BCD equal to ECG. I say that parallelogram AC
παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον λόγον has to parallelogram CF the ratio compounded out of
ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν. (the ratios of) their sides.
Κείσθω γὰρ ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΒΓ τῇ ΓΗ· ἐπ᾿ For let BC be laid down so as to be straight-on to

εὐθείας ἄρα ἐστὶ καὶ ἡ ΔΓ τῇ ΓΕ. καὶ συμπεπληρώσθω τὸ CG. Thus, DC is also straight-on to CE [Prop. 1.14].
ΔΗ παραλληλόγραμμον, καὶ ἐκκείσθω τις εὐθεῖα ἡ Κ, καὶ And let the parallelogram DG have been completed. And
γεγονέτω ὡς μὲν ἡ ΒΓ πρὸς τὴν ΓΗ, οὕτως ἡ Κ πρὸς τὴν let some straight-line K have been laid down. And let it
Λ, ὡς δὲ ἡ ΔΓ πρὸς τὴν ΓΕ, οὕτως ἡ Λ πρὸς τὴν Μ. be contrived that as BC (is) to CG, so K (is) to L, and
Οἱ ἄρα λόγοι τῆς τε Κ πρὸς τὴν Λ καὶ τῆς Λ πρὸς as DC (is) to CE, so L (is) to M [Prop. 6.12].

τὴν Μ οἱ αὐτοί εἰσι τοῖς λόγοις τῶν πλευρῶν, τῆς τε ΒΓ Thus, the ratios of K to L and of L to M are the same
πρὸς τὴν ΓΗ καὶ τῆς ΔΓ πρὸς τὴν ΓΕ. ἀλλ᾿ ὁ τῆς Κ πρὸς as the ratios of the sides, (namely), BC to CG and DC
Μ λόγος σύγκειται ἔκ τε τοῦ τῆς Κ πρὸς Λ λόγου καὶ to CE (respectively). But, the ratio of K to M is com-
τοῦ τῆς Λ πρὸς Μ· ὥστε καὶ ἡ Κ πρὸς τὴν Μ λόγον ἔχει pounded out of the ratio of K to L and (the ratio) of L
τὸν συγκείμενον ἐκ τῶν πλευρῶν. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ to M . Hence, K also has to M the ratio compounded
πρὸς τὴν ΓΗ, οὕτως τὸ ΑΓ παραλληλόγραμμον πρὸς τὸ out of (the ratios of) the sides (of the parallelograms).
ΓΘ, ἀλλ᾿ ὡς ἡ ΒΓ πρὸς τὴν ΓΗ, οὕτως ἡ Κ πρὸς τὴν Λ, And since as BC is to CG, so parallelogram AC (is) to
καὶ ὡς ἄρα ἡ Κ πρὸς τὴν Λ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΘ. CH [Prop. 6.1], but as BC (is) to CG, so K (is) to L,
πάλιν, ἐπεί ἐστιν ὡς ἡ ΔΓ πρὸς τὴν ΓΕ, οὕτως τὸ ΓΘ πα- thus, also, as K (is) to L, so (parallelogram) AC (is) to
ραλληλόγραμμον πρὸς τὸ ΓΖ, ἀλλ᾿ ὡς ἡ ΔΓ πρὸς τὴν ΓΕ, CH . Again, since as DC (is) to CE, so parallelogram

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STOIQEIWN �þ. ELEMENTS BOOK 6
οὕτως ἡ Λ πρὸς τὴν Μ, καὶ ὡς ἄρα ἡ Λ πρὸς τὴν Μ, οὕτως CH (is) to CF [Prop. 6.1], but as DC (is) to CE, so L
τὸ ΓΘ παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον. (is) to M , thus, also, as L (is) to M , so parallelogram
ἐπεὶ οὖν ἐδείχθη, ὡς μὲν ἡ Κ πρὸς τὴν Λ, οὕτως τὸ ΑΓ CH (is) to parallelogram CF . Therefore, since it was
παραλληλόγραμμον πρὸς τὸ ΓΘ παραλληλόγραμμον, ὡς δὲ shown that as K (is) to L, so parallelogram AC (is) to
ἡ Λ πρὸς τὴν Μ, οὕτως τὸ ΓΘ παραλληλόγραμμον πρὸς τὸ parallelogram CH , and as L (is) to M , so parallelogram
ΓΖ παραλληλόγραμμον, δι᾿ ἴσου ἄρα ἐστὶν ὡς ἡ Κ πρὸς τὴν CH (is) to parallelogram CF , thus, via equality, as K is
Μ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΖ παραλληλόγραμμον. ἡ δὲ Κ to M , so (parallelogram) AC (is) to parallelogram CF
πρὸς τὴν Μ λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν· [Prop. 5.22]. And K has to M the ratio compounded out
καὶ τὸ ΑΓ ἄρα πρὸς τὸ ΓΖ λόγον ἔχει τὸν συγκείμενον ἐκ of (the ratios of) the sides (of the parallelograms). Thus,
τῶν πλευρῶν. (parallelogram) AC also has to (parallelogram) CF the

ratio compounded out of (the ratio of) their sides.

Β Γ

∆ Θ

Ζ F

C G

D H

Τὰ ἄρα ἰσογώνια παραλληλόγραμμα πρὸς ἄλληλα λόγον Thus, equiangular parallelograms have to one another
ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν· ὅπερ ἔδει δεῖξαι. the ratio compounded out of (the ratio of) their sides.

(Which is) the very thing it was required to show.

† In modern terminology, if two ratios are “compounded” then they are multiplied together.kdþ. Proposition 24
Παντὸς παραλληλογράμμου τὰ περὶ τὴν διάμετρον πα- In any parallelogram the parallelograms about the di-

ραλληλόγραμμα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις. agonal are similar to the whole, and to one another.
῎Εστω παραλληλόγραμμον τὸ ΑΒΓΔ, διάμετρος δὲ Let ABCD be a parallelogram, and AC its diagonal.

αὐτοῦ ἡ ΑΓ, περὶ δὲ τὴν ΑΓ παραλληλόγραμμα ἔστω τὰ ΕΗ, And let EG and HK be parallelograms about AC. I say
ΘΚ· λέγω, ὅτι ἑκάτερον τῶν ΕΗ, ΘΚ παραλληλογράμμων that the parallelograms EG and HK are each similar to
ὅμοιόν ἐστι ὅλῳ τῷ ΑΒΓΔ καὶ ἀλλήλοις. the whole (parallelogram) ABCD, and to one another.
᾿Επεὶ γὰρ τριγώνου τοῦ ΑΒΓ παρὰ μίαν τῶν πλευρῶν For since EF has been drawn parallel to one of the

τὴν ΒΓ ἦκται ἡ ΕΖ, ἀνάλογόν ἐστιν ὡς ἡ ΒΕ πρὸς τὴν sides BC of triangle ABC, proportionally, as BE is to
ΕΑ, οὕτως ἡ ΓΖ πρὸς τὴν ΖΑ. πάλιν, ἐπεὶ τριγώνου τοῦ EA, so CF (is) to FA [Prop. 6.2]. Again, since FG has
ΑΓΔ παρὰ μίαν τὴν ΓΔ ἦκται ἡ ΖΗ, ἀνάλογόν ἐστιν ὡς ἡ been drawn parallel to one (of the sides) CD of trian-
ΓΖ πρὸς τὴν ΖΑ, οὕτως ἡ ΔΗ πρὸς τὴν ΗΑ. ἀλλ᾿ ὡς ἡ gle ACD, proportionally, as CF is to FA, so DG (is) to
ΓΖ πρὸς τὴν ΖΑ, οὕτως ἐδείχθη καὶ ἡ ΒΕ πρὸς τὴν ΕΑ· GA [Prop. 6.2]. But, as CF (is) to FA, so it was also
καὶ ὡς ἄρα ἡ ΒΕ πρὸς τὴν ΕΑ, οὕτως ἡ ΔΗ πρὸς τὴν shown (is) BE to EA. And thus as BE (is) to EA, so
ΗΑ, καὶ συνθέντι ἄρα ὡς ἡ ΒΑ πρὸς ΑΕ, οὕτως ἡ ΔΑ DG (is) to GA. And, thus, compounding, as BA (is) to
πρὸς ΑΗ, καὶ ἐναλλὰξ ὡς ἡ ΒΑ πρὸς τὴν ΑΔ, οὕτως ἡ AE, so DA (is) to AG [Prop. 5.18]. And, alternately, as
ΕΑ πρὸς τὴν ΑΗ. τῶν ἄρα ΑΒΓΔ, ΕΗ παραλληλογράμμων BA (is) to AD, so EA (is) to AG [Prop. 5.16]. Thus,
ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὴν κοινὴν γωνίαν τὴν ὑπὸ in parallelograms ABCD and EG the sides about the
ΒΑΔ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΗΖ τῇ ΔΓ, ἴση ἐστὶν common angle BAD are proportional. And since GF is
ἡ μὲν ὑπὸ ΑΖΗ γωνία τῇ ὑπὸ ΔΓΑ· καὶ κοινὴ τῶν δύο parallel to DC, angle AFG is equal to DCA [Prop. 1.29].

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STOIQEIWN �þ. ELEMENTS BOOK 6
τριγώνων τῶν ΑΔΓ, ΑΗΖ ἡ ὑπὸ ΔΑΓ γωνία· ἰσογώνιον And angle DAC (is) common to the two triangles ADC
ἄρα ἐστὶ τὸ ΑΔΓ τρίγωνον τῷ ΑΗΖ τριγώνῳ. διὰ τὰ αὐτὰ and AGF . Thus, triangle ADC is equiangular to triangle
δὴ καὶ τὸ ΑΓΒ τρίγωνον ἰσογώνιόν ἐστι τῷ ΑΖΕ τριγώνῳ, AGF [Prop. 1.32]. So, for the same (reasons), triangle
καὶ ὅλον τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΕΗ παραλλη- ACB is equiangular to triangle AFE, and the whole par-
λογράμμῳ ἰσογώνιόν ἐστιν. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΑΔ allelogram ABCD is equiangular to parallelogram EG.
πρὸς τὴν ΔΓ, οὕτως ἡ ΑΗ πρὸς τὴν ΗΖ, ὡς δὲ ἡ ΔΓ πρὸς Thus, proportionally, as AD (is) to DC, so AG (is) to
τὴν ΓΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ, ὡς δὲ ἡ ΑΓ πρὸς τὴν GF , and as DC (is) to CA, so GF (is) to FA, and as AC
ΓΒ, οὕτως ἡ ΑΖ πρὸς τὴν ΖΕ, καὶ ἔτι ὡς ἡ ΓΒ πρὸς τὴν (is) to CB, so AF (is) to FE, and, further, as CB (is)
ΒΑ, οὕτως ἡ ΖΕ πρὸς τὴν ΕΑ. καὶ ἐπεὶ ἐδείχθη ὡς μὲν to BA, so FE (is) to EA [Prop. 6.4]. And since it was
ἡ ΔΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ, ὡς δὲ ἡ shown that as DC is to CA, so GF (is) to FA, and as
ΑΓ πρὸς τὴν ΓΒ, οὕτως ἡ ΑΖ πρὸς τὴν ΖΕ, δι᾿ ἴσου ἄρα AC (is) to CB, so AF (is) to FE, thus, via equality, as
ἐστὶν ὡς ἡ ΔΓ πρὸς τὴν ΓΒ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΕ. DC is to CB, so GF (is) to FE [Prop. 5.22]. Thus, in
τῶν ἄρα ΑΒΓΔ, ΕΗ παραλληλογράμμων ἀνάλογόν εἰσιν parallelograms ABCD and EG the sides about the equal
αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας· ὅμοιον ἄρα ἐστὶ τὸ angles are proportional. Thus, parallelogram ABCD is
ΑΒΓΔ παραλληλογράμμον τῷ ΕΗ παραλληλογράμμῳ. διὰ similar to parallelogram EG [Def. 6.1]. So, for the same
τὰ αὐτὰ δὴ τὸ ΑΒΓΔ παραλληλόγραμμον καὶ τῷ ΚΘ πα- (reasons), parallelogram ABCD is also similar to par-
ραλληλογράμμῳ ὅμοιόν ἐστιν· ἑκάτερον ἄρα τῶν ΕΗ, ΘΚ allelogram KH . Thus, parallelograms EG and HK are
παραλληλογράμμων τῷ ΑΒΓΔ [παραλληλογράμμῳ] ὅμοιόν each similar to [parallelogram] ABCD. And (rectilin-
ἐστιν. τὰ δὲ τῷ αὐτῷ εὐθυγράμμῳ ὅμοια καὶ ἀλλήλοις ἐστὶν ear figures) similar to the same rectilinear figure are also
ὅμοια· καὶ τὸ ΕΗ ἄρα παραλληλόγραμμον τῷ ΘΚ παραλλη- similar to one another [Prop. 6.21]. Thus, parallelogram
λογράμμῳ ὅμοιόν ἐστιν. EG is also similar to parallelogram HK.

∆ Γ

Α Β

Ζ
Η

EA B

G
F

D K C

Παντὸς ἄρα παραλληλογράμμου τὰ περὶ τὴν διάμετρον Thus, in any parallelogram the parallelograms about
παραλληλόγραμμα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις· ὅπερ the diagonal are similar to the whole, and to one another.
ἔδει δεῖξαι. (Which is) the very thing it was required to show.keþ. Proposition 25
Τῷ δοθέντι εὐθυγράμμῳ ὅμοιον καὶ ἄλλῳ τῷ δοθέντι To construct a single (rectilinear figure) similar to a

ἴσον τὸ αὐτὸ συστήσασθαι. given rectilinear figure, and equal to a different given rec-
tilinear figure.

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STOIQEIWN �þ. ELEMENTS BOOK 6
Η Θ

∆

Λ Ε Μ

Ζ F

B
C

G H

L M

῎Εστω τὸ μὲν δοθὲν εὐθύγραμμον, ᾧ δεῖ ὅμοιον Let ABC be the given rectilinear figure to which it is
συστήσασθαι, τὸ ΑΒΓ, ᾧ δὲ δεῖ ἴσον, τὸ Δ· δεῖ δὴ τῷ required to construct a similar (rectilinear figure), and D
μὲν ΑΒΓ ὅμοιον, τῷ δὲ Δ ἴσον τὸ αὐτὸ συστήσασθαι. the (rectilinear figure) to which (the constructed figure)
Παραβεβλήσθω γὰρ παρὰ μὲν τὴν ΒΓ τῷ ΑΒΓ τριγώνῳ is required (to be) equal. So it is required to construct

ἴσον παραλληλόγραμμον τὸ ΒΕ, παρὰ δὲ τὴν ΓΕ τῷ Δ ἴσον a single (rectilinear figure) similar to ABC, and equal to
παραλληλόγραμμον τὸ ΓΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΓΕ, ἥ ἐστιν D.
ἴση τῇ ὑπὸ ΓΒΛ. ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ μὲν ΒΓ τῇ ΓΖ, ἡ For let the parallelogram BE, equal to triangle ABC,
δὲ ΛΕ τῇ ΕΜ. καὶ εἰλήφθω τῶν ΒΓ, ΓΖ μέση ἀνάλογον ἡ have been applied to (the straight-line) BC [Prop. 1.44],
ΗΘ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΗΘ τῷ ΑΒΓ ὅμοιόν τε καὶ and the parallelogram CM , equal to D, (have been ap-
ὁμοίως κείμενον τὸ ΚΗΘ. plied) to (the straight-line) CE, in the angle FCE, which
Καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς τὴν ΗΘ, οὕτως ἡ ΗΘ is equal to CBL [Prop. 1.45]. Thus, BC is straight-on to

πρὸς τὴν ΓΖ, ἐὰν δὲ τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἔστιν CF , and LE to EM [Prop. 1.14]. And let the mean pro-
ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης portion GH have been taken of BC and CF [Prop. 6.13].
εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀνα- And let KGH , similar, and similarly laid out, to ABC
γραφόμενον, ἔστιν ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΖ, οὕτως τὸ have been described on GH [Prop. 6.18].
ΑΒΓ τρίγωνον πρὸς τὸ ΚΗΘ τρίγωνον. ἀλλὰ καὶ ὡς ἡ ΒΓ And since as BC is to GH , so GH (is) to CF , and if
πρὸς τὴν ΓΖ, οὕτως τὸ ΒΕ παραλληλόγραμμον πρὸς τὸ ΕΖ three straight-lines are proportional then as the first is to
παραλληλόγραμμον. καὶ ὡς ἄρα τὸ ΑΒΓ τρίγωνον πρὸς τὸ the third, so the figure (described) on the first (is) to the
ΚΗΘ τρίγωνον, οὕτως τὸ ΒΕ παραλληλόγραμμον πρὸς τὸ similar, and similarly described, (figure) on the second
ΕΖ παραλληλόγραμμον· ἐναλλὰξ ἄρα ὡς τὸ ΑΒΓ τρίγωνον [Prop. 6.19 corr.], thus as BC is to CF , so triangle ABC
πρὸς τὸ ΒΕ παραλληλόγραμμον, οὕτως τὸ ΚΗΘ τρίγωνον (is) to triangle KGH . But, also, as BC (is) to CF , so
πρὸς τὸ ΕΖ παραλληλόγραμμον. ἴσον δὲ τὸ ΑΒΓ τρίγωνον parallelogram BE (is) to parallelogram EF [Prop. 6.1].
τῷ ΒΕ παραλληλογράμμῳ· ἴσον ἄρα καὶ τὸ ΚΗΘ τρίγωνον And, thus, as triangle ABC (is) to triangle KGH , so par-
τῷ ΕΖ παραλληλογράμμῳ. ἀλλὰ τὸ ΕΖ παραλληλόγραμμον allelogram BE (is) to parallelogram EF . Thus, alter-
τῷ Δ ἐστιν ἴσον· καὶ τὸ ΚΗΘ ἄρα τῷ Δ ἐστιν ἴσον. ἔστι nately, as triangle ABC (is) to parallelogram BE, so tri-
δὲ τὸ ΚΗΘ καὶ τῷ ΑΒΓ ὅμοιον. angle KGH (is) to parallelogram EF [Prop. 5.16]. And
Τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓ ὅμοιον καὶ ἄλλῳ triangle ABC (is) equal to parallelogram BE. Thus, tri-

τῷ δοθέντι τῷ Δ ἴσον τὸ αὐτὸ συνέσταται τὸ ΚΗΘ· ὅπερ angle KGH (is) also equal to parallelogram EF . But,
ἔδει ποιῆσαι. parallelogram EF is equal to D. Thus, KGH is also equal

to D. And KGH is also similar to ABC.

Thus, a single (rectilinear figure) KGH has been con-
structed (which is) similar to the given rectilinear figure

ABC, and equal to a different given (rectilinear figure)

D. (Which is) the very thing it was required to do.k�þ. Proposition 26
᾿Εὰν ἀπὸ παραλληλογράμμου παραλληλόγραμμον ἀφαι- If from a parallelogram a(nother) parallelogram is

ρεθῇ ὅμοιόν τε τῷ ὅλῳ καὶ ὁμοίως κείμενον κοινὴν γωνίαν subtracted (which is) similar, and similarly laid out, to
ἔχον αὐτῷ, περὶ τὴν αὐτὴν διάμετρόν ἐστι τῷ ὅλῳ. the whole, having a common angle with it, then (the sub-
Ἀπὸ γὰρ παραλληλογράμμου τοῦ ΑΒΓΔ παραλληλόγρα- tracted parallelogram) is about the same diagonal as the

μμον ἀφῃρήσθω τὸ ΑΖ ὅμοιον τῷ ΑΒΓΔ καὶ ὁμοίως whole.
κείμενον κοινὴν γωνίαν ἔχον αὐτῷ τὴν ὑπὸ ΔΑΒ· λέγω, For, from parallelogram ABCD, let (parallelogram)

183

STOIQEIWN �þ. ELEMENTS BOOK 6
ὅτι περὶ τὴν αὐτὴν διάμετρόν ἐστι τὸ ΑΒΓΔ τῷ ΑΖ. AF have been subtracted (which is) similar, and similarly

laid out, to ABCD, having the common angle DAB with

it. I say that ABCD is about the same diagonal as AF .

ΗΑ

∆

ΖΕ

Κ K

A G D

Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, ἔστω [αὐτῶν] διάμετρος ἡ For (if) not, then, if possible, let AHC be [ABCD’s]
ΑΘΓ, καὶ ἐκβληθεῖσα ἡ ΗΖ διήχθω ἐπὶ τὸ Θ, καὶ ἤχθω diagonal. And producing GF , let it have been drawn
διὰ τοῦ Θ ὁπορέρᾳ τῶν ΑΔ, ΒΓ παράλληλος ἡ ΘΚ. through to (point) H . And let HK have been drawn
᾿Επεὶ οὖν περὶ τὴν αὐτὴν διάμετρόν ἐστι τὸ ΑΒΓΔ τῷ through (point) H , parallel to either of AD or BC

ΚΗ, ἔστιν ἄρα ὡς ἡ ΔΑ πρὸς τὴν ΑΒ, οὕτως ἡ ΗΑ πρὸς [Prop. 1.31].
τὴν ΑΚ. ἔστι δὲ καὶ διὰ τὴν ὁμοιότητα τῶν ΑΒΓΔ, ΕΗ καὶ Therefore, since ABCD is about the same diagonal as
ὡς ἡ ΔΑ πρὸς τὴν ΑΒ, οὕτως ἡ ΗΑ πρὸς τὴν ΑΕ· καὶ ὡς KG, thus as DA is to AB, so GA (is) to AK [Prop. 6.24].
ἄρα ἡ ΗΑ πρὸς τὴν ΑΚ, οὕτως ἡ ΗΑ πρὸς τὴν ΑΕ. ἡ ΗΑ And, on account of the similarity of ABCD and EG, also,
ἄρα πρὸς ἑκατέραν τῶν ΑΚ, ΑΕ τὸν αὐτὸν ἔχει λόγον. ἴση as DA (is) to AB, so GA (is) to AE. Thus, also, as GA
ἄρα ἐστὶν ἡ ΑΕ τῇ ΑΚ ἡ ἐλάττων τῇ μείζονι· ὅπερ ἐστὶν (is) to AK, so GA (is) to AE. Thus, GA has the same
ἀδύνατον. οὐκ ἄρα οὔκ ἐστι περὶ τὴν αὐτὴν διάμετρον τὸ ratio to each of AK and AE. Thus, AE is equal to AK
ΑΒΓΔ τῷ ΑΖ· περὶ τὴν αὐτὴν ἄρα ἐστὶ διάμετρον τὸ ΑΒΓΔ [Prop. 5.9], the lesser to the greater. The very thing is
παραλληλόγραμμον τῷ ΑΖ παραλληλογράμμῳ. impossible. Thus, ABCD is not not about the same di-
᾿Εὰν ἄρα ἀπὸ παραλληλογράμμου παραλληλόγραμμον agonal as AF . Thus, parallelogram ABCD is about the

ἀφαιρεθῇ ὅμοιόν τε τῷ ὅλῳ καὶ ὁμοίως κείμενον κοινὴν same diagonal as parallelogram AF .
γωνίαν ἔχον αὐτῷ, περὶ τὴν αὐτὴν διάμετρόν ἐστι τῷ ὅλῳ· Thus, if from a parallelogram a(nother) parallelogram
ὅπερ ἔδει δεῖξαι. is subtracted (which is) similar, and similarly laid out,

to the whole, having a common angle with it, then (the

subtracted parallelogram) is about the same diagonal as
the whole. (Which is) the very thing it was required to

show.kzþ. Proposition 27
Πάντων τῶν παρὰ τὴν αὐτὴν εὐθεῖαν παραβαλλομένων Of all the parallelograms applied to the same straight-

παραλληλογράμμων καὶ ἐλλειπόντων εἴδεσι παραλληλογράμ- line, and falling short by parallelogrammic figures similar,
μοις ὁμοίοις τε καὶ ὁμοίως κειμένοις τῷ ἀπὸ τῆς ἡμισείας and similarly laid out, to the (parallelogram) described
ἀναγραφομένῳ μέγιστόν ἐστι τὸ ἀπὸ τῆς ἡμισείας παρα- on half (the straight-line), the greatest is the [parallelo-
βαλλόμενον [παραλληλόγραμμον] ὅμοιον ὂν τῷ ἐλλείμμαντι. gram] applied to half (the straight-line) which (is) similar
῎Εστω εὐθεῖα ἡ ΑΒ καὶ τετμήσθω δίχα κατὰ τὸ Γ, to (that parallelogram) by which it falls short.

καὶ παραβεβλήσθω παρὰ τὴν ΑΒ εὐθεῖαν τὸ ΑΔ παραλ- Let AB be a straight-line, and let it have been cut in
ληλόγραμμον ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΔΒ ἀνα- half at (point) C [Prop. 1.10]. And let the parallelogram
γραφέντι ἀπὸ τῆς ἡμισείας τῆς ΑΒ, τουτέστι τῆς ΓΒ· λέγω, AD have been applied to the straight-line AB, falling
ὅτι πάντων τῶν παρὰ τὴν ΑΒ παραβαλλομένων παραλλη- short by the parallelogrammic figure DB (which is) ap-
λογράμμων καὶ ἐλλειπόντων εἴδεσι [παραλληλογράμμοις] plied to half of AB—that is to say, CB. I say that of all
ὁμοίοις τε καὶ ὁμοίως κειμένοις τῷ ΔΒ μέγιστόν ἐστι τὸ the parallelograms applied to AB, and falling short by

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STOIQEIWN �þ. ELEMENTS BOOK 6
ΑΔ. παραβεβλήσθω γὰρ παρὰ τὴν ΑΒ εὐθεῖαν τὸ ΑΖ πα- [parallelogrammic] figures similar, and similarly laid out,
ραλληλόγραμμον ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΖΒ to DB, the greatest is AD. For let the parallelogram AF
ὁμοίῳ τε καὶ ὁμοίως κειμένῳ τῷ ΔΒ· λέγω, ὅτι μεῖζόν ἐστι have been applied to the straight-line AB, falling short by
τὸ ΑΔ τοῦ ΑΖ. the parallelogrammic figure FB (which is) similar, and

similarly laid out, to DB. I say that AD is greater than
AF .

Ζ Θ

Γ Κ

∆

Μ
Η

Ν
F

A C K

᾿Επεὶ γὰρ ὅμοιόν ἐστι τὸ ΔΒ παραλληλόγραμμον τῷ ΖΒ For since parallelogram DB is similar to parallelo-
παραλληλογράμμῳ, περὶ τὴν αὐτήν εἰσι διάμετρον. ἤχθω gram FB, they are about the same diagonal [Prop. 6.26].
αὐτῶν διάμετρος ἡ ΔΒ, καὶ καταγεγράφθω τὸ σχῆμα. Let their (common) diagonal DB have been drawn, and
᾿Επεὶ οὖν ἴσον ἐστὶ τὸ ΓΖ τῷ ΖΕ, κοινὸν δὲ τὸ ΖΒ, let the (rest of the) figure have been described.

ὅλον ἄρα τὸ ΓΘ ὅλῳ τῷ ΚΕ ἐστιν ἴσον. ἀλλὰ τὸ ΓΘ τῷ Therefore, since (complement) CF is equal to (com-
ΓΗ ἐστιν ἴσον, ἐπεὶ καὶ ἡ ΑΓ τῇ ΓΒ. καὶ τὸ ΗΓ ἄρα τῷ ΕΚ plement) FE [Prop. 1.43], and (parallelogram) FB is
ἐστιν ἴσον. κοινὸν προσκείσθω τὸ ΓΖ· ὅλον ἄρα τὸ ΑΖ τῷ common, the whole (parallelogram) CH is thus equal
ΛΜΝ γνώμονί ἐστιν ἴσον· ὥστε τὸ ΔΒ παραλληλόγραμμον, to the whole (parallelogram) KE. But, (parallelogram)
τουτέστι τὸ ΑΔ, τοῦ ΑΖ παραλληλογράμμου μεῖζόν ἐστιν. CH is equal to CG, since AC (is) also (equal) to CB
Πάντων ἄρα τῶν παρὰ τὴν αὐτὴν εὐθεῖαν παραβαλ- [Prop. 6.1]. Thus, (parallelogram) GC is also equal

λομένων παραλληλογράμμων καὶ ἐλλειπόντων εἴδεσι παραλ- to EK. Let (parallelogram) CF have been added to
ληλογράμμοις ὁμοίοις τε καὶ ὁμοίως κειμένοις τῷ ἀπὸ τῆς both. Thus, the whole (parallelogram) AF is equal to
ἡμισείας ἀναγραφομένῳ μέγιστόν ἐστι τὸ ἀπὸ τῆς ἡμισείας the gnomon LMN . Hence, parallelogram DB—that is to
παραβληθέν· ὅπερ ἔδει δεῖξαι. say, AD—is greater than parallelogram AF .

Thus, for all parallelograms applied to the same

straight-line, and falling short by a parallelogrammic
figure similar, and similarly laid out, to the (parallelo-

gram) described on half (the straight-line), the greatest

is the [parallelogram] applied to half (the straight-line).
(Which is) the very thing it was required to show.khþ. Proposition 28†

Παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι εὐθυγράμμῳ To apply a parallelogram, equal to a given rectilin-
ἴσον παραλληλόγραμμον παραβαλεῖν ἐλλεῖπον εἴδει πα- ear figure, to a given straight-line, (the applied parallel-
ραλληλογράμμῳ ὁμοίῳ τῷ δοθέντι· δεῖ δὲ τὸ διδόμενον ogram) falling short by a parallelogrammic figure similar
εὐθύγραμμον [ᾧ δεῖ ἴσον παραβαλεῖν] μὴ μεῖζον εἶναι τοῦ to a given (parallelogram). It is necessary for the given
ἀπὸ τῆς ἡμισείας ἀναγραφομένου ὁμοίου τῷ ἐλλείμματι [τοῦ rectilinear figure [to which it is required to apply an equal
τε ἀπὸ τῆς ἡμισείας καὶ ᾧ δεῖ ὅμοιον ἐλλείπειν]. (parallelogram)] not to be greater than the (parallelo-
῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν gram) described on half (of the straight-line) and similar

εὐθύγραμμον, ᾧ δεῖ ἴσον παρὰ τὴν ΑΒ παραβαλεῖν, τὸ Γ μὴ to the deficit.
μεῖζον [ὂν] τοῦ ἀπὸ τῆς ἡμισείας τῆς ΑΒ ἀναγραφομένου Let AB be the given straight-line, and C the given
ὁμοίου τῷ ἐλλείμματι, ᾧ δὲ δεῖ ὅμοιον ἐλλείπειν, τὸ Δ· δεῖ δὴ rectilinear figure to which the (parallelogram) applied to

185

STOIQEIWN �þ. ELEMENTS BOOK 6
παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ τῷ δοθέντι εὐθυγράμμῳ AB is required (to be) equal, [being] not greater than
τῷ Γ ἴσον παραλληλόγραμμον παραβαλεῖν ἐλλεῖπον εἴδει πα- the (parallelogram) described on half of AB and similar
ραλληλογράμμῳ ὁμοίῳ ὄντι τῷ Δ. to the deficit, and D the (parallelogram) to which the

deficit is required (to be) similar. So it is required to apply

a parallelogram, equal to the given rectilinear figure C, to
the straight-line AB, falling short by a parallelogrammic

figure which is similar to D.

P
G

A S B
R DT

J O

H Z

X L

K N

M
QUF

A BE

FGH

U
V

Τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ε σημεῖον, καὶ ἀνα- Let AB have been cut in half at point E [Prop. 1.10],
γεγράφθω ἀπὸ τῆς ΕΒ τῷ Δ ὅμοιον καὶ ὁμοίως κείμενον and let (parallelogram) EBFG, (which is) similar, and
τὸ ΕΒΖΗ, καὶ συμπεπληρώσθω τὸ ΑΗ παραλληλόγραμμον. similarly laid out, to (parallelogram) D, have been de-
Εἰ μὲν οὖν ἴσον ἐστὶ τὸ ΑΗ τῷ Γ, γεγονὸς ἂν εἴη τὸ ἐπι- scribed on EB [Prop. 6.18]. And let parallelogram AG

ταχθέν· παραβέβληται γὰρ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν have been completed.
ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον Therefore, if AG is equal to C then the thing pre-
τὸ ΑΗ ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΗΒ ὁμοίῳ ὄντι scribed has happened. For a parallelogram AG, equal
τῷ Δ. εἰ δὲ οὔ, μεῖζόν ἔστω τὸ ΘΕ τοῦ Γ. ἴσον δὲ τὸ ΘΕ to the given rectilinear figure C, has been applied to the
τῷ ΗΒ· μεῖζον ἄρα καὶ τὸ ΗΒ τοῦ Γ. ᾧ δὴ μεῖζόν ἐστι given straight-line AB, falling short by a parallelogram-
τὸ ΗΒ τοῦ Γ, ταύτῃ τῇ ὑπεροχῇ ἴσον, τῷ δὲ Δ ὅμοιον καὶ mic figure GB which is similar to D. And if not, let HE
ὁμοίως κείμενον τὸ αὐτὸ συνεστάτω τὸ ΚΛΜΝ. ἀλλὰ τὸ Δ be greater than C. And HE (is) equal to GB [Prop. 6.1].
τῷ ΗΒ [ἐστιν] ὅμοιον· καὶ τὸ ΚΜ ἄρα τῷ ΗΒ ἐστιν ὅμοιον. Thus, GB (is) also greater than C. So, let (parallelo-
ἔστω οὖν ὁμόλογος ἡ μὲν ΚΛ τῂ ΗΕ, ἡ δὲ ΛΜ τῇ ΗΖ. gram) KLMN have been constructed (so as to be) both
καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΗΒ τοῖς Γ, ΚΜ, μεῖζον ἄρα ἐστὶ τὸ similar, and similarly laid out, to D, and equal to the ex-
ΗΒ τοῦ ΚΜ· μείζων ἄρα ἐστὶ καὶ ἡ μὲν ΗΕ τῆς ΚΛ, ἡ δὲ cess by which GB is greater than C [Prop. 6.25]. But,
ΗΖ τῆς ΛΜ. κείσθω τῇ μὲν ΚΛ ἴση ἡ ΗΞ, τῇ δὲ ΛΜ ἴση GB [is] similar to D. Thus, KM is also similar to GB
ἡ ΗΟ, καὶ συμπεπληρώσθω τὸ ΞΗΟΠ παραλληλόγραμμον· [Prop. 6.21]. Therefore, let KL correspond to GE, and
ἴσον ἄρα καὶ ὅμοιον ἐστι [τὸ ΗΠ] τῷ ΚΜ [ἀλλὰ τὸ ΚΜ τῷ LM to GF . And since (parallelogram) GB is equal to
ΗΒ ὅμοιόν ἐστιν]. καὶ τὸ ΗΠ ἄρα τῷ ΗΒ ὅμοιόν ἐστιν· περὶ (figure) C and (parallelogram) KM , GB is thus greater
τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὸ ΗΠ τῷ ΗΒ. ἔστω αὐτῶν than KM . Thus, GE is also greater than KL, and GF
διάμετρος ἡ ΗΠΒ, καὶ καταγεγράφθω τὸ σχῆμα. than LM . Let GO be made equal to KL, and GP to LM
᾿Επεὶ οὖν ἴσον ἐστὶ τὸ ΒΗ τοῖς Γ, ΚΜ, ὧν τὸ ΗΠ τῷ [Prop. 1.3]. And let the parallelogram OGPQ have been

ΚΜ ἐστιν ἴσον, λοιπὸς ἄρα ὁ ΥΧΦ γνόμων λοιπῷ τῷ Γ ἴσος completed. Thus, [GQ] is equal and similar to KM [but,
ἐστίν. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΟΡ τῷ ΞΣ, κοινὸν προσκείσθω KM is similar to GB]. Thus, GQ is also similar to GB
τὸ ΠΒ· ὅλον ἄρα τὸ ΟΒ ὅλῳ τῷ ΞΒ ἴσον ἐστίν. ἀλλὰ τὸ ΞΒ [Prop. 6.21]. Thus, GQ and GB are about the same diag-
τῷ ΤΕ ἐστιν ἴσον, ἐπεὶ καὶ πλευρὰ ἡ ΑΕ πλευρᾷ τῇ ΕΒ ἐστιν onal [Prop. 6.26]. Let GQB be their (common) diagonal,
ἴση· καὶ τὸ ΤΕ ἄρα τῷ ΟΒ ἐστιν ἴσον. κοινὸν προσκείσθω and let the (remainder of the) figure have been described.
τὸ ΞΣ· ὅλον ἄρα τὸ ΤΣ ὅλῳ τῷ ΦΧΥ γνώμονί ἐστιν ἴσον. Therefore, since BG is equal to C and KM , of which
ἀλλ᾿ ὁ ΦΧΥ γνώμων τῷ Γ ἐδείχθη ἴσος· καὶ τὸ ΤΣ ἄρα τῷ GQ is equal to KM , the remaining gnomon UWV is thus
Γ ἐστιν ἴσον. equal to the remainder C. And since (the complement)
Παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν τὴν ΑΒ τῷ δοθέντι PR is equal to (the complement) OS [Prop. 1.43], let

εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον παραβέβληται (parallelogram) QB have been added to both. Thus, the
τὸ ΣΤ ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΠΒ ὁμοίῳ ὄντι whole (parallelogram) PB is equal to the whole (par-

186

STOIQEIWN �þ. ELEMENTS BOOK 6
τῷ Δ [ἐπειδήπερ τὸ ΠΒ τῷ ΗΠ ὅμοιόν ἐστιν]· ὅπερ ἔδει allelogram) OB. But, OB is equal to TE, since side
ποιῆσαι. AE is equal to side EB [Prop. 6.1]. Thus, TE is also

equal to PB. Let (parallelogram) OS have been added
to both. Thus, the whole (parallelogram) TS is equal to

the gnomon V WU . But, gnomon V WU was shown (to
be) equal to C. Therefore, (parallelogram) TS is also

equal to (figure) C.

Thus, the parallelogram ST , equal to the given rec-
tilinear figure C, has been applied to the given straight-

line AB, falling short by the parallelogrammic figure QB,

which is similar to D [inasmuch as QB is similar to GQ
[Prop. 6.24] ]. (Which is) the very thing it was required

to do.

† This proposition is a geometric solution of the quadratic equation x2−αx+β = 0. Here, x is the ratio of a side of the deficit to the corresponding
side of figure D, α is the ratio of the length of AB to the length of that side of figure D which corresponds to the side of the deficit running along

AB, and β is the ratio of the areas of figures C and D. The constraint corresponds to the condition β < α2/4 for the equation to have real roots. Only the smaller root of the equation is found. The larger root can be found by a similar method.kjþ. Proposition 29† Παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι εὐθυγράμμῳ To apply a parallelogram, equal to a given rectilin- ἴσον παραλληλόγραμμον παραβαλεῖν ὑπερβάλλον εἴδει πα- ear figure, to a given straight-line, (the applied parallelo- ραλληλογράμμῳ ὁμοίῳ τῷ δοθέντι. gram) overshooting by a parallelogrammic figure similar to a given (parallelogram). F G JL M OBA N P E Z H K Q D X Y W B C F HKML P D GN E OQ A V X ῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν Let AB be the given straight-line, and C the given εὐθύγραμμον, ᾧ δεῖ ἴσον παρὰ τὴν ΑΒ παραβαλεῖν, τὸ Γ, rectilinear figure to which the (parallelogram) applied to ᾧ δὲ δεῖ ὅμοιον ὑπερβάλλειν, τὸ Δ· δεῖ δὴ παρὰ τὴν ΑΒ AB is required (to be) equal, and D the (parallelogram) εὐθεῖαν τῷ Γ εὐθυγράμμῳ ἴσον παραλληλόγραμμον παρα- to which the excess is required (to be) similar. So it is βαλεῖν ὑπερβάλλον εἴδει παραλληλογράμμῳ ὁμοίῳ τῷ Δ. required to apply a parallelogram, equal to the given rec- Τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ε, καὶ ἀναγεγράθω tilinear figure C, to the given straight-line AB, overshoot- ἀπὸ τὴς ΕΒ τῷ Δ ὅμοιον καὶ ὁμοίως κείμενον παραλ- ing by a parallelogrammic figure similar to D. ληλόγραμμον τὸ ΒΖ, καὶ συναμφοτέροις μὲν τοῖς ΒΖ, Γ Let AB have been cut in half at (point) E [Prop. 1.10], ἴσον, τῷ δὲ Δ ὅμοιον καὶ ὁμοίως κείμενον τὸ αὐτὸ συ- and let the parallelogram BF , (which is) similar, and νεστάτω τὸ ΗΘ. ὁμόλογος δὲ ἔστω ἡ μὲν ΚΘ τῇ ΖΛ, ἡ δὲ similarly laid out, to D, have been described on EB ΚΗ τῇ ΖΕ. καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΗΘ τοῦ ΖΒ, μείζων ἄρα [Prop. 6.18]. And let (parallelogram) GH have been con- ἐστὶ καὶ ἡ μὲν ΚΘ τῆς ΖΛ, ἡ δὲ ΚΗ τῇ ΖΕ. ἐκβεβλήσθωσαν structed (so as to be) both similar, and similarly laid out, αἱ ΖΛ, ΖΕ, καὶ τῇ μὲν ΚΘ ἴση ἔστω ἡ ΖΛΜ, τῇ δὲ ΚΗ ἴση to D, and equal to the sum of BF and C [Prop. 6.25]. ἡ ΖΕΝ, καὶ συμπεπληρώσθω τὸ ΜΝ· τὸ ΜΝ ἄρα τῷ ΗΘ And let KH correspond to FL, and KG to FE. And since ἴσον τέ ἐστι καὶ ὅμοιον. ἀλλὰ τὸ ΗΘ τῷ ΕΛ ἐστιν ὅμοιον· (parallelogram) GH is greater than (parallelogram) FB, 187 STOIQEIWN �þ. ELEMENTS BOOK 6 καὶ τὸ ΜΝ ἄρα τῷ ΕΛ ὅμοιόν ἐστιν· περὶ τὴν αὐτὴν ἄρα KH is thus also greater than FL, and KG than FE. διάμετρόν ἐστι τὸ ΕΛ τῷ ΜΝ. ἤχθω αὐτῶν διάμετρος ἡ Let FL and FE have been produced, and let FLM be ΖΞ, καὶ καταγεγράφθω τὸ σχῆμα. (made) equal to KH , and FEN to KG [Prop. 1.3]. And ᾿Επεὶ ἴσον ἐστὶ τὸ ΗΘ τοῖς ΕΛ, Γ, ἀλλὰ τὸ ΗΘ τῷ ΜΝ let (parallelogram) MN have been completed. Thus, ἴσον ἐστίν, καὶ τὸ ΜΝ ἄρα τοῖς ΕΛ, Γ ἴσον ἐστίν. κοινὸν MN is equal and similar to GH . But, GH is similar to ἀφῃρήσθω τὸ ΕΛ· λοιπὸς ἄρα ὁ ΨΧΦ γνώμων τῷ Γ ἐστιν EL. Thus, MN is also similar to EL [Prop. 6.21]. EL is ἴσος. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΕ τῇ ΕΒ, ἴσον ἐστὶ καὶ τὸ ΑΝ τῷ thus about the same diagonal as MN [Prop. 6.26]. Let ΝΒ, τουτέστι τῷ ΛΟ. κοινὸν προσκείσθω τὸ ΕΞ· ὅλον ἄρα their (common) diagonal FO have been drawn, and let τὸ ΑΞ ἴσον ἐστὶ τῷ ΦΧΨ γνώμονι. ἀλλὰ ὁ ΦΧΨ γνώμων the (remainder of the) figure have been described. τῷ Γ ἴσος ἐστίν· καὶ τὸ ΑΞ ἄρα τῷ Γ ἴσον ἐστίν. And since (parallelogram) GH is equal to (parallel- Παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν τὴν ΑΒ τῷ δοθέντι ogram) EL and (figure) C, but GH is equal to (paral- εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον παραβέβληται lelogram) MN , MN is thus also equal to EL and C. τὸ ΑΞ ὑπερβάλλον εἴδει παραλληλογράμμῳ τῷ ΠΟ ὁμοίῳ Let EL have been subtracted from both. Thus, the re- ὄντι τῷ Δ, ἐπεὶ καὶ τῷ ΕΛ ἐστιν ὅμοιον τὸ ΟΠ· ὅπερ ἔδει maining gnomon XWV is equal to (figure) C. And since ποιῆσαι. AE is equal to EB, (parallelogram) AN is also equal to (parallelogram) NB [Prop. 6.1], that is to say, (parallel- ogram) LP [Prop. 1.43]. Let (parallelogram) EO have been added to both. Thus, the whole (parallelogram) AO is equal to the gnomon V WX . But, the gnomon V WX is equal to (figure) C. Thus, (parallelogram) AO is also equal to (figure) C. Thus, the parallelogram AO, equal to the given rec- tilinear figure C, has been applied to the given straight- line AB, overshooting by the parallelogrammic figure QP which is similar to D, since PQ is also similar to EL [Prop. 6.24]. (Which is) the very thing it was required to do. † This proposition is a geometric solution of the quadratic equation x2+α x−β = 0. Here, x is the ratio of a side of the excess to the corresponding side of figure D, α is the ratio of the length of AB to the length of that side of figure D which corresponds to the side of the excess running along AB, and β is the ratio of the areas of figures C and D. Only the positive root of the equation is found.lþ. Proposition 30† Τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην ἄκρον καὶ μέσον To cut a given finite straight-line in extreme and mean λόγον τεμεῖν. ratio. Ε Β Γ Ζ Α ∆ Θ E F H B C A D ῎Εστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ· δεῖ δὴ τὴν Let AB be the given finite straight-line. So it is re- ΑΒ εὐθεῖαν ἄκρον καὶ μέσον λόγον τεμεῖν. quired to cut the straight-line AB in extreme and mean 188 STOIQEIWN �þ. ELEMENTS BOOK 6 Ἀναγεγράφθω ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΒΓ, καὶ πα- ratio. ραβεβλήσθω παρὰ τὴν ΑΓ τῷ ΒΓ ἴσον παραλληλόγραμμον Let the square BC have been described on AB [Prop. τὸ ΓΔ ὑπερβάλλον εἴδει τῷ ΑΔ ὁμοίῳ τῷ ΒΓ. 1.46], and let the parallelogram CD, equal to BC, have Τετράγωνον δέ ἐστι τὸ ΒΓ· τετράγωνον ἄρα ἐστι καὶ been applied to AC, overshooting by the figure AD τὸ ΑΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΒΓ τῷ ΓΔ, κοινὸν ἀφῃρήσθω (which is) similar to BC [Prop. 6.29]. τὸ ΓΕ· λοιπὸν ἄρα τὸ ΒΖ λοιπῷ τῷ ΑΔ ἐστιν ἴσον. ἔστι And BC is a square. Thus, AD is also a square. δὲ αὐτῷ καὶ ἰσογώνιον· τῶν ΒΖ, ΑΔ ἄρα ἀντιπεπόνθασιν αἱ And since BC is equal to CD, let (rectangle) CE have πλευραὶ αἱ περὶ τὰς ἴσας γωνίας· ἔστιν ἄρα ὡς ἡ ΖΕ πρὸς been subtracted from both. Thus, the remaining (rect- τὴν ΕΔ, οὕτως ἡ ΑΕ πρὸς τὴν ΕΒ. ἴση δὲ ἡ μὲν ΖΕ τῇ ΑΒ, angle) BF is equal to the remaining (square) AD. And ἡ δὲ ΕΔ τῇ ΑΕ. ἔστιν ἄρα ὡς ἡ ΒΑ πρὸς τὴν ΑΕ, οὕτως ἡ it is also equiangular to it. Thus, the sides of BF and ΑΕ πρὸς τὴν ΕΒ. μείζων δὲ ἡ ΑΒ τῆς ΑΕ· μείζων ἄρα καὶ AD about the equal angles are reciprocally proportional ἡ ΑΕ τῆς ΕΒ. [Prop. 6.14]. Thus, as FE is to ED, so AE (is) to EB. ῾Η ἄρα ΑΒ εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ And FE (is) equal to AB, and ED to AE. Thus, as BA is τὸ Ε, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστι τὸ ΑΕ· ὅπερ ἔδει to AE, so AE (is) to EB. And AB (is) greater than AE. ποιῆσαι. Thus, AE (is) also greater than EB [Prop. 5.14]. Thus, the straight-line AB has been cut in extreme and mean ratio at E, and AE is its greater piece. (Which is) the very thing it was required to do. † This method of cutting a straight-line is sometimes called the “Golden Section”—see Prop. 2.11.laþ. Proposition 31 ᾿Εν τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν In right-angled triangles, the figure (drawn) on the γωνίαν ὑποτεινούσης πλευρᾶς εἶδος ἴσον ἐστὶ τοῖς ἀπὸ τῶν side subtending the right-angle is equal to the (sum of τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν εἴδεσι τοῖς ὁμοίοις the) similar, and similarly described, figures on the sides τε καὶ ὁμοίως ἀναγραφομένοις. surrounding the right-angle. Α ∆Β Γ D A CB ῎Εστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν Let ABC be a right-angled triangle having the angle ὑπὸ ΒΑΓ γωνίαν· λέγω, ὅτι τὸ ἀπὸ τῆς ΒΓ εἶδος ἴσον ἐστὶ BAC a right-angle. I say that the figure (drawn) on BC is τοῖς ἀπὸ τῶν ΒΑ, ΑΓ εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως equal to the (sum of the) similar, and similarly described, ἀναγραφομένοις. figures on BA and AC. ῎Ηχθω κάθετος ἡ ΑΔ. Let the perpendicular AD have been drawn [Prop. ᾿Επεὶ οὖν ἐν ὀρθογωνίῳ τριγώνῳ τῷ ΑΒΓ ἀπὸ τῆς πρὸς 1.12]. τῷ Α ὀρθῆς γωνίας ἐπὶ τὴν ΒΓ βάσιν κάθετος ἦκται ἡ ΑΔ, Therefore, since, in the right-angled triangle ABC, τὰ ΑΒΔ, ΑΔΓ πρὸς τῇ καθέτῳ τρίγωνα ὅμοιά ἐστι τῷ τε the (straight-line) AD has been drawn from the right- ὅλῳ τῷ ΑΒΓ καὶ ἀλλήλοις. καὶ ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓ τῷ angle at A perpendicular to the base BC, the trian- ΑΒΔ, ἔστιν ἄρα ὡς ἡ ΓΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΑΒ πρὸς gles ABD and ADC about the perpendicular are sim- τὴν ΒΔ. καὶ ἐπεὶ τρεῖς εὐθεῖαι ἀνάλογόν εἰσιν, ἔστιν ὡς ἡ ilar to the whole (triangle) ABC, and to one another πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς [Prop. 6.8]. And since ABC is similar to ABD, thus 189 STOIQEIWN �þ. ELEMENTS BOOK 6 τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον. as CB is to BA, so AB (is) to BD [Def. 6.1]. And ὡς ἄρα ἡ ΓΒ πρὸς τὴν ΒΔ, οὕτως τὸ ἀπὸ τῆς ΓΒ εἶδος since three straight-lines are proportional, as the first is πρὸς τὸ ἀπὸ τῆς ΒΑ τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον. to the third, so the figure (drawn) on the first is to the διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ ΒΓ πρὸς τὴν ΓΔ, οὕτως τὸ ἀπὸ τῆς similar, and similarly described, (figure) on the second ΒΓ εἶδος πρὸς τὸ ἀπὸ τῆς ΓΑ. ὥστε καὶ ὡς ἡ ΒΓ πρὸς τὰς [Prop. 6.19 corr.]. Thus, as CB (is) to BD, so the fig- ΒΔ, ΔΓ, οὕτως τὸ ἀπὸ τῆς ΒΓ εἶδος πρὸς τὰ ἀπὸ τῶν ΒΑ, ure (drawn) on CB (is) to the similar, and similarly de- ΑΓ τὰ ὅμοια καὶ ὁμοίως ἀναγραφόμενα. ἴση δὲ ἡ ΒΓ ταῖς scribed, (figure) on BA. And so, for the same (reasons), ΒΔ, ΔΓ· ἴσον ἄρα καὶ τὸ ἄπὸ τῆς ΒΓ εἶδος τοῖς ἀπὸ τῶν as BC (is) to CD, so the figure (drawn) on BC (is) to ΒΑ, ΑΓ εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις. the (figure) on CA. Hence, also, as BC (is) to BD and ᾿Εν ἄρα τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν DC, so the figure (drawn) on BC (is) to the (sum of the) ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς εἶδος ἴσον ἐστὶ τοῖς similar, and similarly described, (figures) on BA and AC ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν εἴδεσι τοῖς [Prop. 5.24]. And BC is equal to BD and DC. Thus, the ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις· ὅπερ ἔδει δεῖξαι. figure (drawn) on BC (is) also equal to the (sum of the) similar, and similarly described, figures on BA and AC [Prop. 5.9]. Thus, in right-angled triangles, the figure (drawn) on the side subtending the right-angle is equal to the (sum of the) similar, and similarly described, figures on the sides surrounding the right-angle. (Which is) the very thing it was required to show.lbþ. Proposition 32 ᾿Εὰν δύο τρίγωνα συντεθῇ κατὰ μίαν γωνίαν τὰς δύο If two triangles, having two sides proportional to two πλευρὰς ταῖς δυσὶ πλευραῖς ἀνάλογον ἔχοντα ὥστε τὰς sides, are placed together at a single angle such that the ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι, αἱ λοιπαὶ corresponding sides are also parallel, then the remaining τῶν τριγώνων πλευραὶ ἐπ᾿ εὐθείας ἔσονται. sides of the triangles will be straight-on (with respect to one another). ΓΒ Ε ∆ Α CB E A D ῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΓΕ τὰς δύο πλευρὰς τὰς Let ABC and DCE be two triangles having the two ΒΑ, ΑΓ ταῖς δυσὶ πλευραῖς ταῖς ΔΓ, ΔΕ ἀνάλογον ἔχοντα, sides BA and AC proportional to the two sides DC and ὡς μὲν τὴν ΑΒ πρὸς τὴν ΑΓ, οὕτως τὴν ΔΓ πρὸς τὴν ΔΕ, DE—so that as AB (is) to AC, so DC (is) to DE—and παράλληλον δὲ τὴν μὲν ΑΒ τῇ ΔΓ, τὴν δὲ ΑΓ τῇ ΔΕ· λέγω, (having side) AB parallel to DC, and AC to DE. I say ὅτι ἐπ᾿ εὐθείας ἐστὶν ἡ ΒΓ τῇ ΓΕ. that (side) BC is straight-on to CE. ᾿Επεὶ γὰρ παράλληλός ἐστιν ἡ ΑΒ τῇ ΔΓ, καὶ εἰς αὐτὰς For since AB is parallel to DC, and the straight-line ἐμπέπτωκεν εὐθεῖα ἡ ΑΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ, AC has fallen across them, the alternate angles BAC and ΑΓΔ ἴσαι ἀλλήλαις εἰσίν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΔΕ τῇ ACD are equal to one another [Prop. 1.29]. So, for the ὑπὸ ΑΓΔ ἴση ἐστίν. ὥστε καὶ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΕ ἐστιν same (reasons), CDE is also equal to ACD. And, hence, ἴση. καὶ ἐπεὶ δύο τρίγωνά ἐστι τὰ ΑΒΓ, ΔΓΕ μίαν γωνίαν BAC is equal to CDE. And since ABC and DCE are τὴν πρὸς τῷ Α μιᾷ γωνίᾳ τῇ πρὸς τῷ Δ ἴσην ἔχοντα, περὶ two triangles having the one angle at A equal to the one 190 STOIQEIWN �þ. ELEMENTS BOOK 6 δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ὡς τὴν ΒΑ πρὸς angle at D, and the sides about the equal angles pro- τὴν ΑΓ, οὕτως τὴν ΓΔ πρὸς τὴν ΔΕ, ἰσογώνιον ἄρα ἐστὶ portional, (so that) as BA (is) to AC, so CD (is) to τὸ ΑΒΓ τρίγωνον τῷ ΔΓΕ τριγώνῳ· ἴση ἄρα ἡ ὑπὸ ΑΒΓ DE, triangle ABC is thus equiangular to triangle DCE γωνία τῇ ὑπὸ ΔΓΕ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΒΑΓ [Prop. 6.6]. Thus, angle ABC is equal to DCE. And (an- ἴση· ὅλη ἄρα ἡ ὑπὸ ΑΓΕ δυσὶ ταῖς ὑπὸ ΑΒΓ, ΒΑΓ ἴση ἐστίν. gle) ACD was also shown (to be) equal to BAC. Thus, κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ· αἱ ἄρα ὑπὸ ΑΓΕ, ΑΓΒ ταῖς the whole (angle) ACE is equal to the two (angles) ABC ὑπὸ ΒΑΓ, ΑΓΒ, ΓΒΑ ἴσαι εἰσίν. ἀλλ᾿ αἱ ὑπὸ ΒΑΓ, ΑΒΓ, and BAC. Let ACB have been added to both. Thus, ΑΓΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν· καὶ αἱ ὑπὸ ΑΓΕ, ΑΓΒ ἄρα ACE and ACB are equal to BAC, ACB, and CBA. δυσὶν ὀρθαῖς ἴσαι εἰσίν. πρὸς δή τινι εὐθείᾳ τῇ ΑΓ καὶ τῷ But, BAC, ABC, and ACB are equal to two right-angles πρὸς αὐτῇ σημείῳ τῷ Γ δύο εὐθεῖαι αἱ ΒΓ, ΓΕ μὴ ἐπὶ τὰ [Prop. 1.32]. Thus, ACE and ACB are also equal to two αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνάις τὰς ὑπὸ ΑΓΕ, ΑΓΒ right-angles. Thus, the two straight-lines BC and CE, δυσὶν ὀρθαῖς ἴσας ποιοῦσιν· ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΒΓ τῇ not lying on the same side, make adjacent angles ACE ΓΕ. and ACB (whose sum is) equal to two right-angles with ᾿Εὰν ἄρα δύο τρίγωνα συντεθῇ κατὰ μίαν γωνίαν τὰς some straight-line AC, at the point C on it. Thus, BC is δύο πλευρὰς ταῖς δυσὶ πλευραῖς ἀνάλογον ἔχοντα ὥστε τὰς straight-on to CE [Prop. 1.14]. ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι, αἱ λοιπαὶ Thus, if two triangles, having two sides proportional τῶν τριγώνων πλευραὶ ἐπ᾿ εὐθείας ἔσονται· ὅπερ ἔδει δεῖξαι. to two sides, are placed together at a single angle such that the corresponding sides are also parallel, then the remaining sides of the triangles will be straight-on (with respect to one another). (Which is) the very thing it was required to show.lgþ. Proposition 33 ᾿Εν τοῖς ἴσοις κύκλοις αἱ γωνίαι τὸν αὐτὸν ἔχουσι In equal circles, angles have the same ratio as the (ra- λόγον ταῖς περιφερείαις, ἐφ᾿ ὧν βεβήκασιν, ἐάν τε πρὸς tio of the) circumferences on which they stand, whether τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι. they are standing at the centers (of the circles) or at the circumferences. Z L H K M J A D N B G E H L K M A D N B E G C F ῎Εστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ, καὶ πρὸς μὲν τοῖς Let ABC and DEF be equal circles, and let BGC and κέντροις αὐτῶν τοῖς Η, Θ γωνίαι ἔστωσαν αἱ ὑπὸ ΒΗΓ, EHF be angles at their centers, G and H (respectively), ΕΘΖ, πρὸς δὲ ταῖς περιφερείαις αἱ ὑπὸ ΒΑΓ, ΕΔΖ· λέγω, and BAC and EDF (angles) at their circumferences. I ὅτι ἐστὶν ὡς ἡ ΒΓ περιφέρεια πρὸς τὴν ΕΖ περιφέρειαν, say that as circumference BC is to circumference EF , so οὕτως ἥ τε ὑπὸ ΒΗΓ γωνία πρὸς τὴν ὑπὸ ΕΘΖ καὶ ἡ ὑπὸ angle BGC (is) to EHF , and (angle) BAC to EDF . ΒΑΓ πρὸς τὴν ὑπὸ ΕΔΖ. For let any number whatsoever of consecutive (cir- Κείσθωσαν γὰρ τῇ μὲν ΒΓ περιφερείᾳ ἴσαι κατὰ τὸ ἑξῆς cumferences), CK and KL, be made equal to circumfer- ὁσαιδηποτοῦν αἱ ΓΚ, ΚΛ, τῇ δὲ ΕΖ περιφερείᾳ ἴσαι ὁσαι- ence BC, and any number whatsoever, FM and MN , to δηποτοῦν αἱ ΖΜ, ΜΝ, καὶ ἐπεζεύχθωσαν αἱ ΗΚ, ΗΛ, ΘΜ, circumference EF . And let GK, GL, HM , and HN have ΘΝ. been joined. ᾿Επεὶ οὖν ἴσαι εἰσὶν αἱ ΒΓ, ΓΚ, ΚΛ περιφέρειαι ἀλλήλαις, Therefore, since circumferences BC, CK, and KL are ἴσαι εἰσὶ καὶ αἱ ὑπὸ ΒΗΓ, ΓΗΚ, ΚΗΛ γωνίαι ἀλλήλαις· equal to one another, angles BGC, CGK, and KGL are ὁσαπλασίων ἄρα ἐστὶν ἡ ΒΛ περιφέρεια τῆς ΒΓ, τοσαυτα- also equal to one another [Prop. 3.27]. Thus, as many πλασίων ἐστὶ καὶ ἡ ὑπὸ ΒΗΛ γωνία τῆς ὑπὸ ΒΗΓ. διὰ τὰ times as circumference BL is (divisible) by BC, so many 191 STOIQEIWN �þ. ELEMENTS BOOK 6 αὐτὰ δὴ καὶ ὁσαπλασίων ἐστὶν ἡ ΝΕ περιφέρεια τῆς ΕΖ, το- times is angle BGL also (divisible) by BGC. And so, for σαυταπλασίων ἐστὶ καὶ ἡ ὑπὸ ΝΘΕ γωνία τῆς ὑπὸ ΕΘΖ. εἰ the same (reasons), as many times as circumference NE ἄρα ἴση ἐστὶν ἡ ΒΛ περιφέρεια τῇ ΕΝ περιφερείᾳ, ἴση ἐστὶ is (divisible) by EF , so many times is angle NHE also καὶ γωνία ἡ ὑπὸ ΒΗΛ τῇ ὑπὸ ΕΘΝ, καὶ εἰ μείζων ἐστὶν ἡ ΒΛ (divisible) by EHF . Thus, if circumference BL is equal περιφέρεια τῆς ΕΝ περιφερείας, μείζων ἐστὶ καὶ ἡ ὑπὸ ΒΗΛ to circumference EN then angle BGL is also equal to γωνία τῆς ὑπὸ ΕΘΝ, καὶ εἰ ἐλάσσων, ἐλάσσων. τεσσάρων EHN [Prop. 3.27], and if circumference BL is greater δὴ ὄντων μεγεθῶν, δύο μὲν περιφερειῶν τῶν ΒΓ, ΕΖ, δύο than circumference EN then angle BGL is also greater δὲ γωνιῶν τῶν ὑπὸ ΒΗΓ, ΕΘΖ, εἴληπται τῆς μὲν ΒΓ περι- than EHN ,† and if (BL is) less (than EN then BGL is φερείας καὶ τῆς ὑπὸ ΒΗΓ γωνίας ἰσάκις πολλαπλασίων ἥ τε also) less (than EHN). So there are four magnitudes, ΒΛ περιφέρεια καὶ ἡ ὑπὸ ΒΗΛ γωνία, τῆς δὲ ΕΖ περιφερείας two circumferences BC and EF , and two angles BGC καὶ τῆς ὑπὸ ΕΘΖ γωνίας ἥ τε ΕΝ περιφέρια καὶ ἡ ὑπὸ ΕΘΝ and EHF . And equal multiples have been taken of cir- γωνία. καὶ δέδεικται, ὅτι εἰ ὑπερέχει ἡ ΒΛ περιφέρεια τῆς cumference BC and angle BGC, (namely) circumference ΕΝ περιφερείας, ὑπερέχει καὶ ἡ ὑπὸ ΒΗΛ γωνία τῆς ὑπο BL and angle BGL, and of circumference EF and an- ΕΘΝ γωνίας, καὶ εἰ ἴση, ἴση, καὶ εἰ ἐλάσσων, ἐλάσσων. gle EHF , (namely) circumference EN and angle EHN . ἔστιν ἄρα, ὡς ἡ ΒΓ περιφέρεια πρὸς τὴν ΕΖ, οὕτως ἡ ὑπὸ And it has been shown that if circumference BL exceeds ΒΗΓ γωνία πρὸς τὴν ὑπὸ ΕΘΖ. ἀλλ᾿ ὡς ἡ ὑπὸ ΒΗΓ γωνία circumference EN then angle BGL also exceeds angle πρὸς τὴν ὑπὸ ΕΘΖ, οὕτως ἡ ὑπὸ ΒΑΓ πρὸς τὴν ὑπὸ ΕΔΖ. EHN , and if (BL is) equal (to EN then BGL is also) διπλασία γὰρ ἑκατέρα ἑκατέρας. καὶ ὡς ἄρα ἡ ΒΓ περιφέρεια equal (to EHN), and if (BL is) less (than EN then BGL πρὸς τὴν ΕΖ περιφέρειαν, οὕτως ἥ τε ὑπὸ ΒΗΓ γωνία πρὸς is also) less (than EHN). Thus, as circumference BC τὴν ὑπὸ ΕΘΖ καὶ ἡ ὑπὸ ΒΑΓ πρὸς τὴν ὑπὸ ΕΔΖ. (is) to EF , so angle BGC (is) to EHF [Def. 5.5]. But as ᾿Εν ἄρα τοῖς ἴσοις κύκλοις αἱ γωνίαι τὸν αὐτὸν ἔχουσι angle BGC (is) to EHF , so (angle) BAC (is) to EDF λόγον ταῖς περιφερείαις, ἐφ᾿ ὧν βεβήκασιν, ἐάν τε πρὸς τοῖς [Prop. 5.15]. For the former (are) double the latter (re- κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι· ὅπερ spectively) [Prop. 3.20]. Thus, also, as circumference BC ἔδει δεῖξαι. (is) to circumference EF , so angle BGC (is) to EHF , and BAC to EDF . Thus, in equal circles, angles have the same ratio as the (ratio of the) circumferences on which they stand, whether they are standing at the centers (of the circles) or at the circumferences. (Which is) the very thing it was required to show. † This is a straight-forward generalization of Prop. 3.27 192 ELEMENTS BOOK 7 Elementary Number Theory† †The propositions contained in Books 7–9 are generally attributed to the school of Pythagoras. 193 STOIQEIWN zþ. ELEMENTS BOOK 7VOroi. Definitions αʹ.Μονάς ἐστιν, καθ᾿ ἣν ἕκαστον τῶν ὄντων ἓν λέγεται. 1. A unit is (that) according to which each existing βʹ. Ἀριθμὸς δὲ τὸ ἐκ μονάδων συγκείμενον πλῆθος. (thing) is said (to be) one. γʹ. Μέρος ἐστὶν ἀριθμὸς ἀριθμοῦ ὁ ἐλάσσων τοῦ 2. And a number (is) a multitude composed of units.† μείζονος, ὅταν καταμετρῇ τὸν μείζονα. 3. A number is part of a(nother) number, the lesser of δʹ. Μέρη δέ, ὅταν μὴ καταμετρῇ. the greater, when it measures the greater.‡ εʹ. Πολλαπλάσιος δὲ ὁ μείζων τοῦ ἐλάσσονος, ὅταν κα- 4. But (the lesser is) parts (of the greater) when it ταμετρῆται ὑπὸ τοῦ ἐλάσσονος. does not measure it.§ ϛʹ. ῎Αρτιος ἀριθμός ἐστιν ὁ δίχα διαιρούμενος. 5. And the greater (number is) a multiple of the lesser ζʹ. Περισσὸς δὲ ὁ μὴ διαιρούμενος δίχα ἢ [ὁ] μονάδι when it is measured by the lesser. διαφέρων ἀρτίου ἀριθμοῦ. 6. An even number is one (which can be) divided in ηʹ. Ἀρτιάκις ἄρτιος ἀριθμός ἐστιν ὁ ὑπὸ ἀρτίου ἀριθμοῦ half. μετρούμενος κατὰ ἄρτιον ἀριθμόν. 7. And an odd number is one (which can)not (be) θʹ. ῎Αρτιάκις δὲ περισσός ἐστιν ὁ ὑπὸ ἀρτίου ἀριθμοῦ divided in half, or which differs from an even number by μετρούμενος κατὰ περισσὸν ἀριθμόν. a unit. ιʹ. Περισσάκις δὲ περισσὸς ἀριθμός ἐστιν ὁ ὑπὸ περισσοῦ 8. An even-times-even number is one (which is) mea- ἀριθμοῦ μετρούμενος κατὰ περισσὸν ἀριθμόν. sured by an even number according to an even number.¶ ιαʹ. Πρῶτος ἀριθμός ἐστιν ὁ μονάδι μόνῃ μετρούμενος. 9. And an even-times-odd number is one (which ιβʹ. Πρῶτοι πρὸς ἀλλήλους ἀριθμοί εἰσιν οἱ μονάδι μόνῃ is) measured by an even number according to an odd μετρούμενοι κοινῷ μέτρῳ. number.∗ ιγʹ. Σύνθετος ἀριθμός ἐστιν ὁ ἀριθμῷ τινι μετρούμενος. 10. And an odd-times-odd number is one (which ιδʹ. Σύνθετοι δὲ πρὸς ἀλλήλους ἀριθμοί εἰσιν οἱ ἀριθμῷ is) measured by an odd number according to an odd τινι μετρούμενοι κοινῷ μέτρῳ. number.$ ιεʹ. Ἀριθμὸς ἀριθμὸν πολλαπλασιάζειν λέγεται, ὅταν, 11. A prime‖ number is one (which is) measured by a ὅσαι εἰσὶν ἐν αὐτῷ μονάδες, τοσαυτάκις συντεθῇ ὁ πολ- unit alone. λαπλασιαζόμενος, καὶ γένηταί τις. 12. Numbers prime to one another are those (which ιϛʹ. ῞Οταν δὲ δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους are) measured by a unit alone as a common measure. ποιῶσί τινα, ὁ γενόμενος ἐπίπεδος καλεῖται, πλευραὶ δὲ 13. A composite number is one (which is) measured αὐτοῦ οἱ πολλαπλασιάσαντες ἀλλήλους ἀριθμοί. by some number. ιζʹ. ῞Οταν δὲ τρεῖς ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους 14. And numbers composite to one another are those ποιῶσί τινα, ὁ γενόμενος στερεός ἐστιν, πλευραὶ δὲ αὐτοῦ (which are) measured by some number as a common οἱ πολλαπλασιάσαντες ἀλλήλους ἀριθμοί. measure. ιηʹ. Τετράγωνος ἀριθμός ἐστιν ὁ ἰσάκις ἴσος ἢ [ὁ] ὑπὸ 15. A number is said to multiply a(nother) number δύο ἴσων ἀριθμῶν περιεχόμενος. when the (number being) multiplied is added (to itself) ιθʹ. Κύβος δὲ ὁ ἰσάκις ἴσος ἰσάκις ἢ [ὁ] ὑπὸ τριῶν ἴσων as many times as there are units in the former (number), ἀριθμῶν περιεχόμενος. and (thereby) some (other number) is produced. κʹ. Ἀριθμοὶ ἀνάλογόν εἰσιν, ὅταν ὁ πρῶτος τοῦ δευτέρου 16. And when two numbers multiplying one another καὶ ὁ τρίτος τοῦ τετάρτου ἰσάκις ᾖ πολλαπλάσιος ἢ τὸ αὐτὸ make some (other number) then the (number so) cre- μέρος ἢ τὰ αὐτὰ μέρη ὦσιν. ated is called plane, and its sides (are) the numbers which καʹ. ῞Ομοιοι ἐπίπεδοι καὶ στερεοὶ ἀριθμοί εἰσιν οἱ multiply one another. ανάλογον ἔχοντες τὰς πλευράς. 17. And when three numbers multiplying one another κβʹ. Τέλειος ἀριθμός ἐστιν ὁ τοῖς ἑαυτοῦ μέρεσιν ἴσος make some (other number) then the (number so) created ὤν. is (called) solid, and its sides (are) the numbers which multiply one another. 18. A square number is an equal times an equal, or (a plane number) contained by two equal numbers. 19. And a cube (number) is an equal times an equal times an equal, or (a solid number) contained by three equal numbers. 194 STOIQEIWN zþ. ELEMENTS BOOK 7 20. Numbers are proportional when the first is the same multiple, or the same part, or the same parts, of the second that the third (is) of the fourth. 21. Similar plane and solid numbers are those having proportional sides. 22. A perfect number is that which is equal to its own parts.†† † In other words, a “number” is a positive integer greater than unity. ‡ In other words, a number a is part of another number b if there exists some number n such that n a = b. § In other words, a number a is parts of another number b (where a < b) if there exist distinct numbers, m and n, such that n a = m b. ¶ In other words. an even-times-even number is the product of two even numbers. ∗ In other words, an even-times-odd number is the product of an even and an odd number. $ In other words, an odd-times-odd number is the product of two odd numbers. ‖ Literally, “first”. †† In other words, a perfect number is equal to the sum of its own factors.aþ. Proposition 1 Δύο ἀριθμῶν ἀνίσων ἐκκειμένων, ἀνθυφαιρουμένου δὲ Two unequal numbers (being) laid down, and the ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος, ἐὰν ὁ λειπόμενος lesser being continually subtracted, in turn, from the μηδέποτε καταμετρῇ τὸν πρὸ ἑαυτοῦ, ἕως οὗ λειφθῇ μονάς, greater, if the remainder never measures the (number) οἱ ἐξ ἀρχῆς ἀριθμοὶ πρῶτοι πρὸς ἀλλὴλους ἔσονται. preceding it, until a unit remains, then the original num- bers will be prime to one another. Ε Θ Α Ζ Γ Η ∆Β E G A H F B D C Δύο γὰρ [ἀνίσων] ἀριθμῶν τῶν ΑΒ, ΓΔ ἀνθυφαι- For two [unequal] numbers, AB and CD, the lesser ρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος ὁ λειπόμενος being continually subtracted, in turn, from the greater, μηδέποτε καταμετρείτω τὸν πρὸ ἑαυτοῦ, ἕως οὗ λειφθῇ let the remainder never measure the (number) preceding μονάς· λέγω, ὅτι οἱ ΑΒ, ΓΔ πρῶτοι πρὸς ἀλλήλους εἰσίν, it, until a unit remains. I say that AB and CD are prime τουτέστιν ὅτι τοὺς ΑΒ, ΓΔ μονὰς μόνη μετρεῖ. to one another—that is to say, that a unit alone measures Εἰ γὰρ μή εἰσιν οἱ ΑΒ, ΓΔ πρῶτοι πρὸς ἀλλήλους, (both) AB and CD. μετρήσει τις αὐτοὺς ἀριθμός. μετρείτω, καὶ ἔστω ὁ Ε· καὶ ὁ For if AB and CD are not prime to one another then μὲν ΓΔ τὸν ΒΖ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΖΑ, some number will measure them. Let (some number) ὁ δὲ ΑΖ τὸν ΔΗ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΗΓ, measure them, and let it be E. And let CD measuring ὁ δὲ ΗΓ τὸν ΖΘ μετρῶν λειπέτω μονάδα τὴν ΘΑ. BF leave FA less than itself, and let AF measuring DG ᾿Επεὶ οὖν ὁ Ε τὸν ΓΔ μετρεῖ, ὁ δὲ ΓΔ τὸν ΒΖ μετρεῖ, leave GC less than itself, and let GC measuring FH leave καὶ ὁ Ε ἄρα τὸν ΒΖ μετρεῖ· μετρεῖ δὲ καὶ ὅλον τὸν ΒΑ· a unit, HA. καὶ λοιπὸν ἄρα τὸν ΑΖ μετρήσει. ὁ δὲ ΑΖ τὸν ΔΗ μετρεῖ· In fact, since E measures CD, and CD measures BF , καὶ ὁ Ε ἄρα τὸν ΔΗ μετρεῖ· μετρεῖ δὲ καὶ ὅλον τὸν ΔΓ· E thus also measures BF .† And (E) also measures the καὶ λοιπὸν ἄρα τὸν ΓΗ μετρήσει. ὁ δὲ ΓΗ τὸν ΖΘ μετρεῖ· whole of BA. Thus, (E) will also measure the remainder 195 STOIQEIWN zþ. ELEMENTS BOOK 7 καὶ ὁ Ε ἄρα τὸν ΖΘ μετρεῖ· μετρεῖ δὲ καὶ ὅλον τὸν ΖΑ· AF .‡ And AF measures DG. Thus, E also measures DG. καὶ λοιπὴν ἄρα τὴν ΑΘ μονάδα μετρήσει ἀριθμὸς ὤν· ὅπερ And (E) also measures the whole of DC. Thus, (E) will ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς ΑΒ, ΓΔ ἀριθμοὺς μετρήσει also measure the remainder CG. And CG measures FH . τις ἀριθμός· οἱ ΑΒ, ΓΔ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν· Thus, E also measures FH . And (E) also measures the ὅπερ ἔδει δεῖξαι. whole of FA. Thus, (E) will also measure the remaining unit AH , (despite) being a number. The very thing is impossible. Thus, some number does not measure (both) the numbers AB and CD. Thus, AB and CD are prime to one another. (Which is) the very thing it was required to show. † Here, use is made of the unstated common notion that if a measures b, and b measures c, then a also measures c, where all symbols denote numbers. ‡ Here, use is made of the unstated common notion that if a measures b, and a measures part of b, then a also measures the remainder of b, where all symbols denote numbers. bþ. Proposition 2 Δύο ἀριθμῶν δοθέντων μὴ πρώτων πρὸς ἀλλήλους τὸ To find the greatest common measure of two given μέγιστον αὐτῶν κοινὸν μέτρον εὑρεῖν. numbers (which are) not prime to one another. ∆ Η Α Ε Β Γ Ζ B G F A E C D ῎Εστωσαν οἱ δοθέντες δύο ἀριθμοὶ μὴ πρῶτοι πρὸς Let AB and CD be the two given numbers (which ἀλλήλους οἱ ΑΒ, ΓΔ. δεῖ δὴ τῶν ΑΒ, ΓΔ τὸ μέγιστον κοινὸν are) not prime to one another. So it is required to find μέτρον εὑρεῖν. the greatest common measure of AB and CD. Εἰ μὲν οὖν ὁ ΓΔ τὸν ΑΒ μετρεῖ, μετρεῖ δὲ καὶ ἑαυτόν, ὁ In fact, if CD measures AB, CD is thus a common ΓΔ ἄρα τῶν ΓΔ, ΑΒ κοινὸν μέτρον ἐστίν. καὶ φανερόν, ὅτι measure of CD and AB, (since CD) also measures itself. καὶ μέγιστον· οὐδεὶς γὰρ μείζων τοῦ ΓΔ τὸν ΓΔ μετρήσει. And (it is) manifest that (it is) also the greatest (com- Εἰ δὲ οὐ μετρεῖ ὁ ΓΔ τὸν ΑΒ, τῶν ΑΒ, ΓΔ ἀνθυφαι- mon measure). For nothing greater than CD can mea- ρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος λειφθήσεταί sure CD. τις ἀριθμός, ὃς μετρήσει τὸν πρὸ ἑαυτοῦ. μονὰς μὲν But if CD does not measure AB then some number γὰρ οὐ λειφθήσεται· εἰ δὲ μή, ἔσονται οἱ ΑΒ, ΓΔ πρῶτοι will remain from AB and CD, the lesser being contin- πρὸς ἀλλήλους· ὅπερ οὐχ ὑπόκειται. λειφθήσεταί τις ἄρα ually subtracted, in turn, from the greater, which will ἀριθμὸς, ὃς μετρήσει τὸν πρὸ ἑαυτοῦ. καὶ ὁ μὲν ΓΔ τὸν measure the (number) preceding it. For a unit will not be ΒΕ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΕΑ, ὁ δὲ ΕΑ τὸν left. But if not, AB and CD will be prime to one another ΔΖ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΖΓ, ὁ δὲ ΓΖ τὸν [Prop. 7.1]. The very opposite thing was assumed. Thus, ΑΕ μετρείτω. ἐπεὶ οὖν ὁ ΓΖ τὸν ΑΕ μετρεῖ, ὁ δὲ ΑΕ τὸν some number will remain which will measure the (num- ΔΖ μετρεῖ, καὶ ὁ ΓΖ ἄρα τὸν ΔΖ μετρήσει. μετρεῖ δὲ καὶ ber) preceding it. And let CD measuring BE leave EA ἑαυτόν· καὶ ὅλον ἄρα τὸν ΓΔ μετρήσει. ὁ δὲ ΓΔ τὸν ΒΕ less than itself, and let EA measuring DF leave FC less μετρεῖ· καὶ ὁ ΓΖ ἄρα τὸν ΒΕ μετρεῖ· μετρεῖ δὲ καὶ τὸν ΕΑ· than itself, and let CF measure AE. Therefore, since CF καὶ ὅλον ἄρα τὸν ΒΑ μετρήσει· μετρεῖ δὲ καὶ τὸν ΓΔ· ὁ ΓΖ measures AE, and AE measures DF , CF will thus also ἄρα τοὺς ΑΒ, ΓΔ μετρεῖ. ὁ ΓΖ ἄρα τῶν ΑΒ, ΓΔ κοινὸν measure DF . And it also measures itself. Thus, it will 196 STOIQEIWN zþ. ELEMENTS BOOK 7 μέτρον ἐστίν. λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ μή ἐστιν ὁ also measure the whole of CD. And CD measures BE. ΓΖ τῶν ΑΒ, ΓΔ μέγιστον κοινὸν μέτρον, μετρήσει τις τοὺς Thus, CF also measures BE. And it also measures EA. ΑΒ, ΓΔ ἀριθμοὺς ἀριθμὸς μείζων ὢν τοῦ ΓΖ. μετρείτω, Thus, it will also measure the whole of BA. And it also καὶ ἔστω ὁ Η. καὶ ἐπεὶ ὁ Η τὸν ΓΔ μετρεῖ, ὁ δὲ ΓΔ τὸν measures CD. Thus, CF measures (both) AB and CD. ΒΕ μετρεῖ, καὶ ὁ Η ἄρα τὸν ΒΕ μετρεῖ· μετρεῖ δὲ καὶ ὅλον Thus, CF is a common measure of AB and CD. So I say τὸν ΒΑ· καὶ λοιπὸν ἄρα τὸν ΑΕ μετρήσει. ὁ δὲ ΑΕ τὸν that (it is) also the greatest (common measure). For if ΔΖ μετρεῖ· καὶ ὁ Η ἄρα τὸν ΔΖ μετρήσει· μετρεῖ δὲ καὶ CF is not the greatest common measure of AB and CD ὅλον τὸν ΔΓ· καὶ λοιπὸν ἄρα τὸν ΓΖ μετρήσει ὁ μείζων then some number which is greater than CF will mea- τὸν ἐλάσσονα· ὅπερ ἐστὶν ἀδύνατον· οὐκ ἄρα τοὺς ΑΒ, ΓΔ sure the numbers AB and CD. Let it (so) measure (AB ἀριθμοὺς ἀριθμός τις μετρήσει μείζων ὢν τοῦ ΓΖ· ὁ ΓΖ ἄρα and CD), and let it be G. And since G measures CD, τῶν ΑΒ, ΓΔ μέγιστόν ἐστι κοινὸν μέτρον [ὅπερ ἔδει δεῖξαι]. and CD measures BE, G thus also measures BE. And it also measures the whole of BA. Thus, it will also mea- sure the remainder AE. And AE measures DF . Thus, G will also measure DF . And it also measures the whole of DC. Thus, it will also measure the remainder CF , the greater (measuring) the lesser. The very thing is im- possible. Thus, some number which is greater than CF cannot measure the numbers AB and CD. Thus, CF is the greatest common measure of AB and CD. [(Which is) the very thing it was required to show].Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν ἀριθμὸς δύο ἀριθμοὺς So it is manifest, from this, that if a number measures μετρῇ, καὶ τὸ μέγιστον αὐτῶν κοινὸν μέτρον μετρήσει· ὅπερ two numbers then it will also measure their greatest com- ἔδει δεῖξαι. mon measure. (Which is) the very thing it was required to show.gþ. Proposition 3 Τριῶν ἀριθμῶν δοθέντων μὴ πρώτων πρὸς ἀλλήλους τὸ To find the greatest common measure of three given μέγιστον αὐτῶν κοινὸν μέτρον εὑρεῖν. numbers (which are) not prime to one another. ΖΑ Β Γ ∆ Ε FA B C D E ῎Εστωσαν οἱ δοθέντες τρεῖς ἀριθμοὶ μὴ πρῶτοι πρὸς Let A, B, and C be the three given numbers (which ἀλλήλους οἱ Α, Β, Γ· δεῖ δὴ τῶν Α, Β, Γ τὸ μέγιστον κοινὸν are) not prime to one another. So it is required to find μέτρον εὑρεῖν. the greatest common measure of A, B, and C. Εἰλήφθω γὰρ δύο τῶν Α, Β τὸ μέγιστον κοινὸν μέτρον ὁ For let the greatest common measure, D, of the two Δ· ὁ δὴ Δ τὸν Γ ἤτοι μετρεῖ ἢ οὐ μετρεῖ. μετρείτω πρότερον· (numbers) A and B have been taken [Prop. 7.2]. So D μετρεῖ δέ καὶ τοὺς Α, Β· ὁ Δ ἄρα τοὺς Α, Β, Γ μετρεῖ· ὁ either measures, or does not measure, C. First of all, let Δ ἄρα τῶν Α, Β, Γ κοινὸν μέτρον ἐστίν. λέγω δή, ὅτι καὶ it measure (C). And it also measures A and B. Thus, D 197 STOIQEIWN zþ. ELEMENTS BOOK 7 μέγιστον. εἰ γὰρ μή ἐστιν ὁ Δ τῶν Α, Β, Γ μέγιστον κοινὸν measures A, B, and C. Thus, D is a common measure μέτρον, μετρήσει τις τοὺς Α, Β, Γ ἀριθμοὺς ἀριθμὸς μείζων of A, B, and C. So I say that (it is) also the greatest ὢν τοῦ Δ. μετρείτω, καὶ ἔστω ὁ Ε. ἐπεὶ οὖν ὁ Ε τοὺς Α, Β, (common measure). For if D is not the greatest common Γ μετρεῖ, καὶ τοὺς Α, Β ἄρα μετρήσει· καὶ τὸ τῶν Α, Β ἄρα measure of A, B, and C then some number greater than μέγιστον κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Α, Β μέγιστον D will measure the numbers A, B, and C. Let it (so) κοινὸν μέτρον ἐστὶν ὁ Δ· ὁ Ε ἄρα τὸν Δ μετρεῖ ὁ μείζων measure (A, B, and C), and let it be E. Therefore, since τὸν ἐλάσσονα· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Α, Β, Γ E measures A, B, and C, it will thus also measure A and ἀριθμοὺς ἀριθμός τις μετρήσει μείζων ὢν τοῦ Δ· ὁ Δ ἄρα B. Thus, it will also measure the greatest common mea- τῶν Α, Β, Γ μέγιστόν ἐστι κοινὸν μέτρον. sure of A and B [Prop. 7.2 corr.]. And D is the greatest Μὴ μετρείτω δὴ ὁ Δ τὸν Γ· λέγω πρῶτον, ὅτι οἱ Γ, Δ common measure of A and B. Thus, E measures D, the οὔκ εἰσι πρῶτοι πρὸς ἀλλήλους. ἐπεὶ γὰρ οἱ Α, Β, Γ οὔκ greater (measuring) the lesser. The very thing is impossi- εἰσι πρῶτοι πρὸς ἀλλήλους, μετρήσει τις αὐτοὺς ἀριθμός. ὁ ble. Thus, some number which is greater than D cannot δὴ τοὺς Α, Β, Γ μετρῶν καὶ τοὺς Α, Β μετρήσει, καὶ τὸ measure the numbers A, B, and C. Thus, D is the great- τῶν Α, Β μέγιστον κοινὸν μέτρον τὸν Δ μετρήσει· μετρεῖ est common measure of A, B, and C. δὲ καὶ τὸν Γ· τοὺς Δ, Γ ἄρα ἀριθμοὺς ἀριθμός τις μετρήσει· So let D not measure C. I say, first of all, that C οἱ Δ, Γ ἄρα οὔκ εἰσι πρῶτοι πρὸς ἀλλήλους. εἰλήφθω οὖν and D are not prime to one another. For since A, B, C αὐτῶν τὸ μέγιστον κοινὸν μέτρον ὁ Ε. καὶ ἐπεὶ ὁ Ε τὸν Δ are not prime to one another, some number will measure μετρεῖ, ὁ δὲ Δ τοὺς Α, Β μετρεῖ, καὶ ὁ Ε ἄρα τοὺς Α, Β them. So the (number) measuring A, B, and C will also μετρεῖ· μετρεῖ δὲ καὶ τὸν Γ· ὁ Ε ἄρα τοὺς Α, Β, Γ μετρεῖ. measure A and B, and it will also measure the greatest ὁ Ε ἄρα τῶν Α, Β, Γ κοινόν ἐστι μέτρον. λέγω δή, ὅτι καὶ common measure, D, of A and B [Prop. 7.2 corr.]. And μέγιστον. εἰ γὰρ μή ἐστιν ὁ Ε τῶν Α, Β, Γ τὸ μέγιστον it also measures C. Thus, some number will measure the κοινὸν μέτρον, μετρήσει τις τοὺς Α, Β, Γ ἀριθμοὺς ἀριθμὸς numbers D and C. Thus, D and C are not prime to one μείζων ὢν τοῦ Ε. μετρείτω, καὶ ἔστω ὁ Ζ. καὶ ἐπεὶ ὁ Ζ τοὺς another. Therefore, let their greatest common measure, Α, Β, Γ μετρεῖ, καὶ τοὺς Α, Β μετρεῖ· καὶ τὸ τῶν Α, Β ἄρα E, have been taken [Prop. 7.2]. And since E measures μέγιστον κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Α, Β μέγιστον D, and D measures A and B, E thus also measures A κοινὸν μέτρον ἐστὶν ὁ Δ· ὁ Ζ ἄρα τὸν Δ μετρεῖ· μετρεῖ δὲ and B. And it also measures C. Thus, E measures A, B, καὶ τὸν Γ· ὁ Ζ ἄρα τοὺς Δ, Γ μετρεῖ· καὶ τὸ τῶν Δ, Γ ἄρα and C. Thus, E is a common measure of A, B, and C. So μέγιστον κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Δ, Γ μέγιστον I say that (it is) also the greatest (common measure). For κοινὸν μέτρον ἐστὶν ὁ Ε· ὁ Ζ ἄρα τὸν Ε μετρεῖ ὁ μείζων if E is not the greatest common measure of A, B, and C τὸν ἐλάσσονα· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Α, Β, Γ then some number greater than E will measure the num- ἀριθμοὺς ἀριθμός τις μετρήσει μείζων ὢν τοῦ Ε· ὁ Ε ἄρα bers A, B, and C. Let it (so) measure (A, B, and C), and τῶν Α, Β, Γ μέγιστόν ἐστι κοινὸν μέτρον· ὅπερ ἔδει δεῖξαι. let it be F . And since F measures A, B, and C, it also measures A and B. Thus, it will also measure the great- est common measure of A and B [Prop. 7.2 corr.]. And D is the greatest common measure of A and B. Thus, F measures D. And it also measures C. Thus, F measures D and C. Thus, it will also measure the greatest com- mon measure of D and C [Prop. 7.2 corr.]. And E is the greatest common measure of D and C. Thus, F measures E, the greater (measuring) the lesser. The very thing is impossible. Thus, some number which is greater than E does not measure the numbers A, B, and C. Thus, E is the greatest common measure of A, B, and C. (Which is) the very thing it was required to show.dþ. Proposition 4 ῞Απας ἀριθμὸς παντὸς ἀριθμοῦ ὁ ἐλάσσων τοῦ μείζονος Any number is either part or parts of any (other) num- ἤτοι μέρος ἐστὶν ἢ μέρη. ber, the lesser of the greater. ῎Εστωσαν δύο ἀριθμοὶ οἱ Α, ΒΓ, καὶ ἔστω ἐλάσσων ὁ Let A and BC be two numbers, and let BC be the ΒΓ· λέγω, ὅτι ὁ ΒΓ τοῦ Α ἤτοι μέρος ἐστὶν ἢ μέρη. lesser. I say that BC is either part or parts of A. 198 STOIQEIWN zþ. ELEMENTS BOOK 7 Οἱ Α, ΒΓ γὰρ ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν ἢ οὔ. For A and BC are either prime to one another, or not. ἔστωσαν πρότερον οἱ Α, ΒΓ πρῶτοι πρὸς ἀλλήλους. διαι- Let A and BC, first of all, be prime to one another. So ρεθέντος δὴ τοῦ ΒΓ εἰς τὰς ἐν αὐτῷ μονάδας ἔσται ἑκάστη separating BC into its constituent units, each of the units μονὰς τῶν ἐν τῷ ΒΓ μέρος τι τοῦ Α· ὥστε μέρη ἐστὶν ὁ ΒΓ in BC will be some part of A. Hence, BC is parts of A. τοῦ Α. Ε Ζ Γ Β Α ∆ D C F E B A Μὴ ἔστωσαν δὴ οἱ Α, ΒΓ πρῶτοι πρὸς ἀλλήλους· ὁ δὴ So let A and BC be not prime to one another. So BC ΒΓ τὸν Α ἤτοι μετρεῖ ἢ οὐ μετρεῖ. εἰ μὲν οὖν ὁ ΒΓ τὸν either measures, or does not measure, A. Therefore, if Α μετρεῖ, μέρος ἐστὶν ὁ ΒΓ τοῦ Α. εἰ δὲ οὔ, εἰλήφθω τῶν BC measures A then BC is part of A. And if not, let the Α, ΒΓ μέγιστον κοινὸν μέτρον ὁ Δ, καὶ διῃρήσθω ὁ ΒΓ εἰς greatest common measure, D, of A and BC have been τοὺς τῷ Δ ἴσους τοὺς ΒΕ, ΕΖ, ΖΓ. καὶ ἐπεὶ ὁ Δ τὸν Α taken [Prop. 7.2], and let BC have been divided into BE, μετρεῖ, μέρος ἐστὶν ὁ Δ τοῦ Α· ἴσος δὲ ὁ Δ ἑκάστῳ τῶν EF , and FC, equal to D. And since D measures A, D is ΒΕ, ΕΖ, ΖΓ· καὶ ἕκαστος ἄρα τῶν ΒΕ, ΕΖ, ΖΓ τοῦ Α μέρος a part of A. And D is equal to each of BE, EF , and FC. ἐστίν· ὥστε μέρη ἐστὶν ὁ ΒΓ τοῦ Α. Thus, BE, EF , and FC are also each part of A. Hence, ῞Απας ἄρα ἀριθμὸς παντὸς ἀριθμοῦ ὁ ἐλάσσων τοῦ BC is parts of A. μείζονος ἤτοι μέρος ἐστὶν ἢ μέρη· ὅπερ ἔδει δεῖξαι. Thus, any number is either part or parts of any (other) number, the lesser of the greater. (Which is) the very thing it was required to show.eþ. Proposition 5† ᾿Εὰν ἀριθμὸς ἀριθμοῦ μέρος ᾖ, καὶ ἕτερος ἑτέρου τὸ If a number is part of a number, and another (num- αὐτὸ μέρος ᾖ, καὶ συναμφότερος συναμφοτέρου τὸ αὐτὸ ber) is the same part of another, then the sum (of the μέρος ἔσται, ὅπερ ὁ εἷς τοῦ ἑνός. leading numbers) will also be the same part of the sum (of the following numbers) that one (number) is of an- other. ∆ Γ Η Β Ζ Θ Ε Α D C G B F H E A Ἀριθμὸς γὰρ ὁ Α [ἀριθμοῦ] τοῦ ΒΓ μέρος ἔστω, καὶ For let a number A be part of a [number] BC, and 199 STOIQEIWN zþ. ELEMENTS BOOK 7 ἕτερος ὁ Δ ἑτέρου τοῦ ΕΖ τὸ αὐτὸ μέρος, ὅπερ ὁ Α τοῦ another (number) D (be) the same part of another (num- ΒΓ· λέγω, ὅτι καὶ συναμφότερος ὁ Α, Δ συναμφοτέρου τοῦ ber) EF that A (is) of BC. I say that the sum A, D is also ΒΓ, ΕΖ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὁ Α τοῦ ΒΓ. the same part of the sum BC, EF that A (is) of BC. ᾿Επεὶ γάρ, ὃ μέρος ἐστὶν ὁ Α τοῦ ΒΓ, τὸ αὐτὸ μέρος ἐστὶ For since which(ever) part A is of BC, D is the same καὶ ὁ Δ τοῦ ΕΖ, ὅσοι ἄρα εἰσὶν ἐν τῷ ΒΓ ἀριθμοὶ ἴσοι τῷ part of EF , thus as many numbers as are in BC equal Α, τοσοῦτοί εἰσι καὶ ἐν τῷ ΕΖ ἀριθμοὶ ἴσοι τῷ Δ. διῇρήσθω to A, so many numbers are also in EF equal to D. Let ὁ μὲν ΒΓ εἰς τοὺς τῷ Α ἴσους τοὺς ΒΗ, ΗΓ, ὁ δὲ ΕΖ εἰς BC have been divided into BG and GC, equal to A, and τοὺς τῷ Δ ἴσους τοὺς ΕΘ, ΘΖ· ἔσται δὴ ἴσον τὸ πλῆθος EF into EH and HF , equal to D. So the multitude of τῶν ΒΗ, ΗΓ τῷ πλήθει τῶν ΕΘ, ΘΖ. καὶ ἐπεὶ ἴσος ἐστὶν (divisions) BG, GC will be equal to the multitude of (di- ὁ μὲν ΒΗ τῷ Α, ὁ δὲ ΕΘ τῷ Δ, καὶ οἱ ΒΗ, ΕΘ ἄρα τοῖς visions) EH , HF . And since BG is equal to A, and EH Α, Δ ἴσοι. διὰ τὰ αὐτὰ δὴ καὶ οἱ ΗΓ, ΘΖ τοῖς Α, Δ. ὅσοι to D, thus BG, EH (is) also equal to A, D. So, for the ἄρα [εἰσὶν] ἐν τῷ ΒΓ ἀριθμοὶ ἴσοι τῷ Α, τοσοῦτοί εἰσι καὶ same (reasons), GC, HF (is) also (equal) to A, D. Thus, ἐν τοῖς ΒΓ, ΕΖ ἴσοι τοῖς Α, Δ. ὁσαπλασίων ἄρα ἐστὶν ὁ ΒΓ as many numbers as [are] in BC equal to A, so many are τοῦ Α, τοσαυταπλασίων ἐστὶ καὶ συναμφότερος ὁ ΒΓ, ΕΖ also in BC, EF equal to A, D. Thus, as many times as συναμφοτέρου τοῦ Α, Δ. ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ ΒΓ, τὸ BC is (divisible) by A, so many times is the sum BC, EF αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ Α, Δ συναμφοτέρου also (divisible) by the sum A, D. Thus, which(ever) part τοῦ ΒΓ, ΕΖ· ὅπερ ἔδει δεῖξαι. A is of BC, the sum A, D is also the same part of the sum BC, EF . (Which is) the very thing it was required to show. † In modern notation, this proposition states that if a = (1/n) b and c = (1/n) d then (a+ c) = (1/n) (b+d), where all symbols denote numbers.�þ. Proposition 6† ᾿Εὰν ἀριθμὸς ἀριθμοῦ μέρη ᾖ, καὶ ἕτερος ἑτέρου τὰ If a number is parts of a number, and another (num- αὐτὰ μέρη ᾖ, καὶ συναμφότερος συναμφοτέρου τὰ αὐτὰ ber) is the same parts of another, then the sum (of the μέρη ἔσται, ὅπερ ὁ εἷς τοῦ ἑνός. leading numbers) will also be the same parts of the sum (of the following numbers) that one (number) is of an- other. Ε Θ ∆ Η Α Β Γ Ζ E G A C D H F B Ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ Γ μέρη ἔστω, καὶ ἕτερος For let a number AB be parts of a number C, and an- ὁ ΔΕ ἑτέρου τοῦ Ζ τὰ αὐτὰ μέρη, ἅπερ ὁ ΑΒ τοῦ Γ· λέγω, other (number) DE (be) the same parts of another (num- ὅτι καὶ συναμφότερος ὁ ΑΒ, ΔΕ συναμφοτέρου τοῦ Γ, Ζ ber) F that AB (is) of C. I say that the sum AB, DE is τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὁ ΑΒ τοῦ Γ. also the same parts of the sum C, F that AB (is) of C. ᾿Επεὶ γάρ, ἃ μέρη ἐστὶν ὁ ΑΒ τοῦ Γ, τὰ αὐτὰ μέρη καὶ For since which(ever) parts AB is of C, DE (is) also ὁ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μέρη τοῦ Γ, τοσαῦτά the same parts of F , thus as many parts of C as are in AB, ἐστι καὶ ἐν τῷ ΔΕ μέρη τοῦ Ζ. διῃρήσθω ὁ μὲν ΑΒ εἰς τὰ so many parts of F are also in DE. Let AB have been τοῦ Γ μέρη τὰ ΑΗ, ΗΒ, ὁ δὲ ΔΕ εἰς τὰ τοῦ Ζ μέρη τὰ divided into the parts of C, AG and GB, and DE into the ΔΘ, ΘΕ· ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΒ τῷ πλήθει parts of F , DH and HE. So the multitude of (divisions) τῶν ΔΘ, ΘΕ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ Γ, τὸ AG, GB will be equal to the multitude of (divisions) DH , 200 STOIQEIWN zþ. ELEMENTS BOOK 7 αὑτὸ μέρος ἐστὶ καὶ ὁ ΔΘ τοῦ Ζ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΗ HE. And since which(ever) part AG is of C, DH is also τοῦ Γ, τὸ αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ ΑΗ, ΔΘ the same part of F , thus which(ever) part AG is of C, συναμφοτέρου τοῦ Γ, Ζ. διὰ τὰ αὐτὰ δὴ καὶ ὃ μέρος ἐστὶν ὁ the sum AG, DH is also the same part of the sum C, F ΗΒ τοῦ Γ, τὸ αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ ΗΒ, ΘΕ [Prop. 7.5]. And so, for the same (reasons), which(ever) συναμφοτέρου τοῦ Γ, Ζ. ἃ ἄρα μέρη ἐστὶν ὁ ΑΒ τοῦ Γ, τὰ part GB is of C, the sum GB, HE is also the same part αὐτὰ μέρη ἐστὶ καὶ συναμφότερος ὁ ΑΒ, ΔΕ συναμφοτέρου of the sum C, F . Thus, which(ever) parts AB is of C, τοῦ Γ, Ζ· ὅπερ ἔδει δεῖξαι. the sum AB, DE is also the same parts of the sum C, F . (Which is) the very thing it was required to show. † In modern notation, this proposition states that if a = (m/n) b and c = (m/n) d then (a + c) = (m/n) (b + d), where all symbols denote numbers. zþ. Proposition 7† ᾿Εὰν ἀριθμὸς ἀριθμοῦ μέρος ᾖ, ὅπερ ἀφαιρεθεὶς ἀφαι- If a number is that part of a number that a (part) ρεθέντος, καὶ ὁ λοιπὸς τοῦ λοιποῦ τὸ αὐτὸ μέρος ἔσται, taken away (is) of a (part) taken away then the remain- ὅπερ ὁ ὅλος τοῦ ὅλου. der will also be the same part of the remainder that the whole (is) of the whole. Η Α Γ Ζ ∆ ΒΕ EA G C DF B Ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ ΓΔ μέρος ἔστω, ὅπερ For let a number AB be that part of a number CD ἀφαιρεθεὶς ὁ ΑΕ ἀφαιρεθέντος τοῦ ΓΖ· λέγω, ὅτι καὶ λοιπὸς that a (part) taken away AE (is) of a part taken away ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὅλος ὁ ΑΒ CF . I say that the remainder EB is also the same part of ὅλου τοῦ ΓΔ. the remainder FD that the whole AB (is) of the whole ῝Ο γὰρ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἔστω CD. καὶ ὁ ΕΒ τοῦ ΓΗ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ For which(ever) part AE is of CF , let EB also be the αὐτὸ μέρος ἐστὶ καὶ ὁ ΕΒ τοῦ ΓΗ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΕ same part of CG. And since which(ever) part AE is of τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΒ τοῦ ΗΖ. ὃ δὲ μέρος CF , EB is also the same part of CG, thus which(ever) ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ὑπόκειται καὶ ὁ ΑΒ τοῦ part AE is of CF , AB is also the same part of GF ΓΔ· ὃ ἄρα μέρος ἐστὶ καὶ ὁ ΑΒ τοῦ ΗΖ, τὸ αὐτὸ μέρος ἐστὶ [Prop. 7.5]. And which(ever) part AE is of CF , AB is καὶ τοῦ ΓΔ· ἴσος ἄρα ἐστὶν ὁ ΗΖ τῷ ΓΔ. κοινὸς ἀφῃρήσθω also assumed (to be) the same part of CD. Thus, also, ὁ ΓΖ· λοιπὸς ἄρα ὁ ΗΓ λοιπῷ τῷ ΖΔ ἐστιν ἴσος. καὶ ἐπεί, which(ever) part AB is of GF , (AB) is also the same ὃ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος [ἐστὶ] καὶ ὁ ΕΒ part of CD. Thus, GF is equal to CD. Let CF have been τοῦ ΗΓ, ἴσος δὲ ὁ ΗΓ τῷ ΖΔ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΕ τοῦ subtracted from both. Thus, the remainder GC is equal ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΕΒ τοῦ ΖΔ. ἀλλὰ ὃ μέρος to the remainder FD. And since which(ever) part AE is ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΒ τοῦ ΓΔ· of CF , EB [is] also the same part of GC, and GC (is) καὶ λοιπὸς ἄρα ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστίν, equal to FD, thus which(ever) part AE is of CF , EB is ὅπερ ὅλος ὁ ΑΒ ὅλου τοῦ ΓΔ· ὅπερ ἔδει δεῖξαι. also the same part of FD. But, which(ever) part AE is of CF , AB is also the same part of CD. Thus, the remain- der EB is also the same part of the remainder FD that the whole AB (is) of the whole CD. (Which is) the very thing it was required to show. † In modern notation, this proposition states that if a = (1/n) b and c = (1/n) d then (a− c) = (1/n) (b−d), where all symbols denote numbers.hþ. Proposition 8† ᾿Εὰν ἀριθμὸς ἀριθμοῦ μέρη ᾖ, ἅπερ ἀφαιρεθεὶς ἀφαι- If a number is those parts of a number that a (part) ρεθέντος, καὶ ὁ λοιπὸς τοῦ λοιποῦ τὰ αὐτὰ μέρη ἔσται, taken away (is) of a (part) taken away then the remain- ἅπερ ὁ ὅλος τοῦ ὅλου. der will also be the same parts of the remainder that the 201 STOIQEIWN zþ. ELEMENTS BOOK 7 whole (is) of the whole. Β Η Α Λ Ε Ν Γ Ζ ∆ Μ Κ Θ C A E F G KM N L B H D Ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ ΓΔ μέρη ἔστω, ἅπερ For let a number AB be those parts of a number CD ἀφαιρεθεὶς ὁ ΑΕ ἀφαιρεθέντος τοῦ ΓΖ· λέγω, ὅτι καὶ λοιπὸς that a (part) taken away AE (is) of a (part) taken away ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὅλος ὁ ΑΒ CF . I say that the remainder EB is also the same parts ὅλου τοῦ ΓΔ. of the remainder FD that the whole AB (is) of the whole Κείσθω γὰρ τῷ ΑΒ ἴσος ὁ ΗΘ, ἃ ἄρα μέρη ἐστὶν ὁ ΗΘ CD. τοῦ ΓΔ, τὰ αὐτὰ μέρη ἐστὶ καὶ ὁ ΑΕ τοῦ ΓΖ. διῃρήσθω ὁ For let GH be laid down equal to AB. Thus, μὲν ΗΘ εἰς τὰ τοῦ ΓΔ μέρη τὰ ΗΚ, ΚΘ, ὁ δὲ ΑΕ εἰς τὰ τοῦ which(ever) parts GH is of CD, AE is also the same ΓΖ μέρη τὰ ΑΛ, ΛΕ· ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΗΚ, ΚΘ parts of CF . Let GH have been divided into the parts τῷ πλήθει τῶν ΑΛ, ΛΕ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΗΚ τοῦ of CD, GK and KH , and AE into the part of CF , AL ΓΔ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΛ τοῦ ΓΖ, μείζων δὲ ὁ ΓΔ and LE. So the multitude of (divisions) GK, KH will be τοῦ ΓΖ, μείζων ἄρα καὶ ὁ ΗΚ τοῦ ΑΛ. κείσθω τῷ ΑΛ ἴσος equal to the multitude of (divisions) AL, LE. And since ὁ ΗΜ. ὃ ἄρα μέρος ἐστὶν ὁ ΗΚ τοῦ ΓΔ, τὸ αὐτὸ μέρος ἐστὶ which(ever) part GK is of CD, AL is also the same part καὶ ὁ ΗΜ τοῦ ΓΖ· καὶ λοιπὸς ἄρα ὁ ΜΚ λοιποῦ τοῦ ΖΔ of CF , and CD (is) greater than CF , GK (is) thus also τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὅλος ὁ ΗΚ ὅλου τοῦ ΓΔ. πάλιν greater than AL. Let GM be made equal to AL. Thus, ἐπεί, ὃ μέρος ἐστὶν ὁ ΚΘ τοῦ ΓΔ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ which(ever) part GK is of CD, GM is also the same part ΕΛ τοῦ ΓΖ, μείζων δὲ ὁ ΓΔ τοῦ ΓΖ, μείζων ἄρα καὶ ὁ ΘΚ of CF . Thus, the remainder MK is also the same part of τοῦ ΕΛ. κείσθω τῷ ΕΛ ἴσος ὁ ΚΝ. ὃ ἄρα μέρος ἐστὶν ὁ ΚΘ the remainder FD that the whole GK (is) of the whole τοῦ ΓΔ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΚΝ τοῦ ΓΖ· καὶ λοιπὸς CD [Prop. 7.5]. Again, since which(ever) part KH is of ἄρα ὁ ΝΘ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὅλος ὁ CD, EL is also the same part of CF , and CD (is) greater ΚΘ ὅλου τοῦ ΓΔ. ἐδείχθη δὲ καὶ λοιπὸς ὁ ΜΚ λοιποῦ τοῦ than CF , HK (is) thus also greater than EL. Let KN be ΖΔ τὸ αὐτὸ μέρος ὤν, ὅπερ ὅλος ὁ ΗΚ ὅλου τοῦ ΓΔ· καὶ made equal to EL. Thus, which(ever) part KH (is) of συναμφότερος ἄρα ὁ ΜΚ, ΝΘ τοῦ ΔΖ τὰ αὐτὰ μέρη ἐστίν, CD, KN is also the same part of CF . Thus, the remain- ἅπερ ὅλος ὁ ΘΗ ὅλου τοῦ ΓΔ. ἴσος δὲ συναμφότερος μὲν der NH is also the same part of the remainder FD that ὁ ΜΚ, ΝΘ τῷ ΕΒ, ὁ δὲ ΘΗ τῷ ΒΑ· καὶ λοιπὸς ἄρα ὁ ΕΒ the whole KH (is) of the whole CD [Prop. 7.5]. And the λοιποῦ τοῦ ΖΔ τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὅλος ὁ ΑΒ ὅλου remainder MK was also shown to be the same part of τοῦ ΓΔ· ὅπερ ἔδει δεῖξαι. the remainder FD that the whole GK (is) of the whole CD. Thus, the sum MK, NH is the same parts of DF that the whole HG (is) of the whole CD. And the sum MK, NH (is) equal to EB, and HG to BA. Thus, the remainder EB is also the same parts of the remainder FD that the whole AB (is) of the whole CD. (Which is) the very thing it was required to show. † In modern notation, this proposition states that if a = (m/n) b and c = (m/n) d then (a − c) = (m/n) (b − d), where all symbols denote numbers. jþ. Proposition 9† ᾿Εὰν ἀριθμὸς ἀριθμοῦ μέρος ᾖ, καὶ ἕτερος ἑτέρου τὸ If a number is part of a number, and another (num- αὐτὸ μέρος ᾖ, καὶ ἐναλλάξ, ὃ μέρος ἐστὶν ἢ μέρη ὁ πρῶτος ber) is the same part of another, also, alternately, τοῦ τρίτου, τὸ αὐτὸ μέρος ἔσται ἢ τὰ αὐτὰ μέρη καὶ ὁ which(ever) part, or parts, the first (number) is of the δεύτερος τοῦ τετάρτου. third, the second (number) will also be the same part, or 202 STOIQEIWN zþ. ELEMENTS BOOK 7 the same parts, of the fourth. Β Ε Θ Ζ ∆Α Γ Η F A C G B D E H Ἀριθμὸς γὰρ ὁ Α ἀριθμοῦ τοῦ ΒΓ μέρος ἔστω, καὶ ἕτε- For let a number A be part of a number BC, and an- ρος ὁ Δ ἑτέρου τοῦ ΕΖ τὸ αὐτὸ μέρος, ὅπερ ὁ Α τοῦ ΒΓ· other (number) D (be) the same part of another EF that λέγω, ὅτι καὶ ἐναλλάξ, ὃ μέρος ἐστὶν ὁ Α τοῦ Δ ἢ μέρη, τὸ A (is) of BC. I say that, also, alternately, which(ever) αὐτὸ μέρος ἐστὶ καὶ ὁ ΒΓ τοῦ ΕΖ ἢ μέρη. part, or parts, A is of D, BC is also the same part, or ᾿Επεὶ γὰρ ὃ μέρος ἐστὶν ὁ Α τοῦ ΒΓ, τὸ αὐτὸ μέρος ἐστὶ parts, of EF . καὶ ὁ Δ τοῦ ΕΖ, ὅσοι ἄρα εἰσὶν ἐν τῷ ΒΓ ἀριθμοὶ ἴσοι τῷ For since which(ever) part A is of BC, D is also the Α, τοσοῦτοί εἰσι καὶ ἐν τῷ ΕΖ ἴσοι τῷ Δ. διῃρήσθω ὁ μὲν same part of EF , thus as many numbers as are in BC ΒΓ εἰς τοὺς τῷ Α ἴσους τοὺς ΒΗ, ΗΓ, ὁ δὲ ΕΖ εἰς τοὺς τῷ equal to A, so many are also in EF equal to D. Let BC Δ ἴσους τοὺς ΕΘ, ΘΖ· ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΒΗ, have been divided into BG and GC, equal to A, and EF ΗΓ τῷ πλήθει τῶν ΕΘ, ΘΖ. into EH and HF , equal to D. So the multitude of (di- Καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΒΗ, ΗΓ ἀριθμοὶ ἀλλήλοις, εἰσὶ visions) BG, GC will be equal to the multitude of (divi- δὲ καὶ οἱ ΕΘ, ΘΖ ἀριθμοὶ ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ sions) EH , HF . πλῆθος τῶν ΒΗ, ΗΓ τῷ πλήθει τῶν ΕΘ, ΘΖ, ὃ ἄρα μέρος And since the numbers BG and GC are equal to one ἐστὶν ὁ ΒΗ τοῦ ΕΘ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΗΓ another, and the numbers EH and HF are also equal to τοῦ ΘΖ ἢ τὰ αὐτὰ μέρη· ὥστε καὶ ὃ μέρος ἐστὶν ὁ ΒΗ τοῦ one another, and the multitude of (divisions) BG, GC ΕΘ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ ΒΓ is equal to the multitude of (divisions) EH , HC, thus συναμφοτέρου τοῦ ΕΖ ἢ τὰ αὐτὰ μέρη. ἴσος δὲ ὁ μὲν ΒΗ which(ever) part, or parts, BG is of EH , GC is also τῷ Α, ὁ δὲ ΕΘ τῷ Δ· ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ Δ ἢ μέρη, the same part, or the same parts, of HF . And hence, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΒΓ τοῦ ΕΖ ἢ τὰ αὐτὰ μέρη· ὅπερ which(ever) part, or parts, BG is of EH , the sum BC ἔδει δεῖξαι. is also the same part, or the same parts, of the sum EF [Props. 7.5, 7.6]. And BG (is) equal to A, and EH to D. Thus, which(ever) part, or parts, A is of D, BC is also the same part, or the same parts, of EF . (Which is) the very thing it was required to show. † In modern notation, this proposition states that if a = (1/n) b and c = (1/n) d then if a = (k/l) c then b = (k/l) d, where all symbols denote numbers. iþ. Proposition 10† ᾿Εὰν ἀριθμὸς ἀριθμοῦ μέρη ᾖ, καὶ ἕτερος ἑτέρου τὰ αὐτὰ If a number is parts of a number, and another (num- μέρη ᾖ, καὶ ἐναλλάξ, ἃ μέρη ἐστὶν ὁ πρῶτος τοῦ τρίτου ἢ ber) is the same parts of another, also, alternately, μέρος, τὰ αὐτὰ μέρη ἔσται καὶ ὁ δεύτερος τοῦ τετάρτου ἢ which(ever) parts, or part, the first (number) is of the τὸ αὐτὸ μέρος. third, the second will also be the same parts, or the same Ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ Γ μέρη ἔστω, καὶ ἕτερος part, of the fourth. ὁ ΔΕ ἑτέρου τοῦ Ζ τὰ αὐτὰ μέρη· λέγω, ὅτι καὶ ἐναλλάξ, For let a number AB be parts of a number C, and ἃ μέρη ἐστὶν ὁ ΑΒ τοῦ ΔΕ ἢ μέρος, τὰ αὐτὰ μέρη ἐστὶ καὶ another (number) DE (be) the same parts of another F . ὁ Γ τοῦ Ζ ἢ τὸ αὐτὸ μέρος. I say that, also, alternately, which(ever) parts, or part, 203 STOIQEIWN zþ. ELEMENTS BOOK 7 AB is of DE, C is also the same parts, or the same part, of F . Ζ Η Ε Θ ∆ Α Β Γ A B G C E H D F ᾿Επεὶ γάρ, ἃ μέρη ἐστὶν ὁ ΑΒ τοῦ Γ, τὰ αὐτὰ μέρη ἐστὶ For since which(ever) parts AB is of C, DE is also καὶ ὁ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μέρη τοῦ Γ, the same parts of F , thus as many parts of C as are in τοσαῦτα καὶ ἐν τῷ ΔΕ μέρη τοῦ Ζ. διῃρήσθω ὁ μὲν ΑΒ εἰς AB, so many parts of F (are) also in DE. Let AB have τὰ τοῦ Γ μέρη τὰ ΑΗ, ΗΒ, ὁ δὲ ΔΕ εἰς τὰ τοῦ Ζ μέρη τὰ been divided into the parts of C, AG and GB, and DE ΔΘ, ΘΕ· ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΒ τῷ πλήθει into the parts of F , DH and HE. So the multitude of τῶν ΔΘ, ΘΕ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ Γ, τὸ αὐτὸ (divisions) AG, GB will be equal to the multitude of (di- μέρος ἐστὶ καὶ ὁ ΔΘ τοῦ Ζ, καὶ ἐναλλάξ, ὃ μέρος ἐστὶν ὁ visions) DH , HE. And since which(ever) part AG is ΑΗ τοῦ ΔΘ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Ζ ἢ of C, DH is also the same part of F , also, alternately, τὰ αὐτὰ μέρη. διὰ τὰ αὐτὰ δὴ καί, ὃ μέρος ἐστὶν ὁ ΗΒ τοῦ which(ever) part, or parts, AG is of DH , C is also the ΘΕ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Ζ ἢ τὰ αὐτὰ same part, or the same parts, of F [Prop. 7.9]. And so, μέρη· ὥστε καί [ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ ΔΘ ἢ μέρη, τὸ for the same (reasons), which(ever) part, or parts, GB is αὐτὸ μέρος ἐστὶ καὶ ὁ ΗΒ τοῦ ΘΕ ἢ τὰ αὐτὰ μέρη· καὶ ὃ of HE, C is also the same part, or the same parts, of F ἄρα μέρος ἐστὶν ὁ ΑΗ τοῦ ΔΘ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ [Prop. 7.9]. And so [which(ever) part, or parts, AG is of καὶ ὁ ΑΒ τοῦ ΔΕ ἢ τὰ αὐτὰ μέρη· ἀλλ᾿ ὃ μέρος ἐστὶν ὁ ΑΗ DH , GB is also the same part, or the same parts, of HE. τοῦ ΔΘ ἢ μέρη, τὸ αὐτὸ μέρος ἐδείχθη καὶ ὁ Γ τοῦ Ζ ἢ τὰ And thus, which(ever) part, or parts, AG is of DH , AB is αὐτὰ μέρη, καὶ] ἃ [ἄρα] μέρη ἐστὶν ὁ ΑΒ τοῦ ΔΕ ἢ μέρος, also the same part, or the same parts, of DE [Props. 7.5, τὰ αὐτὰ μέρη ἐστὶ καὶ ὁ Γ τοῦ Ζ ἢ τὸ αὐτὸ μέρος· ὅπερ ἔδει 7.6]. But, which(ever) part, or parts, AG is of DH , C δεῖξαι. was also shown (to be) the same part, or the same parts, of F . And, thus] which(ever) parts, or part, AB is of DE, C is also the same parts, or the same part, of F . (Which is) the very thing it was required to show. † In modern notation, this proposition states that if a = (m/n) b and c = (m/n) d then if a = (k/l) c then b = (k/l) d, where all symbols denote numbers. iaþ. Proposition 11 ᾿Εαν ᾖ ὡς ὅλος πρὸς ὅλον, οὕτως ἀφαιρεθεὶς πρὸς ἀφαι- If as the whole (of a number) is to the whole (of an- ρεθέντα, καὶ ὁ λοιπὸς πρὸς τὸν λοιπὸν ἔσται, ὡς ὅλος πρὸς other), so a (part) taken away (is) to a (part) taken away, ὅλον. then the remainder will also be to the remainder as the ῎Εστω ὡς ὅλος ὁ ΑΒ πρὸς ὅλον τὸν ΓΔ, οὕτως ἀφαι- whole (is) to the whole. ρεθεὶς ὁ ΑΕ πρὸς ἀφαιρεθέντα τὸν ΓΖ· λέγω, ὅτι καὶ λοιπὸς Let the whole AB be to the whole CD as the (part) ὁ ΕΒ πρὸς λοιπὸν τὸν ΖΔ ἐστιν, ὡς ὅλος ὁ ΑΒ πρὸς ὅλον taken away AE (is) to the (part) taken away CF . I say τὸν ΓΔ. that the remainder EB is to the remainder FD as the whole AB (is) to the whole CD. 204 STOIQEIWN zþ. ELEMENTS BOOK 7 ∆Β Ε Α Γ Ζ A F E C B D ᾿Επεί ἐστιν ὡς ὁ ΑΒ πρὸς τὸν ΓΔ, οὕτως ὁ ΑΕ πρὸς (For) since as AB is to CD, so AE (is) to CF , thus τὸν ΓΖ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΒ τοῦ ΓΔ ἢ μέρη, τὸ αὐτὸ which(ever) part, or parts, AB is of CD, AE is also the μέρος ἐστὶ καὶ ὁ ΑΕ τοῦ ΓΖ ἢ τὰ αὐτὰ μέρη. καὶ λοιπὸς same part, or the same parts, of CF [Def. 7.20]. Thus, ἄρα ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστὶν ἢ μέρη, ἅπερ the remainder EB is also the same part, or parts, of the ὁ ΑΒ τοῦ ΓΔ. ἔστιν ἄρα ὡς ὁ ΕΒ πρὸς τὸν ΖΔ, οὕτως ὁ remainder FD that AB (is) of CD [Props. 7.7, 7.8]. ΑΒ πρὸς τὸν ΓΔ· ὅπερ ἔδει δεῖξαι. Thus, as EB is to FD, so AB (is) to CD [Def. 7.20]. (Which is) the very thing it was required to show. † In modern notation, this proposition states that if a : b :: c : d then a : b :: a − c : b − d, where all symbols denote numbers.ibþ. Proposition 12† ᾿Εὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ ἀνάλογον, ἔσται ὡς εἷς If any multitude whatsoever of numbers are propor- τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως ἅπαντες οἱ tional then as one of the leading (numbers is) to one of ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους. the following so (the sum of) all of the leading (numbers) will be to (the sum of) all of the following. ∆ΒΑ Γ DA B C ῎Εστωσαν ὁποσοιοῦν ἀριθμοὶ ἀνάλογον οἱ Α, Β, Γ, Δ, Let any multitude whatsoever of numbers, A, B, C, ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ· λέγω, ὅτι ἐστὶν D, be proportional, (such that) as A (is) to B, so C (is) ὡς ὁ Α πρὸς τὸν Β, οὕτως οἱ Α, Γ πρὸς τοὺς Β, Δ. to D. I say that as A is to B, so A, C (is) to B, D. ᾿Επεὶ γάρ ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς For since as A is to B, so C (is) to D, thus which(ever) τὸν Δ, ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ Β ἢ μέρη, τὸ αὐτὸ μέρος part, or parts, A is of B, C is also the same part, or parts, ἐστὶ καὶ ὁ Γ τοῦ Δ ἢ μέρη. καὶ συναμφότερος ἄρα ὁ Α, of D [Def. 7.20]. Thus, the sum A, C is also the same Γ συναμφοτέρου τοῦ Β, Δ τὸ αὐτὸ μέρος ἐστὶν ἢ τὰ αὐτὰ part, or the same parts, of the sum B, D that A (is) of B μέρη, ἅπερ ὁ Α τοῦ Β. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως [Props. 7.5, 7.6]. Thus, as A is to B, so A, C (is) to B, D οἱ Α, Γ πρὸς τοὺς Β, Δ· ὅπερ ἔδει δεῖξαι. [Def. 7.20]. (Which is) the very thing it was required to show. 205 STOIQEIWN zþ. ELEMENTS BOOK 7 † In modern notation, this proposition states that if a : b :: c : d then a : b :: a + c : b + d, where all symbols denote numbers.igþ. Proposition 13† ᾿Εὰν τέσσαρες ἀριθμοὶ ἀνάλογον ὦσιν, καὶ ἐναλλὰξ If four numbers are proportional then they will also ἀνάλογον ἔσονται. be proportional alternately. ∆ΒΑ Γ DA B C ῎Εστωσαν τέσσαρες ἀριθμοὶ ἀνάλογον οἱ Α, Β, Γ, Δ, Let the four numbers A, B, C, and D be proportional, ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ· λέγω, ὅτι καὶ (such that) as A (is) to B, so C (is) to D. I say that they ἐναλλὰξ ἀνάλογον ἔσονται, ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Β will also be proportional alternately, (such that) as A (is) πρὸς τὸν Δ. to C, so B (is) to D. ᾿Επεὶ γάρ ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν For since as A is to B, so C (is) to D, thus which(ever) Δ, ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ Β ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ part, or parts, A is of B, C is also the same part, καὶ ὁ Γ τοῦ Δ ἢ τὰ αὐτὰ μέρη. ἐναλλὰξ ἄρα, ὃ μέρος ἐστὶν or the same parts, of D [Def. 7.20]. Thus, alterately, ὁ Α τοῦ Γ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Β τοῦ Δ ἢ τὰ which(ever) part, or parts, A is of C, B is also the same αὐτὰ μέρη. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Β πρὸς part, or the same parts, of D [Props. 7.9, 7.10]. Thus, as τὸν Δ· ὅπερ ἔδει δεῖξαι. A is to C, so B (is) to D [Def. 7.20]. (Which is) the very thing it was required to show. † In modern notation, this proposition states that if a : b :: c : d then a : c :: b : d, where all symbols denote numbers.idþ. Proposition 14† ᾿Εὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ καὶ ἄλλοι αὐτοῖς ἴσοι τὸ If there are any multitude of numbers whatsoever, πλῆθος σύνδυο λαμβανόμενοι καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ δι᾿ and (some) other (numbers) of equal multitude to them, ἴσου ἐν τῷ αὐτῷ λόγῷ ἔσονται. (which are) also in the same ratio taken two by two, then they will also be in the same ratio via equality. Ζ Α ∆ Β Γ Ε F A D B C E ῎Εστωσαν ὁποσοιοῦν ἀριθμοὶ οἱ Α, Β, Γ καὶ ἄλλοι αὐτοῖς Let there be any multitude of numbers whatsoever, A, ἴσοι τὸ πλῆθος σύνδυο λαμβανόμενοι ἐν τῷ αὐτῷ λόγῳ οἱ B, C, and (some) other (numbers), D, E, F , of equal Δ, Ε, Ζ, ὡς μὲν ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε, multitude to them, (which are) in the same ratio taken ὡς δὲ ὁ Β πρὸς τὸν Γ, οὕτως ὁ Ε πρὸς τὸν Ζ· λέγω, ὅτι two by two, (such that) as A (is) to B, so D (is) to E, καὶ δι᾿ ἴσου ἐστὶν ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Δ πρὸς τὸν and as B (is) to C, so E (is) to F . I say that also, via Ζ. equality, as A is to C, so D (is) to F . ᾿Επεὶ γάρ ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς For since as A is to B, so D (is) to E, thus, alternately, τὸν Ε, ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, οὕτως ὁ Β as A is to D, so B (is) to E [Prop. 7.13]. Again, since πρὸς τὸν Ε. πάλιν, ἐπεί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, οὕτως ὁ as B is to C, so E (is) to F , thus, alternately, as B is 206 STOIQEIWN zþ. ELEMENTS BOOK 7 Ε πρὸς τὸν Ζ, ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Β πρὸς τὸν Ε, οὕτως to E, so C (is) to F [Prop. 7.13]. And as B (is) to E, ὁ Γ πρὸς τὸν Ζ. ὡς δὲ ὁ Β πρὸς τὸν Ε, οὕτως ὁ Α πρὸς so A (is) to D. Thus, also, as A (is) to D, so C (is) to F . τὸν Δ· καὶ ὡς ἄρα ὁ Α πρὸς τὸν Δ, οὕτως ὁ Γ πρὸς τὸν Thus, alternately, as A is to C, so D (is) to F [Prop. 7.13]. Ζ· ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Δ πρὸς (Which is) the very thing it was required to show. τὸν Ζ· ὅπερ ἔδει δεῖξαι. † In modern notation, this proposition states that if a : b :: d : e and b : c :: e : f then a : c :: d : f , where all symbols denote numbers.ieþ. Proposition 15 ᾿Εὰν μονὰς ἀριθμόν τινα μετρῇ, ἰσακις δὲ ἕτερος ἀριθμὸς If a unit measures some number, and another num- ἄλλον τινὰ ἀριθμὸν μετρῇ, καὶ ἐναλλὰξ ἰσάκις ἡ μονὰς τὸν ber measures some other number as many times, then, τρίτον ἀριθμὸν μετρήσει καὶ ὁ δεύτερος τὸν τέταρτον. also, alternately, the unit will measure the third num- ber as many times as the second (number measures) the fourth. Ζ Α ∆ Β Η Θ Γ Ε Κ Λ F H D A B E G C K L Μονὰς γὰρ ἡ Α ἀριθμόν τινα τὸν ΒΓ μετρείτω, ἰσάκις δὲ For let a unit A measure some number BC, and let ἕτερος ἀριθμὸς ὁ Δ ἄλλον τινὰ ἀριθμὸν τὸν ΕΖ μετρείτω· another number D measure some other number EF as λέγω, ὅτι καὶ ἐναλλὰξ ἰσάκις ἡ Α μονὰς τὸν Δ ἀριθμὸν many times. I say that, also, alternately, the unit A also μετρεῖ καὶ ὁ ΒΓ τὸν ΕΖ. measures the number D as many times as BC (measures) ᾿Επεὶ γὰρ ἰσάκις ἡ Α μονὰς τὸν ΒΓ ἀριθμὸν μετρεῖ καὶ ὁ EF . Δ τὸν ΕΖ, ὅσαι ἄρα εἰσὶν ἐν τῷ ΒΓ μονάδες, τοσοῦτοί εἰσι For since the unit A measures the number BC as καὶ ἐν τῷ ΕΖ ἀριθμοὶ ἴσοι τῷ Δ. διῃρήσθω ὁ μὲν ΒΓ εἰς τὰς many times as D (measures) EF , thus as many units as ἐν ἑαυτῷ μονάδας τὰς ΒΗ, ΗΘ, ΘΓ, ὁ δὲ ΕΖ εἰς τοὺς τῷ Δ are in BC, so many numbers are also in EF equal to ἴσους τοὺς ΕΚ, ΚΛ, ΛΖ. ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΒΗ, D. Let BC have been divided into its constituent units, ΗΘ, ΘΓ τῷ πλήθει τῶν ΕΚ, ΚΛ, ΛΖ. καὶ ἐπεὶ ἴσαι εἰσὶν αἱ BG, GH , and HC, and EF into the (divisions) EK, KL, ΒΗ, ΗΘ, ΘΓ μονάδες ἀλλήλαις, εἰσὶ δὲ καὶ οἱ ΕΚ, ΚΛ, ΛΖ and LF , equal to D. So the multitude of (units) BG, ἀριθμοὶ ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ πλῆθος τῶν ΒΗ, GH , HC will be equal to the multitude of (divisions) ΗΘ, ΘΓ μονάδων τῷ πλήθει τῶν ΕΚ, ΚΛ, ΛΖ ἀριθμῶν, EK, KL, LF . And since the units BG, GH , and HC ἔσται ἄρα ὡς ἡ ΒΗ μονὰς πρὸς τὸν ΕΚ ἀριθμόν, οὕτως ἡ are equal to one another, and the numbers EK, KL, and ΗΘ μονὰς πρὸς τὸν ΚΛ ἀριθμὸν καὶ ἡ ΘΓ μονὰς πρὸς τὸν LF are also equal to one another, and the multitude of ΛΖ ἀριθμόν. ἔσται ἄρα καὶ ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα the (units) BG, GH , HC is equal to the multitude of the τῶν ἑπομένων, οὕτως ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας numbers EK, KL, LF , thus as the unit BG (is) to the τοὺς ἑπομένους· ἔστιν ἄρα ὡς ἡ ΒΗ μονὰς πρὸς τὸν ΕΚ number EK, so the unit GH will be to the number KL, ἀριθμόν, οὕτως ὁ ΒΓ πρὸς τὸν ΕΖ. ἴση δὲ ἡ ΒΗ μονὰς τῇ and the unit HC to the number LF . And thus, as one of Α μονάδι, ὁ δὲ ΕΚ ἀριθμὸς τῷ Δ ἀριθμῷ. ἔστιν ἄρα ὡς ἡ the leading (numbers is) to one of the following, so (the Α μονὰς πρὸς τὸν Δ ἀριθμόν, οὕτως ὁ ΒΓ πρὸς τὸν ΕΖ. sum of) all of the leading will be to (the sum of) all of ἰσάκις ἄρα ἡ Α μονὰς τὸν Δ ἀριθμὸν μετρεῖ καὶ ὁ ΒΓ τὸν the following [Prop. 7.12]. Thus, as the unit BG (is) to ΕΖ· ὅπερ ἔδει δεῖξαι. the number EK, so BC (is) to EF . And the unit BG (is) equal to the unit A, and the number EK to the number D. Thus, as the unit A is to the number D, so BC (is) to EF . Thus, the unit A measures the number D as many times as BC (measures) EF [Def. 7.20]. (Which is) the very thing it was required to show. † This proposition is a special case of Prop. 7.9. 207 STOIQEIWN zþ. ELEMENTS BOOK 7i�þ. Proposition 16† Εὰν δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί If two numbers multiplying one another make some τινας, οἱ γενόμενοι ἐξ αὐτῶν ἴσοι ἀλλήλοις ἔσονται. (numbers) then the (numbers) generated from them will be equal to one another. Ε Β Α Γ ∆ E A B C D ῎Εστωσαν δύο ἀριθμοὶ οἱ Α, Β, καὶ ὁ μὲν Α τὸν Β πολ- Let A and B be two numbers. And let A make C (by) λαπλασιάσας τὸν Γ ποιείτω, ὁ δὲ Β τὸν Α πολλαπλασιάσας multiplying B, and let B make D (by) multiplying A. I τὸν Δ ποιείτω· λέγω, ὅτι ἴσος ἐστὶν ὁ Γ τῷ Δ. say that C is equal to D. ᾿Επεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, For since A has made C (by) multiplying B, B thus ὁ Β ἄρα τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. μετρεῖ δὲ measures C according to the units in A [Def. 7.15]. And καὶ ἡ Ε μονὰς τὸν Α ἀριθμὸν κατὰ τὰς ἐν αὐτῷ μονάδας· the unit E also measures the number A according to the ἰσάκις ἄρα ἡ Ε μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν units in it. Thus, the unit E measures the number A as Γ. ἐναλλὰξ ἄρα ἰσάκις ἡ Ε μονὰς τὸν Β ἀριθμὸν μετρεῖ καὶ many times as B (measures) C. Thus, alternately, the ὁ Α τὸν Γ. πάλιν, ἐπεὶ ὁ Β τὸν Α πολλαπλασιάσας τὸν Δ unit E measures the number B as many times as A (mea- πεποίηκεν, ὁ Α ἄρα τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Β μονάδας. sures) C [Prop. 7.15]. Again, since B has made D (by) μετρεῖ δὲ καὶ ἡ Ε μονὰς τὸν Β κατὰ τὰς ἐν αὐτῷ μονάδας· multiplying A, A thus measures D according to the units ἰσάκις ἄρα ἡ Ε μονὰς τὸν Β ἀριθμὸν μετρεῖ καὶ ὁ Α τὸν Δ. in B [Def. 7.15]. And the unit E also measures B ac- ἰσάκις δὲ ἡ Ε μονὰς τὸν Β ἀριθμὸν ἐμέτρει καὶ ὁ Α τὸν Γ· cording to the units in it. Thus, the unit E measures the ἰσάκις ἄρα ὁ Α ἑκάτερον τῶν Γ, Δ μετρεῖ. ἴσος ἄρα ἐστὶν number B as many times as A (measures) D. And the ὁ Γ τῷ Δ· ὅπερ ἔδει δεῖξαι. unit E was measuring the number B as many times as A (measures) C. Thus, A measures each of C and D an equal number of times. Thus, C is equal to D. (Which is) the very thing it was required to show. † In modern notation, this proposition states that a b = b a, where all symbols denote numbers.izþ. Proposition 17† ᾿Εὰν ἀριθμὸς δύο ἀριθμοὺς πολλαπλασιάσας ποιῇ τινας, If a number multiplying two numbers makes some οἱ γενόμενοι ἐξ αὐτῶν τὸν αὐτὸν ἕξουσι λόγον τοῖς πολλα- (numbers) then the (numbers) generated from them will πλασιασθεῖσιν. have the same ratio as the multiplied (numbers). Ε Α Β ∆ Ζ Γ E A B D F C Ἀριθμὸς γὰρ ὁ Α δύο ἀριθμοὺς τοὺς Β, Γ πολλα- For let the number A make (the numbers) D and πλασιάσας τοὺς Δ, Ε ποιείτω· λέγω, ὅτι ἐστὶν ὡς ὁ Β πρὸς E (by) multiplying the two numbers B and C (respec- τὸν Γ, οὕτως ὁ Δ πρὸς τὸν Ε. tively). I say that as B is to C, so D (is) to E. ᾿Επεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Δ πεποίηκεν, For since A has made D (by) multiplying B, B thus ὁ Β ἄρα τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. μετρεῖ measures D according to the units in A [Def. 7.15]. And 208 STOIQEIWN zþ. ELEMENTS BOOK 7 δὲ καὶ ἡ Ζ μονὰς τὸν Α ἀριθμὸν κατὰ τὰς ἐν αὐτῷ μονάδας· the unit F also measures the number A according to the ἰσάκις ἄρα ἡ Ζ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Δ. units in it. Thus, the unit F measures the number A as ἔστιν ἄρα ὡς ἡ Ζ μονὰς πρὸς τὸν Α ἀριθμόν, οὕτως ὁ Β many times as B (measures) D. Thus, as the unit F is πρὸς τὸν Δ. διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ Ζ μονὰς πρὸς τὸν Α to the number A, so B (is) to D [Def. 7.20]. And so, for ἀριθμόν, οὕτως ὁ Γ πρὸς τὸν Ε· καὶ ὡς ἄρα ὁ Β πρὸς τὸν the same (reasons), as the unit F (is) to the number A, Δ, οὕτως ὁ Γ πρὸς τὸν Ε. ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Β πρὸς so C (is) to E. And thus, as B (is) to D, so C (is) to E. τὸν Γ, οὕτως ὁ Δ πρὸς τὸν Ε· ὅπερ ἔδει δεῖξαι. Thus, alternately, as B is to C, so D (is) to E [Prop. 7.13]. (Which is) the very thing it was required to show. † In modern notation, this proposition states that if d = a b and e = a c then d : e :: b : c, where all symbols denote numbers.ihþ. Proposition 18† ᾿Εὰν δύο ἀριθμοὶ ἀριθμόν τινα πολλαπλασιάσαντες If two numbers multiplying some number make some ποιῶσί τινας, οἱ γενόμενοι ἐξ αὐτῶν τὸν αὐτὸν ἕξουσι λόγον (other numbers) then the (numbers) generated from τοῖς πολλαπλασιάσασιν. them will have the same ratio as the multiplying (num- bers). Ε Β Α Γ ∆ E B A C D Δύο γὰρ ἀριθμοὶ οἱ Α, Β ἀριθμόν τινα τὸν Γ πολλα- For let the two numbers A and B make (the numbers) πλασιάσαντες τοὺς Δ, Ε ποιείτωσαν· λέγω, ὅτι ἐστὶν ὡς ὁ D and E (respectively, by) multiplying some number C. Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε. I say that as A is to B, so D (is) to E. ᾿Επεὶ γὰρ ὁ Α τὸν Γ πολλαπλασιάσας τὸν Δ πεποίηκεν, For since A has made D (by) multiplying C, C has καὶ ὁ Γ ἄρα τὸν Α πολλαπλασιάσας τὸν Δ πεποίηκεν. διὰ thus also made D (by) multiplying A [Prop. 7.16]. So, for τὰ αὐτὰ δὴ καὶ ὁ Γ τὸν Β πολλαπλασιάσας τὸν Ε πεποίηκεν. the same (reasons), C has also made E (by) multiplying ἀριθμὸς δὴ ὁ Γ δύο ἀριθμοὺς τοὺς Α, Β πολλαπλασιάσας B. So the number C has made D and E (by) multiplying τοὺς Δ, Ε πεποίηκεν. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως the two numbers A and B (respectively). Thus, as A is to ὁ Δ πρὸς τὸν Ε· ὅπερ ἔδει δεῖξαι. B, so D (is) to E [Prop. 7.17]. (Which is) the very thing it was required to show. † In modern notation, this propositions states that if a c = d and b c = e then a : b :: d : e, where all symbols denote numbers.ijþ. Proposition 19† ᾿Εὰν τέσσαρες ἀριθμοὶ ἀνάλογον ὦσιν, ὁ ἐκ πρώτου καὶ If four number are proportional then the number cre- τετάρτου γενόμενος ἀριθμὸς ἴσος ἔσται τῷ ἐκ δευτέρου καὶ ated from (multiplying) the first and fourth will be equal τρίτου γενομένῳ ἀριθμῷ· καὶ ἐὰν ὁ ἐκ πρώτου καὶ τετάρτου to the number created from (multiplying) the second and γενόμενος ἀριθμὸς ἴσος ᾖ τῷ ἐκ δευτέρου καὶ τρίτου, οἱ third. And if the number created from (multiplying) the τέσσασρες ἀριθμοὶ ἀνάλογον ἔσονται. first and fourth is equal to the (number created) from ῎Εστωσαν τέσσαρες ἀριθμοὶ ἀνάλογον οἱ Α, Β, Γ, Δ, (multiplying) the second and third then the four num- ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ, καὶ ὁ μὲν Α bers will be proportional. τὸν Δ πολλαπλασιάσας τὸν Ε ποιείτω, ὁ δὲ Β τὸν Γ πολ- Let A, B, C, and D be four proportional numbers, λαπλασιάσας τὸν Ζ ποιείτω· λέγω, ὅτι ἴσος ἐστὶν ὁ Ε τῷ Ζ. (such that) as A (is) to B, so C (is) to D. And let A make E (by) multiplying D, and let B make F (by) multiplying C. I say that E is equal to F . 209 STOIQEIWN zþ. ELEMENTS BOOK 7 ΗΑ Β Γ ∆ Ε Ζ GA B C D E F ῾Ο γὰρ Α τὸν Γ πολλαπλασιάσας τὸν Η ποιείτω. ἐπεὶ For let A make G (by) multiplying C. Therefore, since οὖν ὁ Α τὸν Γ πολλαπλασιάσας τὸν Η πεποίηκεν, τὸν δὲ A has made G (by) multiplying C, and has made E (by) Δ πολλαπλασιάσας τὸν Ε πεποίηκεν, ἀριθμὸς δὴ ὁ Α δύο multiplying D, the number A has made G and E by mul- ἀριθμοὺς τοὺς Γ, Δ πολλαπλασιάσας τούς Η, Ε πεποίηκεν. tiplying the two numbers C and D (respectively). Thus, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Η πρὸς τὸν Ε. ἀλλ᾿ as C is to D, so G (is) to E [Prop. 7.17]. But, as C (is) to ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α πρὸς τὸν Β· καὶ ὡς ἄρα D, so A (is) to B. Thus, also, as A (is) to B, so G (is) to ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς τὸν Ε. πάλιν, ἐπεὶ ὁ Α E. Again, since A has made G (by) multiplying C, but, τὸν Γ πολλαπλασιάσας τὸν Η πεποίηκεν, ἀλλὰ μὴν καὶ ὁ in fact, B has also made F (by) multiplying C, the two Β τὸν Γ πολλαπλασιάσας τὸν Ζ πεποίηκεν, δύο δὴ ἀριθμοὶ numbers A and B have made G and F (respectively, by) οἱ Α, Β ἀριθμόν τινα τὸν Γ πολλαπλασιάσαντες τοὺς Η, Ζ multiplying some number C. Thus, as A is to B, so G (is) πεποιήκασιν. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς to F [Prop. 7.18]. But, also, as A (is) to B, so G (is) to τὸν Ζ. ἀλλὰ μὴν καὶ ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς E. And thus, as G (is) to E, so G (is) to F . Thus, G has τὸν Ε· καὶ ὡς ἄρα ὁ Η πρὸς τὸν Ε, οὕτως ὁ Η πρὸς τὸν the same ratio to each of E and F . Thus, E is equal to F Ζ. ὁ Η ἄρα πρὸς ἑκάτερον τῶν Ε, Ζ τὸν αὐτὸν ἔχει λόγον· [Prop. 5.9]. ἴσος ἄρα ἐστὶν ὁ Ε τῷ Ζ. So, again, let E be equal to F . I say that as A is to B, ῎Εστω δὴ πάλιν ἴσος ὁ Ε τῷ Ζ· λέγω, ὅτι ἐστὶν ὡς ὁ Α so C (is) to D. πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ. For, with the same construction, since E is equal to F , Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ ἴσος ἐστὶν ὁ thus as G is to E, so G (is) to F [Prop. 5.7]. But, as G Ε τῷ Ζ, ἔστιν ἄρα ὡς ὁ Η πρὸς τὸν Ε, οὕτως ὁ Η πρὸς τὸν (is) to E, so C (is) to D [Prop. 7.17]. And as G (is) to F , Ζ. ἀλλ᾿ ὡς μὲν ὁ Η πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς τὸν Δ, ὡς so A (is) to B [Prop. 7.18]. And, thus, as A (is) to B, so δὲ ὁ Η πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν Β. καὶ ὡς ἄρα ὁ Α C (is) to D. (Which is) the very thing it was required to πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ· ὅπερ ἔδει δεῖξαι. show. † In modern notation, this proposition reads that if a : b :: c : d then a d = b c, and vice versa, where all symbols denote numbers.kþ. Proposition 20 Οἱ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων The least numbers of those (numbers) having the αὐτοῖς μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ same ratio measure those (numbers) having the same ra- τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα. tio as them an equal number of times, the greater (mea- ῎Εστωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον suring) the greater, and the lesser the lesser. ἐχόντων τοῖς Α, Β οἱ ΓΔ, ΕΖ· λέγω, ὅτι ἰσάκις ὁ ΓΔ τὸν For let CD and EF be the least numbers having the Α μετρεῖ καὶ ὁ ΕΖ τὸν Β. same ratio as A and B (respectively). I say that CD mea- sures A the same number of times as EF (measures) B. 210 STOIQEIWN zþ. ELEMENTS BOOK 7 Ε Α Β Γ Η ∆ Ζ Θ F A B C G D E H ῾Ο ΓΔ γὰρ τοῦ Α οὔκ ἐστι μέρη. εἰ γὰρ δυνατόν, ἔστω· For CD is not parts of A. For, if possible, let it be καὶ ὁ ΕΖ ἄρα τοῦ Β τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὁ ΓΔ τοῦ (parts of A). Thus, EF is also the same parts of B that Α. ὅσα ἄρα ἐστὶν ἐν τῷ ΓΔ μέρη τοῦ Α, τοσαῦτά ἐστι καὶ CD (is) of A [Def. 7.20, Prop. 7.13]. Thus, as many parts ἐν τῷ ΕΖ μέρη τοῦ Β. διῃρήσθω ὁ μὲν ΓΔ εἰς τὰ τοῦ Α of A as are in CD, so many parts of B are also in EF . Let μέρη τὰ ΓΗ, ΗΔ, ὁ δὲ ΕΖ εἰς τὰ τοῦ Β μέρη τὰ ΕΘ, ΘΖ· CD have been divided into the parts of A, CG and GD, ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΓΗ, ΗΔ τῷ πλήθει τῶν ΕΘ, and EF into the parts of B, EH and HF . So the multi- ΘΖ. καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΓΗ, ΗΔ ἀριθμοὶ ἀλλήλοις, εἰσὶ tude of (divisions) CG, GD will be equal to the multitude δὲ καὶ οἱ ΕΘ, ΘΖ ἀριθμοὶ ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ of (divisions) EH , HF . And since the numbers CG and πλῆθος τῶν ΓΗ, ΗΔ τῷ πλήθει τῶν ΕΘ, ΘΖ, ἔστιν ἄρα ὡς GD are equal to one another, and the numbers EH and ὁ ΓΗ πρὸς τὸν ΕΘ, οὕτως ὁ ΗΔ πρὸς τὸν ΘΖ. ἔσται ἄρα HF are also equal to one another, and the multitude of καὶ ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως (divisions) CG, GD is equal to the multitude of (divi- ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους. ἔστιν sions) EH , HF , thus as CG is to EH , so GD (is) to HF . ἄρα ὡς ὁ ΓΗ πρὸς τὸν ΕΘ, οὕτως ὁ ΓΔ πρὸς τὸν ΕΖ· οἱ Thus, as one of the leading (numbers is) to one of the ΓΗ, ΕΘ ἄρα τοῖς ΓΔ, ΕΖ ἐν τῷ αὐτῷ λόγῳ εἰσὶν ἐλάσσονες following, so will (the sum of) all of the leading (num- ὄντες αὐτῶν· ὅπερ ἐστὶν ἀδύνατον· ὑπόκεινται γὰρ οἱ ΓΔ, bers) be to (the sum of) all of the following [Prop. 7.12]. ΕΖ ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. οὐκ Thus, as CG is to EH , so CD (is) to EF . Thus, CG ἄρα μέρη ἐστὶν ὁ ΓΔ τοῦ Α· μέρος ἄρα. καὶ ὁ ΕΖ τοῦ Β τὸ and EH are in the same ratio as CD and EF , being less αὐτὸ μέρος ἐστίν, ὅπερ ὁ ΓΔ τοῦ Α· ἰσάκις ἄρα ὁ ΓΔ τὸν than them. The very thing is impossible. For CD and Α μετρεῖ καὶ ὁ ΕΖ τὸν Β· ὅπερ ἔδει δεῖξαι. EF were assumed (to be) the least of those (numbers) having the same ratio as them. Thus, CD is not parts of A. Thus, (it is) a part (of A) [Prop. 7.4]. And EF is the same part of B that CD (is) of A [Def. 7.20, Prop 7.13]. Thus, CD measures A the same number of times that EF (measures) B. (Which is) the very thing it was required to show.kaþ. Proposition 21 Οἱ πρῶτοι πρὸς ἀλλήλους ἀριθμοὶ ἐλάχιστοί εἰσι τῶν τὸν Numbers prime to one another are the least of those αὐτὸν λόγον ἐχόντων αὐτοῖς. (numbers) having the same ratio as them. ῎Εστωσαν πρῶτοι πρὸς ἀλλήλους ἀριθμοὶ οἱ Α, Β· λέγω, Let A and B be numbers prime to one another. I say ὅτι οἱ Α, Β ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων that A and B are the least of those (numbers) having the αὐτοῖς. same ratio as them. Εἰ γὰρ μή, ἔσονταί τινες τῶν Α, Β ἐλάσσονες ἀριθμοὶ For if not then there will be some numbers less than A ἐν τῷ αὐτῷ λόγῳ ὄντες τοῖς Α, Β. ἔστωσαν οἱ Γ, Δ. and B which are in the same ratio as A and B. Let them be C and D. 211 STOIQEIWN zþ. ELEMENTS BOOK 7 ΕΑ Β Γ ∆ EA B C D ᾿Επεὶ οὖν οἱ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον Therefore, since the least numbers of those (num- ἐχόντων μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις bers) having the same ratio measure those (numbers) ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάττων τὸν ἐλάττονα, having the same ratio (as them) an equal number of τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος times, the greater (measuring) the greater, and the lesser τὸν ἑπόμενον, ἰσάκις ἄρα ὁ Γ τὸν Α μετρεῖ καὶ ὁ Δ τὸν Β. the lesser—that is to say, the leading (measuring) the ὁσάκις δὴ ὁ Γ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν leading, and the following the following—C thus mea- τῷ Ε. καὶ ὁ Δ ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας. sures A the same number of times that D (measures) B καὶ ἐπεὶ ὁ Γ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, καί ὁ [Prop. 7.20]. So as many times as C measures A, so many Ε ἄρα τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Γ μονάδας. διὰ τὰ αὐτὰ units let there be in E. Thus, D also measures B accord- δὴ ὁ Ε καὶ τὸν Β μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας. ὁ Ε ing to the units in E. And since C measures A according ἄρα τοὺς Α, Β μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους· ὅπερ to the units in E, E thus also measures A according to ἐστὶν ἀδύνατον. οὐκ ἄρα ἔσονταί τινες τῶν Α, Β ἐλάσσονες the units in C [Prop. 7.16]. So, for the same (reasons), E ἀριθμοὶ ἐν τῷ αὐτῷ λόγῳ ὄντες τοῖς Α, Β. οἱ Α, Β ἄρα also measures B according to the units in D [Prop. 7.16]. ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς· ὅπερ Thus, E measures A and B, which are prime to one an- ἔδει δεῖξαι. other. The very thing is impossible. Thus, there cannot be any numbers less than A and B which are in the same ratio as A and B. Thus, A and B are the least of those (numbers) having the same ratio as them. (Which is) the very thing it was required to show.kbþ. Proposition 22 Οἱ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων The least numbers of those (numbers) having the αὐτοῖς πρῶτοι πρὸς ἀλλήλους εἰσίν. same ratio as them are prime to one another. Ε Α Β Γ ∆ E A B C D ῎Εστωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον Let A and B be the least numbers of those (numbers) ἐχόντων αὐτοῖς οἱ Α, Β· λέγω, ὅτι οἱ Α, Β πρῶτοι πρὸς having the same ratio as them. I say that A and B are ἀλλήλους εἰσίν. prime to one another. Εἰ γὰρ μή εἰσι πρῶτοι πρὸς ἀλλήλους, μετρήσει τις For if they are not prime to one another then some αὐτοὺς ἀριθμός. μετρείτω, καὶ ἔστω ὁ Γ. καὶ ὁσάκις μὲν number will measure them. Let it (so measure them), ὁ Γ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Δ, and let it be C. And as many times as C measures A, so 212 STOIQEIWN zþ. ELEMENTS BOOK 7 ὁσάκις δὲ ὁ Γ τὸν Β μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν many units let there be in D. And as many times as C τῷ Ε. measures B, so many units let there be in E. ᾿Επεὶ ὁ Γ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας, ὁ Since C measures A according to the units in D, C Γ ἄρα τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ has thus made A (by) multiplying D [Def. 7.15]. So, for αὐτὰ δὴ καὶ ὁ Γ τὸν Ε πολλαπλασιάσας τὸν Β πεποίηκεν. the same (reasons), C has also made B (by) multiplying ἀριθμὸς δὴ ὁ Γ δύο ἀριθμοὺς τοὺς Δ, Ε πολλαπλασιάσας E. So the number C has made A and B (by) multiplying τοὺς Α, Β πεποίηκεν· ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ε, οὕτως the two numbers D and E (respectively). Thus, as D is ὁ Α πρὸς τὸν Β· οἱ Δ, Ε ἄρα τοῖς Α, Β ἐν τῷ αὐτῷ λόγῳ to E, so A (is) to B [Prop. 7.17]. Thus, D and E are in εἰσὶν ἐλάσσονες ὄντες αὐτῶν· ὅπερ ἐστὶν ἀδύνατον. οὐκ the same ratio as A and B, being less than them. The ἄρα τοὺς Α, Β ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ Α, Β ἄρα very thing is impossible. Thus, some number does not πρῶτοι πρὸς ἀλλήλους εἰσίν· ὅπερ ἔδει δεῖξαι. measure the numbers A and B. Thus, A and B are prime to one another. (Which is) the very thing it was required to show.kgþ. Proposition 23 ᾿Εὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, ὁ τὸν ἕνα If two numbers are prime to one another then a num- αὐτῶν μετρῶν ἀριθμὸς πρὸς τὸν λοιπὸν πρῶτος ἔσται. ber measuring one of them will be prime to the remaining (one). ΒΑ Γ ∆ . A B C D ῎Εστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Β, Let A and B be two numbers (which are) prime to τὸν δὲ Α μετρείτω τις ἀριθμὸς ὁ Γ· λέγω, ὅτι καὶ οἱ Γ, Β one another, and let some number C measure A. I say πρῶτοι πρὸς ἀλλήλους εἰσίν. that C and B are also prime to one another. Εἰ γὰρ μή εἰσιν οἱ Γ, Β πρῶτοι πρὸς ἀλλήλους, μετρήσει For if C and B are not prime to one another then [τις] τοὺς Γ, Β ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. ἐπεὶ ὁ [some] number will measure C and B. Let it (so) mea- Δ τὸν Γ μετρεῖ, ὁ δὲ Γ τὸν Α μετρεῖ, καὶ ὁ Δ ἄρα τὸν Α sure (them), and let it be D. Since D measures C, and C μετρεῖ. μετρεῖ δὲ καὶ τὸν Β· ὁ Δ ἄρα τοὺς Α, Β μετρεῖ measures A, D thus also measures A. And (D) also mea- πρώτους ὄντας πρὸς ἀλλήλους· ὅπερ ἐστὶν ἀδύνατον. οὐκ sures B. Thus, D measures A and B, which are prime ἄρα τοὺς Γ, Β ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ Γ, Β ἄρα to one another. The very thing is impossible. Thus, some πρῶτοι πρὸς ἀλλήλους εἰσίν· ὅπερ ἔδει δεῖξαι. number does not measure the numbers C and B. Thus, C and B are prime to one another. (Which is) the very thing it was required to show.kdþ. Proposition 24 ᾿Εὰν δύο ἀριθμοὶ πρός τινα ἀριθμὸν πρῶτοι ὦσιν, καὶ ὁ If two numbers are prime to some number then the ἐξ αὐτῶν γενόμενος πρὸς τὸν αὐτὸν πρῶτος ἔσται. number created from (multiplying) the former (two num- bers) will also be prime to the latter (number). 213 STOIQEIWN zþ. ELEMENTS BOOK 7 ΖΑ Β Γ ∆ Ε FA B C D E Δύο γὰρ ἀριθμοὶ οἱ Α, Β πρός τινα ἀριθμὸν τὸν Γ πρῶτοι For let A and B be two numbers (which are both) ἔστωσαν, καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Δ ποιείτω· prime to some number C. And let A make D (by) multi- λέγω, ὅτι οἱ Γ, Δ πρῶτοι πρὸς ἀλλήλους εἰσίν. plying B. I say that C and D are prime to one another. Εἰ γὰρ μή εἰσιν οἱ Γ, Δ πρῶτοι πρὸς ἀλλήλους, μετρήσει For if C and D are not prime to one another then [τις] τοὺς Γ, Δ ἀριθμός. μετρείτω, καὶ ἔστω ὁ Ε. καὶ ἐπεὶ [some] number will measure C and D. Let it (so) mea- οἱ Γ, Δ πρῶτοι πρὸς ἀλλήλους εἰσίν, τὸν δὲ Γ μετρεῖ τις sure them, and let it be E. And since C and A are prime ἀριθμὸς ὁ Ε, οἱ Α, Ε ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. to one another, and some number E measures C, A and ὁσάκις δὴ ὁ Ε τὸν Δ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν E are thus prime to one another [Prop. 7.23]. So as τῷ Ζ· καὶ ὁ Ζ ἄρα τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας. many times as E measures D, so many units let there ὁ Ε ἄρα τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ be in F . Thus, F also measures D according to the units μὴν καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Δ πεποίηκεν· ἴσος in E [Prop. 7.16]. Thus, E has made D (by) multiply- ἄρα ἐστὶν ὁ ἑκ τῶν Ε, Ζ τῷ ἐκ τῶν Α, Β. ἐὰν δὲ ὁ ὑπὸ ing F [Def. 7.15]. But, in fact, A has also made D (by) τῶν ἄκρων ἴσος ᾖ τῷ ὑπὸ τῶν μέσων, οἱ τέσσαρες ἀριθμοὶ multiplying B. Thus, the (number created) from (multi- ἀνάλογόν εἰσιν· ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Α, οὕτως ὁ Β plying) E and F is equal to the (number created) from πρὸς τὸν Ζ. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ (multiplying) A and B. And if the (rectangle contained) δὲ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς by the (two) outermost is equal to the (rectangle con- μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων tained) by the middle (two) then the four numbers are τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε proportional [Prop. 6.15]. Thus, as E is to A, so B (is) ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον· ὁ to F . And A and E (are) prime (to one another). And Ε ἄρα τὸν Β μετρεῖ. μετρεῖ δὲ καὶ τὸν Γ· ὁ Ε ἄρα τοὺς Β, Γ (numbers) prime (to one another) are also the least (of μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους· ὅπερ ἐστὶν ἀδύνατον. those numbers having the same ratio) [Prop. 7.21]. And οὐκ ἄρα τοὺς Γ, Δ ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ Γ, Δ the least numbers of those (numbers) having the same ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν· ὅπερ ἔδει δεῖξαι. ratio measure those (numbers) having the same ratio as them an equal number of times, the greater (measuring) the greater, and the lesser the lesser—that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures B. And it also measures C. Thus, E measures B and C, which are prime to one another. The very thing is impossible. Thus, some number cannot measure the numbers C and D. Thus, C and D are prime to one another. (Which is) the very thing it was required to show.keþ. Proposition 25 ᾿Εὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, ὁ ἐκ τοῦ If two numbers are prime to one another then the ἑνὸς αὐτῶν γενόμενος πρὸς τὸν λοιπὸν πρῶτος ἔσται. number created from (squaring) one of them will be ῎Εστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Β, prime to the remaining (number). καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι Let A and B be two numbers (which are) prime to 214 STOIQEIWN zþ. ELEMENTS BOOK 7 οἱ Β, Γ πρῶτοι πρὸς ἀλλὴλους εἰσίν. one another. And let A make C (by) multiplying itself. I say that B and C are prime to one another. ∆Α Β Γ DA B C Κείσθω γὰρ τῷ Α ἴσος ὁ Δ. ἐπεὶ οἱ Α, Β πρῶτοι πρὸς For let D be made equal to A. Since A and B are ἀλλήλους εἰσίν, ἴσος δὲ ὁ Α τῷ Δ, καί οἱ Δ, Β ἄρα πρῶτοι prime to one another, and A (is) equal to D, D and B are πρὸς ἀλλήλους εἰσίν· ἑκάτερος ἄρα τῶν Δ, Α πρὸς τὸν Β thus also prime to one another. Thus, D and A are each πρῶτός ἐστιν· καὶ ὁ ἐκ τῶν Δ, Α ἄρα γενόμενος πρὸς τὸν Β prime to B. Thus, the (number) created from (multily- πρῶτος ἔσται. ὁ δὲ ἐκ τῶν Δ, Α γενόμενος ἀριθμός ἐστιν ὁ ing) D and A will also be prime to B [Prop. 7.24]. And C Γ. οἱ Γ, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν· ὅπερ ἔδει δεῖξαι. is the number created from (multiplying) D and A. Thus, C and B are prime to one another. (Which is) the very thing it was required to show.k�þ. Proposition 26 ᾿Εὰν δύο ἀριθμοὶ πρὸς δύο ἀριθμοὺς ἀμφότεροι πρὸς If two numbers are both prime to each of two numbers ἑκάτερον πρῶτοι ὦσιν, καὶ οἱ ἐξ αὐτῶν γενόμενοι πρῶτοι then the (numbers) created from (multiplying) them will πρὸς ἀλλήλους ἔσονται. also be prime to one another. Ζ Γ Β Α ∆ Ε F A C B D E Δύο γὰρ ἀριθμοὶ οἱ Α, Β πρὸς δύο ἀριθμοὺς τοὺς Γ, For let two numbers, A and B, both be prime to each Δ ἀμφότεροι πρὸς ἑκάτερον πρῶτοι ἔστωσαν, καὶ ὁ μὲν of two numbers, C and D. And let A make E (by) mul- Α τὸν Β πολλαπλασιάσας τὸν Ε ποιείτω, ὁ δὲ Γ τὸν Δ tiplying B, and let C make F (by) multiplying D. I say πολλαπλασιάσας τὸν Ζ ποιείτω· λέγω, ὅτι οἱ Ε, Ζ πρῶτοι that E and F are prime to one another. πρὸς ἀλλήλους εἰσίν. For since A and B are each prime to C, the (num- ᾿Επεὶ γὰρ ἑκάτερος τῶν Α, Β πρὸς τὸν Γ πρῶτός ἐστιν, ber) created from (multiplying) A and B will thus also καὶ ὁ ἐκ τῶν Α, Β ἄρα γενόμενος πρὸς τὸν Γ πρῶτος ἔσται. be prime to C [Prop. 7.24]. And E is the (number) cre- ὁ δὲ ἐκ τῶν Α, Β γενόμενός ἐστιν ὁ Ε· οἱ Ε, Γ ἄρα πρῶτοι ated from (multiplying) A and B. Thus, E and C are πρὸς ἀλλήλους εἰσίν. διὰ τὰ αὐτὰ δὴ καὶ οἱ Ε, Δ πρῶτοι prime to one another. So, for the same (reasons), E and πρὸς ἀλλήλους εἰσίν. ἑκάτερος ἄρα τῶν Γ, Δ πρὸς τὸν Ε D are also prime to one another. Thus, C and D are each πρῶτός ἐστιν. καὶ ὁ ἐκ τῶν Γ, Δ ἄρα γενόμενος πρὸς τὸν prime to E. Thus, the (number) created from (multiply- Ε πρῶτος ἔσται. ὁ δὲ ἐκ τῶν Γ, Δ γενόμενός ἐστιν ὁ Ζ. οἱ ing) C and D will also be prime to E [Prop. 7.24]. And Ε, Ζ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν· ὅπερ ἔδει δεῖξαι. F is the (number) created from (multiplying) C and D. Thus, E and F are prime to one another. (Which is) the very thing it was required to show. 215 STOIQEIWN zþ. ELEMENTS BOOK 7kzþ. Proposition 27† ᾿Εὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ πολ- If two numbers are prime to one another and each λαπλασιάσας ἑκάτερος ἑαυτὸν ποιῇ τινα, οἱ γενόμενοι ἐξ makes some (number by) multiplying itself then the num- αὐτῶν πρῶτοι πρὸς ἀλλήλους ἔσονται, κἂν οἱ ἐξ ἀρχῆς bers created from them will be prime to one another, and τοὺς γενομένους πολλαπλασιάσαντες ποιῶσί τινας, κἀκεῖνοι if the original (numbers) make some (more numbers by) πρῶτοι πρὸς ἀλλήλους ἔσονται [καὶ ἀεὶ περὶ τοὺς ἄκρους multiplying the created (numbers) then these will also be τοῦτο συμβαίνει]. prime to one another [and this always happens with the extremes]. ΖΑ Β Γ ∆ Ε FA B C D E ῎Εστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Β, Let A and B be two numbers prime to one another, καὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Γ ποιείτω, τὸν and let A make C (by) multiplying itself, and let it make δὲ Γ πολλαπλασιάσας τὸν Δ ποιείτω, ὁ δὲ Β ἑαυτὸν μὲν D (by) multiplying C. And let B make E (by) multiplying πολλαπλασιάσας τὸν Ε ποιείτω, τὸν δὲ Ε πολλαπλασιάσας itself, and let it make F by multiplying E. I say that C τὸν Ζ ποιείτω· λέγω, ὅτι οἵ τε Γ, Ε καὶ οἱ Δ, Ζ πρῶτοι πρὸς and E, and D and F , are prime to one another. ἀλλήλους εἰσίν. For since A and B are prime to one another, and A has ᾿Επεὶ γὰρ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, καὶ ὁ made C (by) multiplying itself, C and B are thus prime Α ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν, οἱ Γ, Β ἄρα to one another [Prop. 7.25]. Therefore, since C and B πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐπεὶ οὖν οἱ Γ, Β πρῶτοι πρὸς are prime to one another, and B has made E (by) mul- ἀλλήλους εἰσίν, καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Ε tiplying itself, C and E are thus prime to one another πεποίηκεν, οἱ Γ, Ε ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. πάλιν, [Prop. 7.25]. Again, since A and B are prime to one an- ἐπεὶ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, καὶ ὁ Β ἑαυτὸν other, and B has made E (by) multiplying itself, A and πολλαπλασιάσας τὸν Ε πεποίηκεν, οἱ Α, Ε ἄρα πρῶτοι πρὸς E are thus prime to one another [Prop. 7.25]. Therefore, ἀλλήλους εἰσίν. ἐπεὶ οὖν δύο ἀριθμοὶ οἱ Α, Γ πρὸς δύο since the two numbers A and C are both prime to each ἀριθμοὺς τοὺς Β, Ε ἀμφότεροι πρὸς ἑκάτερον πρῶτοί εἰσιν, of the two numbers B and E, the (number) created from καὶ ὁ ἐκ τῶν Α, Γ ἄρα γενόμενος πρὸς τὸν ἐκ τῶν Β, Ε (multiplying) A and C is thus prime to the (number cre- πρῶτός ἐστιν. καί ἐστιν ὁ μὲν ἐκ τῶν Α, Γ ὁ Δ, ὁ δὲ ἐκ ated) from (multiplying) B and E [Prop. 7.26]. And D is τῶν Β, Ε ὁ Ζ. οἱ Δ, Ζ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν· the (number created) from (multiplying) A and C, and F ὅπερ ἔδει δεῖξαι. the (number created) from (multiplying) B and E. Thus, D and F are prime to one another. (Which is) the very thing it was required to show. † In modern notation, this proposition states that if a is prime to b, then a2 is also prime to b2, as well as a3 to b3, etc., where all symbols denote numbers. khþ. Proposition 28 ᾿Εὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ συ- If two numbers are prime to one another then their ναμφότερος πρὸς ἑκάτερον αὐτῶν πρῶτος ἔσται· καὶ ἐὰν sum will also be prime to each of them. And if the sum συναμφότερος πρὸς ἕνα τινὰ αὐτῶν πρῶτος ᾖ, καὶ οἱ ἐξ (of two numbers) is prime to any one of them then the ἀρχῆς ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ἔσονται. original numbers will also be prime to one another. 216 STOIQEIWN zþ. ELEMENTS BOOK 7 Γ ∆ Α Β C D A B Συγκείσθωσαν γὰρ δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλ- For let the two numbers, AB and BC, (which are) ους οἱ ΑΒ, ΒΓ· λέγω, ὅτι καὶ συναμφότερος ὁ ΑΓ πρὸς prime to one another, be laid down together. I say that ἑκάτερον τῶν ΑΒ, ΒΓ πρῶτός ἐστιν. their sum AC is also prime to each of AB and BC. Εἰ γὰρ μή εἰσιν οἱ ΓΑ, ΑΒ πρῶτοι πρὸς ἀλλήλους, For if CA and AB are not prime to one another then μετρήσει τις τοὺς ΓΑ, ΑΒ ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. some number will measure CA and AB. Let it (so) mea- ἐπεὶ οὖν ὁ Δ τοὺς ΓΑ, ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸν ΒΓ sure (them), and let it be D. Therefore, since D measures μετρήσει. μετρεῖ δὲ καὶ τὸν ΒΑ· ὁ Δ ἄρα τοὺς ΑΒ, ΒΓ με- CA and AB, it will thus also measure the remainder BC. τρεῖ πρώτους ὄντας πρὸς ἀλλήλους· ὅπερ ἐστὶν ἀδύνατον. And it also measures BA. Thus, D measures AB and οὐκ ἄρα τοὺς ΓΑ, ΑΒ ἀριθμοὺς ἀριθμός τις μετρήσει· οἱ BC, which are prime to one another. The very thing is ΓΑ, ΑΒ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. διὰ τὰ αὐτὰ δὴ impossible. Thus, some number cannot measure (both) καὶ οἱ ΑΓ, ΓΒ πρῶτοι πρὸς ἀλλήλους εἰσίν. ὁ ΓΑ ἄρα πρὸς the numbers CA and AB. Thus, CA and AB are prime ἑκάτερον τῶν ΑΒ, ΒΓ πρῶτός ἐστιν. to one another. So, for the same (reasons), AC and CB ῎Εστωσαν δὴ πάλιν οἱ ΓΑ, ΑΒ πρῶτοι πρὸς ἀλλήλους· are also prime to one another. Thus, CA is prime to each λέγω, ὅτι καὶ οἱ ΑΒ, ΒΓ πρῶτοι πρὸς ἀλλήλους εἰσίν. of AB and BC. Εἰ γὰρ μή εἰσιν οἱ ΑΒ, ΒΓ πρῶτοι πρὸς ἀλλήλους, So, again, let CA and AB be prime to one another. I μετρήσει τις τοὺς ΑΒ, ΒΓ ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. say that AB and BC are also prime to one another. καὶ ἐπεὶ ὁ Δ ἑκάτερον τῶν ΑΒ, ΒΓ μετρεῖ, καὶ ὅλον ἄρα τὸν For if AB and BC are not prime to one another then ΓΑ μετρήσει. μετρεῖ δὲ καὶ τὸν ΑΒ· ὁ Δ ἄρα τοὺς ΓΑ, ΑΒ some number will measure AB and BC. Let it (so) mea- μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους· ὅπερ ἐστὶν ἀδύνατον. sure (them), and let it be D. And since D measures each οὐκ ἄρα τοὺς ΑΒ, ΒΓ ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ of AB and BC, it will thus also measure the whole of ΑΒ, ΒΓ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν· ὅπερ ἔδει δεῖξαι. CA. And it also measures AB. Thus, D measures CA and AB, which are prime to one another. The very thing is impossible. Thus, some number cannot measure (both) the numbers AB and BC. Thus, AB and BC are prime to one another. (Which is) the very thing it was required to show.kjþ. Proposition 29 ῞Απας πρῶτος ἀριθμὸς πρὸς ἅπαντα ἀριθμόν, ὃν μὴ με- Every prime number is prime to every number which τρεῖ, πρῶτός ἐστιν. it does not measure. Γ Α Β C A B ῎Εστω πρῶτος ἀριθμὸς ὁ Α καὶ τὸν Β μὴ μετρείτω· λέγω, Let A be a prime number, and let it not measure B. I ὅτι οἱ Β, Α πρῶτοι πρὸς ἀλλήλους εἰσίν. say that B and A are prime to one another. For if B and Εἰ γὰρ μή εἰσιν οἱ Β, Α πρῶτοι πρὸς ἀλλήλους, μετρήσει A are not prime to one another then some number will τις αὐτοὺς ἀριθμός. μετρείτω ὁ Γ. ἐπεὶ ὁ Γ τὸν Β μετρεῖ, measure them. Let C measure (them). Since C measures ὁ δὲ Α τὸν Β οὐ μετρεῖ, ὁ Γ ἄρα τῷ Α οὔκ ἐστιν ὁ αὐτός. B, and A does not measure B, C is thus not the same as καὶ ἐπεὶ ὁ Γ τοὺς Β, Α μετρεῖ, καὶ τὸν Α ἄρα μετρεῖ πρῶτον A. And since C measures B and A, it thus also measures ὄντα μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα A, which is prime, (despite) not being the same as it. τοὺς Β, Α μετρήσει τις ἀριθμός. οἱ Α, Β ἄρα πρῶτοι πρὸς The very thing is impossible. Thus, some number cannot ἀλλήλους εἰσίν· ὅπερ ἔδει δεῖξαι. measure (both) B and A. Thus, A and B are prime to one another. (Which is) the very thing it was required to 217 STOIQEIWN zþ. ELEMENTS BOOK 7 show.lþ. Proposition 30 ᾿Εὰν δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί If two numbers make some (number by) multiplying τινα, τὸν δὲ γενόμενον ἐξ αὐτῶν μετρῇ τις πρῶτος ἀριθμός, one another, and some prime number measures the num- καὶ ἕνα τῶν ἐξ ἀρχῆς μετρήσει. ber (so) created from them, then it will also measure one of the original (numbers). Ε Α Β Γ ∆ E A B C D Δύο γὰρ ἀριθμοὶ οἱ Α, Β πολλαπλασιάσαντες ἀλλήλους For let two numbers A and B make C (by) multiplying τὸν Γ ποιείτωσαν, τὸν δὲ Γ μετρείτω τις πρῶτος ἀριθμὸς ὁ one another, and let some prime number D measure C. I Δ· λέγω, ὅτι ὁ Δ ἕνα τῶν Α, Β μετρεῖ. say that D measures one of A and B. Τὸν γὰρ Α μὴ μετρείτω· καί ἐστι πρῶτος ὁ Δ· οἱ Α, For let it not measure A. And since D is prime, A Δ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ὁσάκις ὁ Δ τὸν Γ and D are thus prime to one another [Prop. 7.29]. And μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ οὖν ὁ Δ as many times as D measures C, so many units let there τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, ὁ Δ ἄρα τὸν Ε be in E. Therefore, since D measures C according to πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α τὸν the units E, D has thus made C (by) multiplying E Β πολλαπλασιάσας τὸν Γ πεποίηκεν· ἴσος ἄρα ἐστὶν ὁ ἐκ [Def. 7.15]. But, in fact, A has also made C (by) multi- τῶν Δ, Ε τῷ ἐκ τῶν Α, Β. ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Α, plying B. Thus, the (number created) from (multiplying) οὕτως ὁ Β πρὸς τὸν Ε. οἱ δὲ Δ, Α πρῶτοι, οἱ δὲ πρῶτοι καὶ D and E is equal to the (number created) from (mul- ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον tiplying) A and B. Thus, as D is to A, so B (is) to E ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν [Prop. 7.19]. And D and A (are) prime (to one another), ἐλάσσονα, τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ and (numbers) prime (to one another are) also the least ἐπόμενος τὸν ἑπόμενον· ὁ Δ ἄρα τὸν Β μετρεῖ. ὁμοίως δὴ (of those numbers having the same ratio) [Prop. 7.21], δείξομεν, ὅτι καὶ ἐὰν τὸν Β μὴ μετρῇ, τὸν Α μετρήσει. ὁ Δ and the least (numbers) measure those (numbers) hav- ἄρα ἕνα τῶν Α, Β μετρεῖ· ὅπερ ἔδει δεῖξαι. ing the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser—that is to say, the leading (measuring) the lead- ing, and the following the following [Prop. 7.20]. Thus, D measures B. So, similarly, we can also show that if (D) does not measure B then it will measure A. Thus, D measures one of A and B. (Which is) the very thing it was required to show.laþ. Proposition 31 ῞Απας σύνθεντος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ με- Every composite number is measured by some prime τρεῖται. number. Let A be a composite number. I say that A is measured ῎Εστω σύνθεντος ἀριθμὸς ὁ Α· λέγω, ὅτι ὁ Α ὑπὸ πρώτου by some prime number. τινὸς ἀριθμοῦ μετρεῖται. For since A is composite, some number will measure ᾿Επεὶ γὰρ σύνθετός ἐστιν ὁ Α, μετρήσει τις αὐτὸν it. Let it (so) measure (A), and let it be B. And if B 218 STOIQEIWN zþ. ELEMENTS BOOK 7 ἀριθμός. μετρείτω, καὶ ἔστω ὁ Β. καὶ εἰ μὲν πρῶτός ἐστιν ὁ is prime then that which was prescribed has happened. Β, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. εἰ δὲ σύνθετος, μετρήσει And if (B is) composite then some number will measure τις αὐτὸν ἀριθμός. μετρείτω, καὶ ἔστω ὁ Γ. καὶ ἐπεὶ ὁ Γ it. Let it (so) measure (B), and let it be C. And since τὸν Β μετρεῖ, ὁ δὲ Β τὸν Α μετρεῖ, καὶ ὁ Γ ἄρα τὸν Α C measures B, and B measures A, C thus also measures μετρεῖ. καὶ εἰ μὲν πρῶτός ἐστιν ὁ Γ, γεγονὸς ἂν εἴη τὸ A. And if C is prime then that which was prescribed has ἐπιταχθέν. εἰ δὲ σύνθετος, μετρήσει τις αὐτὸν ἀριθμός. happened. And if (C is) composite then some number τοιαύτης δὴ γινομένης ἐπισκέψεως ληφθήσεταί τις πρῶτος will measure it. So, in this manner of continued inves- ἀριθμός, ὃς μετρήσει. εἰ γὰρ οὐ ληφθήσεται, μετρήσουσι tigation, some prime number will be found which will τὸν Α ἀριθμὸν ἄπειροι ἀριθμοί, ὧν ἕτερος ἑτέρου ἐλάσσων measure (the number preceding it, which will also mea- ἐστίν· ὅπερ ἐστὶν ἀδύνατον ἐν ἀριθμοῖς. ληφθήσεταί τις ἄρα sure A). And if (such a number) cannot be found then an πρῶτος ἀριθμός, ὃς μετρήσει τὸν πρὸ ἑαυτοῦ, ὃς καὶ τὸν Α infinite (series of) numbers, each of which is less than the μετρήσει. preceding, will measure the number A. The very thing is impossible for numbers. Thus, some prime number will (eventually) be found which will measure the (number) preceding it, which will also measure A. Γ Α Β C A B ῞Απας ἄρα σύνθεντος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ Thus, every composite number is measured by some μετρεῖται· ὅπερ ἔδει δεῖξαι. prime number. (Which is) the very thing it was required to show.lbþ. Proposition 32 ῞Απας ἀριθμὸς ἤτοι πρῶτός ἐστιν ἢ ὑπὸ πρώτου τινὸς Every number is either prime or is measured by some ἀριθμοῦ μετρεῖται. prime number. Α A ῎Εστω ἀριθμὸς ὁ Α· λέγω, ὅτι ὁ Α ἤτοι πρῶτός ἐστιν ἢ Let A be a number. I say that A is either prime or is ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. measured by some prime number. Εἰ μὲν οὖν πρῶτός ἐστιν ὁ Α, γεγονὸς ἂν εἴη τό In fact, if A is prime then that which was prescribed ἐπιταχθέν. εἰ δὲ σύνθετος, μετρήσει τις αὐτὸν πρῶτος has happened. And if (it is) composite then some prime ἀριθμός. number will measure it [Prop. 7.31]. ῞Απας ἄρα ἀριθμὸς ἤτοι πρῶτός ἐστιν ἢ ὑπὸ πρώτου Thus, every number is either prime or is measured τινὸς ἀριθμοῦ μετρεῖται· ὅπερ ἔδει δεῖξαι. by some prime number. (Which is) the very thing it was required to show.lgþ. Proposition 33 Ἀριθμῶν δοθέντων ὁποσωνοῦν εὑρεῖν τοὺς ἐλαχίστους To find the least of those (numbers) having the same τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. ratio as any given multitude of numbers. ῎Εστωσαν οἱ δοθέντες ὁποσοιοῦν ἀριθμοὶ οἱ Α, Β, Γ· δεῖ Let A, B, and C be any given multitude of numbers. δὴ εὑρεῖν τοὺς ἐλαχίστους τῶν τὸν αὐτὸν λόγον ἐχόντων So it is required to find the least of those (numbers) hav- τοῖς Α, Β, Γ. ing the same ratio as A, B, and C. Οἱ Α, Β, Γ γὰρ ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν ἢ οὔ. For A, B, and C are either prime to one another, or εἰ μὲν οὖν οἱ Α, Β, Γ πρῶτοι πρὸς ἀλλήλους εἰσίν, ἐλάχιστοί not. In fact, if A, B, and C are prime to one another then εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. they are the least of those (numbers) having the same ratio as them [Prop. 7.22]. 219 STOIQEIWN zþ. ELEMENTS BOOK 7 ΜΑ Β Γ ∆ Ε Ζ Η Θ Κ Λ L MA B C D E F G H K Εἰ δὲ οὔ, εἰλήφθω τῶν Α, Β, Γ τὸ μέγιστον κοινὸν And if not, let the greatest common measure, D, of μέτρον ὁ Δ, καὶ ὁσάκις ὁ Δ ἕκαστον τῶν Α, Β, Γ μετρεῖ, A, B, and C have be taken [Prop. 7.3]. And as many τοσαῦται μονάδες ἔστωσαν ἐν ἑκάστῳ τῶν Ε, Ζ, Η. καὶ times as D measures A, B, C, so many units let there ἕκαστος ἄρα τῶν Ε, Ζ, Η ἕκαστον τῶν Α, Β, Γ μετρεῖ κατὰ be in E, F , G, respectively. And thus E, F , G mea- τὰς ἐν τῷ Δ μονάδας. οἱ Ε, Ζ, Η ἄρα τοὺς Α, Β, Γ ἰσάκις sure A, B, C, respectively, according to the units in D μετροῦσιν· οἱ Ε, Ζ, Η ἄρα τοῖς Α, Β, Γ ἐν τῷ αὐτῷ λόγῳ [Prop. 7.15]. Thus, E, F , G measure A, B, C (respec- εἰσίν. λέγω δή, ὅτι καὶ ἐλάχιστοι. εἰ γὰρ μή εἰσιν οἱ Ε, Ζ, tively) an equal number of times. Thus, E, F , G are in Η ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β, Γ, the same ratio as A, B, C (respectively) [Def. 7.20]. So I ἔσονται [τινες] τῶν Ε, Ζ, Η ἐλάσσονες ἀριθμοὶ ἐν τῷ αὐτῷ say that (they are) also the least (of those numbers hav- λόγῳ ὄντες τοῖς Α, Β, Γ. ἔστωσαν οἱ Θ, Κ, Λ· ἰσάκις ἄρα ὁ ing the same ratio as A, B, C). For if E, F , G are not Θ τὸν Α μετρεῖ καὶ ἑκάτερος τῶν Κ, Λ ἑκάτερον τῶν Β, Γ. the least of those (numbers) having the same ratio as A, ὁσάκις δὲ ὁ Θ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν B, C (respectively), then there will be [some] numbers τῷ Μ· καὶ ἑκάτερος ἄρα τῶν Κ, Λ ἑκάτερον τῶν Β, Γ μετρεῖ less than E, F , G which are in the same ratio as A, B, C κατὰ τὰς ἐν τῷ Μ μονάδας. καὶ ἐπεὶ ὁ Θ τὸν Α μετρεῖ κατὰ (respectively). Let them be H , K, L. Thus, H measures τὰς ἐν τῷ Μ μονάδας, καὶ ὁ Μ ἄρα τὸν Α μετρεῖ κατὰ τὰς A the same number of times that K, L also measure B, ἐν τῷ Θ μονάδας. διὰ τὰ αὐτὰ δὴ ὁ Μ καὶ ἑκάτερον τῶν Β, C, respectively. And as many times as H measures A, so Γ μετρεῖ κατὰ τὰς ἐν ἑκατέρῳ τῶν Κ, Λ μονάδας· ὁ Μ ἄρα many units let there be in M . Thus, K, L measure B, τοὺς Α, Β, Γ μετρεῖ. καὶ ἐπεὶ ὁ Θ τὸν Α μετρεῖ κατὰ τὰς C, respectively, according to the units in M . And since ἐν τῷ Μ μονάδας, ὁ Θ ἄρα τὸν Μ πολλαπλασιάσας τὸν Α H measures A according to the units in M , M thus also πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ε τὸν Δ πολλαπλασιάσας measures A according to the units in H [Prop. 7.15]. So, τὸν Α πεποίηκεν. ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Ε, Δ τῷ ἐκ for the same (reasons), M also measures B, C accord- τῶν Θ, Μ. ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Μ πρὸς ing to the units in K, L, respectively. Thus, M measures τὸν Δ. μείζων δὲ ὁ Ε τοῦ Θ· μείζων ἄρα καὶ ὁ Μ τοῦ Δ. A, B, and C. And since H measures A according to the καὶ μετρεῖ τοὺς Α, Β, Γ· ὅπερ ἐστὶν ἀδύνατον· ὑπόκειται units in M , H has thus made A (by) multiplying M . So, γὰρ ὁ Δ τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον. οὐκ ἄρα for the same (reasons), E has also made A (by) multiply- ἔσονταί τινες τῶν Ε, Ζ, Η ἐλάσσονες ἀριθμοὶ ἐν τῷ αὐτῷ ing D. Thus, the (number created) from (multiplying) λόγῳ ὄντες τοῖς Α, Β, Γ. οἱ Ε, Ζ, Η ἄρα ἐλάχιστοί εἰσι τῶν E and D is equal to the (number created) from (multi- τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β, Γ· ὅπερ ἔδει δεῖξαι. plying) H and M . Thus, as E (is) to H , so M (is) to D [Prop. 7.19]. And E (is) greater than H . Thus, M (is) also greater than D [Prop. 5.13]. And (M) measures A, B, and C. The very thing is impossible. For D was assumed (to be) the greatest common measure of A, B, and C. Thus, there cannot be any numbers less than E, F , G which are in the same ratio as A, B, C (respec- tively). Thus, E, F , G are the least of (those numbers) having the same ratio as A, B, C (respectively). (Which is) the very thing it was required to show.ldþ. Proposition 34 Δύο ἀριθμῶν δοθέντων εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν To find the least number which two given numbers ἀριθμόν. (both) measure. ῎Εστωσαν οἱ δοθέντες δύο ἀριθμοὶ οἱ Α, Β· δεῖ δὴ εὑρεῖν, Let A and B be the two given numbers. So it is re- 220 STOIQEIWN zþ. ELEMENTS BOOK 7 ὃν ἐλάχιστον μετροῦσιν ἀριθμόν. quired to find the least number which they (both) mea- sure. Ζ Α Β Γ ∆ Ε F A B C D E Οἱ Α, Β γὰρ ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν ἢ οὔ. For A and B are either prime to one another, or not. ἔστωσαν πρότερον οἱ Α, Β πρῶτοι πρὸς ἀλλήλους, καὶ ὁ Α Let them, first of all, be prime to one another. And let A τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω· καὶ ὁ Β ἄρα τὸν Α make C (by) multiplying B. Thus, B has also made C πολλαπλασιάσας τὸν Γ πεποίηκεν. οἱ Α, Β ἄρα τὸν Γ με- (by) multiplying A [Prop. 7.16]. Thus, A and B (both) τροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ μή, μετρήσουσί measure C. So I say that (C) is also the least (num- τινα ἀριθμὸν οἱ Α, Β ἐλάσσονα ὄντα τοῦ Γ. μετρείτωσαν ber which they both measure). For if not, A and B will τὸν Δ. καὶ ὁσάκις ὁ Α τὸν Δ μετρεῖ, τοσαῦται μονάδες (both) measure some (other) number which is less than ἔστωσαν ἐν τῷ Ε, ὁσάκις δὲ ὁ Β τὸν Δ μετρεῖ, τοσαῦται C. Let them (both) measure D (which is less than C). μονάδες ἔστωσαν ἐν τῷ Ζ. ὁ μὲν Α ἄρα τὸν Ε πολλα- And as many times as A measures D, so many units let πλασιάσας τὸν Δ πεποίηκεν, ὁ δὲ Β τὸν Ζ πολλαπλασιάσας there be in E. And as many times as B measures D, τὸν Δ πεποίηκεν· ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Α, Ε τῷ ἐκ τῶν so many units let there be in F . Thus, A has made D Β, Ζ. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Ζ πρὸς τὸν (by) multiplying E, and B has made D (by) multiply- Ε. οἱ δὲ Α, Β πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ing F . Thus, the (number created) from (multiplying) ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ A and E is equal to the (number created) from (multi- τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα· ὁ Β plying) B and F . Thus, as A (is) to B, so F (is) to E ἄρα τὸν Ε μετρεῖ, ὡς ἑπόμενος ἑπόμενον. καὶ ἐπεὶ ὁ Α τοὺς [Prop. 7.19]. And A and B are prime (to one another), Β, Ε πολλαπλασιάσας τοὺς Γ, Δ πεποίηκεν, ἔστιν ἄρα ὡς and prime (numbers) are the least (of those numbers ὁ Β πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς τὸν Δ. μετρεῖ δὲ ὁ Β τὸν having the same ratio) [Prop. 7.21], and the least (num- Ε· μετρεῖ ἄρα καὶ ὁ Γ τὸν Δ ὁ μείζων τὸν ἐλάσσονα· ὅπερ bers) measure those (numbers) having the same ratio (as ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Α, Β μετροῦσί τινα ἀριθμὸν them) an equal number of times, the greater (measuring) ἐλάσσονα ὄντα τοῦ Γ. ὁ Γ ἄρα ἐλάχιστος ὢν ὑπὸ τῶν Α, Β the greater, and the lesser the lesser [Prop. 7.20]. Thus, μετρεῖται. B measures E, as the following (number measuring) the following. And since A has made C and D (by) multi- plying B and E (respectively), thus as B is to E, so C (is) to D [Prop. 7.17]. And B measures E. Thus, C also measures D, the greater (measuring) the lesser. The very thing is impossible. Thus, A and B do not (both) mea- sure some number which is less than C. Thus, C is the least (number) which is measured by (both) A and B. Θ Α Ζ Γ ∆ Η Β Ε H A D B EF C G Μὴ ἔστωσαν δὴ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους, So let A and B be not prime to one another. And καὶ εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον let the least numbers, F and E, have been taken having ἐχόντων τοῖς Α, Β οἱ Ζ, Ε· ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Α, Ε τῷ the same ratio as A and B (respectively) [Prop. 7.33]. 221 STOIQEIWN zþ. ELEMENTS BOOK 7 ἐκ τῶν Β, Ζ. καὶ ὁ Α τὸν Ε πολλαπλασιάσας τὸν Γ ποιείτω· Thus, the (number created) from (multiplying) A and E καὶ ὁ Β ἄρα τὸν Ζ πολλαπλασιάσας τὸν Γ πεποίηκεν· οἱ Α, is equal to the (number created) from (multiplying) B Β ἄρα τὸν Γ μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ and F [Prop. 7.19]. And let A make C (by) multiplying μή, μετρήσουσί τινα ἀριθμὸν οἱ Α, Β ἐλάσσονα ὄντα τοῦ E. Thus, B has also made C (by) multiplying F . Thus, Γ. μετρείτωσαν τὸν Δ. καὶ ὁσάκις μὲν ὁ Α τὸν Δ μετρεῖ, A and B (both) measure C. So I say that (C) is also the τοσαῦται μονάδες ἔστωσαν ἐν τῷ Η, ὁσάκις δὲ ὁ Β τὸν Δ least (number which they both measure). For if not, A μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Θ. ὁ μὲν Α ἄρα and B will (both) measure some number which is less τὸν Η πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ δὲ Β τὸν Θ than C. Let them (both) measure D (which is less than πολλαπλασιάσας τὸν Δ πεποίηκεν. ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν C). And as many times as A measures D, so many units Α, Η τῷ ἐκ τῶν Β, Θ· ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως let there be in G. And as many times as B measures D, ὁ Θ πρὸς τὸν Η. ὡς δὲ ὁ Α πρὸς τὸν Β, οὕτως ὁ Ζ πρὸς so many units let there be in H . Thus, A has made D τὸν Ε· καὶ ὡς ἄρα ὁ Ζ πρὸς τὸν Ε, οὕτως ὁ Θ πρὸς τὸν (by) multiplying G, and B has made D (by) multiplying Η. οἱ δὲ Ζ, Ε ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν H . Thus, the (number created) from (multiplying) A and αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ G is equal to the (number created) from (multiplying) B ἐλάσσων τὸν ἐλάσσονα· ὁ Ε ἄρα τὸν Η μετρεῖ. καὶ ἐπεὶ ὁ and H . Thus, as A is to B, so H (is) to G [Prop. 7.19]. Α τοὺς Ε, Η πολλαπλασιάσας τοὺς Γ, Δ πεποίηκεν, ἔστιν And as A (is) to B, so F (is) to E. Thus, also, as F (is) ἄρα ὡς ὁ Ε πρὸς τὸν Η, οὕτως ὁ Γ πρὸς τὸν Δ. ὁ δὲ Ε τὸν to E, so H (is) to G. And F and E are the least (num- Η μετρεῖ· καὶ ὁ Γ ἄρα τὸν Δ μετρεῖ ὁ μείζων τὸν ἐλάσσονα· bers having the same ratio as A and B), and the least ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Α, Β μετρήσουσί τινα (numbers) measure those (numbers) having the same ra- ἀριθμὸν ἐλάσσονα ὄντα τοῦ Γ. ὁ Γ ἄρα ἐλάχιστος ὢν ὑπὸ tio an equal number of times, the greater (measuring) τῶν Α, Β μετρεῖται· ὅπερ ἔπει δεῖξαι. the greater, and the lesser the lesser [Prop. 7.20]. Thus, E measures G. And since A has made C and D (by) mul- tiplying E and G (respectively), thus as E is to G, so C (is) to D [Prop. 7.17]. And E measures G. Thus, C also measures D, the greater (measuring) the lesser. The very thing is impossible. Thus, A and B do not (both) mea- sure some (number) which is less than C. Thus, C (is) the least (number) which is measured by (both) A and B. (Which is) the very thing it was required to show.leþ. Proposition 35 ᾿Εὰν δύο ἀριθμοὶ ἀριθμόν τινα μετρῶσιν, καὶ ὁ ἐλάχιστος If two numbers (both) measure some number then the ὑπ᾿ αὐτῶν μετρούμενος τὸν αὐτὸν μετρήσει. least (number) measured by them will also measure the same (number). ∆ Α Β Ε Γ Ζ D A B E C F Δύο γὰρ ἀριθμοὶ οἱ Α, Β ἀριθμόν τινα τὸν ΓΔ με- For let two numbers, A and B, (both) measure some τρείτωσαν, ἐλάχιστον δὲ τὸν Ε· λέγω, ὅτι καὶ ὁ Ε τὸν ΓΔ number CD, and (let) E (be the) least (number mea- μετρεῖ. sured by both A and B). I say that E also measures CD. Εἰ γὰρ οὐ μετρεῖ ὁ Ε τὸν ΓΔ, ὁ Ε τὸν ΔΖ μετρῶν For if E does not measure CD then let E leave CF λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΓΖ. καὶ ἐπεὶ οἱ Α, Β τὸν Ε less than itself (in) measuring DF . And since A and B μετροῦσιν, ὁ δὲ Ε τὸν ΔΖ μετρεῖ, καὶ οἱ Α, Β ἄρα τὸν (both) measure E, and E measures DF , A and B will ΔΖ μετρήσουσιν. μετροῦσι δὲ καὶ ὅλον τὸν ΓΔ· καὶ λοιπὸν thus also measure DF . And (A and B) also measure the ἄρα τὸν ΓΖ μετρήσουσιν ἐλάσσονα ὄντα τοῦ Ε· ὅπερ ἐστὶν whole of CD. Thus, they will also measure the remainder ἀδύνατον. οὐκ ἄρα οὐ μετρεῖ ὁ Ε τὸν ΓΔ· μετρεῖ ἄρα· ὅπερ CF , which is less than E. The very thing is impossible. ἔδει δεῖξαι. Thus, E cannot not measure CD. Thus, (E) measures 222 STOIQEIWN zþ. ELEMENTS BOOK 7 (CD). (Which is) the very thing it was required to show.l�þ. Proposition 36 Τριῶν ἀριθμῶν δοθέντων εὑρεῖν, ὃν ἐλάχιστον με- To find the least number which three given numbers τροῦσιν ἀριθμόν. (all) measure. ῎Εστωσαν οἱ δοθέντες τρεῖς ἀριθμοὶ οἱ Α, Β, Γ· δεῖ δὴ Let A, B, and C be the three given numbers. So it is εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν ἀριθμόν. required to find the least number which they (all) mea- sure. Ζ Α Β Γ ∆ Ε F A B C D E Εἰλήφθω γὰρ ὑπὸ δύο τῶν Α, Β ἐλάχιστος μετρούμενος For let the least (number), D, measured by the two ὁ Δ. ὁ δὴ Γ τὸν Δ ἤτοι μετρεῖ ἢ οὐ μετρεῖ. μετρείτω (numbers) A and B have been taken [Prop. 7.34]. So C πρότερον. μετροῦσι δὲ καὶ οἱ Α, Β τὸν Δ· οἱ Α, Β, Γ either measures, or does not measure, D. Let it, first of ἄρα τὸν Δ μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ all, measure (D). And A and B also measure D. Thus, μή, μετρήσουσιν [τινα] ἀριθμὸν οἱ Α, Β, Γ ἐλάσσονα ὄντα A, B, and C (all) measure D. So I say that (D is) also τοῦ Δ. μετρείτωσαν τὸν Ε. ἐπεὶ οἱ Α, Β, Γ τὸν Ε μετροῦσιν, the least (number measured by A, B, and C). For if not, καὶ οἱ Α, Β ἄρα τὸν Ε μετροῦσιν. καὶ ὁ ἐλάχιστος ἄρα ὑπὸ A, B, and C will (all) measure [some] number which τῶν Α, Β μετρούμενος [τὸν Ε] μετρήσει. ἐλάχιστος δὲ ὑπὸ is less than D. Let them measure E (which is less than τῶν Α, Β μετρούμενός ἐστιν ὁ Δ· ὁ Δ ἄρα τὸν Ε μετρήσει D). Since A, B, and C (all) measure E then A and B ὁ μείζων τὸν ἐλάσσονα· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ thus also measure E. Thus, the least (number) measured Α, Β, Γ μετρήσουσί τινα ἀριθμὸν ἐλάσσονα ὄντα τοῦ Δ· οἱ by A and B will also measure [E] [Prop. 7.35]. And D Α, Β, Γ ἄρα ἐλάχιστον τὸν Δ μετροῦσιν. is the least (number) measured by A and B. Thus, D Μὴ μετρείτω δὴ πάλιν ὁ Γ τὸν Δ, καὶ εἰλήφθω ὑπὸ τῶν will measure E, the greater (measuring) the lesser. The Γ, Δ ἐλάχιστος μετρούμενος ἀριθμὸς ὁ Ε. ἐπεὶ οἱ Α, Β very thing is impossible. Thus, A, B, and C cannot (all) τὸν Δ μετροῦσιν, ὁ δὲ Δ τὸν Ε μετρεῖ, καὶ οἱ Α, Β ἄρα measure some number which is less than D. Thus, A, B, τὸν Ε μετροῦσιν. μετρεῖ δὲ καὶ ὁ Γ [τὸν Ε· καὶ] οἱ Α, Β, and C (all) measure the least (number) D. Γ ἄρα τὸν Ε μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ So, again, let C not measure D. And let the least γὰρ μή, μετρήσουσί τινα οἱ Α, Β, Γ ἐλάσσονα ὄντα τοῦ Ε. number, E, measured by C and D have been taken μετρείτωσαν τὸν Ζ. ἐπεὶ οἱ Α, Β, Γ τὸν Ζ μετροῦσιν, καὶ οἱ [Prop. 7.34]. Since A and B measure D, and D measures Α, Β ἄρα τὸν Ζ μετροῦσιν· καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν E, A and B thus also measure E. And C also measures Α, Β μετρούμενος τὸν Ζ μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν [E]. Thus, A, B, and C [also] measure E. So I say that Α, Β μετρούμενός ἐστιν ὁ Δ· ὁ Δ ἄρα τὸν Ζ μετρεῖ. μετρεῖ (E is) also the least (number measured by A, B, and C). δὲ καὶ ὁ Γ τὸν Ζ· οἱ Δ, Γ ἄρα τὸν Ζ μετροῦσιν· ὥστε καὶ ὁ For if not, A, B, and C will (all) measure some (number) ἐλάχιστος ὑπὸ τῶν Δ, Γ μετρούμενος τὸν Ζ μετρήσει. ὁ δὲ which is less than E. Let them measure F (which is less ἐλάχιστος ὑπὸ τῶν Γ, Δ μετρούμενός ἐστιν ὁ Ε· ὁ Ε ἄρα than E). Since A, B, and C (all) measure F , A and B τὸν Ζ μετρεῖ ὁ μείζων τὸν ἐλάσσονα· ὅπερ ἐστὶν ἀδύνατον. thus also measure F . Thus, the least (number) measured οὐκ ἄρα οἱ Α, Β, Γ μετρήσουσί τινα ἀριθμὸν ἐλάσσονα ὄντα by A and B will also measure F [Prop. 7.35]. And D τοῦ Ε. ὁ Ε ἄρα ἐλάχιστος ὢν ὑπὸ τῶν Α, Β, Γ μετρεῖται· is the least (number) measured by A and B. Thus, D ὅπερ ἔδει δεῖξαι. measures F . And C also measures F . Thus, D and C (both) measure F . Hence, the least (number) measured by D and C will also measure F [Prop. 7.35]. And E 223 STOIQEIWN zþ. ELEMENTS BOOK 7 is the least (number) measured by C and D. Thus, E measures F , the greater (measuring) the lesser. The very thing is impossible. Thus, A, B, and C cannot measure some number which is less than E. Thus, E (is) the least (number) which is measured by A, B, and C. (Which is) the very thing it was required to show.lzþ. Proposition 37 ᾿Εὰν ἀριθμὸς ὑπό τινος ἀριθμοῦ μετρῆται, ὁ μετρούμενος If a number is measured by some number then the ὁμώνυμον μέρος ἕξει τῷ μετροῦντι. (number) measured will have a part called the same as the measuring (number). Α Β Γ ∆ D B A C Ἀριθμὸς γάρ ὁ Α ὑπό τινος ἀριθμοῦ τοῦ Β μετρείσθω· For let the number A be measured by some number λέγω, ὅτι ὁ Α ὁμώνυμον μέρος ἔχει τῷ Β. B. I say that A has a part called the same as B. ῾Οσάκις γὰρ ὁ Β τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστω- For as many times as B measures A, so many units σαν ἐν τῷ Γ. ἐπεὶ ὁ Β τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Γ let there be in C. Since B measures A according to the μονάδας, μετρεῖ δὲ καὶ ἡ Δ μονὰς τὸν Γ ἀριθμὸν κατὰ τὰς units in C, and the unit D also measures C according ἐν αὐτῷ μονάδας, ἰσάκις ἄρα ἡ Δ μονὰς τὸν Γ ἀριθμὸν με- to the units in it, the unit D thus measures the number τρεῖ καὶ ὁ Β τὸν Α. ἐναλλὰξ ἄρα ἰσάκις ἡ Δ μονὰς τὸν Β C as many times as B (measures) A. Thus, alternately, ἀριθμὸν μετρεῖ καὶ ὁ Γ τὸν Α· ὃ ἄρα μέρος ἐστὶν ἡ Δ μονὰς the unit D measures the number B as many times as C τοῦ Β ἀριθμοῦ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Α. ἡ δὲ Δ (measures) A [Prop. 7.15]. Thus, which(ever) part the μονὰς τοῦ Β ἀριθμοῦ μέρος ἐστὶν ὁμώνυμον αὐτῷ· καὶ ὁ Γ unit D is of the number B, C is also the same part of A. ἄρα τοῦ Α μέρος ἐστὶν ὁμώνυμον τῷ Β. ὥστε ὁ Α μέρος And the unit D is a part of the number B called the same ἔχει τὸν Γ ὁμώνυμον ὄντα τῷ Β· ὅπερ ἔδει δεῖξαι. as it (i.e., a Bth part). Thus, C is also a part of A called the same as B (i.e., C is the Bth part of A). Hence, A has a part C which is called the same as B (i.e., A has a Bth part). (Which is) the very thing it was required to show.lhþ. Proposition 38 ᾿Εὰν ἀριθμος μέρος ἔχῃ ὁτιοῦν, ὑπὸ ὁμωνύμου ἀριθμοῦ If a number has any part whatever then it will be mea- μετρηθήσεται τῷ μέρει. sured by a number called the same as the part. ∆ Α Β Γ D A B C Ἀριθμὸς γὰρ ὁ Α μέρος ἐχέτω ὁτιοῦν τὸν Β, καὶ τῷ Β For let the number A have any part whatever, B. And μέρει ὁμώνυμος ἔστω [ἀριθμὸς] ὁ Γ· λέγω, ὅτι ὁ Γ τὸν Α let the [number] C be called the same as the part B (i.e., μετρεῖ. B is the Cth part of A). I say that C measures A. ᾿Επεὶ γὰρ ὁ Β τοῦ Α μέρος ἐστὶν ὁμώνυμον τῷ Γ, ἔστι For since B is a part of A called the same as C, and δὲ καὶ ἡ Δ μονὰς τοῦ Γ μέρος ὁμώνυμον αὐτῷ, ὃ ἄρα μέρος the unit D is also a part of C called the same as it (i.e., 224 STOIQEIWN zþ. ELEMENTS BOOK 7 ἐστὶν ἡ Δ μονὰς τοῦ Γ ἀριθμοῦ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Β D is the Cth part of C), thus which(ever) part the unit D τοῦ Α· ἰσάκις ἄρα ἡ Δ μονὰς τὸν Γ ἀριθμὸν μετρεῖ καὶ ὁ Β is of the number C, B is also the same part of A. Thus, τὸν Α. ἐναλλὰξ ἄρα ἰσάκις ἡ Δ μονὰς τὸν Β ἀριθμὸν μετρεῖ the unit D measures the number C as many times as B καὶ ὁ Γ τὸν Α. ὁ Γ ἄρα τὸν Α μετρεῖ· ὅπερ ἔδει δεὶξαι. (measures) A. Thus, alternately, the unit D measures the number B as many times as C (measures) A [Prop. 7.15]. Thus, C measures A. (Which is) the very thing it was required to show.ljþ. Proposition 39 Ἀριθμὸν εὐρεῖν, ὃς ἐλάχιστος ὢν ἕξει τὰ δοθέντα μέρη. To find the least number that will have given parts. Η Θ Α Β Γ ∆ Ε Ζ H A B C D E F G ῎Εστω τὰ δοθέντα μέρη τὰ Α, Β, Γ· δεῖ δὴ ἀριθμὸν Let A, B, and C be the given parts. So it is required εὑρεῖν, ὃς ἐλάχιστος ὢν ἕξει τὰ Α, Β, Γ μέρη. to find the least number which will have the parts A, B, ῎Εστωσαν γὰρ τοῖς Α, Β, Γ μέρεσιν ὁμώνυμοι ἀριθμοὶ and C (i.e., an Ath part, a Bth part, and a Cth part). οἱ Δ, Ε, Ζ, καὶ εἰλήφθω ὑπὸ τῶν Δ, Ε, Ζ ἐλάχιστος με- For let D, E, and F be numbers having the same τρούμενος ἀριθμὸς ὁ Η. names as the parts A, B, and C (respectively). And let ῾Ο Η ἄρα ὁμώνυμα μέρη ἔχει τοῖς Δ, Ε, Ζ. τοῖς δὲ Δ, the least number, G, measured by D, E, and F , have Ε, Ζ ὁμώνυμα μέρη ἐστὶ τὰ Α, Β, Γ· ὁ Η ἄρα ἔχει τὰ Α, Β, been taken [Prop. 7.36]. Γ μέρη. λέγω δή, ὅτι καὶ ἐλάχιστος ὤν, εἰ γὰρ μή, ἔσται τις Thus, G has parts called the same as D, E, and F τοῦ Η ἐλάσσων ἀριθμός, ὃς ἕξει τὰ Α, Β, Γ μέρη. ἔστω ὁ [Prop. 7.37]. And A, B, and C are parts called the same Θ. ἐπεὶ ὁ Θ ἔχει τὰ Α, Β, Γ μέρη, ὁ Θ ἄρα ὑπὸ ὁμωνύμων as D, E, and F (respectively). Thus, G has the parts A, ἀριθμῶν μετρηθήσεται τοῖς Α, Β, Γ μέρεσιν. τοῖς δὲ Α, Β, B, and C. So I say that (G) is also the least (number Γ μέρεσιν ὁμώνυμοι ἀριθμοί εἰσιν οἱ Δ, Ε, Ζ· ὁ Θ ἄρα ὑπὸ having the parts A, B, and C). For if not, there will be τῶν Δ, Ε, Ζ μετρεῖται. καί ἐστιν ἐλάσσων τοῦ Η· ὅπερ some number less than G which will have the parts A, ἐστὶν ἀδύνατον. οὐκ ἄρα ἔσται τις τοῦ Η ἐλάσσων ἀριθμός, B, and C. Let it be H . Since H has the parts A, B, and ὃς ἕξει τὰ Α, Β, Γ μέρη· ὅπερ ἔδει δεῖξαι. C, H will thus be measured by numbers called the same as the parts A, B, and C [Prop. 7.38]. And D, E, and F are numbers called the same as the parts A, B, and C (respectively). Thus, H is measured by D, E, and F . And (H) is less than G. The very thing is impossible. Thus, there cannot be some number less than G which will have the parts A, B, and C. (Which is) the very thing it was required to show. 225 226 ELEMENTS BOOK 8 Continued Proportion† †The propositions contained in Books 7–9 are generally attributed to the school of Pythagoras. 227 STOIQEIWN hþ. ELEMENTS BOOK 8aþ. Proposition 1 ᾿Εὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, οἱ δὲ If there are any multitude whatsoever of continuously ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους ὦσιν, ἐλάχιστοί εἰσι proportional numbers, and the outermost of them are τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. prime to one another, then the (numbers) are the least of those (numbers) having the same ratio as them. Θ Α Β Γ ∆ Ε Ζ Η H EA B C D F G ῎Εστωσαν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Let A, B, C, D be any multitude whatsoever of con- Γ, Δ, οἱ δὲ ἄκροι αὐτῶν οἱ Α, Δ, πρῶτοι πρὸς ἀλλήλους tinuously proportional numbers. And let the outermost ἔστωσαν· λέγω, ὅτι οἱ Α, Β, Γ, Δ ἐλάχιστοί εἰσι τῶν τὸν of them, A and D, be prime to one another. I say that αὐτὸν λόγον ἐχόντων αὐτοῖς. A, B, C, D are the least of those (numbers) having the Εἰ γὰρ μή, ἔστωσαν ἐλάττονες τῶν Α, Β, Γ, Δ οἱ Ε, same ratio as them. Ζ, Η, Θ ἐν τῷ αὐτῷ λόγῳ ὄντες αὐτοῖς. καὶ ἐπεὶ οἱ Α, For if not, let E, F , G, H be less than A, B, C, D Β, Γ, Δ ἐν τῷ αὐτῷ λόγῳ εἰσὶ τοῖς Ε, Ζ, Η, Θ, καί ἐστιν (respectively), being in the same ratio as them. And since ἴσον τὸ πλῆθος [τῶν Α, Β, Γ, Δ] τῷ πλήθει [τῶν Ε, Ζ, Η, A, B, C, D are in the same ratio as E, F , G, H , and the Θ], δι᾿ ἴσου ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, ὁ Ε πρὸς τὸν multitude [of A, B, C, D] is equal to the multitude [of E, Θ. οἱ δὲ Α, Δ πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ F , G, H], thus, via equality, as A is to D, (so) E (is) to H ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας [Prop. 7.14]. And A and D (are) prime (to one another). ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, And prime (numbers are) also the least of those (numbers τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος having the same ratio as them) [Prop. 7.21]. And the τὸν ἑπόμενον. μετρεῖ ἄρα ὁ Α τὸν Ε ὁ μείζων τὸν ἐλάσσονα· least numbers measure those (numbers) having the same ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Ε, Ζ, Η, Θ ἐλάσσονες ratio (as them) an equal number of times, the greater ὄντες τῶν Α, Β, Γ, Δ ἐν τῷ αὐτῷ λόγῳ εἰσὶν αὐτοῖς. οἱ Α, (measuring) the greater, and the lesser the lesser—that Β, Γ, Δ ἄρα ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων is to say, the leading (measuring) the leading, and the αὐτοῖς· ὅπερ ἔδει δεῖξαι. following the following [Prop. 7.20]. Thus, A measures E, the greater (measuring) the lesser. The very thing is impossible. Thus, E, F , G, H , being less than A, B, C, D, are not in the same ratio as them. Thus, A, B, C, D are the least of those (numbers) having the same ratio as them. (Which is) the very thing it was required to show.bþ. Proposition 2 Αριθμοὺς εὑρεῖν ἑξῆς ἀνάλογον ἐλαχίστους, ὅσους ἂν To find the least numbers, as many as may be pre- ἐπιτάξῃ τις, ἐν τῷ δοθέντι λόγῳ. scribed, (which are) continuously proportional in a given ῎Εστω ὁ δοθεὶς λόγος ἐν ἐλάχίστοις ἀριθμοῖς ὁ τοῦ ratio. Α πρὸς τὸν Β· δεῖ δὴ ἀριθμοὺς εὑρεῖν ἑξῆς ἀνάλογον Let the given ratio, (expressed) in the least numbers, ἐλαχίστους, ὅσους ἄν τις ἐπιτάξῃ, ἐν τῷ τοῦ Α πρὸς τὸν Β be that of A to B. So it is required to find the least num- λόγῳ. bers, as many as may be prescribed, (which are) in the ᾿Επιτετάχθωσαν δὴ τέσσαρες, καὶ ὁ Α ἑαυτὸν πολλα- ratio of A to B. πλασιάσας τὸν Γ ποιείτω, τὸν δὲ Β πολλαπλασιάσας τὸν Δ Let four (numbers) have been prescribed. And let A ποιείτω, καὶ ἔτι ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Ε ποιείτω, make C (by) multiplying itself, and let it make D (by) καὶ ἔτι ὁ Α τοὺς Γ, Δ, Ε πολλαπλασιάσας τοὺς Ζ, Η, Θ multiplying B. And, further, let B make E (by) multiply- ποιείτω, ὁ δὲ Β τὸν Ε πολλαπλασιάσας τὸν Κ ποιείτω. ing itself. And, further, let A make F , G, H (by) mul- tiplying C, D, E. And let B make K (by) multiplying E. 228 STOIQEIWN hþ. ELEMENTS BOOK 8 Κ Α Β Γ ∆ Ε Ζ Η Θ K A B C D E F G H Καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Γ And since A has made C (by) multiplying itself, and πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Δ πεποίηκεν, has made D (by) multiplying B, thus as A is to B, [so] C ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, [οὕτως] ὁ Γ πρὸς τὸν Δ. (is) to D [Prop. 7.17]. Again, since A has made D (by) πάλιν, ἐπεὶ ὁ μὲν Α τὸν Β πολλαπλασιάσας τὸν Δ πεποίηκεν, multiplying B, and B has made E (by) multiplying itself, ὁ δὲ Β ἑαυτὸν πολλαπλασιάσας τὸν Ε πεποίηκεν, ἑκάτερος A, B have thus made D, E, respectively, (by) multiplying ἄρα τῶν Α, Β τὸν Β πολλαπλασιάσας ἑκάτερον τῶν Δ, Ε B. Thus, as A is to B, so D (is) to E [Prop. 7.18]. But, as πεποίηκεν. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς A (is) to B, (so) C (is) to D. And thus as C (is) to D, (so) τὸν Ε. ἀλλ᾿ ὡς ὁ Α πρὸς τὸν Β, ὁ Γ πρὸς τὸν Δ· καὶ ὡς D (is) to E. And since A has made F , G (by) multiplying ἄρα ὁ Γ πρὸς τὸν Δ, ὁ Δ πρὸς τὸν Ε. καὶ ἐπεὶ ὁ Α τοὺς Γ, C, D, thus as C is to D, [so] F (is) to G [Prop. 7.17]. Δ πολλαπλασιάσας τοὺς Ζ, Η πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ And as C (is) to D, so A was to B. And thus as A (is) πρὸς τὸν Δ, [οὕτως] ὁ Ζ πρὸς τὸν Η. ὡς δὲ ὁ Γ πρὸς τὸν to B, (so) F (is) to G. Again, since A has made G, H Δ, οὕτως ἦν ὁ Α πρὸς τὸν Β· καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, ὁ (by) multiplying D, E, thus as D is to E, (so) G (is) to Ζ πρὸς τὸν Η. πάλιν, ἐπεὶ ὁ Α τοὺς Δ, Ε πολλαπλασιάσας H [Prop. 7.17]. But, as D (is) to E, (so) A (is) to B. τοὺς Η, Θ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ε, ὁ Η And thus as A (is) to B, so G (is) to H . And since A, B πρὸς τὸν Θ. ἀλλ᾿ ὡς ὁ Δ πρὸς τὸν Ε, ὁ Α πρὸς τὸν Β. καὶ have made H , K (by) multiplying E, thus as A is to B, ὡς ἄρα ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς τὸν Θ. καὶ ἐπεὶ so H (is) to K. But, as A (is) to B, so F (is) to G, and οἱ Α, Β τὸν Ε πολλαπλασιάσαντες τοὺς Θ, Κ πεποιήκασιν, G to H . And thus as F (is) to G, so G (is) to H , and H ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Θ πρὸς τὸν Κ. ἀλλ᾿ to K. Thus, C, D, E and F , G, H , K are (both continu- ὡς ὁ Α πρὸς τὸν Β, οὕτως ὅ τε Ζ πρὸς τὸν Η καὶ ὁ Η πρὸς ously) proportional in the ratio of A to B. So I say that τὸν Θ. καὶ ὡς ἄρα ὁ Ζ πρὸς τὸν Η, οὕτως ὅ τε Η πρὸς (they are) also the least (sets of numbers continuously τὸν Θ καὶ ὁ Θ πρὸς τὸν Κ· οἱ Γ, Δ, Ε ἄρα καὶ οἱ Ζ, Η, proportional in that ratio). For since A and B are the Θ, Κ ἀνάλογόν εἰσιν ἐν τῷ τοῦ Α πρὸς τὸν Β λόγῳ. λέγω least of those (numbers) having the same ratio as them, δή, ὅτι καὶ ἐλάχιστοι. ἐπεὶ γὰρ οἱ Α, Β ἐλάχιστοί εἰσι τῶν and the least of those (numbers) having the same ratio τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, οἱ δὲ ἐλάχιστοι τῶν τὸν are prime to one another [Prop. 7.22], A and B are thus αὐτὸν λόγον ἐχόντων πρῶτοι πρὸς ἀλλήλους εἰσίν, οἱ Α, Β prime to one another. And A, B have made C, E, respec- ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἑκάτερος μὲν τῶν Α, tively, (by) multiplying themselves, and have made F , K Β ἑαυτὸν πολλαπλασιάσας ἑκάτερον τῶν Γ, Ε πεποίηκεν, by multiplying C, E, respectively. Thus, C, E and F , K ἑκάτερον δὲ τῶν Γ, Ε πολλαπλασιάσας ἑκάτερον τῶν Ζ, Κ are prime to one another [Prop. 7.27]. And if there are πεποίηκεν· οἱ Γ, Ε ἄρα καὶ οἱ Ζ, Κ πρῶτοι πρὸς ἀλλήλους any multitude whatsoever of continuously proportional εἰσίν. ἐὰν δὲ ὦσιν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, οἱ numbers, and the outermost of them are prime to one δὲ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους ὦσιν, ἐλάχιστοί εἰσι another, then the (numbers) are the least of those (num- τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. οἱ Γ, Δ, Ε ἄρα καὶ bers) having the same ratio as them [Prop. 8.1]. Thus, C, οἱ Ζ, Η, Θ, Κ ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων D, E and F , G, H , K are the least of those (continuously τοῖς Α, Β· ὅπερ ἔδει δεῖξαι. proportional sets of numbers) having the same ratio as A and B. (Which is) the very thing it was required to show. 229 STOIQEIWN hþ. ELEMENTS BOOK 8Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν τρεῖς ἀριθμοὶ ἑξῆς So it is clear, from this, that if three continuously pro- ἀνάλογον ἐλάχιστοι ὦσι τῶν τὸν αὐτὸν λόγον ἐχόντων portional numbers are the least of those (numbers) hav- αὐτοῖς, οἱ ἄκρον αὐτῶν τετράγωνοί εἰσιν, ἐὰν δὲ τέσσαρες, ing the same ratio as them then the outermost of them κύβοι. are square, and, if four (numbers), cube.gþ. Proposition 3 ᾿Εὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ἐλάχιστ- If there are any multitude whatsoever of continu- οι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, οἱ ἄκροι αὐτῶν ously proportional numbers (which are) the least of those πρῶτοι πρὸς ἀλλήλους εἰσίν. (numbers) having the same ratio as them then the outer- most of them are prime to one another. Κ Α Β Γ ∆ Λ Μ Ν Ξ Ε Ζ Η Θ O A B C D E F G H K L M N ῎Εστωσαν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ἐλάχιστοι Let A, B, C, D be any multitude whatsoever of con- τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς οἱ Α, Β, Γ, Δ· λέγω, tinuously proportional numbers (which are) the least of ὅτι οἱ ἄκροι αὐτῶν οἱ Α, Δ πρῶτοι πρὸς ἀλλήλους εἰσίν. those (numbers) having the same ratio as them. I say Εἰλήφθωσαν γὰρ δύο μὲν ἀριθμοὶ ἐλάχιστοι ἐν τῷ τῶν that the outermost of them, A and D, are prime to one Α, Β, Γ, Δ λόγῳ οἱ Ε, Ζ, τρεῖς δὲ οἱ Η, Θ, Κ, καὶ ἑξῆς another. ἑνὶ πλείους, ἕως τὸ λαμβανόμενον πλῆθος ἴσον γένηται τῷ For let the two least (numbers) E, F (which are) πλήθει τῶν Α, Β, Γ, Δ. εἰλήφθωσαν καὶ ἔστωσαν οἱ Λ, Μ, in the same ratio as A, B, C, D have been taken Ν, Ξ. [Prop. 7.33]. And the three (least numbers) G, H , K Καὶ ἐπεὶ οἱ Ε, Ζ ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον [Prop. 8.2]. And (so on), successively increasing by one, ἐχόντων αὐτοῖς, πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ until the multitude of (numbers) taken is made equal to ἑκάτερος τῶν Ε, Ζ ἑαυτὸν μὲν πολλαπλασιάσας ἑκάτερον the multitude of A, B, C, D. Let them have been taken, τῶν Η, Κ πεποίηκεν, ἑκάτερον δὲ τῶν Η, Κ πολλα- and let them be L, M , N , O. πλασιάσας ἑκάτερον τῶν Λ, Ξ πεποίηκεν, καὶ οἱ Η, Κ ἄρα And since E and F are the least of those (numbers) καὶ οἱ Λ, Ξ πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ οἱ Α, Β, having the same ratio as them they are prime to one an- Γ, Δ ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, other [Prop. 7.22]. And since E, F have made G, K, re- εἰσὶ δὲ καὶ οἱ Λ, Μ, Ν, Ξ ἐλάχιστοι ἐν τῷ αὐτῷ λόγῳ ὄντες spectively, (by) multiplying themselves [Prop. 8.2 corr.], τοῖς Α, Β, Γ, Δ, καί ἐστιν ἴσον τὸ πλῆθος τῶν Α, Β, Γ, and have made L, O (by) multiplying G, K, respec- Δ τῷ πλήθει τῶν Λ, Μ, Ν, Ξ, ἕκαστος ἄρα τῶν Α, Β, Γ, tively, G, K and L, O are thus also prime to one another Δ ἑκάστῳ τῶν Λ, Μ, Ν, Ξ ἴσος ἐστίν· ἴσος ἄρα ἐστὶν ὁ [Prop. 7.27]. And since A, B, C, D are the least of those μὲν Α τῷ Λ, ὁ δὲ Δ τῷ Ξ. καί εἰσιν οἱ Λ, Ξ πρῶτοι πρὸς (numbers) having the same ratio as them, and L, M , N , ἀλλήλους. καὶ οἱ Α, Δ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν· O are also the least (of those numbers having the same ὅπερ ἔδει δεῖξαι. ratio as them), being in the same ratio as A, B, C, D, and the multitude of A, B, C, D is equal to the multitude of L, M , N , O, thus A, B, C, D are equal to L, M , N , O, respectively. Thus, A is equal to L, and D to O. And L and O are prime to one another. Thus, A and D are also prime to one another. (Which is) the very thing it was 230 STOIQEIWN hþ. ELEMENTS BOOK 8 required to show.dþ. Proposition 4 Λόγων δοθέντων ὁποσωνοῦν ἐν ἐλαχίστοις ἀριθμοῖς For any multitude whatsoever of given ratios, (ex- ἀριθμοὺς εὑρεῖν ἑξῆς ἀνάλογον ἐλαχίστους ἐν τοῖς δοθεῖσι pressed) in the least numbers, to find the least numbers λόγοις. continuously proportional in these given ratios. Λ Α Γ Ε Ν Ξ Μ Ο Β ∆ Ζ Θ Η Κ L A C E B D F N O M P H G K ῎Εστωσαν οἱ δοθέντες λόγοι ἐν ἐλαχίστοις ἀριθμοῖς ὅ Let the given ratios, (expressed) in the least numbers, τε τοῦ Α πρὸς τὸν Β καὶ ὁ τοῦ Γ πρὸς τὸν Δ καὶ ἔτι ὁ be the (ratios) of A to B, and of C to D, and, further, τοῦ Ε πρὸς τὸν Ζ· δεῖ δὴ ἀριθμοὺς εὑρεῖν ἑξῆς ἀνάλογον of E to F . So it is required to find the least numbers ἐλαχίστους ἔν τε τῷ τοῦ Α πρὸς τὸν Β λόγῳ καὶ ἐν τῷ τοῦ continuously proportional in the ratio of A to B, and of Γ πρὸς τὸν Δ καὶ ἔτι τῷ τοῦ Ε πρὸς τὸν Ζ. C to B, and, further, of E to F . Εἰλήφθω γὰρ ὁ ὑπὸ τῶν Β, Γ ἐλάχιστος μετρούμενος For let the least number, G, measured by (both) B and ἀριθμὸς ὁ Η. καὶ ὁσάκις μὲν ὁ Β τὸν Η μετρεῖ, τοσαυτάκις C have be taken [Prop. 7.34]. And as many times as B καὶ ὁ Α τὸν Θ μετρείτω, ὁσάκις δὲ ὁ Γ τὸν Η μετρεῖ, το- measures G, so many times let A also measure H . And as σαυτάκις καὶ ὁ Δ τὸν Κ μετρείτω. ὁ δὲ Ε τὸν Κ ἤτοι μετρεῖ many times as C measures G, so many times let D also ἢ οὐ μετρεῖ. μετρείτω πρότερον. καὶ ὁσάκις ὁ Ε τὸν Κ με- measure K. And E either measures, or does not measure, τρεῖ, τοσαυτάκις καὶ ὁ Ζ τὸν Λ μετρείτω. καὶ ἐπεὶ ἰσάκις ὁ K. Let it, first of all, measure (K). And as many times as Α τὸν Θ μετρεῖ καὶ ὁ Β τὸν Η, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν E measures K, so many times let F also measure L. And Β, οὕτως ὁ Θ πρὸς τὸν Η. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Γ πρὸς since A measures H the same number of times that B also τὸν Δ, οὕτως ὁ Η πρὸς τὸν Κ, καὶ ἔτι ὡς ὁ Ε πρὸς τὸν Ζ, (measures) G, thus as A is to B, so H (is) to G [Def. 7.20, οὕτως ὁ Κ πρὸς τὸν Λ· οἱ Θ, Η, Κ, Λ ἄρα ἑξῆς ἀνάλογόν Prop. 7.13]. And so, for the same (reasons), as C (is) to εἰσιν ἔν τε τῷ τοῦ Α πρὸς τὸν Β καὶ ἐν τῷ τοῦ Γ πρὸς τὸν D, so G (is) to K, and, further, as E (is) to F , so K (is) Δ καὶ ἔτι ἐν τῷ τοῦ Ε πρὸς τὸν Ζ λόγῳ. λέγω δή, ὅτι καὶ to L. Thus, H , G, K, L are continuously proportional in ἐλάχιστοι. εἰ γὰρ μή εἰσιν οἱ Θ, Η, Κ, Λ ἑξῆς ἀνάλογον the ratio of A to B, and of C to D, and, further, of E to ἐλάχιστοι ἔν τε τοῖς τοῦ Α πρὸς τὸν Β καὶ τοῦ Γ πρὸς τὸν F . So I say that (they are) also the least (numbers con- Δ καὶ ἐν τῷ τοῦ Ε πρὸς τὸν Ζ λόγοις, ἔστωσαν οἱ Ν, Ξ, tinuously proportional in these ratios). For if H , G, K, Μ, Ο. καὶ ἐπεί ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Ν πρὸς L are not the least numbers continuously proportional in τὸν Ξ, οἱ δὲ Α, Β ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς the ratios of A to B, and of C to D, and of E to F , let N , τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα O, M , P be (the least such numbers). And since as A is καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε ἡγούμενος to B, so N (is) to O, and A and B are the least (numbers τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον, ὁ Β ἄρα τὸν which have the same ratio as them), and the least (num- Ξ μετρεῖ. διὰ τὰ αὐτὰ δὴ καὶ ὁ Γ τὸν Ξ μετρεῖ· οἱ Β, Γ bers) measure those (numbers) having the same ratio (as ἄρα τὸν Ξ μετροῦσιν· καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν Β, Γ them) an equal number of times, the greater (measur- μετρούμενος τὸν Ξ μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν Β, Γ ing) the greater, and the lesser the lesser—that is to say, μετρεῖται ὁ Η· ὁ Η ἄρα τὸν Ξ μετρεῖ ὁ μείζων τὸν ἐλάσσονα· the leading (measuring) the leading, and the following ὅπερ ἐστὶν ἀδύντατον. οὐκ ἄρα ἔσονταί τινες τῶν Θ, Η, Κ, the following [Prop. 7.20], B thus measures O. So, for Λ ἐλάσσονες ἀριθμοὶ ἑξῆς ἔν τε τῷ τοῦ Α πρὸς τὸν Β καὶ the same (reasons), C also measures O. Thus, B and C τῷ τοῦ Γ πρὸς τὸν Δ καὶ ἔτι τῷ τοῦ Ε πρὸς τὸν Ζ λόγῷ. (both) measure O. Thus, the least number measured by (both) B and C will also measure O [Prop. 7.35]. And G (is) the least number measured by (both) B and C. 231 STOIQEIWN hþ. ELEMENTS BOOK 8 Thus, G measures O, the greater (measuring) the lesser. The very thing is impossible. Thus, there cannot be any numbers less than H , G, K, L (which are) continuously (proportional) in the ratio of A to B, and of C to D, and, further, of E to F . Τ Μ Ξ Ν Ο Ε Γ Α Β ∆ Ζ Θ Η Κ Π Ρ Σ T A C E B D H G K N O M P F Q R S Μὴ μετρείτω δὴ ὁ Ε τὸν Κ, καὶ εἰλήφθω ὑπὸ τῶν Ε, So let E not measure K. And let the least num- Κ ἐλάχιστος μετρούμενος ἀριθμὸς ὁ Μ. καὶ ὁσάκις μὲν ber, M , measured by (both) E and K have been taken ὁ Κ τὸν Μ μετρεῖ, τοσαυτάκις καὶ ἑκάτερος τῶν Θ, Η [Prop. 7.34]. And as many times as K measures M , so ἑκάτερον τῶν Ν, Ξ μετρείτω, ὁσάακις δὲ ὁ Ε τὸν Μ με- many times let H , G also measure N , O, respectively. τρεῖ, τοσαυτάκις καὶ ὁ Ζ τὸν Ο μετρείτω. ἐπεὶ ἰσάκις ὁ Θ And as many times as E measures M , so many times let τὸν Ν μετρεῖ καὶ ὁ Η τὸν Ξ, ἔστιν ἄρα ὡς ὁ Θ πρὸς τὸν F also measure P . Since H measures N the same num- Η, οὕτως ὁ Ν πρὸς τὸν Ξ. ὡς δὲ ὁ Θ πρὸς τὸν Η, οὕτως ber of times as G (measures) O, thus as H is to G, so ὁ Α πρὸς τὸν Β· καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, οὕτως ὁ Ν N (is) to O [Def. 7.20, Prop. 7.13]. And as H (is) to G, πρὸς τὸν Ξ. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Γ πρὸς τὸν Δ, οὕτως so A (is) to B. And thus as A (is) to B, so N (is) to ὁ Ξ πρὸς τὸν Μ. πάλιν, ἐπεὶ ἰσάκις ὁ Ε τὸν Μ μετρεῖ καὶ O. And so, for the same (reasons), as C (is) to D, so O ὁ Ζ τὸν Ο, ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Ζ, οὕτως ὁ Μ πρὸς (is) to M . Again, since E measures M the same num- τὸν Ο· οἱ Ν, Ξ, Μ, Ο ἄρα ἑξῆς ἀνάλογόν εἰσιν ἐν τοῖς τοῦ ber of times as F (measures) P , thus as E is to F , so τε Α πρὸς τὸν Β καὶ τοῦ Γ πρὸς τὸν Δ καὶ ἔτι τοῦ Ε πρὸς M (is) to P [Def. 7.20, Prop. 7.13]. Thus, N , O, M , P τὸν Ζ λόγοις. λέγω δή, ὅτι καὶ ἐλάχιστοι ἐν τοῖς Α Β, Γ are continuously proportional in the ratios of A to B, and Δ, Ε Ζ λόγοις. εἰ γὰρ μή, ἔσονταί τινες τῶν Ν, Ξ, Μ, Ο of C to D, and, further, of E to F . So I say that (they ἐλάσσονες ἀριθμοὶ ἑξῆς ἀνάλογον ἐν τοῖς Α Β, Γ Δ, Ε Ζ are) also the least (numbers) in the ratios of A B, C D, λόγοις. ἔστωσαν οἱ Π, Ρ, Σ, Τ. καὶ ἐπεί ἐστιν ὡς ὁ Π πρὸς E F . For if not, then there will be some numbers less τὸν Ρ, οὕτως ὁ Α πρὸς τὸν Β, οἱ δὲ Α, Β ἐλάχιστοι, οἱ δὲ than N , O, M , P (which are) continuously proportional ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας αὐτοῖς in the ratios of A B, C D, E F . Let them be Q, R, S, ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν T . And since as Q is to R, so A (is) to B, and A and B ἑπόμενον, ὁ Β ἄρα τὸν Ρ μετρεῖ. διὰ τὰ αὐτὰ δὴ καὶ ὁ Γ (are) the least (numbers having the same ratio as them), τὸν Ρ μετρεῖ· οἱ Β, Γ ἄρα τὸν Ρ μετροῦσιν. καὶ ὁ ἐλάχιστος and the least (numbers) measure those (numbers) hav- ἄρα ὑπὸ τῶν Β, Γ μετούμενος τὸν Ρ μετρήσει. ἐλάχιστος ing the same ratio as them an equal number of times, δὲ ὑπὸ τῶν Β, Γ μετρούμενος ἐστιν ὁ Η· ὁ Η ἄρα τὸν Ρ the leading (measuring) the leading, and the following μετρεῖ. καί ἐστιν ὡς ὁ Η πρὸς τὸν Ρ, οὕτως ὁ Κ πρὸς τὸν the following [Prop. 7.20], B thus measures R. So, for Σ· καὶ ὁ Κ ἄρα τὸν Σ μετρεῖ. μετρεῖ δὲ καὶ ὁ Ε τὸν Σ· οἱ Ε, the same (reasons), C also measures R. Thus, B and C Κ ἄρα τὸν Σ μετροῦσιν. καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν Ε, Κ (both) measure R. Thus, the least (number) measured by μετρούμενος τὸν Σ μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν Ε, Κ (both) B and C will also measure R [Prop. 7.35]. And G μετρούμενός ἐστιν ὁ Μ· ὁ Μ ἄρα τὸν Σ μετρεῖ ὁ μείζων τὸν is the least number measured by (both) B and C. Thus, ἐλάσσονα· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἔσονταί τινες τῶν G measures R. And as G is to R, so K (is) to S. Thus, 232 STOIQEIWN hþ. ELEMENTS BOOK 8 Ν, Ξ, Μ, Ο ἐλάσσονες ἀριθμοὶ ἑξῆς ἀνάλογον ἔν τε τοῖς K also measures S [Def. 7.20]. And E also measures τοῦ Α πρὸς τὸν Β καὶ τοῦ Γ πρὸς τὸν Δ καὶ ἔτι τοῦ Ε πρὸς S [Prop. 7.20]. Thus, E and K (both) measure S. Thus, τὸν Ζ λόγοις· οἱ Ν, Ξ, Μ, Ο ἄρα ἑξῆς ἀνάλογον ἐλάχιστοί the least (number) measured by (both) E and K will also εἰσιν ἐν τοῖς Α Β, Γ Δ, Ε Ζ λόγοις· ὅπερ ἔδει δεῖξαι. measure S [Prop. 7.35]. And M is the least (number) measured by (both) E and K. Thus, M measures S, the greater (measuring) the lesser. The very thing is impos- sible. Thus there cannot be any numbers less than N , O, M , P (which are) continuously proportional in the ratios of A to B, and of C to D, and, further, of E to F . Thus, N , O, M , P are the least (numbers) continuously propor- tional in the ratios of A B, C D, E F . (Which is) the very thing it was required to show.eþ. Proposition 5 Οἱ ἐπίπεδοι ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχουσι τὸν Plane numbers have to one another the ratio compoun- συγκείμενον ἐκ τῶν πλευρῶν. ded† out of (the ratios of) their sides. Λ Α Β Γ Ε ∆ Ζ Η Θ Κ L A B C E D F G H K ῎Εστωσαν ἐπίπεδοι ἀριθμοὶ οἱ Α, Β, καὶ τοῦ μὲν Α Let A and B be plane numbers, and let the numbers πλευραὶ ἔστωσαν οἱ Γ, Δ ἀριθμοί, τοῦ δὲ Β οἱ Ε, Ζ· λέγω, C, D be the sides of A, and (the numbers) E, F (the ὅτι ὁ Α πρὸς τὸν Β λόγον ἔχει τὸν συγκείμενον ἐκ τῶν sides) of B. I say that A has to B the ratio compounded πλευρῶν. out of (the ratios of) their sides. Λόγων γὰρ δοθέντων τοῦ τε ὃν ἔχει ὁ Γ πρὸς τὸν Ε καὶ For given the ratios which C has to E, and D (has) to ὁ Δ πρὸς τὸν Ζ εἰλήφθωσαν ἀριθμοὶ ἑξῆς ἐλάχιστοι ἐν τοῖς F , let the least numbers, G, H , K, continuously propor- Γ Ε, Δ Ζ λόγοις, οἱ Η, Θ, Κ, ὥστε εἶναι ὡς μὲν τὸν Γ πρὸς tional in the ratios C E, D F have been taken [Prop. 8.4], τὸν Ε, οὕτως τὸν Η πρὸς τὸν Θ, ὡς δὲ τὸν Δ πρὸς τὸν Ζ, so that as C is to E, so G (is) to H , and as D (is) to F , so οὕτως τὸν Θ πρὸς τὸν Κ. καὶ ὁ Δ τὸν Ε πολλαπλασιάσας H (is) to K. And let D make L (by) multiplying E. τὸν Λ ποιείτω. And since D has made A (by) multiplying C, and has Καὶ ἐπεὶ ὁ Δ τὸν μὲν Γ πολλαπλασιάσας τὸν Α made L (by) multiplying E, thus as C is to E, so A (is) to πεποίηκεν, τὸν δὲ Ε πολλαπλασιάσας τὸν Λ πεποίηκεν, L [Prop. 7.17]. And as C (is) to E, so G (is) to H . And ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Ε, οὕτως ὁ Α πρὸς τὸν Λ. ὡς thus as G (is) to H , so A (is) to L. Again, since E has δὲ ὁ Γ πρὸς τὸν Ε, οὕτως ὁ Η πρὸς τὸν Θ· καὶ ὡς ἄρα ὁ made L (by) multiplying D [Prop. 7.16], but, in fact, has Η πρὸς τὸν Θ, οὕτως ὁ Α πρὸς τὸν Λ. πάλιν, ἐπεὶ ὁ Ε τὸν also made B (by) multiplying F , thus as D is to F , so L Δ πολλαπλασιάσας τὸν Λ πεποίηκεν, ἀλλὰ μὴν καὶ τὸν Ζ (is) to B [Prop. 7.17]. But, as D (is) to F , so H (is) to πολλαπλασιάσας τὸν Β πεποίηκεν, ἔστιν ἄρα ὡς ὁ Δ πρὸς K. And thus as H (is) to K, so L (is) to B. And it was τὸν Ζ, οὕτως ὁ Λ πρὸς τὸν Β. ἀλλ᾿ ὡς ὁ Δ πρὸς τὸν Ζ, also shown that as G (is) to H , so A (is) to L. Thus, via οὕτως ὁ Θ πρὸς τὸν Κ· καὶ ὡς ἄρα ὁ Θ πρὸς τὸν Κ, οὕτως equality, as G is to K, [so] A (is) to B [Prop. 7.14]. And ὁ Λ πρὸς τὸν Β. ἐδείχθη δὲ καὶ ὡς ὁ Η πρὸς τὸν Θ, οὕτως G has to K the ratio compounded out of (the ratios of) ὁ Α πρὸς τὸν Λ· δι᾿ ἴσου ἄρα ἐστὶν ὡς ὁ Η πρὸς τὸν Κ, the sides (of A and B). Thus, A also has to B the ratio [οὕτως] ὁ Α πρὸς τὸν Β. ὁ δὲ Η πρὸς τὸν Κ λόγον ἔχει compounded out of (the ratios of) the sides (of A and B). 233 STOIQEIWN hþ. ELEMENTS BOOK 8 τὸν συγκείμενον ἐκ τῶν πλευρῶν· καὶ ὁ Α ἄρα πρὸς τὸν (Which is) the very thing it was required to show. Β λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν· ὅπερ ἔδει δεῖξαι. † i.e., multiplied. �þ. Proposition 6 ᾿Εὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, ὁ δὲ If there are any multitude whatsoever of continuously πρῶτος τὸν δεύτερον μὴ μετρῇ, οὐδὲ ἄλλος οὐδεὶς οὐδένα proportional numbers, and the first does not measure the μετρήσει. second, then no other (number) will measure any other (number) either. Θ Α Β Γ ∆ Ε Ζ Η E F G H A B C D ῎Εστωσαν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Let A, B, C, D, E be any multitude whatsoever of Γ, Δ, Ε, ὁ δὲ Α τὸν Β μὴ μετρείτω· λέγω, ὅτι οὐδὲ ἄλλος continuously proportional numbers, and let A not mea- οὐδεὶς οὐδένα μετρήσει. sure B. I say that no other (number) will measure any ῞Οτι μὲν οὖν οἱ Α, Β, Γ, Δ, Ε ἑξῆς ἀλλήλους οὐ με- other (number) either. τροῦσιν, φανερόν· οὐδὲ γὰρ ὁ Α τὸν Β μετρεῖ. λέγω Now, (it is) clear that A, B, C, D, E do not succes- δή, ὅτι οὐδὲ ἄλλος οὐδεὶς οὐδένα μετρήσει. εἰ γὰρ δυ- sively measure one another. For A does not even mea- νατόν, μετρείτω ὁ Α τὸν Γ. καὶ ὅσοι εἰσὶν οἱ Α, Β, Γ, sure B. So I say that no other (number) will measure τοσοῦτοι εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν any other (number) either. For, if possible, let A measure λόγον ἐχόντων τοῖς Α, Β, Γ οἱ Ζ, Η, Θ. καὶ ἐπεὶ οἱ Ζ, C. And as many (numbers) as are A, B, C, let so many Η, Θ ἐν τῷ αὐτῷ λόγῳ εἰσὶ τοῖς Α, Β, Γ, καί ἐστιν ἴσον τὸ of the least numbers, F , G, H , have been taken of those πλῆθος τῶν Α, Β, Γ τῷ πλήθει τῶν Ζ, Η, Θ, δι᾿ ἴσου ἄρα (numbers) having the same ratio as A, B, C [Prop. 7.33]. ἐστὶν ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Ζ πρὸς τὸν Θ. καὶ ἐπεί And since F , G, H are in the same ratio as A, B, C, and ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Ζ πρὸς τὸν Η, οὐ μετρεῖ the multitude of A, B, C is equal to the multitude of F , δὲ ὁ Α τὸν Β, οὐ μετρεῖ ἄρα οὐδὲ ὁ Ζ τὸν Η· οὐκ ἄρα μονάς G, H , thus, via equality, as A is to C, so F (is) to H ἐστιν ὁ Ζ· ἡ γὰρ μονὰς πάντα ἀριθμὸν μετρεῖ. καί εἰσιν οἱ [Prop. 7.14]. And since as A is to B, so F (is) to G, Ζ, Θ πρῶτοι πρὸς ἀλλήλους [οὐδὲ ὁ Ζ ἄρα τὸν Θ μετρεῖ]. and A does not measure B, F does not measure G either καί ἐστιν ὡς ὁ Ζ πρὸς τὸν Θ, οὕτως ὁ Α πρὸς τὸν Γ· οὐδὲ [Def. 7.20]. Thus, F is not a unit. For a unit measures ὁ Α ἄρα τὸν Γ μετρεῖ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἄλλος all numbers. And F and H are prime to one another οὐδεὶς οὐδένα μετρήσει· ὅπερ ἔδει δεῖξαι. [Prop. 8.3] [and thus F does not measure H]. And as F is to H , so A (is) to C. And thus A does not measure C either [Def. 7.20]. So, similarly, we can show that no other (number) can measure any other (number) either. (Which is) the very thing it was required to show.zþ. Proposition 7 ᾿Εὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ [ἑξῆς] ἀνάλογον, ὁ δὲ If there are any multitude whatsoever of [continu- πρῶτος τὸν ἔσχατον μετρῇ, καὶ τὸν δεύτερον μετρήσει. ously] proportional numbers, and the first measures the 234 STOIQEIWN hþ. ELEMENTS BOOK 8 last, then (the first) will also measure the second. Α Β Γ ∆ D A B C ῎Εστωσαν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, Let A, B, C, D be any number whatsoever of continu- Δ, ὁ δὲ Α τὸν Δ μετρείτω· λέγω, ὅτι καὶ ὁ Α τὸν Β μετρεῖ. ously proportional numbers. And let A measure D. I say Εἰ γὰρ οὐ μετρεῖ ὁ Α τὸν Β, οὐδὲ ἄλλος οὐδεὶς οὐδένα that A also measures B. μετρήσει· μετρεῖ δὲ ὁ Α τὸν Δ. μετρεῖ ἄρα καὶ ὁ Α τὸν Β· For if A does not measure B then no other (number) ὅπερ ἔδει δεῖξαι. will measure any other (number) either [Prop. 8.6]. But A measures D. Thus, A also measures B. (Which is) the very thing it was required to show.hþ. Proposition 8 ᾿Εὰν δύο ἀριθμῶν μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον If between two numbers there fall (some) numbers in ἐμπίπτωσιν ἀριθμοί, ὅσοι εἰς αὐτοὺς μεταξὺ κατὰ τὸ συ- continued proportion then, as many numbers as fall in νεχὲς ἀνόλογον ἐμπίπτουσιν ἀριθμοί, τοσοῦτοι καὶ εἰς τοὺς between them in continued proportion, so many (num- τὸν αὐτὸν λόγον ἔχοντας [αὐτοῖς] μεταξὺ κατὰ τὸ συνὲχες bers) will also fall in between (any two numbers) having ἀνάλογον ἐμπεσοῦνται the same ratio [as them] in continued proportion. ΕΑ Γ ∆ Β Η Θ Κ Λ Ζ Ν Μ F G K L H A C D B E M N Δύο γὰρ ἀριθμῶν τῶν Α, Β μεταξὺ κατὰ τὸ συνεχὲς For let the numbers, C and D, fall between two num- ἀνάλογον ἐμπιπτέτωσαν ἀριθμοὶ οἱ Γ, Δ, καὶ πεποιήσθω ὡς bers, A and B, in continued proportion, and let it have ὁ Α πρὸς τὸν Β, οὕτως ὁ Ε πρὸς τὸν Ζ· λέγω, ὅτι ὅσοι εἰς been contrived (that) as A (is) to B, so E (is) to F . I say τοὺς Α, Β μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεπτώκασιν that, as many numbers as have fallen in between A and ἀριθμοί, τοσοῦτοι καὶ εἰς τοὺς Ε, Ζ μεταξὺ κατὰ τὸ συνεχὲς B in continued proportion, so many (numbers) will also ἀνάλογον ἐμπεσοῦνται. fall in between E and F in continued proportion. ῞Οσοι γάρ εἰσι τῷ πλήθει οἱ Α, Β, Γ, Δ, τοσοῦτοι For as many as A, B, C, D are in multitude, let so εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον many of the least numbers, G, H , K, L, having the same ἐχόντων τοῖς Α, Γ, Δ, Β οἱ Η, Θ, Κ, Λ· οἱ ἄρα ἄκροι ratio as A, B, C, D, have been taken [Prop. 7.33]. Thus, αὐτῶν οἱ Η, Λ πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ οἱ Α, the outermost of them, G and L, are prime to one another Γ, Δ, Β τοῖς Η, Θ, Κ, Λ ἐν τῷ αὐτῷ λόγῳ εἰσίν, καί ἐστιν [Prop. 8.3]. And since A, B, C, D are in the same ratio ἴσον τὸ πλῆθος τῶν Α, Γ, Δ, Β τῷ πλήθει τῶν Η, Θ, Κ, as G, H , K, L, and the multitude of A, B, C, D is equal Λ, δι᾿ ἴσου ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς to the multitude of G, H , K, L, thus, via equality, as A is τὸν Λ. ὡς δὲ ὁ Α πρὸς τὸν Β, οὕτως ὁ Ε πρὸς τὸν Ζ· καὶ to B, so G (is) to L [Prop. 7.14]. And as A (is) to B, so 235 STOIQEIWN hþ. ELEMENTS BOOK 8 ὡς ἄρα ὁ Η πρὸς τὸν Λ, οὕτως ὁ Ε πρὸς τὸν Ζ. οἱ δὲ Η, Λ E (is) to F . And thus as G (is) to L, so E (is) to F . And πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ G and L (are) prime (to one another). And (numbers) μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων prime (to one another are) also the least (numbers hav- τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε ing the same ratio as them) [Prop. 7.21]. And the least ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον. numbers measure those (numbers) having the same ratio ἰσάκις ἄρα ὁ Η τὸν Ε μετρεῖ καὶ ὁ Λ τὸν Ζ. ὁσάκις δὴ ὁ Η (as them) an equal number of times, the greater (measur- τὸν Ε μετρεῖ, τοσαυτάκις καὶ ἑκάτερος τῶν Θ, Κ ἑκάτερον ing) the greater, and the lesser the lesser—that is to say, τῶν Μ, Ν μετρείτω· οἱ Η, Θ, Κ, Λ ἄρα τοὺς Ε, Μ, Ν, Ζ the leading (measuring) the leading, and the following ἰσάκις μετροῦσιν. οἱ Η, Θ, Κ, Λ ἄρα τοῖς Ε, Μ, Ν, Ζ ἐν τῷ the following [Prop. 7.20]. Thus, G measures E the same αὐτῷ λόγῳ εἰσίν. ἀλλὰ οἱ Η, Θ, Κ, Λ τοῖς Α, Γ, Δ, Β ἐν τῷ number of times as L (measures) F . So as many times as αὐτῷ λόγῳ εἰσίν· καὶ οἱ Α, Γ, Δ, Β ἄρα τοῖς Ε, Μ, Ν, Ζ ἐν G measures E, so many times let H , K also measure M , τῷ αὐτῷ λόγῳ εἰσίν. οἱ δὲ Α, Γ, Δ, Β ἑξῆς ἀνάλογόν εἰσιν· N , respectively. Thus, G, H , K, L measure E, M , N , καὶ οἱ Ε, Μ, Ν, Ζ ἄρα ἑξῆς ἀνάλογόν εἰσιν. ὅσοι ἄρα εἰς F (respectively) an equal number of times. Thus, G, H , τοὺς Α, Β μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεπτώκασιν K, L are in the same ratio as E, M , N , F [Def. 7.20]. ἀριθμοί, τοσοῦτοι καὶ εἰς τοὺς Ε, Ζ μεταξὺ κατὰ τὸ συνεχὲς But, G, H , K, L are in the same ratio as A, C, D, B. ἀνάλογον ἐμπεπτώκασιν ἀριθμοί· ὅπερ ἔδει δεῖξαι. Thus, A, C, D, B are also in the same ratio as E, M , N , F . And A, C, D, B are continuously proportional. Thus, E, M , N , F are also continuously proportional. Thus, as many numbers as have fallen in between A and B in continued proportion, so many numbers have also fallen in between E and F in continued proportion. (Which is) the very thing it was required to show.jþ. Proposition 9 ᾿Εὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ If two numbers are prime to one another and there εἰς αὐτοὺς μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτωσιν fall in between them (some) numbers in continued pro- ἀριθμοί, ὅσοι εἰς αὐτοὺς μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον portion then, as many numbers as fall in between them ἐμπίπτουσιν ἀριθμοί, τοσοῦτοι καὶ ἑκατέρου αὐτῶν καὶ in continued proportion, so many (numbers) will also fall μονάδος μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεσοῦνται. between each of them and a unit in continued proportion. ΘΑ Γ ∆ Β Ε Ζ Η Ο Ξ Ν Μ Λ Κ G A C D B H K L M N O P E F ῎Εστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Let A and B be two numbers (which are) prime to Β, καὶ εἰς αὐτοὺς μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον one another, and let the (numbers) C and D fall in be- ἐμπιπτέτωσαν οἱ Γ, Δ, καὶ ἐκκείσθω ἡ Ε μονάς· λέγω, tween them in continued proportion. And let the unit E ὅτι ὅσοι εἰς τοὺς Α, Β μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον be set out. I say that, as many numbers as have fallen ἐμπεπτώκασιν ἀριθμοί, τοσοῦτοι καὶ ἑκατέρου τῶν Α, in between A and B in continued proportion, so many Β καὶ τῆς μονάδος μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον (numbers) will also fall between each of A and B and ἐμπεσοῦνται. the unit in continued proportion. Εἰλήφθωσαν γὰρ δύο μὲν ἀριθμοὶ ἐλάχιστοι ἐν τῷ τῶν For let the least two numbers, F and G, which are in Α, Γ, Δ, Β λόγῳ ὄντες οἱ Ζ, Η, τρεῖς δὲ οἱ Θ, Κ, Λ, καὶ ἀεὶ the ratio of A, C, D, B, have been taken [Prop. 7.33]. 236 STOIQEIWN hþ. ELEMENTS BOOK 8 ἑξῆς ἑνὶ πλείους, ἕως ἂν ἴσον γένηται τὸ πλῆθος αὐτῶν τῷ And the (least) three (numbers), H , K, L. And so on, πλήθει τῶν Α, Γ, Δ, Β. εἰλήφθωσαν, καὶ ἔστωσαν οἱ Μ, Ν, successively increasing by one, until the multitude of the Ξ, Ο. φανερὸν δή, ὅτι ὁ μὲν Ζ ἑαυτὸν πολλαπλασιάσας τὸν (least numbers taken) is made equal to the multitude of Θ πεποίηκεν, τὸν δὲ Θ πολλαπλασιάσας τὸν Μ πεποίηκεν, A, C, D, B [Prop. 8.2]. Let them have been taken, and καὶ ὁ Η ἑαυτὸν μὲν πολλαπλασιάσας τὸν Λ πεποίηκεν, τὸν let them be M , N , O, P . So (it is) clear that F has made δὲ Λ πολλαπλασιάσας τὸν Ο πεποίηκεν. καὶ ἐπεὶ οἱ Μ, Ν, H (by) multiplying itself, and has made M (by) multi- Ξ, Ο ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Ζ, plying H . And G has made L (by) multiplying itself, and Η, εἰσὶ δὲ καὶ οἱ Α, Γ, Δ, Β ἐλάχιστοι τῶν τὸν αὐτὸν λόγον has made P (by) multiplying L [Prop. 8.2 corr.]. And ἐχόντων τοῖς Ζ, Η, καί ἐστιν ἴσον τὸ πλῆθος τῶν Μ, Ν, Ξ, since M , N , O, P are the least of those (numbers) hav- Ο τῷ πλήθει τῶν Α, Γ, Δ, Β, ἕκαστος ἄρα τῶν Μ, Ν, Ξ, Ο ing the same ratio as F , G, and A, C, D, B are also the ἑκάστῳ τῶν Α, Γ, Δ, Β ἴσος ἐστίν· ἴσος ἄρα ἐστὶν ὁ μὲν Μ least of those (numbers) having the same ratio as F , G τῷ Α, ὁ δὲ Ο τῷ Β. καὶ ἐπεὶ ὁ Ζ ἑαυτὸν πολλαπλασιάσας [Prop. 8.2], and the multitude of M , N , O, P is equal τὸν Θ πεποίηκεν, ὁ Ζ ἄρα τὸν Θ μετρεῖ κατὰ τὰς ἐν τῷ Ζ to the multitude of A, C, D, B, thus M , N , O, P are μονάδας. μετρεῖ δὲ καὶ ἡ Ε μονὰς τὸν Ζ κατὰ τὰς ἐν αὐτῷ equal to A, C, D, B, respectively. Thus, M is equal to μονάδας· ἰσάκις ἄρα ἡ Ε μονὰς τὸν Ζ ἀριθμὸν μετρεῖ καὶ ὁ Ζ A, and P to B. And since F has made H (by) multiply- τὸν Θ. ἔστιν ἄρα ὡς ἡ Ε μονὰς πρὸς τὸν Ζ ἀριθμόν, οὕτως ing itself, F thus measures H according to the units in F ὁ Ζ πρὸς τὸν Θ. πάλιν, ἐπεὶ ὁ Ζ τὸν Θ πολλαπλασιάσας [Def. 7.15]. And the unit E also measures F according to τὸν Μ πεποίηκεν, ὁ Θ ἄρα τὸν Μ μετρεῖ κατὰ τὰς ἐν τῷ Ζ the units in it. Thus, the unit E measures the number F μονάδας. μετρεῖ δὲ καὶ ἡ Ε μονὰς τὸν Ζ ἀριθμὸν κατὰ τὰς ἐν as many times as F (measures) H . Thus, as the unit E is αὐτῷ μονάδας· ἰσάκις ἄρα ἡ Ε μονὰς τὸν Ζ ἀριθμὸν μετρεῖ to the number F , so F (is) to H [Def. 7.20]. Again, since καὶ ὁ Θ τὸν Μ. ἔστιν ἄρα ὡς ἡ Ε μονὰς πρὸς τὸν Ζ ἀριθμόν, F has made M (by) multiplying H , H thus measures M οὕτως ὁ Θ πρὸς τὸν Μ. ἐδείχθη δὲ καὶ ὡς ἡ Ε μονὰς πρὸς according to the units in F [Def. 7.15]. And the unit E τὸν Ζ ἀριθμόν, οὕτως ὁ Ζ πρὸς τὸν Θ· καὶ ὡς ἄρα ἡ Ε μονὰς also measures the number F according to the units in it. πρὸς τὸν Ζ ἀριθμόν, οὕτως ὁ Ζ πρὸς τὸν Θ καὶ ὁ Θ πρὸς Thus, the unit E measures the number F as many times τὸν Μ. ἴσος δὲ ὁ Μ τῷ Α· ἔστιν ἄρα ὡς ἡ Ε μονὰς πρὸς τὸν as H (measures) M . Thus, as the unit E is to the number Ζ ἀριθμόν, οὕτως ὁ Ζ πρὸς τὸν Θ καὶ ὁ Θ πρὸς τὸν Α. διὰ F , so H (is) to M [Prop. 7.20]. And it was shown that as τὰ αὐτὰ δὴ καὶ ὡς ἡ Ε μονὰς πρὸς τὸν Η ἀριθμόν, οὕτως ὁ the unit E (is) to the number F , so F (is) to H . And thus Η πρὸς τὸν Λ καὶ ὁ Λ πρὸς τὸν Β. ὅσοι ἄρα εἰς τοὺς Α, Β as the unit E (is) to the number F , so F (is) to H , and H μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεπτώκασιν ἀριθμοί, (is) to M . And M (is) equal to A. Thus, as the unit E is τοσοῦτοι καὶ ἑκατέρου τῶν Α, Β καὶ μονάδος τῆς Ε μεταξὺ to the number F , so F (is) to H , and H to A. And so, for κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεπτώκασιν ἀριθμοί· ὅπερ ἔδει the same (reasons), as the unit E (is) to the number G, δεῖξαι. so G (is) to L, and L to B. Thus, as many (numbers) as have fallen in between A and B in continued proportion, so many numbers have also fallen between each of A and B and the unit E in continued proportion. (Which is) the very thing it was required to show.iþ. Proposition 10 ᾿Εάν δύο ἀριθμῶν ἑκατέρου καὶ μονάδος μεταξὺ κατὰ If (some) numbers fall between each of two numbers τὸ συνεχὲς ἀνάλογον ἐμπίπτωσιν ἀριθμοί, ὅσοι ἑκατέρου and a unit in continued proportion then, as many (num- αὐτῶν καὶ μονάδος μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον bers) as fall between each of the (two numbers) and the ἐμπίπτουσιν ἀριθμοί, τοσοῦτοι καὶ εἰς αὐτοὺς μεταξὺ κατὰ unit in continued proportion, so many (numbers) will τὸ συνεχὲς ἀνάλογον ἐμπεσοῦνται. also fall in between the (two numbers) themselves in con- Δύο γὰρ ἀριθμῶν τῶν Α, Β καὶ μονάδος τῆς Γ με- tinued proportion. ταξύ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπιπτέτωσαν ἀριθμοὶ οἵ For let the numbers D, E and F , G fall between the τε Δ, Ε καὶ οἱ Ζ, Η· λέγω, ὅτι ὅσοι ἑκατέρου τῶν Α, numbers A and B (respectively) and the unit C in con- Β καὶ μονάδος τῆς Γ μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον tinued proportion. I say that, as many numbers as have ἐμπεπτώκασιν ἀριθμοί, τοσοῦτοι καὶ εἰς τοὺς Α, Β μεταξὺ fallen between each of A and B and the unit C in contin- κατὰ τὸ συνεχὲς ἀνάλογον ἐμπεσοῦνται. ued proportion, so many will also fall in between A and B in continued proportion. 237 STOIQEIWN hþ. ELEMENTS BOOK 8 Ε Γ ∆ Α Θ Κ Λ Β Η Ζ Γ L C C D E A B G F H K ῾Ο Δ γὰρ τὸν Ζ πολλαπλασιάσας τὸν Θ ποιείτω, For let D make H (by) multiplying F . And let D, F ἑκάτερος δὲ τῶν Δ, Ζ τὸν Θ πολλαπλασιάσας ἑκάτερον make K, L, respectively, by multiplying H . τῶν Κ, Λ ποιείτω. As since as the unit C is to the number D, so D (is) to Καὶ ἐπεί ἐστιν ὡς ἡ Γ μονὰς πρὸς τὸν Δ ἀριθμόν, οὕτως E, the unit C thus measures the number D as many times ὁ Δ πρὸς τὸν Ε, ἰσάκις ἄρα ἡ Γ μονὰς τὸν Δ ἀριθμὸν μετρεῖ as D (measures) E [Def. 7.20]. And the unit C measures καὶ ὁ Δ τὸν Ε. ἡ δὲ Γ μονὰς τὸν Δ ἀριθμὸν μετρεῖ κατὰ τὰς the number D according to the units in D. Thus, the ἐν τῷ Δ μονάδας· καὶ ὁ Δ ἄρα ἀριθμὸς τὸν Ε μετρεῖ κατὰ number D also measures E according to the units in D. τὰς ἐν τῷ Δ μονάδας· ὁ Δ ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Thus, D has made E (by) multiplying itself. Again, since Ε πεποίηκεν. πάλιν, ἐπεί ἐστιν ὡς ἡ Γ [μονὰς] πρὸς τὸν as the [unit] C is to the number D, so E (is) to A, the Δ ἀριθμὸν, οὕτως ὁ Ε πρὸς τὸν Α, ἰσάκις ἄρα ἡ Γ μονὰς unit C thus measures the number D as many times as E τὸν Δ ἀριθμὸν μετρεῖ καὶ ὁ Ε τὸν Α. ἡ δὲ Γ μονὰς τὸν (measures) A [Def. 7.20]. And the unit C measures the Δ ἀριθμὸν μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας· καὶ ὁ Ε ἄρα number D according to the units in D. Thus, E also mea- τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας· ὁ Δ ἄρα τὸν Ε sures A according to the units in D. Thus, D has made πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ A (by) multiplying E. And so, for the same (reasons), F μὲν Ζ ἑαυτὸν πολλαπλασιάσας τὸν Η πεποίηκεν, τὸν δὲ Η has made G (by) multiplying itself, and has made B (by) πολλαπλασιάσας τὸν Β πεποίηκεν. καὶ ἐπεὶ ὁ Δ ἑαυτὸν μὲν multiplying G. And since D has made E (by) multiplying πολλαπλασιάσας τὸν Ε πεποίηκεν, τὸν δὲ Ζ πολλαπλασιάσας itself, and has made H (by) multiplying F , thus as D is to τὸν Θ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Ε F , so E (is) to H [Prop 7.17]. And so, for the same rea- πρὸς τὸν Θ. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ sons, as D (is) to F , so H (is) to G [Prop. 7.18]. And thus Θ πρὸς τὸν Η. καὶ ὡς ἄρα ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Θ πρὸς as E (is) to H , so H (is) to G. Again, since D has made τὸν Η. πάλιν, ἐπεὶ ὁ Δ ἑκάτερον τῶν Ε, Θ πολλαπλασιάσας A, K (by) multiplying E, H , respectively, thus as E is to ἑκάτερον τῶν Α, Κ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Θ, H , so A (is) to K [Prop 7.17]. But, as E (is) to H , so D οὕτως ὁ Α πρὸς τὸν Κ. ἀλλ᾿ ὡς ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Δ (is) to F . And thus as D (is) to F , so A (is) to K. Again, πρὸς τὸν Ζ· καὶ ὡς ἄρα ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν since D, F have made K, L, respectively, (by) multiply- Κ. πάλιν, ἐπεὶ ἑκάτερος τῶν Δ, Ζ τὸν Θ πολλαπλασιάσας ing H , thus as D is to F , so K (is) to L [Prop. 7.18]. But, ἑκάτερον τῶν Κ, Λ πεποίηκεν, ἕστιν ἄρα ὡς ὁ Δ πρὸς τὸν as D (is) to F , so A (is) to K. And thus as A (is) to K, Ζ, οὕτως ὁ Κ πρὸς τὸν Λ. ἀλλ᾿ ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ so K (is) to L. Further, since F has made L, B (by) mul- Α πρὸς τὸν Κ· καὶ ὡς ἄρα ὁ Α πρὸς τὸν Κ, οὕτως ὁ Κ πρὸς tiplying H , G, respectively, thus as H is to G, so L (is) to τὸν Λ. ἔτι ἐπεὶ ὁ Ζ ἑκάτερον τῶν Θ, Η πολλαπλασιάσας B [Prop 7.17]. And as H (is) to G, so D (is) to F . And ἑκάτερον τῶν Λ, Β πεποίηκεν, ἔστιν ἄρα ὡς ὁ Θ πρὸς τὸν thus as D (is) to F , so L (is) to B. And it was also shown Η, οὕτως ὁ Λ πρὸς τὸν Β. ὡς δὲ ὁ Θ πρὸς τὸν Η, οὕτως that as D (is) to F , so A (is) to K, and K to L. And thus ὁ Δ πρὸς τὸν Ζ· καὶ ὡς ἄρα ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Λ as A (is) to K, so K (is) to L, and L to B. Thus, A, K, πρὸς τὸν Β. ἐδείχθη δὲ καὶ ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὅ τε L, B are successively in continued proportion. Thus, as Α πρὸς τὸν Κ καὶ ὁ Κ πρὸς τὸν Λ· καὶ ὡς ἄρα ὁ Α πρὸς many numbers as fall between each of A and B and the τὸν Κ, οὕτως ὁ Κ πρὸς τὸν Λ καὶ ὁ Λ πρὸς τὸν Β. οἱ Α, unit C in continued proportion, so many will also fall in Κ, Λ, Β ἄρα κατὰ τὸ συνεχὲς ἑξῆς εἰσιν ἀνάλογον. ὅσοι between A and B in continued proportion. (Which is) ἄρα ἑκατέρου τῶν Α, Β καὶ τῆς Γ μονάδος μεταξὺ κατὰ the very thing it was required to show. τὸ συνεχὲς ἀνάλογον ἐμπίπτουσιν ἀριθμοί, τοσοῦτοι καὶ εἰς τοὺς Α, Β μεταξὺ κατὰ τὸ συνεχὲς ἐμπεσοῦνται· ὅπερ ἔδει 238 STOIQEIWN hþ. ELEMENTS BOOK 8 δεῖξαι. iaþ. Proposition 11 Δύο τετραγώνων ἀριθμῶν εἷς μέσος ἀνάλογόν ἐστιν There exists one number in mean proportion to two ἀριθμός, καὶ ὁ τετράγωνος πρὸς τὸν τετράγωνον δι- (given) square numbers.† And (one) square (number) πλασίονα λόγον ἔχει ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν. has to the (other) square (number) a squared‡ ratio with respect to (that) the side (of the former has) to the side (of the latter). ∆ Α Β Γ Ε D A B C E ῎Εστωσαν τετράγωνοι ἀριθμοὶ οἱ Α, Β, καὶ τοῦ μὲν Α Let A and B be square numbers, and let C be the side πλευρὰ ἔστω ὁ Γ, τοῦ δὲ Β ὁ Δ· λέγω, ὅτι τῶν Α, Β εἷς of A, and D (the side) of B. I say that there exists one μέσος ἀνάλογόν ἐστιν ἀριθμός, καὶ ὁ Α πρὸς τὸν Β δι- number in mean proportion to A and B, and that A has πλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ. to B a squared ratio with respect to (that) C (has) to D. ῾Ο Γ γὰρ τὸν Δ πολλαπλασιάσας τὸν Ε ποιείτω. καὶ For let C make E (by) multiplying D. And since A is ἐπεὶ τετράγωνός ἐστιν ὁ Α, πλευρὰ δὲ αὐτοῦ ἐστιν ὁ Γ, ὁ Γ square, and C is its side, C has thus made A (by) multi- ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ plying itself. And so, for the same (reasons), D has made δὴ καὶ ὁ Δ ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν. ἐπεὶ B (by) multiplying itself. Therefore, since C has made A, οὖν ὁ Γ ἑκάτερον τῶν Γ, Δ πολλαπλασιάσας ἑκάτερον τῶν E (by) multiplying C, D, respectively, thus as C is to D, Α, Ε πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α so A (is) to E [Prop. 7.17]. And so, for the same (rea- πρὸς τὸν Ε. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ sons), as C (is) to D, so E (is) to B [Prop. 7.18]. And Ε πρὸς τὸν Β. καὶ ὡς ἄρα ὁ Α πρὸς τὸν Ε, οὕτως ὁ Ε πρὸς thus as A (is) to E, so E (is) to B. Thus, one number τὸν Β. τῶν Α, Β ἄρα εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός. (namely, E) is in mean proportion to A and B. Λέγω δή, ὅτι καὶ ὁ Α πρὸς τὸν Β διπλασίονα λόγον ἔχει So I say that A also has to B a squared ratio with ἤπερ ὁ Γ πρὸς τὸν Δ. ἐπεὶ γὰρ τρεῖς ἀριθμοὶ ἀνάλογόν εἰσιν respect to (that) C (has) to D. For since A, E, B are οἱ Α, Ε, Β, ὁ Α ἄρα πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ three (continuously) proportional numbers, A thus has ὁ Α πρὸς τὸν Ε. ὡς δὲ ὁ Α πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς to B a squared ratio with respect to (that) A (has) to E τὸν Δ. ὁ Α ἄρα πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ἡ [Def. 5.9]. And as A (is) to E, so C (is) to D. Thus, A has Γ πλευρὰ πρὸς τὴν Δ· ὅπερ ἔδει δεῖξαι. to B a squared ratio with respect to (that) side C (has) to (side) D. (Which is) the very thing it was required to show. † In other words, between two given square numbers there exists a number in continued proportion. ‡ Literally, “double”. ibþ. Proposition 12 Δύο κύβων ἀριθμῶν δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί, There exist two numbers in mean proportion to two καὶ ὁ κύβος πρὸς τὸν κύβον τριπλασίονα λόγον ἔχει ἤπερ ἡ (given) cube numbers.† And (one) cube (number) has to πλευρὰ πρὸς τὴν πλευράν. the (other) cube (number) a cubed‡ ratio with respect ῎Εστωσαν κύβοι ἀριθμοὶ οἱ Α, Β καὶ τοῦ μὲν Α πλευρὰ to (that) the side (of the former has) to the side (of the ἔστω ὁ Γ, τοῦ δὲ Β ὁ Δ· λέγω, ὅτι τῶν Α, Β δύο μέσοι latter). ἀνάλογόν εἰσιν ἀριθμοί, καὶ ὁ Α πρὸς τὸν Β τριπλασίονα Let A and B be cube numbers, and let C be the side λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ. of A, and D (the side) of B. I say that there exist two numbers in mean proportion to A and B, and that A has 239 STOIQEIWN hþ. ELEMENTS BOOK 8 to B a cubed ratio with respect to (that) C (has) to D. Θ Α Β Γ ∆ Ε Ζ Η Κ K A B C D E F G H ῾Ο γὰρ Γ ἑαυτὸν μὲν πολλαπλασιάσας τὸν Ε ποιείτω, For let C make E (by) multiplying itself, and let it τὸν δὲ Δ πολλαπλασιάσας τὸν Ζ ποιείτω, ὁ δὲ Δ ἑαυτὸν make F (by) multiplying D. And let D make G (by) mul- πολλαπλασιάσας τὸν Η ποιείτω, ἑκάτερος δὲ τῶν Γ, Δ τὸν tiplying itself, and let C, D make H , K, respectively, (by) Ζ πολλαπλασιάσας ἑκάτερον τῶν Θ, Κ ποιείτω. multiplying F . Καὶ ἐπεὶ κύβος ἐστὶν ὁ Α, πλευρὰ δὲ αὐτοῦ ὁ Γ, καὶ ὁ And since A is cube, and C (is) its side, and C has Γ ἑαυτὸν μὲν πολλαπλασιάσας τὸν Ε πεποίηκεν, ὁ Γ ἄρα made E (by) multiplying itself, C has thus made E (by) ἑαυτὸν μὲν πολλαπλασιάσας τὸν Ε πεποίηκεν, τὸν δὲ Ε multiplying itself, and has made A (by) multiplying E. πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ And so, for the same (reasons), D has made G (by) mul- Δ ἑαυτὸν μὲν πολλαπλασιάσας τὸν Η πεποίηκεν, τὸν δὲ Η tiplying itself, and has made B (by) multiplying G. And πολλαπλασιάσας τὸν Β πεποίηκεν. καὶ ἐπεὶ ὁ Γ ἑκάτερον since C has made E, F (by) multiplying C, D, respec- τῶν Γ, Δ πολλαπλασιάσας ἑκάτερον τῶν Ε, Ζ πεποίηκεν, tively, thus as C is to D, so E (is) to F [Prop. 7.17]. And ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς τὸν Ζ. so, for the same (reasons), as C (is) to D, so F (is) to G διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ζ πρὸς [Prop. 7.18]. Again, since C has made A, H (by) multi- τὸν Η. πάλιν, ἐπεὶ ὁ Γ ἑκάτερον τῶν Ε, Ζ πολλαπλασιάσας plying E, F , respectively, thus as E is to F , so A (is) to ἑκάτερον τῶν Α, Θ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν H [Prop. 7.17]. And as E (is) to F , so C (is) to D. And Ζ, οὕτως ὁ Α πρὸς τὸν Θ. ὡς δὲ ὁ Ε πρὸς τὸν Ζ, οὕτως ὁ Γ thus as C (is) to D, so A (is) to H . Again, since C, D πρὸς τὸν Δ· καὶ ὡς ἄρα ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α πρὸς τὸν have made H , K, respectively, (by) multiplying F , thus Θ. πάλιν, ἐπεὶ ἑκάτερος τῶν Γ, Δ τὸν Ζ πολλαπλασιάσας as C is to D, so H (is) to K [Prop. 7.18]. Again, since D ἑκάτερον τῶν Θ, Κ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν has made K, B (by) multiplying F , G, respectively, thus Δ, οὕτως ὁ Θ πρὸς τὸν Κ. πάλιν, ἐπεὶ ὁ Δ ἑκάτερον τῶν as F is to G, so K (is) to B [Prop. 7.17]. And as F (is) Ζ, Η πολλαπλασιάσας ἑκάτερον τῶν Κ, Β πεποίηκεν, ἔστιν to G, so C (is) to D. And thus as C (is) to D, so A (is) ἄρα ὡς ὁ Ζ πρὸς τὸν Η, οὕτως ὁ Κ πρὸς τὸν Β. ὡς δὲ ὁ Ζ to H , and H to K, and K to B. Thus, H and K are two πρὸς τὸν Η, οὕτως ὁ Γ πρὸς τὸν Δ· καὶ ὡς ἄρα ὁ Γ πρὸς (numbers) in mean proportion to A and B. τὸν Δ, οὕτως ὅ τε Α πρὸς τὸν Θ καὶ ὁ Θ πρὸς τὸν Κ καὶ So I say that A also has to B a cubed ratio with re- ὁ Κ πρὸς τὸν Β. τῶν Α, Β ἄρα δύο μέσοι ἀνάλογόν εἰσιν spect to (that) C (has) to D. For since A, H , K, B are οἱ Θ, Κ. four (continuously) proportional numbers, A thus has Λέγω δή, ὅτι καὶ ὁ Α πρὸς τὸν Β τριπλασίονα λόγον ἔχει to B a cubed ratio with respect to (that) A (has) to H ἤπερ ὁ Γ πρὸς τὸν Δ. ἐπεὶ γὰρ τέσσαρες ἀριθμοὶ ἀνάλογόν [Def. 5.10]. And as A (is) to H , so C (is) to D. And εἰσιν οἱ Α, Θ, Κ, Β, ὁ Α ἄρα πρὸς τὸν Β τριπλασίονα λόγον [thus] A has to B a cubed ratio with respect to (that) C ἔχει ἤπερ ὁ Α πρὸς τὸν Θ. ὡς δὲ ὁ Α πρὸς τὸν Θ, οὕτως ὁ (has) to D. (Which is) the very thing it was required to Γ πρὸς τὸν Δ· καὶ ὁ Α [ἄρα] πρὸς τὸν Β τριπλασίονα λόγον show. ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ· ὅπερ ἔδει δεῖξαι. † In other words, between two given cube numbers there exist two numbers in continued proportion. ‡ Literally, “triple”. igþ. Proposition 13 ᾿Εὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, καὶ If there are any multitude whatsoever of continuously πολλαπλασιάσας ἕκαστος ἑαυτὸν ποιῇ τινα, οἱ γενόμενοι proportional numbers, and each makes some (number ἐξ αὐτῶν ἀνάλογον ἔσονται· καὶ ἐὰν οἱ ἐξ ἀρχῆς τοὺς by) multiplying itself, then the (numbers) created from γενομένους πολλαπλασιάσαντες ποιῶσί τινας, καὶ αὐτοὶ them will (also) be (continuously) proportional. And if ἀνάλογον ἔσονται [καὶ ἀεὶ περὶ τοὺς ἄκρους τοῦτο συμβαίνει]. the original (numbers) make some (more numbers by) ῎Εστωσαν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, οἱ Α, Β, multiplying the created (numbers) then these will also 240 STOIQEIWN hþ. ELEMENTS BOOK 8 Γ, ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Β πρὸς τὸν Γ, καὶ οἱ be (continuously) proportional [and this always happens Α, Β, Γ ἑαυτοὺς μὲν πολλαπλασιάσαντες τοὺς Δ, Ε, Ζ with the extremes]. ποιείτωσαν, τοὺς δὲ Δ, Ε, Ζ πολλαπλασιάσαντες τοὺς Η, Let A, B, C be any multitude whatsoever of contin- Θ, Κ ποιείτωσαν· λέγω, ὅτι οἵ τε Δ, Ε, Ζ καὶ οἱ Η, Θ, Κ uously proportional numbers, (such that) as A (is) to B, ἑξῆς ἀνάλογον εἰσιν. so B (is) to C. And let A, B, C make D, E, F (by) multiplying themselves, and let them make G, H , K (by) multiplying D, E, F . I say that D, E, F and G, H , K are continuously proportional. Π Α Β Γ ∆ Ε Ζ Κ Θ Η Λ Ξ Μ Ν Ο P A B C D E F G H K L O M N Q ῾Ο μὲν γὰρ Α τὸν Β πολλαπλασιάσας τὸν Λ ποιείτω, For let A make L (by) multiplying B. And let A, B ἑκάτερος δὲ τῶν Α, Β τὸν Λ πολλαπλασιάσας ἑκάτερον τῶν make M , N , respectively, (by) multiplying L. And, again, Μ, Ν ποιείτω. καὶ πάλιν ὁ μὲν Β τὸν Γ πολλαπλασιάσας τὸν let B make O (by) multiplying C. And let B, C make P , Ξ ποιείτω, ἑκάτερος δὲ τῶν Β, Γ τὸν Ξ πολλαπλασιάσας Q, respectively, (by) multplying O. ἑκάτερον τῶν Ο, Π ποιείτω. So, similarly to the above, we can show that D, L, ῾Ομοίως δὴ τοῖς ἐπάνω δεῖξομεν, ὅτι οἱ Δ, Λ, Ε καὶ οἱ E and G, M , N , H are continuously proportional in the Η, Μ, Ν, Θ ἑξῆς εἰσιν ἀνάλογον ἐν τῷ τοῦ Α πρὸς τὸν ratio of A to B, and, further, (that) E, O, F and H , P , Q, Β λόγῳ, καὶ ἔτι οἱ Ε, Ξ, Ζ καὶ οἱ Θ, Ο, Π, Κ ἑξῆς εἰσιν K are continuously proportional in the ratio of B to C. ἀνάλογον ἐν τῷ τοῦ Β πρὸς τὸν Γ λόγῳ. καί ἐστιν ὡς ὁ Α And as A is to B, so B (is) to C. And thus D, L, E are in πρὸς τὸν Β, οὕτως ὁ Β πρὸς τὸν Γ· καὶ οἱ Δ, Λ, Ε ἄρα τοῖς the same ratio as E, O, F , and, further, G, M , N , H (are Ε, Ξ, Ζ ἐν τῷ αὐτῷ λόγῳ εἰσὶ καὶ ἔτι οἱ Η, Μ, Ν, Θ τοῖς in the same ratio) as H , P , Q, K. And the multitude of Θ, Ο, Π, Κ. καί ἐστιν ἴσον τὸ μὲν τῶν Δ, Λ, Ε πλῆθος τῷ D, L, E is equal to the multitude of E, O, F , and that of τῶν Ε, Ξ, Ζ πλήθει, τὸ δὲ τῶν Η, Μ, Ν, Θ τῷ τῶν Θ, Ο, G, M , N , H to that of H , P , Q, K. Thus, via equality, as Π, Κ· δι᾿ ἴσου ἄρα ἐστὶν ὡς μὲν ὁ Δ πρὸς τὸν Ε, οὕτως ὁ D is to E, so E (is) to F , and as G (is) to H , so H (is) to Ε πρὸς τὸν Ζ, ὡς δὲ ὁ Η πρὸς τὸν Θ, οὕτως ὁ Θ πρὸς τὸν K [Prop. 7.14]. (Which is) the very thing it was required Κ· ὅπερ ἔδει δεῖξαι. to show.idþ. Proposition 14 ᾿Εὰν τετράγωνος τετράγωνον μετρῇ, καὶ ἡ πλευρὰ τὴν If a square (number) measures a(nother) square πλευρὰν μετρήσει· καὶ ἐὰν ἡ πλευρὰ τὴν πλευρὰν μετρῇ, καὶ (number) then the side (of the former) will also mea- ὁ τετράγωνος τὸν τετράγωνον μετρήσει. sure the side (of the latter). And if the side (of a square ῎Εστωσαν τετράγωνοι ἀριθμοὶ οἱ Α, Β, πλευραὶ δὲ αὐτῶν number) measures the side (of another square number) ἔστωσαν οἱ Γ, Δ, ὁ δὲ Α τὸν Β μετρείτω· λέγω, ὅτι καὶ ὁ then the (former) square (number) will also measure the Γ τὸν Δ μετρεῖ. (latter) square (number). Let A and B be square numbers, and let C and D be their sides (respectively). And let A measure B. I say that C also measures D. 241 STOIQEIWN hþ. ELEMENTS BOOK 8 ∆ Α Β Ε Γ D A B E C ῾Ο Γ γὰρ τὸν Δ πολλαπλασιάσας τὸν Ε ποιείτω· οἱ Α, Ε, For let C make E (by) multiplying D. Thus, A, E, Β ἄρα ἑξῆς ἀνάλογόν εἰσιν ἐν τῷ τοῦ Γ πρὸς τὸν Δ λόγῳ. B are continuously proportional in the ratio of C to D καὶ ἐπεὶ οἱ Α, Ε, Β ἐξῆς ἀνάλογόν εἰσιν, καὶ μετρεῖ ὁ Α τὸν [Prop. 8.11]. And since A, E, B are continuously pro- Β, μετρεῖ ἄρα καὶ ὁ Α τὸν Ε. καί ἐστιν ὡς ὁ Α πρὸς τὸν Ε, portional, and A measures B, A thus also measures E οὕτως ὁ Γ πρὸς τὸν Δ· μετρεῖ ἄρα καὶ ὁ Γ τὸν Δ. [Prop. 8.7]. And as A is to E, so C (is) to D. Thus, C Πάλιν δὴ ὁ Γ τὸν Δ μετρείτω· λέγω, ὅτι καὶ ὁ Α τὸν Β also measures D [Def. 7.20]. μετρεῖ. So, again, let C measure D. I say that A also measures Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, B. ὅτι οἱ Α, Ε, Β ἑξῆς ἀνάλογόν εἰσιν ἐν τῷ τοῦ Γ πρὸς τὸν Δ For similarly, with the same construction, we can λόγῳ. καὶ ἐπεί ἐστιν ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α πρὸς show that A, E, B are continuously proportional in the τὸν Ε, μετρεῖ δὲ ὁ Γ τὸν Δ, μετρεῖ ἄρα καὶ ὁ Α τὸν Ε. καί ratio of C to D. And since as C is to D, so A (is) to E, εἰσιν οἱ Α, Ε, Β ἑξῆς ἀνάλογον· μετρεῖ ἄρα καὶ ὁ Α τὸν Β. and C measures D, A thus also measures E [Def. 7.20]. ᾿Εὰν ἄρα τετράγωνος τετράγωνον μετρῇ, καὶ ἡ πλευρὰ And A, E, B are continuously proportional. Thus, A also τὴν πλευρὰν μετρήσει· καὶ ἐὰν ἡ πλευρὰ τὴν πλευρὰν μετρῇ, measures B. καὶ ὁ τετράγωνος τὸν τετράγωνον μετρήσει· ὅπερ ἔδει Thus, if a square (number) measures a(nother) square δεῖξαι. (number) then the side (of the former) will also measure the side (of the latter). And if the side (of a square num- ber) measures the side (of another square number) then the (former) square (number) will also measure the (lat- ter) square (number). (Which is) the very thing it was required to show.ieþ. Proposition 15 ᾿Εὰν κύβος ἀριθμὸς κύβον ἀριθμὸν μετρῇ, καὶ ἡ πλευρὰ If a cube number measures a(nother) cube number τὴν πλευρὰν μετρήσει· καὶ ἐὰν ἡ πλευρὰ τὴν πλευρὰν μετρῇ, then the side (of the former) will also measure the side καὶ ὁ κύβος τὸν κύβον μετρήσει. (of the latter). And if the side (of a cube number) mea- Κύβος γὰρ ἀριθμὸς ὁ Α κύβον τὸν Β μετρείτω, καὶ τοῦ sures the side (of another cube number) then the (for- μὲν Α πλευρὰ ἔστω ὁ Γ, τοῦ δὲ Β ὁ Δ· λέγω, ὅτι ὁ Γ τὸν mer) cube (number) will also measure the (latter) cube Δ μετρεῖ. (number). For let the cube number A measure the cube (num- ber) B, and let C be the side of A, and D (the side) of B. I say that C measures D. ΓΑ Β Ε Η Ζ Κ Θ ∆ D A B E G F H K C ῾Ο Γ γὰρ ἑαυτὸν πολλαπλασιάσας τὸν Ε ποιείτω, ὁ δὲ Δ For let C make E (by) multiplying itself. And let 242 STOIQEIWN hþ. ELEMENTS BOOK 8 ἑαυτὸν πολλαπλασιάσας τὸν Η ποιείτω, καὶ ἔτι ὁ Γ τὸν Δ D make G (by) multiplying itself. And, further, [let] C πολλαπλασιάσας τὸν Ζ [ποιείτω], ἑκάτερος δὲ τῶν Γ, Δ τὸν [make] F (by) multiplying D, and let C, D make H , K, Ζ πολλαπλασιάσας ἑκάτερον τῶν Θ, Κ ποιείτω. φανερὸν respectively, (by) multiplying F . So it is clear that E, F , δή, ὅτι οἱ Ε, Ζ, Η καὶ οἱ Α, Θ, Κ, Β ἑξῆς ἀνάλογόν εἰσιν G and A, H , K, B are continuously proportional in the ἐν τῷ τοῦ Γ πρὸς τὸν Δ λόγῳ. καὶ ἐπεὶ οἱ Α, Θ, Κ, Β ἑξῆς ratio of C to D [Prop. 8.12]. And since A, H , K, B are ἀνάλογόν εἰσιν, καὶ μετρεῖ ὁ Α τὸν Β, μετρεῖ ἄρα καὶ τὸν continuously proportional, and A measures B, (A) thus Θ. καί ἐστιν ὡς ὁ Α πρὸς τὸν Θ, οὕτως ὁ Γ πρὸς τὸν Δ· also measures H [Prop. 8.7]. And as A is to H , so C (is) μετρεῖ ἄρα καὶ ὁ Γ τὸν Δ. to D. Thus, C also measures D [Def. 7.20]. Ἀλλὰ δὴ μετρείτω ὁ Γ τὸν Δ· λέγω, ὅτι καὶ ὁ Α τὸν Β And so let C measure D. I say that A will also mea- μετρήσει. sure B. Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δὴ δείξομεν, For similarly, with the same construction, we can ὅτι οἱ Α, Θ, Κ, Β ἑξῆς ἀνάλογόν εἰσιν ἐν τῷ τοῦ Γ πρὸς show that A, H , K, B are continuously proportional in τὸν Δ λόγῳ. καὶ ἐπεὶ ὁ Γ τὸν Δ μετρεῖ, καί ἐστιν ὡς ὁ Γ the ratio of C to D. And since C measures D, and as C is πρὸς τὸν Δ, οὕτως ὁ Α πρὸς τὸν Θ, καὶ ὁ Α ἄρα τὸν Θ to D, so A (is) to H , A thus also measures H [Def. 7.20]. μετρεῖ· ὥστε καὶ τὸν Β μετρεῖ ὁ Α· ὅπερ ἔδει δεῖξαι. Hence, A also measures B. (Which is) the very thing it was required to show.i�þ. Proposition 16 ᾿Εὰν τετράγωνος ἀριθμὸς τετράγωνον ἀριθμὸν μὴ If a square number does not measure a(nother) μετρῇ, οὐδὲ ἡ πλευρὰ τὴν πλευρὰν μετρήσει· κἂν ἡ πλευρὰ square number then the side (of the former) will not τὴν πλευρὰν μὴ μετρῇ, οὐδὲ ὁ τετράγωνος τὸν τετράγωνον measure the side (of the latter) either. And if the side (of μετρήσει. a square number) does not measure the side (of another square number) then the (former) square (number) will not measure the (latter) square (number) either. ∆ Α Β Γ . A B C D ῎Εστωσαν τετρὰγωνοι ἀριθμοὶ οἱ Α, Β, πλευραὶ δὲ αὐτῶν Let A and B be square numbers, and let C and D be ἔστωσαν οἱ Γ, Δ, καὶ μὴ μετρείτω ὁ Α τὸν Β· λὲγω, ὅτι οὐδὲ their sides (respectively). And let A not measure B. I say ὁ Γ τὸν Δ μετρεῖ. that C does not measure D either. Εἰ γὰρ μετρεῖ ὁ Γ τὸν Δ, μετρήσει καὶ ὁ Α τὸν Β. οὐ For if C measures D then A will also measure B μετρεῖ δὲ ὁ Α τὸν Β· οὐδὲ ἄρα ὁ Γ τὸν Δ μετρήσει. [Prop. 8.14]. And A does not measure B. Thus, C will Μὴ μετρείτω [δὴ] πάλιν ὁ Γ τὸν Δ· λέγω, ὅτι οὐδὲ ὁ Α not measure D either. τὸν Β μετρήσει. [So], again, let C not measure D. I say that A will not Εἰ γὰρ μετρεῖ ὁ Α τὸν Β, μετρήσει καὶ ὁ Γ τὸν Δ. οὐ measure B either. μετρεῖ δὲ ὁ Γ τὸν Δ· οὐδ᾿ ἄρα ὁ Α τὸν Β μετρήσει· ὅπερ For if A measures B then C will also measure D ἔδει δεῖξαι. [Prop. 8.14]. And C does not measure D. Thus, A will not measure B either. (Which is) the very thing it was required to show.izþ. Proposition 17 ᾿Εὰν κύβος ἀριθμὸς κύβον ἀριθμὸν μὴ μετρῇ, οὐδὲ ἡ If a cube number does not measure a(nother) cube πλευρὰ τὴν πλευρὰν μετρήσει· κἂν ἡ πλευρὰ τὴν πλευρὰν number then the side (of the former) will not measure the μὴ μετρῇ, οὐδὲ ὁ κύβος τὸν κύβον μετρήσει. side (of the latter) either. And if the side (of a cube num- ber) does not measure the side (of another cube number) then the (former) cube (number) will not measure the (latter) cube (number) either. 243 STOIQEIWN hþ. ELEMENTS BOOK 8 ∆ Α Β Γ . A B C D Κύβος γὰρ ἀριθμὸς ὁ Α κύβον ἀριθμὸν τὸν Β μὴ με- For let the cube number A not measure the cube num- τρείτω, καὶ τοῦ μὲν Α πλευρὰ ἔστω ὁ Γ, τοῦ δὲ Β ὁ Δ· ber B. And let C be the side of A, and D (the side) of B. λέγω, ὅτι ὁ Γ τὸν Δ οὐ μετρήσει. I say that C will not measure D. Εἰ γὰρ μετρεῖ ὁ Γ τὸν Δ, καὶ ὁ Α τὸν Β μετρήσει. οὐ For if C measures D then A will also measure B μετρεῖ δὲ ὁ Α τὸν Β· οὐδ᾿ ἄρα ὁ Γ τὸν Δ μετρεῖ. [Prop. 8.15]. And A does not measure B. Thus, C does Ἀλλὰ δὴ μὴ μετρείτω ὁ Γ τὸν Δ· λέγω, ὅτι οὐδὲ ὁ Α not measure D either. τὸν Β μετρήσει. And so let C not measure D. I say that A will not Εἰ γὰρ ὁ Α τὸν Β μετρεῖ, καὶ ὁ Γ τὸν Δ μετρήσει. οὐ measure B either. μετρεῖ δὲ ὁ Γ τὸν Δ· οὐδ᾿ ἄρα ὁ Α τὸν Β μετρήσει· ὅπερ For if A measures B then C will also measure D ἔδει δεῖξαι. [Prop. 8.15]. And C does not measure D. Thus, A will not measure B either. (Which is) the very thing it was required to show.ihþ. Proposition 18 Δύο ὁμοίων ἐπιπέδων ἀριθμῶν εἷς μέσος ἀνάλογόν ἐστιν There exists one number in mean proportion to two ἀριθμός· καὶ ὁ ἐπίπεδος πρὸς τὸν ἐπίπεδον διπλασίονα λόγον similar plane numbers. And (one) plane (number) has to ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν. the (other) plane (number) a squared† ratio with respect to (that) a corresponding side (of the former has) to a corresponding side (of the latter). ΒΑ Γ ∆ Η Ζ Ε E A C G D B F ῎Εστωσαν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ οἱ Α, Β, καὶ τοῦ Let A and B be two similar plane numbers. And let μὲν Α πλευραὶ ἔστωσαν οἱ Γ, Δ ἀριθμοί, τοῦ δὲ Β οἱ Ε, the numbers C, D be the sides of A, and E, F (the sides) Ζ. καὶ ἐπεὶ ὅμοιοι ἐπίπεδοί εἰσιν οἱ ἀνάλογον ἔχοντες τὰς of B. And since similar numbers are those having pro- πλευράς, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς portional sides [Def. 7.21], thus as C is to D, so E (is) to τὸν Ζ. λέγω οὖν, ὅτι τῶν Α, Β εἷς μέσος ἀνάλογόν ἐστιν F . Therefore, I say that there exists one number in mean ἀριθμός, καὶ ὁ Α πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ὁ proportion to A and B, and that A has to B a squared Γ πρὸς τὸν Ε ἢ ὁ Δ πρὸς τὸν Ζ, τουτέστιν ἤπερ ἡ ὁμόλογος ratio with respect to that C (has) to E, or D to F—that is πλευρὰ πρὸς τὴν ὁμόλογον [πλευράν]. to say, with respect to (that) a corresponding side (has) Καὶ ἐπεί ἐστιν ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς τὸν to a corresponding [side]. Ζ, ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Γ πρὸς τὸν Ε, ὁ Δ πρὸς τὸν Ζ. For since as C is to D, so E (is) to F , thus, alternately, καὶ ἐπεὶ ἐπίπεδός ἐστιν ὁ Α, πλευραὶ δὲ αὐτοῦ οἱ Γ, Δ, ὁ Δ as C is to E, so D (is) to F [Prop. 7.13]. And since A is ἄρα τὸν Γ πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ plane, and C, D its sides, D has thus made A (by) mul- δὴ καὶ ὁ Ε τὸν Ζ πολλαπλασιάσας τὸν Β πεποίηκεν. ὁ Δ tiplying C. And so, for the same (reasons), E has made δὴ τὸν Ε πολλαπλασιάσας τὸν Η ποιείτω. καὶ ἐπεὶ ὁ Δ τὸν B (by) multiplying F . So let D make G (by) multiplying μὲν Γ πολλαπλασιάσας τὸν Α πεποίηκεν, τὸν δὲ Ε πολλα- E. And since D has made A (by) multiplying C, and has πλασιάσας τὸν Η πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Ε, made G (by) multiplying E, thus as C is to E, so A (is) to οὕτως ὁ Α πρὸς τὸν Η. ἀλλ᾿ ὡς ὁ Γ πρὸς τὸν Ε, [οὕτως] G [Prop. 7.17]. But as C (is) to E, [so] D (is) to F . And ὁ Δ πρὸς τὸν Ζ· καὶ ὡς ἄρα ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Α thus as D (is) to F , so A (is) to G. Again, since E has πρὸς τὸν Η. πάλιν, ἐπεὶ ὁ Ε τὸν μὲν Δ πολλαπλασιάσας τὸν made G (by) multiplying D, and has made B (by) multi- Η πεποίηκεν, τὸν δὲ Ζ πολλαπλασιάσας τὸν Β πεποίηκεν, plying F , thus as D is to F , so G (is) to B [Prop. 7.17]. ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Η πρὸς τὸν Β. And it was also shown that as D (is) to F , so A (is) to G. ἐδείχθη δὲ καὶ ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν And thus as A (is) to G, so G (is) to B. Thus, A, G, B are 244 STOIQEIWN hþ. ELEMENTS BOOK 8 Η· καὶ ὡς ἄρα ὁ Α πρὸς τὸν Η, οὕτως ὁ Η πρὸς τὸν Β. οἱ continously proportional. Thus, there exists one number Α, Η, Β ἄρα ἑξῆς ἀνάλογόν εἰσιν. τῶν Α, Β ἄρα εἷς μέσος (namely, G) in mean proportion to A and B. ἀνάλογόν ἐστιν ἀριθμός. So I say that A also has to B a squared ratio with Λέγω δή, ὅτι καὶ ὁ Α πρὸς τὸν Β διπλασίονα λόγον respect to (that) a corresponding side (has) to a corre- ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, sponding side—that is to say, with respect to (that) C τουτέστιν ἤπερ ὁ Γ πρὸς τὸν Ε ἢ ὁ Δ πρὸς τὸν Ζ. ἐπεὶ γὰρ (has) to E, or D to F . For since A, G, B are continuously οἱ Α, Η, Β ἑξῆς ἀνάλογόν εἰσιν, ὁ Α πρὸς τὸν Β διπλασίονα proportional, A has to B a squared ratio with respect to λόγον ἔχει ἤπερ πρὸς τὸν Η. καί ἐστιν ὡς ὁ Α πρὸς τὸν Η, (that A has) to G [Prop. 5.9]. And as A is to G, so C (is) οὕτως ὅ τε Γ πρὸς τὸν Ε καὶ ὁ Δ πρὸς τὸν Ζ. καὶ ὁ Α ἄρα to E, and D to F . And thus A has to B a squared ratio πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Ε ἢ with respect to (that) C (has) to E, or D to F . (Which ὁ Δ πρὸς τὸν Ζ· ὅπερ ἔδει δεῖξαι. is) the very thing it was required to show. † Literally, “double”. ijþ. Proposition 19 Δύο ὁμοίων στερεῶν ἀριθμῶν δύο μέσοι ἀνάλογον Two numbers fall (between) two similar solid num- ἐμπίπτουσιν ἀριθμοί· καὶ ὁ στερεὸς πρὸς τὸν ὅμοιον στερεὸν bers in mean proportion. And a solid (number) has to τριπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν a similar solid (number) a cubed† ratio with respect to ὁμόλογον πλευράν. (that) a corresponding side (has) to a corresponding side. Ξ Α Β Κ Μ Λ Γ ∆ Ε Ζ Η Θ Ν O A B K M L D E F G H C N ῎Εστωσαν δύο ὅμοιοι στερεοὶ οἱ Α, Β, καὶ τοῦ μὲν Α Let A and B be two similar solid numbers, and let πλευραὶ ἔστωσαν οἱ Γ, Δ, Ε, τοῦ δὲ Β οἱ Ζ, Η, Θ. καὶ ἐπεὶ C, D, E be the sides of A, and F , G, H (the sides) of ὅμοιοι στερεοί εἰσιν οἱ ἀνάλογον ἔχοντες τὰς πλευράς, ἔστιν B. And since similar solid (numbers) are those having ἄρα ὡς μὲν ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ζ πρὸς τὸν Η, ὡς δὲ proportional sides [Def. 7.21], thus as C is to D, so F ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Η πρὸς τὸν Θ. λέγω, ὅτι τῶν Α, (is) to G, and as D (is) to E, so G (is) to H . I say that Β δύο μέσοι ἀνάλογόν ἐμπίπτουσιν ἀριθμοί, καὶ ὁ Α πρὸς two numbers fall (between) A and B in mean proportion, τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Ζ καὶ ὁ and (that) A has to B a cubed ratio with respect to (that) Δ πρὸς τὸν Η καὶ ἔτι ὁ Ε πρὸς τὸν Θ. C (has) to F , and D to G, and, further, E to H . ῾Ο Γ γὰρ τὸν Δ πολλαπλασιάσας τὸν Κ ποιείτω, ὁ δὲ For let C make K (by) multiplying D, and let F make Ζ τὸν Η πολλαπλασιάσας τὸν Λ ποιείτω. καὶ ἐπεὶ οἱ Γ, L (by) multiplying G. And since C, D are in the same Δ τοὶς Ζ, Η ἐν τῷ αὐτῷ λόγῳ εἰσίν, καὶ ἐκ μὲν τῶν Γ, ratio as F , G, and K is the (number created) from (mul- Δ ἐστιν ὁ Κ, ἐκ δὲ τῶν Ζ, Η ὁ Λ, οἱ Κ, Λ [ἄρα] ὅμοιοι tiplying) C, D, and L the (number created) from (multi- ἐπίπεδοί εἰσιν ἀριθμοί· τῶν Κ, Λ ἄρα εἷς μέσος ἀνάλογόν plying) F , G, [thus] K and L are similar plane numbers ἐστιν ἀριθμός. ἔστω ὁ Μ. ὁ Μ ἄρα ἐστὶν ὁ ἐκ τῶν Δ, [Def. 7.21]. Thus, there exits one number in mean pro- Ζ, ὡς ἐν τῷ πρὸ τούτου θεωρήματι ἐδείχθη. καὶ ἐπεὶ ὁ portion to K and L [Prop. 8.18]. Let it be M . Thus, M is Δ τὸν μὲν Γ πολλαπλασιάσας τὸν Κ πεποίηκεν, τὸν δὲ Ζ the (number created) from (multiplying) D, F , as shown πολλαπλασιάσας τὸν Μ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς in the theorem before this (one). And since D has made τὸν Ζ, οὕτως ὁ Κ πρὸς τὸν Μ. ἀλλ᾿ ὡς ὁ Κ πρὸς τὸν Μ, K (by) multiplying C, and has made M (by) multiplying ὁ Μ πρὸς τὸν Λ. οἱ Κ, Μ, Λ ἄρα ἑξῆς εἰσιν ἀνάλογον ἐν F , thus as C is to F , so K (is) to M [Prop. 7.17]. But, as 245 STOIQEIWN hþ. ELEMENTS BOOK 8 τῷ τοῦ Γ πρὸς τὸν Ζ λόγῷ. καὶ ἐπεί ἐστιν ὡς ὁ Γ πρὸς τὸν K (is) to M , (so) M (is) to L. Thus, K, M , L are contin- Δ, οὕτως ὁ Ζ πρὸς τὸν Η, ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Γ πρὸς uously proportional in the ratio of C to F . And since as τὸν Ζ, οὕτως ὁ Δ πρὸς τὸν Η. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Δ C is to D, so F (is) to G, thus, alternately, as C is to F , so πρὸς τὸν Η, οὕτως ὁ Ε πρὸς τὸν Θ. οἱ Κ, Μ, Λ ἄρα ἑξῆς D (is) to G [Prop. 7.13]. And so, for the same (reasons), εἰσιν ἀνάλογον ἔν τε τῷ τοῦ Γ πρὸς τὸν Ζ λόγῳ καὶ τῷ as D (is) to G, so E (is) to H . Thus, K, M , L are contin- τοῦ Δ πρὸς τὸν Η καὶ ἔτι τῷ τοῦ Ε πρὸς τὸν Θ. ἑκατερος uously proportional in the ratio of C to F , and of D to G, δὴ τῶν Ε, Θ τὸν Μ πολλαπλασιάσας ἑκάτερον τῶν Ν, Ξ and, further, of E to H . So let E, H make N , O, respec- ποιείτω. καὶ ἐπεὶ στερεός ἐστιν ὁ Α, πλευραὶ δὲ αὐτοῦ εἰσιν tively, (by) multiplying M . And since A is solid, and C, οἱ Γ, Δ, Ε, ὁ Ε ἄρα τὸν ἐκ τῶν Γ, Δ πολλαπλασιάσας τὸν D, E are its sides, E has thus made A (by) multiplying Α πεποίηκεν. ὁ δὲ ἐκ τῶν Γ, Δ ἐστιν ὁ Κ· ὁ Ε ἄρα τὸν Κ the (number created) from (multiplying) C, D. And K πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Θ is the (number created) from (multiplying) C, D. Thus, τὸν Λ πολλαπλασιάσας τὸν Β πεποίηκεν. καὶ ἐπεὶ ὁ Ε τὸν E has made A (by) multiplying K. And so, for the same Κ πολλαπλασιάσας τὸν Α πεποίηκεν, ἀλλὰ μὴν καὶ τὸν Μ (reasons), H has made B (by) multiplying L. And since πολλαπλασιάσας τὸν Ν πεποίηκεν, ἔστιν ἄρα ὡς ὁ Κ πρὸς E has made A (by) multiplying K, but has, in fact, also τὸν Μ, οὕτως ὁ Α πρὸς τὸν Ν. ὡς δὲ ὁ Κ πρὸς τὸν Μ, made N (by) multiplying M , thus as K is to M , so A (is) οὕτως ὅ τε Γ πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν Η καὶ ἔτι ὁ Ε to N [Prop. 7.17]. And as K (is) to M , so C (is) to F , πρὸς τὸν Θ· καὶ ὡς ἄρα ὁ Γ πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν and D to G, and, further, E to H . And thus as C (is) to Η καὶ ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Α πρὸς τὸν Ν. πάλιν, ἐπεὶ F , and D to G, and E to H , so A (is) to N . Again, since ἑκάτερος τῶν Ε, Θ τὸν Μ πολλαπλασιάσας ἑκάτερον τῶν E, H have made N , O, respectively, (by) multiplying M , Ν, Ξ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Ν thus as E is to H , so N (is) to O [Prop. 7.18]. But, as πρὸς τὸν Ξ. ἀλλ᾿ ὡς ὁ Ε πρὸς τὸν Θ, οὕτως ὅ τε Γ πρὸς τὸν E (is) to H , so C (is) to F , and D to G. And thus as C Ζ καὶ ὁ Δ πρὸς τὸν Η· καὶ ὡς ἄρα ὁ Γ πρὸς τὸν Ζ καὶ ὁ Δ (is) to F , and D to G, and E to H , so (is) A to N , and πρὸς τὸν Η καὶ ὁ Ε πρὸς τὸν Θ, οὕτως ὅ τε Α πρὸς τὸν Ν N to O. Again, since H has made O (by) multiplying M , καὶ ὁ Ν πρὸς τὸν Ξ. πάλιν, ἐπεὶ ὁ Θ τὸν Μ πολλαπλασιάσας but has, in fact, also made B (by) multiplying L, thus as τὸν Ξ πεποίηκεν, ἀλλὰ μὴν καὶ τὸν Λ πολλαπλασιάσας τὸν M (is) to L, so O (is) to B [Prop. 7.17]. But, as M (is) Β πεποίηκεν, ἔστιν ἄρα ὡς ὁ Μ πρὸς τὸν Λ, οὕτως ὁ Ξ πρὸς to L, so C (is) to F , and D to G, and E to H . And thus τὸν Β. ἀλλ᾿ ὡς ὁ Μ πρὸς τὸν Λ, οὕτως ὅ τε Γ πρὸς τὸν Ζ as C (is) to F , and D to G, and E to H , so not only (is) καὶ ὁ Δ πρὸς τὸν Η καὶ ὁ Ε πρὸς τὸν Θ. καὶ ὡς ἄρα ὁ Γ O to B, but also A to N , and N to O. Thus, A, N , O, πρὸς τὸν Ζ καὶ ὁ Δ πρὸς τὸν Η καὶ ὁ Ε πρὸς τὸν Θ, οὕτως B are continuously proportional in the aforementioned οὐ μόνον ὁ Ξ πρὸς τὸν Β, ἀλλὰ καὶ ὁ Α πρὸς τὸν Ν καὶ ὁ ratios of the sides. Ν πρὸς τὸν Ξ. οἱ Α, Ν, Ξ, Β ἄρα ἑξῆς εἰσιν ἀνάλογον ἐν So I say that A also has to B a cubed ratio with respect τοῖς εἰρημένοις τῶν πλευρῶν λόγοις. to (that) a corresponding side (has) to a corresponding Λέγω, ὅτι καὶ ὁ Α πρὸς τὸν Β τριπλασίονα λόγον side—that is to say, with respect to (that) the number C ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, (has) to F , or D to G, and, further, E to H . For since A, τουτέστιν ἤπερ ὁ Γ ἀριθμὸς πρὸς τὸν Ζ ἢ ὁ Δ πρὸς τὸν N , O, B are four continuously proportional numbers, A Η καὶ ἔτι ὁ Ε πρὸς τὸν Θ. ἐπεὶ γὰρ τέσσαρες ἀριθμοὶ ἑξῆς thus has to B a cubed ratio with respect to (that) A (has) ἀνάλογόν εἰσιν οἱ Α, Ν, Ξ, Β, ὁ Α ἄρα πρὸς τὸν Β τρι- to N [Def. 5.10]. But, as A (is) to N , so it was shown (is) πλασίονα λόγον ἔχει ἤπερ ὁ Α πρὸς τὸν Ν. ἀλλ᾿ ὡς ὁ Α C to F , and D to G, and, further, E to H . And thus A has πρὸς τὸν Ν, οὕτως ἐδείχθη ὅ τε Γ πρὸς τὸν Ζ καὶ ὁ Δ πρὸς to B a cubed ratio with respect to (that) a corresponding τὸν Η καὶ ἔτι ὁ Ε πρὸς τὸν Θ. καὶ ὁ Α ἄρα πρὸς τὸν Β side (has) to a corresponding side—that is to say, with τριπλασίονα λόγον ἔχει ἤπερ ἡ ομόλογος πλευρὰ πρὸς τὴν respect to (that) the number C (has) to F , and D to G, ὁμόλογον πλευράν, τουτέστιν ἤπερ ὁ Γ ἀριθμὸς πρὸς τὸν and, further, E to H . (Which is) the very thing it was Ζ καὶ ὁ Δ πρὸς τὸν Η καὶ ἔτι ὁ Ε πρὸς τὸν Θ· ὅπερ ἔδει required to show. δεῖξαι. † Literally, “triple”. kþ. Proposition 20 ᾿Εὰν δύο ἀριθμῶν εἷς μέσος ἀνάλογον ἐμπίπτῇ ἀριθμός, If one number falls between two numbers in mean ὅμοιοι ἐπίπεδοι ἔσονται οἱ ἀριθμοί. proportion then the numbers will be similar plane (num- 246 STOIQEIWN hþ. ELEMENTS BOOK 8 Δύο γὰρ ἀριθμῶν τῶν Α, Β εἷς μέσος ἀνάλογον bers). ἐμπιπτέτω ἀριθμὸς ὁ Γ· λέγω, ὅτι οἱ Α, Β ὅμοιοι ἐπίπεδοί For let one number C fall between the two numbers A εἰσιν ἀριθμοί. and B in mean proportion. I say that A and B are similar plane numbers. Η Α Γ Β ∆ Ζ Ε G A C B D F E Εἰλήφθωσαν [γὰρ] ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν [For] let the least numbers, D and E, having the λόγον ἐχόντων τοῖς Α, Γ οἱ Δ, Ε· ἰσάκις ἄρα ὁ Δ τὸν Α same ratio as A and C have been taken [Prop. 7.33]. μετρεῖ καὶ ὁ Ε τὸν Γ. ὁσάκις δὴ ὁ Δ τὸν Α μετρεῖ, τοσαῦται Thus, D measures A as many times as E (measures) C μονάδες ἔστωσαν ἐν τῷ Ζ· ὁ Ζ ἄρα τὸν Δ πολλαπλασιάσας [Prop. 7.20]. So as many times as D measures A, so τὸν Α πεποίηκεν. ὥστε ὁ Α ἐπίπεδός ἐστιν, πλευραὶ δὲ many units let there be in F . Thus, F has made A (by) αὐτοῦ οἱ Δ, Ζ. πάλιν, ἐπεὶ οἱ Δ, Ε ἐλάχιστοί εἰσι τῶν τὸν multiplying D [Def. 7.15]. Hence, A is plane, and D, αὐτὸν λόγον ἐχόντων τοῖς Γ, Β, ἰσάκις ἄρα ὁ Δ τὸν Γ με- F (are) its sides. Again, since D and E are the least τρεῖ καὶ ὁ Ε τὸν Β. ὁσάκις δὴ ὁ Ε τὸν Β μετρεῖ, τοσαῦται of those (numbers) having the same ratio as C and B, μονάδες ἔστωσαν ἐν τῷ Η. ὁ Ε ἄρα τὸν Β μετρεῖ κατὰ τὰς D thus measures C as many times as E (measures) B ἐν τῷ Η μονάδας· ὁ Η ἄρα τὸν Ε πολλαπλασιάσας τὸν Β [Prop. 7.20]. So as many times as E measures B, so πεποίηκεν. ὁ Β ἄρα ἐπίπεδος ἐστι, πλευραὶ δὲ αὐτοῦ εἰσιν many units let there be in G. Thus, E measures B ac- οἱ Ε, Η. οἱ Α, Β ἄρα ἐπίπεδοί εἰσιν ἀριθμοί. λέγω δή, ὅτι cording to the units in G. Thus, G has made B (by) mul- καὶ ὅμοιοι. ἐπεὶ γὰρ ὁ Ζ τὸν μὲν Δ πολλαπλασιάσας τὸν tiplying E [Def. 7.15]. Thus, B is plane, and E, G are Α πεποίηκεν, τὸν δὲ Ε πολλαπλασιάσας τὸν Γ πεποίηκεν, its sides. Thus, A and B are (both) plane numbers. So I ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Α πρὸς τὸν Γ, say that (they are) also similar. For since F has made A τουτέστιν ὁ Γ πρὸς τὸν Β. πάλιν, ἐπεὶ ὁ Ε ἑκάτερον τῶν Ζ, (by) multiplying D, and has made C (by) multiplying E, Η πολλαπλασιάσας τοὺς Γ, Β πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ζ thus as D is to E, so A (is) to C—that is to say, C to B πρὸς τὸν Η, οὕτως ὁ Γ πρὸς τὸν Β. ὡς δὲ ὁ Γ πρὸς τὸν Β, [Prop. 7.17].† Again, since E has made C, B (by) multi- οὕτως ὁ Δ πρὸς τὸν Ε· καὶ ὡς ἄρα ὁ Δ πρὸς τὸν Ε, οὕτως plying F , G, respectively, thus as F is to G, so C (is) to ὁ Ζ πρὸς τὸν Η· καὶ ἐναλλὰξ ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ B [Prop. 7.17]. And as C (is) to B, so D (is) to E. And Ε πρὸς τὸν Η. οἱ Α, Β ἄρα ὅμοιοι ἐπίπεδοι ἀριθμοί εἰσιν· αἱ thus as D (is) to E, so F (is) to G. And, alternately, as D γὰρ πλευραὶ αὐτῶν ἀνάλογόν εἰσιν· ὅπερ ἔδει δεῖξαι. (is) to F , so E (is) to G [Prop. 7.13]. Thus, A and B are similar plane numbers. For their sides are proportional [Def. 7.21]. (Which is) the very thing it was required to show. † This part of the proof is defective, since it is not demonstrated that F × E = C. Furthermore, it is not necessary to show that D : E :: A : C, because this is true by hypothesis.kaþ. Proposition 21 ᾿Εὰν δύο ἀριθμῶν δύο μέσοι ἀνάλογον ἐμπίπτωσιν If two numbers fall between two numbers in mean ἀριθμοί, ὅμοιοι στερεοί εἰσιν οἱ ἀριθμοί. proportion then the (latter) are similar solid (numbers). Δύο γὰρ ἀριθμῶν τῶν Α, Β δύο μέσοι ἀνάλογον For let the two numbers C and D fall between the two ἐμπιπτέτωσαν ἀριθμοὶ οἱ Γ, Δ· λέγω, ὅτι οἱ Α, Β ὅμοιοι numbers A and B in mean proportion. I say that A and στερεοί εἰσιν. B are similar solid (numbers). Εἰλήφθωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν For let the three least numbers E, F , G having the λόγον ἐχόντων τοῖς Α, Γ, Δ τρεῖς οἱ Ε, Ζ, Η· οἱ ἄρα ἄκροι same ratio as A, C, D have been taken [Prop. 8.2]. Thus, αὐτῶν οἱ Ε, Η πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ τῶν the outermost of them, E and G, are prime to one an- Ε, Η εἷς μέσος ἀνάλογον ἐμπέπτωκεν ἀριθμὸς ὁ Ζ, οἱ Ε, other [Prop. 8.3]. And since one number, F , has fallen Η ἄρα ἀριθμοὶ ὅμοιοι ἐπίπεδοί εἰσιν. ἔστωσαν οὖν τοῦ μὲν (between) E and G in mean proportion, E and G are 247 STOIQEIWN hþ. ELEMENTS BOOK 8 Ε πλευραὶ οἱ Θ, Κ, τοῦ δὲ Η οἱ Λ, Μ. φανερὸν ἄρα ἐστὶν thus similar plane numbers [Prop. 8.20]. Therefore, let ἐκ τοῦ πρὸ τούτου, ὅτι οἱ Ε, Ζ, Η ἑξῆς εἰσιν ἀνάλογον ἔν H , K be the sides of E, and L, M (the sides) of G. Thus, τε τῷ τοῦ Θ πρὸς τὸν Λ λόγῳ καὶ τῷ τοῦ Κ πρὸς τὸν Μ. it is clear from the (proposition) before this (one) that E, καὶ ἐπεὶ οἱ Ε, Ζ, Η ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον F , G are continuously proportional in the ratio of H to ἐχόντων τοῖς Α, Γ, Δ, καί ἐστιν ἴσον τὸ πλῆθος τῶν Ε, L, and of K to M . And since E, F , G are the least (num- Ζ, Η τῷ πλήθει τῶν Α, Γ, Δ, δι᾿ ἴσου ἄρα ἐστὶν ὡς ὁ Ε bers) having the same ratio as A, C, D, and the multitude πρὸς τὸν Η, οὕτως ὁ Α πρὸς τὸν Δ. οἱ δὲ Ε, Η πρῶτοι, οἱ of E, F , G is equal to the multitude of A, C, D, thus, δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν via equality, as E is to G, so A (is) to D [Prop. 7.14]. αὐτὸν λόγον ἔχοντας αὐτοῖς ἰσάκις ὅ τε μείζων τὸν μείζονα And E and G (are) prime (to one another), and prime καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε ἡγούμενος (numbers) are also the least (of those numbers having τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον· ἰσάκις ἄρα the same ratio as them) [Prop. 7.21], and the least (num- ὁ Ε τὸν Α μετρεῖ καὶ ὁ Η τὸν Δ. ὁσάκις δὴ ὁ Ε τὸν Α bers) measure those (numbers) having the same ratio as μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ν. ὁ Ν ἄρα τὸν Ε them an equal number of times, the greater (measuring) πολλαπλασιάσας τὸν Α πεποίηκεν. ὁ δὲ Ε ἐστιν ὁ ἐκ τῶν the greater, and the lesser the lesser—that is to say, the Θ, Κ· ὁ Ν ἄρα τὸν ἐκ τῶν Θ, Κ πολλαπλασιάσας τὸν Α leading (measuring) the leading, and the following the πεποίηκεν. στερεὸς ἄρα ἐστὶν ὁ Α, πλευραὶ δὲ αὐτοῦ εἰσιν following [Prop. 7.20]. Thus, E measures A the same οἱ Θ, Κ, Ν. πάλιν, ἐπεὶ οἱ Ε, Ζ, Η ἐλάχιστοί εἰσι τῶν τὸν number of times as G (measures) D. So as many times as αὐτὸν λόγον ἐχόντων τοῖς Γ, Δ, Β, ἰσάκις ἄρα ὁ Ε τὸν Γ E measures A, so many units let there be in N . Thus, N μετρεῖ καὶ ὁ Η τὸν Β. ὁσάκις δὴ ὁ Ε τὸν Γ μετρεῖ, τοσαῦται has made A (by) multiplying E [Def. 7.15]. And E is the μονάδες ἔστωσαν ἐν τῷ Ξ. ὁ Η ἄρα τὸν Β μετρεῖ κατὰ τὰς (number created) from (multiplying) H and K. Thus, N ἐν τῷ Ξ μονάδας· ὁ Ξ ἄρα τὸν Η πολλαπλασιάσας τὸν Β has made A (by) multiplying the (number created) from πεποίηκεν. ὁ δὲ Η ἐστιν ὁ ἐκ τῶν Λ, Μ· ὁ Ξ ἄρα τὸν ἐκ (multiplying) H and K. Thus, A is solid, and its sides are τῶν Λ, Μ πολλαπλασιάσας τὸν Β πεποίηκεν. στερεὸς ἄρα H , K, N . Again, since E, F , G are the least (numbers) ἐστὶν ὁ Β, πλευραὶ δὲ αὐτοῦ εἰσιν οἱ Λ, Μ, Ξ· οἱ Α, Β ἄρα having the same ratio as C, D, B, thus E measures C the στερεοί εἰσιν. same number of times as G (measures) B [Prop. 7.20]. So as many times as E measures C, so many units let there be in O. Thus, G measures B according to the units in O. Thus, O has made B (by) multiplying G. And G is the (number created) from (multiplying) L and M . Thus, O has made B (by) multiplying the (number cre- ated) from (multiplying) L and M . Thus, B is solid, and its sides are L, M , O. Thus, A and B are (both) solid. Ξ Α Γ ∆ Β Ε Ζ Η Θ Κ Ν Λ Μ HA C D B E F G L M O N K Λέγω [δή], ὅτι καὶ ὅμοιοι. ἐπεὶ γὰρ οἱ Ν, Ξ τὸν Ε πολ- [So] I say that (they are) also similar. For since N , O λαπλασιάσαντες τοὺς Α, Γ πεποιήκασιν, ἔστιν ἄρα ὡς ὁ Ν have made A, C (by) multiplying E, thus as N is to O, so πρὸς τὸν Ξ, ὁ Α πρὸς τὸν Γ, τουτέστιν ὁ Ε πρὸς τὸν Ζ. A (is) to C—that is to say, E to F [Prop. 7.18]. But, as ἀλλ᾿ ὡς ὁ Ε πρὸς τὸν Ζ, ὁ Θ πρὸς τὸν Λ καὶ ὁ Κ πρὸς τὸν E (is) to F , so H (is) to L, and K to M . And thus as H Μ· καὶ ὡς ἄρα ὁ Θ πρὸς τὸν Λ, οὕτως ὁ Κ πρὸς τὸν Μ καὶ (is) to L, so K (is) to M , and N to O. And H , K, N are ὁ Ν πρὸς τὸν Ξ. καί εἰσιν οἱ μὲν Θ, Κ, Ν πλευραὶ τοῦ Α, the sides of A, and L, M , O the sides of B. Thus, A and 248 STOIQEIWN hþ. ELEMENTS BOOK 8 οἱ δὲ Ξ, Λ, Μ πλευραὶ τοῦ Β. οἱ Α, Β ἄρα ἀριθμοὶ ὅμοιοι B are similar solid numbers [Def. 7.21]. (Which is) the στερεοί εἰσιν· ὅπερ ἔδει δεῖξαι. very thing it was required to show. † The Greek text has “O, L, M”, which is obviously a mistake.kbþ. Proposition 22 ᾿Εὰν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ δὲ πρῶτος If three numbers are continuously proportional, and τετράγωνος ᾖ, καὶ ὁ τρίτος τετράγωνος ἔσται. the first is square, then the third will also be square. Γ Α Β C A B ῎Εστωσαν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, ὁ δὲ Let A, B, C be three continuously proportional num- πρῶτος ὁ Α τετράγωνος ἔστω· λέγω, ὅτι καὶ ὁ τρίτος ὁ Γ bers, and let the first A be square. I say that the third C τετράγωνός ἐστιν. is also square. ᾿Επεὶ γὰρ τῶν Α, Γ εἷς μέσος ἀνάλογόν ἐστιν ἀριθμὸς For since one number, B, is in mean proportion to ὁ Β, οἱ Α, Γ ἄρα ὅμοιοι ἐπίπεδοί εἰσιν. τετράγωνος δὲ ὁ Α· A and C, A and C are thus similar plane (numbers) τετράγωνος ἄρα καὶ ὁ Γ· ὅπερ ἔδει δεῖξαι. [Prop. 8.20]. And A is square. Thus, C is also square [Def. 7.21]. (Which is) the very thing it was required to show.kgþ. Proposition 23 ᾿Εὰν τέσσαρες ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ δὲ πρῶτος If four numbers are continuously proportional, and κύβος ᾖ, καὶ ὁ τέταρτος κύβος ἔσται. the first is cube, then the fourth will also be cube. ∆ Α Β Γ . A B C D ῎Εστωσαν τέσσαρες ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, Let A, B, C, D be four continuously proportional Δ, ὁ δὲ Α κύβος ἔστω· λέγω, ὅτι καὶ ὁ Δ κύβος ἐστίν. numbers, and let A be cube. I say that D is also cube. ᾿Επεὶ γὰρ τῶν Α, Δ δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοὶ For since two numbers, B and C, are in mean propor- οἱ Β, Γ, οἱ Α, Δ ἄρα ὅμοιοί εἰσι στερεοὶ ἀριθμοί. κύβος δὲ tion to A and D, A and D are thus similar solid numbers ὁ Α· κύβος ἄρα καὶ ὁ Δ· ὅπερ ἔδει δεῖξαι. [Prop. 8.21]. And A (is) cube. Thus, D (is) also cube [Def. 7.21]. (Which is) the very thing it was required to show.kdþ. Proposition 24 ᾿Εὰν δύο ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχωσιν, ὃν If two numbers have to one another the ratio which τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, ὁ δὲ a square number (has) to a(nother) square number, and πρῶτος τετράγωνος ᾖ, καὶ ὁ δεύτερος τετράγωνος ἔσται. the first is square, then the second will also be square. ∆ Α Β Γ D A B C Δύο γὰρ ἀριθμοὶ οἱ Α, Β πρὸς ἀλλήλους λόγον For let two numbers, A and B, have to one another 249 STOIQEIWN hþ. ELEMENTS BOOK 8 ἐχέτωσαν, ὃν τετράγωνος ἀριθμὸς ὁ Γ πρὸς τετράγωνον the ratio which the square number C (has) to the square ἀριθμὸν τὸν Δ, ὁ δὲ Α τετράγωνος ἔστω· λέγω, ὅτι καὶ ὁ number D. And let A be square. I say that B is also Β τετράγωνός ἐστιν. square. ᾿Επεὶ γὰρ οἱ Γ, Δ τετράγωνοί εἰσιν, οἱ Γ, Δ ἄρα ὅμοιοι For since C and D are square, C and D are thus sim- ἐπίπεδοί εἰσιν. τῶν Γ, Δ ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει ilar plane (numbers). Thus, one number falls (between) ἀριθμός. καί ἐστιν ὡς ὁ Γ πρὸς τὸν Δ, ὁ Α πρὸς τὸν Β· C and D in mean proportion [Prop. 8.18]. And as C is καὶ τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. καί to D, (so) A (is) to B. Thus, one number also falls (be- ἐστιν ὁ Α τετράγωνος· καὶ ὁ Β ἄρα τετράγωνός ἐστιν· ὅπερ tween) A and B in mean proportion [Prop. 8.8]. And A ἔδει δεῖξαι. is square. Thus, B is also square [Prop. 8.22]. (Which is) the very thing it was required to show.keþ. Proposition 25 ᾿Εὰν δύο ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχωσιν, ὃν If two numbers have to one another the ratio which κύβος ἀριθμὸς πρὸς κύβον ἀριθμόν, ὁ δὲ πρῶτος κύβος ᾖ, a cube number (has) to a(nother) cube number, and the καὶ ὁ δεύτερος κύβος ἔσται. first is cube, then the second will also be cube. ∆ Α Ε Ζ Β Γ D A E F B C Δύο γὰρ ἀριθμοὶ οἱ Α, Β πρὸς ἀλλήλους λόγον For let two numbers, A and B, have to one another ἐχέτωσαν, ὃν κύβος ἀριθμὸς ὁ Γ πρὸς κύβον ἀριθμὸν τὸν the ratio which the cube number C (has) to the cube Δ, κύβος δὲ ἔστω ὁ Α· λέγω [δή], ὅτι καὶ ὁ Β κύβος ἐστίν. number D. And let A be cube. [So] I say that B is also ᾿Επεὶ γὰρ οἱ Γ, Δ κύβοι εἰσίν, οἱ Γ, Δ ὅμοιοι στε- cube. ρεοί εἰσιν· τῶν Γ, Δ ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν For since C and D are cube (numbers), C and D are ἀριθμοί. ὅσοι δὲ εἰς τοὺς Γ, Δ μεταξὺ κατὰ τὸ συνεχὲς (thus) similar solid (numbers). Thus, two numbers fall ἀνάλογον ἐμπίπτουσιν, τοσοῦτοι καὶ εἰς τοὺς τὸν αὐτὸν (between) C and D in mean proportion [Prop. 8.19]. λόγον ἔχοντας αὐτοῖς· ὥστε καὶ τῶν Α, Β δύο μέσοι And as many (numbers) as fall in between C and D in ἀνάλογον ἐμπίπτουσιν ἀριθμοί. ἐμπιπτέτωσαν οἱ Ε, Ζ. ἐπεὶ continued proportion, so many also (fall) in (between) οὖν τέσσαρες ἀριθμοὶ οἱ Α, Ε, Ζ, Β ἑξῆς ἀνάλογόν εἰσιν, those (numbers) having the same ratio as them (in con- καί ἐστι κύβος ὁ Α, κύβος ἄρα καὶ ὁ Β· ὅπερ ἔδει δεῖξαι. tinued proportion) [Prop. 8.8]. And hence two numbers fall (between) A and B in mean proportion. Let E and F (so) fall. Therefore, since the four numbers A, E, F , B are continuously proportional, and A is cube, B (is) thus also cube [Prop. 8.23]. (Which is) the very thing it was required to show.k�þ. Proposition 26 Οἱ ὅμοιοι ἐπίπεδοι ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχου- Similar plane numbers have to one another the ratio σιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. which (some) square number (has) to a(nother) square number. ∆Α Γ Β Ζ Ε F A C B D E ῎Εστωσαν ὅμοιοι ἐπίπεδοι ἀριθμοὶ οἱ Α, Β· λέγω, ὅτι Let A and B be similar plane numbers. I say that A ὁ Α πρὸς τὸν Β λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς has to B the ratio which (some) square number (has) to 250 STOIQEIWN hþ. ELEMENTS BOOK 8 τετράγωνον ἀριθμόν. a(nother) square number. ᾿Επεὶ γὰρ οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν, τῶν Α, Β ἄρα For since A and B are similar plane numbers, one εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. ἐμπιπτέτω καὶ ἔστω ὁ number thus falls (between) A and B in mean propor- Γ, καὶ εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον tion [Prop. 8.18]. Let it (so) fall, and let it be C. And ἐχόντων τοῖς Α, Γ, Β οἱ Δ, Ε, Ζ· οἱ ἄρα ἄκροι αὐτῶν οἱ let the least numbers, D, E, F , having the same ratio Δ, Ζ τετράγωνοί εἰσιν. καὶ ἐπεί ἐστιν ὡς ὁ Δ πρὸς τὸν Ζ, as A, C, B have been taken [Prop. 8.2]. The outermost οὕτως ὁ Α πρὸς τὸν Β, καί εἰσιν οἱ Δ, Ζ τετράγωνοι, ὁ Α of them, D and F , are thus square [Prop. 8.2 corr.]. And ἄρα πρὸς τὸν Β λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς since as D is to F , so A (is) to B, and D and F are square, τετράγωνον ἀριθμόν· ὅπερ ἔδει δεῖξαι. A thus has to B the ratio which (some) square number (has) to a(nother) square number. (Which is) the very thing it was required to show.kzþ. Proposition 27 Οἱ ὅμοιοι στερεοὶ ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχου- Similar solid numbers have to one another the ratio σιν, ὃν κύβος ἀριθμὸς πρὸς κύβον ἀριθμόν. which (some) cube number (has) to a(nother) cube num- ber. Θ Α Γ ∆ Β Ε Ζ Η H A C D B E F G ῎Εστωσαν ὅμοιοι στερεοὶ ἀριθμοὶ οἱ Α, Β· λέγω, ὅτι ὁ Let A and B be similar solid numbers. I say that A Α πρὸς τὸν Β λόγον ἔχει, ὃν κύβος ἀριθμὸς πρὸς κύβον has to B the ratio which (some) cube number (has) to ἀριθμόν. a(nother) cube number. ᾿Επεὶ γὰρ οἱ Α, Β ὅμοιοι στερεοί εἰσιν, τῶν Α, Β ἄρα δύο For since A and B are similar solid (numbers), two μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. ἐμπιπτέτωσαν οἱ Γ, numbers thus fall (between) A and B in mean proportion Δ, καὶ εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον [Prop. 8.19]. Let C and D have (so) fallen. And let the ἐχόντων τοῖς Α, Γ, Δ, Β ἴσοι αὐτοῖς τὸ πλῆθος οἱ Ε, Ζ, Η, least numbers, E, F , G, H , having the same ratio as A, Θ· οἱ ἄρα ἄκροι αὐτῶν οἱ Ε, Θ κύβοι εἰσίν. καί ἐστιν ὡς ὁ C, D, B, (and) equal in multitude to them, have been Ε πρὸς τὸν Θ, οὕτως ὁ Α πρὸς τὸν Β· καὶ ὁ Α ἄρα πρὸς taken [Prop. 8.2]. Thus, the outermost of them, E and τὸν Β λόγον ἔχει, ὃν κύβος ἀριθμὸς πρὸς κύβον ἀριθμόν· H , are cube [Prop. 8.2 corr.]. And as E is to H , so A (is) ὅπερ ἔδει δεῖξαι. to B. And thus A has to B the ratio which (some) cube number (has) to a(nother) cube number. (Which is) the very thing it was required to show. 251 252 ELEMENTS BOOK 9 Applications of Number Theory† †The propositions contained in Books 7–9 are generally attributed to the school of Pythagoras. 253 STOIQEIWN jþ. ELEMENTS BOOK 9aþ. Proposition 1 ᾿Εὰν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ πολλαπλασιάσαντες If two similar plane numbers make some (number by) ἀλλήλους ποιῶσί τινα, ὁ γενόμενος τετράγωνος ἔσται. multiplying one another then the created (number) will be square. ∆ Α Β Γ D A B C ῎Εστωσαν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ οἱ Α, Β, καὶ ὁ Let A and B be two similar plane numbers, and let A Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ make C (by) multiplying B. I say that C is square. τετράγωνός ἐστιν. For let A make D (by) multiplying itself. D is thus ῾Ο γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω. ὁ Δ square. Therefore, since A has made D (by) multiply- ἄρα τετράγωνός ἐστιν. ἐπεὶ οὖν ὁ Α ἑαυτὸν μὲν πολλα- ing itself, and has made C (by) multiplying B, thus as A πλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν is to B, so D (is) to C [Prop. 7.17]. And since A and Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ B are similar plane numbers, one number thus falls (be- πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί, tween) A and B in mean proportion [Prop. 8.18]. And if τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. ἐὰν δὲ (some) numbers fall between two numbers in continued δύο ἀριθμῶν μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτωσιν proportion then, as many (numbers) as fall in (between) ἀριθμοί, ὅσοι εἰς αὐτοὺς ἐμπίπτουσι, τοσοῦτοι καὶ εἰς τοὺς them (in continued proportion), so many also (fall) in τὸν αὐτὸν λόγον ἔχοντας· ὥστε καὶ τῶν Δ, Γ εἷς μέσος (between numbers) having the same ratio (as them in ἀνάλογον ἐμπίπτει ἀριθμός. καί ἐστι τετράγωνος ὁ Δ· continued proportion) [Prop. 8.8]. And hence one num- τετράγωνος ἄρα καὶ ὁ Γ· ὅπερ ἔδει δεῖξαι. ber falls (between) D and C in mean proportion. And D is square. Thus, C (is) also square [Prop. 8.22]. (Which is) the very thing it was required to show.bþ. Proposition 2 ᾿Εὰν δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσι If two numbers make a square (number by) multiply- τετράγωνον, ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί. ing one another then they are similar plane numbers. ∆ Α Β Γ D A B C ῎Εστωσαν δύο ἀριθμοὶ οἱ Α, Β, καὶ ὁ Α τὸν Β πολλα- Let A and B be two numbers, and let A make the πλασιάσας τετράγωνον τὸν Γ ποιείτω· λέγω, ὅτι οἱ Α, Β square (number) C (by) multiplying B. I say that A and ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί. B are similar plane numbers. ῾Ο γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω· ὁ Δ For let A make D (by) multiplying itself. Thus, D is ἄρα τετράγωνός ἐστιν. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλα- square. And since A has made D (by) multiplying itself, πλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν and has made C (by) multiplying B, thus as A is to B, so Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, ὁ Δ πρὸς τὸν D (is) to C [Prop. 7.17]. And since D is square, and C Γ. καὶ ἐπεὶ ὁ Δ τετράγωνός ἐστιν, ἀλλὰ καὶ ὁ Γ, οἱ Δ, Γ (is) also, D and C are thus similar plane numbers. Thus, ἄρα ὅμοιοι ἐπίπεδοί εἰσιν. τῶν Δ, Γ ἄρα εἷς μέσος ἀνάλογον one (number) falls (between) D and C in mean propor- 254 STOIQEIWN jþ. ELEMENTS BOOK 9 ἐμπίπτει. καί ἐστιν ὡς ὁ Δ πρὸς τὸν Γ, οὕτως ὁ Α πρὸς τὸν tion [Prop. 8.18]. And as D is to C, so A (is) to B. Thus, Β· καὶ τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει. ἐὰν δὲ one (number) also falls (between) A and B in mean pro- δύο ἀριθμῶν εἷς μέσος ἀνάλογον ἐμπίπτῃ, ὅμοιοι ἐπίπεδοί portion [Prop. 8.8]. And if one (number) falls (between) εἰσιν [οἱ] ἀριθμοί· οἱ ἄρα Α, Β ὅμοιοί εἰσιν ἐπίπεδοι· ὅπερ two numbers in mean proportion then [the] numbers are ἔδει δεῖξαι. similar plane (numbers) [Prop. 8.20]. Thus, A and B are similar plane (numbers). (Which is) the very thing it was required to show.gþ. Proposition 3 ᾿Εὰν κύβος ἀριθμὸς ἑαυτὸν πολλαπλασιάσας ποιῇ τινα, If a cube number makes some (number by) multiply- ὁ γενόμενος κύβος ἔσται. ing itself then the created (number) will be cube. ∆ Α Β Γ D A B C Κύβος γὰρ ἀριθμὸς ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β For let the cube number A make B (by) multiplying ποιείτω· λέγω, ὅτι ὁ Β κύβος ἐστίν. itself. I say that B is cube. Εἰλήφθω γὰρ τοῦ Α πλευρὰ ὁ Γ, καὶ ὁ Γ ἑαυτὸν πολλα- For let the side C of A have been taken. And let C πλασιάσας τὸν Δ ποιείτω. φανερὸν δή ἐστιν, ὅτι ὁ Γ τὸν Δ make D by multiplying itself. So it is clear that C has πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ ἐπεὶ ὁ Γ ἑαυτὸν πολ- made A (by) multiplying D. And since C has made D λαπλασιάσας τὸν Δ πεποίηκεν, ὁ Γ ἄρα τὸν Δ μετρεῖ κατὰ (by) multiplying itself, C thus measures D according to τὰς ἐν αὑτῷ μονάδας. ἀλλὰ μὴν καὶ ἡ μονὰς τὸν Γ μετρεῖ the units in it [Def. 7.15]. But, in fact, a unit also mea- κατὰ τὰς ἐν αὐτῷ μονάδας· ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν sures C according to the units in it [Def. 7.20]. Thus, as Γ, ὁ Γ πρὸς τὸν Δ. πάλιν, ἐπεὶ ὁ Γ τὸν Δ πολλαπλασιάσας a unit is to C, so C (is) to D. Again, since C has made A τὸν Α πεποίηκεν, ὁ Δ ἄρα τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Γ (by) multiplying D, D thus measures A according to the μονάδας. μετρεῖ δὲ καὶ ἡ μονὰς τὸν Γ κατὰ τὰς ἐν αὐτῷ units in C. And a unit also measures C according to the μονάδας· ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Γ, ὁ Δ πρὸς τὸν units in it. Thus, as a unit is to C, so D (is) to A. But, Α. ἀλλ᾿ ὡς ἡ μονὰς πρὸς τὸν Γ, ὁ Γ πρὸς τὸν Δ· καὶ ὡς as a unit (is) to C, so C (is) to D. And thus as a unit (is) ἄρα ἡ μονὰς πρὸς τὸν Γ, οὕτως ὁ Γ πρὸς τὸν Δ καὶ ὁ Δ to C, so C (is) to D, and D to A. Thus, two numbers, C πρὸς τὸν Α. τῆς ἄρα μονάδος καὶ τοῦ Α ἀριθμοῦ δύο μέσοι and D, have fallen (between) a unit and the number A ἀνάλογον κατὰ τὸ συνεχὲς ἐμπεπτώκασιν ἀριθμοὶ οἱ Γ, Δ. in continued mean proportion. Again, since A has made πάλιν, ἐπεὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν, B (by) multiplying itself, A thus measures B according ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας· μετρεῖ δὲ to the units in it. And a unit also measures A according καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας· ἔστιν ἄρα ὡς to the units in it. Thus, as a unit is to A, so A (is) to B. ἡ μονὰς πρὸς τὸν Α, ὁ Α πρὸς τὸν Β. τῆς δὲ μονάδος καὶ And two numbers have fallen (between) a unit and A in τοῦ Α δύο μέσοι ἀνάλογον ἐμπεπτώκασιν ἀριθμοί· καὶ τῶν mean proportion. Thus two numbers will also fall (be- Α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπεσοῦνται ἀριθμοί. ἐὰν δὲ tween) A and B in mean proportion [Prop. 8.8]. And if δύο ἀριθμῶν δύο μέσοι ἀνάλογον ἐμπίπτωσιν, ὁ δὲ πρῶτος two (numbers) fall (between) two numbers in mean pro- κύβος ᾖ, καὶ ὁ δεύτερος κύβος ἔσται. καί ἐστιν ὁ Α κύβος· portion, and the first (number) is cube, then the second καὶ ὁ Β ἄρα κύβος ἐστίν· ὅπερ ἔδει δεῖξαι. will also be cube [Prop. 8.23]. And A is cube. Thus, B is also cube. (Which is) the very thing it was required to show.dþ. Proposition 4 ᾿Εὰν κύβος ἀριθμὸς κύβον ἀριθμὸν πολλαπλασιάσας If a cube number makes some (number by) multiply- ποιῇ τινα, ὁ γενόμενος κύβος ἔσται. ing a(nother) cube number then the created (number) 255 STOIQEIWN jþ. ELEMENTS BOOK 9 will be cube. ∆ Α Β Γ D A B C Κύβος γὰρ ἀριθμὸς ὁ Α κύβον ἀριθμὸν τὸν Β πολλα- For let the cube number A make C (by) multiplying πλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ κύβος ἐστίν. the cube number B. I say that C is cube. ῾Ο γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω· ὁ For let A make D (by) multiplying itself. Thus, D is Δ ἄρα κύβος ἐστίν. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλα- cube [Prop. 9.3]. And since A has made D (by) multi- πλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν plying itself, and has made C (by) multiplying B, thus Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς as A is to B, so D (is) to C [Prop. 7.17]. And since A τὸν Γ. καὶ ἐπεὶ οἱ Α, Β κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν οἱ Α, and B are cube, A and B are similar solid (numbers). Β. τῶν Α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί· Thus, two numbers fall (between) A and B in mean pro- ὥστε καὶ τῶν Δ, Γ δύο μέσοι ἀνάλογον ἐμπεσοῦνται portion [Prop. 8.19]. Hence, two numbers will also fall ἀριθμοί. καί ἐστι κύβος ὁ Δ· κύβος ἄρα καὶ ὁ Γ· ὅπερ (between) D and C in mean proportion [Prop. 8.8]. And ἔδει δεῖξαι. D is cube. Thus, C (is) also cube [Prop. 8.23]. (Which is) the very thing it was required to show.eþ. Proposition 5 ᾿Εὰν κύβος ἀριθμὸς ἀριθμόν τινα πολλαπλασιάσας κύβον If a cube number makes a(nother) cube number (by) ποιῇ, καὶ ὁ πολλαπλασιασθεὶς κύβος ἔσται. multiplying some (number) then the (number) multi- plied will also be cube. ∆ Α Β Γ C A B D Κύβος γὰρ ἀριθμὸς ὁ Α ἀριθμόν τινα τὸν Β πολλα- For let the cube number A make the cube (number) πλασιάσας κύβον τὸν Γ ποιείτω· λέγω, ὅτι ὁ Β κύβος ἐστίν. C (by) multiplying some number B. I say that B is cube. ῾Ο γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω· κύβος For let A make D (by) multiplying itself. D is thus ἄρα ἐστίν ὁ Δ. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν cube [Prop. 9.3]. And since A has made D (by) multiply- Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ing itself, and has made C (by) multiplying B, thus as A ἔστιν ἀρα ὡς ὁ Α πρὸς τὸν Β, ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ is to B, so D (is) to C [Prop. 7.17]. And since D and C οἱ Δ, Γ κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν. τῶν Δ, Γ ἄρα are (both) cube, they are similar solid (numbers). Thus, δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. καί ἐστιν ὡς ὁ Δ two numbers fall (between) D and C in mean proportion πρὸς τὸν Γ, οὕτως ὁ Α πρὸς τὸν Β· καὶ τῶν Α, Β ἄρα δύο [Prop. 8.19]. And as D is to C, so A (is) to B. Thus, μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. καί ἐστι κύβος ὁ Α· two numbers also fall (between) A and B in mean pro- κύβος ἄρα ἐστὶ καὶ ὁ Β· ὅπερ ἔδει δεῖξαι. portion [Prop. 8.8]. And A is cube. Thus, B is also cube [Prop. 8.23]. (Which is) the very thing it was required to show.�þ. Proposition 6 ᾿Εὰν ἀριθμὸς ἑαυτὸν πολλαπλασιάσας κύβον ποιῇ, καὶ If a number makes a cube (number by) multiplying 256 STOIQEIWN jþ. ELEMENTS BOOK 9 αὐτὸς κύβος ἔσται. itself then it itself will also be cube. Γ Α Β C A B Ἀριθμὸς γὰρ ὁ Α ἑαυτὸν πολλαπλασιάσας κύβον τὸν Β For let the number A make the cube (number) B (by) ποιείτω· λέγω, ὅτι καὶ ὁ Α κύβος ἐστίν. multiplying itself. I say that A is also cube. ῾Ο γὰρ Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω. ἐπεὶ οὖν For let A make C (by) multiplying B. Therefore, since ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Β πεποίηκεν, τὸν δὲ A has made B (by) multiplying itself, and has made C Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Γ ἄρα κύβος ἐστίν. (by) multiplying B, C is thus cube. And since A has made καὶ ἐπεὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν, ὁ B (by) multiplying itself, A thus measures B according to Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας. μετρεῖ δὲ the units in (A). And a unit also measures A according καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας. ἔστιν ἄρα to the units in it. Thus, as a unit is to A, so A (is) to ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β. καὶ ἐπεὶ B. And since A has made C (by) multiplying B, B thus ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Β ἄρα τὸν measures C according to the units in A. And a unit also Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. μετρεὶ δὲ καὶ ἡ μονὰς measures A according to the units in it. Thus, as a unit is τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας. ἔστιν ἄρα ὡς ἡ μονὰς to A, so B (is) to C. But, as a unit (is) to A, so A (is) to πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Γ. ἀλλ᾿ ὡς ἡ μονὰς πρὸς B. And thus as A (is) to B, (so) B (is) to C. And since B τὸν Α, οὕτως ὁ Α πρὸς τὸν Β· καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, and C are cube, they are similar solid (numbers). Thus, ὁ Β πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Β, Γ κύβοι εἰσίν, ὅμοιοι στερεοί there exist two numbers in mean proportion (between) εἰσιν. τῶν Β, Γ ἄρα δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί. καί B and C [Prop. 8.19]. And as B is to C, (so) A (is) to B. ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Α πρὸς τὸν Β. καὶ τῶν Α, Β Thus, there also exist two numbers in mean proportion ἄρα δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί. καί ἐστιν κύβος ὁ (between) A and B [Prop. 8.8]. And B is cube. Thus, A Β· κύβος ἄρα ἐστὶ καὶ ὁ Α· ὅπερ ἔδει δεὶξαι. is also cube [Prop. 8.23]. (Which is) the very thing it was required to show.zþ. Proposition 7 ᾿Εὰν σύνθετος ἀριθμὸς ἀριθμόν τινα πολλαπλασιάσας If a composite number makes some (number by) mul- ποιῇ τινα, ὁ γενόμενος στερεὸς ἔσται. tiplying some (other) number then the created (number) will be solid. Ε Α Β Γ ∆ E A B C D Σύνθετος γὰρ ἀριθμὸς ὁ Α ἀριθμόν τινα τὸν Β πολλα- For let the composite number A make C (by) multi- πλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ στερεός ἐστιν. plying some number B. I say that C is solid. ᾿Επεὶ γὰρ ὁ Α σύνθετός ἐστιν, ὑπὸ ἀριθμοῦ τινος με- For since A is a composite (number), it will be mea- τρηθήσεται. μετρείσθω ὑπὸ τοῦ Δ, καὶ ὁσάκις ὁ Δ τὸν Α sured by some number. Let it be measured by D. And, μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ οὖν ὁ Δ as many times as D measures A, so many units let there τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, ὁ Ε ἄρα τὸν Δ be in E. Therefore, since D measures A according to πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ ἐπεὶ ὁ Α τὸν Β πολ- the units in E, E has thus made A (by) multiplying D λαπλασιάσας τὸν Γ πεποίηκεν, ὁ δὲ Α ἐστιν ὁ ἐκ τῶν Δ, Ε, [Def. 7.15]. And since A has made C (by) multiplying B, ὁ ἄρα ἐκ τῶν Δ, Ε τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν. and A is the (number created) from (multiplying) D, E, ὁ Γ ἄρα στερεός ἐστιν, πλευραὶ δὲ αὐτοῦ εἰσιν οἱ Δ, Ε, Β· the (number created) from (multiplying) D, E has thus 257 STOIQEIWN jþ. ELEMENTS BOOK 9 ὅπερ ἔδει δεῖξαι. made C (by) multiplying B. Thus, C is solid, and its sides are D, E, B. (Which is) the very thing it was required to show.hþ. Proposition 8 ᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu- ὦσιν, ὁ μὲν τρίτος ἀπὸ τῆς μονάδος τετράγωνος ἔσται ously proportional, (starting) from a unit, then the third καὶ οἱ ἕνα διαλείποντες, ὁ δὲ τέταρτος κύβος καὶ οἱ from the unit will be square, and (all) those (numbers δύο διαλείποντες πάντες, ὁ δὲ ἕβδομος κύβος ἅμα καὶ after that) which leave an interval of one (number), and τετράγωνος καὶ οἱ πέντε διαλείποντες. the fourth (will be) cube, and all those (numbers after that) which leave an interval of two (numbers), and the seventh (will be) both cube and square, and (all) those (numbers after that) which leave an interval of five (num- bers). Ζ Α Β Γ ∆ Ε F A B C D E ῎Εστωσαν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογ- Let any multitude whatsoever of numbers, A, B, C, ον οἱ Α, Β, Γ, Δ, Ε, Ζ· λέγω, ὅτι ὁ μὲν τρίτος ἀπὸ D, E, F , be continuously proportional, (starting) from τῆς μονάδος ὁ Β τετράγωνός ἐστι καὶ οἱ ἕνα διαλείποντες a unit. I say that the third from the unit, B, is square, πάντες, ὁ δὲ τέταρτος ὁ Γ κύβος καὶ οἱ δύο διαλείποντες and all those (numbers after that) which leave an inter- πάντες, ὁ δὲ ἕβδομος ὁ Ζ κύβος ἅμα καὶ τετράγωνος καὶ οἱ val of one (number). And the fourth (from the unit), C, πέντε διαλείποντες πάντες. (is) cube, and all those (numbers after that) which leave ᾿Επεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α an interval of two (numbers). And the seventh (from the πρὸς τὸν Β, ἰσάκις ἄρα ἡ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ unit), F , (is) both cube and square, and all those (num- ὁ Α τὸν Β. ἡ δὲ μονὰς τὸν Α ἀριθμὸν μετρεῖ κατὰ τὰς bers after that) which leave an interval of five (numbers). ἐν αὐτῷ μονάδας· καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν For since as the unit is to A, so A (is) to B, the unit τῷ Α μονάδας. ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Β thus measures the number A the same number of times πεποίηκεν· τετράγωνος ἄρα ἐστὶν ὁ Β. καὶ ἐπεὶ οἱ Β, Γ, Δ as A (measures) B [Def. 7.20]. And the unit measures ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ Β τετράγωνός ἐστιν, καὶ ὁ Δ ἄρα the number A according to the units in it. Thus, A also τετράγωνός ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ζ τετράγωνός measures B according to the units in A. A has thus made ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ ἕνα διαλείποντες B (by) multiplying itself [Def. 7.15]. Thus, B is square. πάντες τετράγωνοί εἰσιν. λέγω δή, ὅτι καὶ ὁ τέταρτος ἀπὸ And since B, C, D are continuously proportional, and B τῆς μονάδος ὁ Γ κύβος ἐστὶ καὶ οἱ δύο διαλείποντες πάντες. is square, D is thus also square [Prop. 8.22]. So, for the ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν same (reasons), F is also square. So, similarly, we can Γ, ἰσάκις ἄρα ἡ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Γ. ἡ also show that all those (numbers after that) which leave δὲ μονὰς τὸν Α ἀριθμὸν μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας· an interval of one (number) are square. So I also say that καὶ ὁ Β ἄρα τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας· ὁ Α the fourth (number) from the unit, C, is cube, and all ἄρα τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν. ἐπεὶ οὖν ὁ those (numbers after that) which leave an interval of two Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Β πεποίηκεν, τὸν δὲ Β (numbers). For since as the unit is to A, so B (is) to C, πολλαπλασιάσας τὸν Γ πεποίηκεν, κύβος ἄρα ἐστὶν ὁ Γ. καὶ the unit thus measures the number A the same number ἐπεὶ οἱ Γ, Δ, Ε, Ζ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ Γ κύβος ἐστίν, of times that B (measures) C. And the unit measures the 258 STOIQEIWN jþ. ELEMENTS BOOK 9 καὶ ὁ Ζ ἄρα κύβος ἐστίν. ἐδείχθη δὲ καὶ τετράγωνος· ὁ ἄρα number A according to the units in A. And thus B mea- ἕβδομος ἀπὸ τῆς μονάδος κύβος τέ ἐστι καὶ τετράγωνος. sures C according to the units in A. A has thus made C ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ πέντε διαλείποντες πάντες (by) multiplying B. Therefore, since A has made B (by) κύβοι τέ εἰσι καὶ τετράγωνοι· ὅπερ ἔδει δεῖξαι. multiplying itself, and has made C (by) multiplying B, C is thus cube. And since C, D, E, F are continuously pro- portional, and C is cube, F is thus also cube [Prop. 8.23]. And it was also shown (to be) square. Thus, the seventh (number) from the unit is (both) cube and square. So, similarly, we can show that all those (numbers after that) which leave an interval of five (numbers) are (both) cube and square. (Which is) the very thing it was required to show.jþ. Proposition 9 ᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἑξῆς κατὰ τὸ συνεχὲς If any multitude whatsoever of numbers is continu- ἀριθμοὶ ἀνάλογον ὦσιν, ὁ δὲ μετὰ τὴν μονάδα τετράγωνος ously proportional, (starting) from a unit, and the (num- ᾖ, καὶ οἱ λοιποὶ πάντες τετράγωνοι ἔσονται. καὶ ἐὰν ὁ μετὰ ber) after the unit is square, then all the remaining (num- τὴν μονάδα κύβος ᾖ, καὶ οἱ λοιποὶ πάντες κύβοι ἔσονται. bers) will also be square. And if the (number) after the unit is cube, then all the remaining (numbers) will also be cube. Ζ Α Β Γ ∆ Ε F A B C D E ῎Εστωσαν ἀπὸ μονάδος ἑξῆς ἀνάλογον ὁσοιδηποτοῦν Let any multitude whatsoever of numbers, A, B, C, ἀριθμοὶ οἱ Α, Β, Γ, Δ, Ε, Ζ, ὁ δὲ μετὰ τὴν μονάδα D, E, F , be continuously proportional, (starting) from a ὁ Α τετράγωνος ἔστω· λέγω, ὅτι καὶ οἱ λοιποὶ πάντες unit. And let the (number) after the unit, A, be square. I τετράγωνοι ἔσονται. say that all the remaining (numbers) will also be square. ῞Οτι μὲν οὖν ὁ τρίτος ἀπὸ τῆς μονάδος ὁ Β τετράγωνός In fact, it has (already) been shown that the third ἐστι καὶ οἱ ἕνα διαπλείποντες πάντες, δέδεικται· λέγω [δή], (number) from the unit, B, is square, and all those (num- ὅτι καὶ οἱ λοιποὶ πάντες τετράγωνοί εἰσιν. ἐπεὶ γὰρ οἱ Α, bers after that) which leave an interval of one (number) Β, Γ ἑξῆς ἀνάλογόν εἰσιν, καί ἐστιν ὁ Α τετράγωνος, καὶ ὁ [Prop. 9.8]. [So] I say that all the remaining (num- Γ [ἄρα] τετράγωνος ἐστιν. πάλιν, ἐπεὶ [καὶ] οἱ Β, Γ, Δ ἑξῆς bers) are also square. For since A, B, C are continu- ἀνάλογόν εἰσιν, καί ἐστιν ὁ Β τετράγωνος, καὶ ὁ Δ [ἄρα] ously proportional, and A (is) square, C is [thus] also τετράγωνός ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ λοιποὶ square [Prop. 8.22]. Again, since B, C, D are [also] con- πάντες τετράγωνοί εἰσιν. tinuously proportional, and B is square, D is [thus] also Ἀλλὰ δὴ ἔστω ὁ Α κύβος· λέγω, ὅτι καὶ οἱ λοιποὶ πάντες square [Prop. 8.22]. So, similarly, we can show that all κύβοι εἰσίν. the remaining (numbers) are also square. ῞Οτι μὲν οὖν ὁ τέταρτος ἀπὸ τῆς μονάδος ὁ Γ κύβος ἐστὶ And so let A be cube. I say that all the remaining καὶ οἱ δύο διαλείποντες πάντες, δέδεικται· λέγω [δή], ὅτι καὶ (numbers) are also cube. οἱ λοιποὶ πάντες κύβοι εἰσίν. ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς In fact, it has (already) been shown that the fourth τὸν Α, οὕτως ὁ Α πρὸς τὸν Β, ἰσάκις ἀρα ἡ μονὰς τὸν Α (number) from the unit, C, is cube, and all those (num- μετρεῖ καὶ ὁ Α τὸν Β. ἡ δὲ μονὰς τὸν Α μετρεῖ κατὰ τὰς ἐν bers after that) which leave an interval of two (numbers) 259 STOIQEIWN jþ. ELEMENTS BOOK 9 αὐτῷ μονάδας· καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ [Prop. 9.8]. [So] I say that all the remaining (numbers) μονάδας· ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν. are also cube. For since as the unit is to A, so A (is) to καί ἐστιν ὁ Α κύβος. ἐὰν δὲ κύβος ἀριθμὸς ἑαυτὸν πολλα- B, the unit thus measures A the same number of times πλασιάσας ποιῇ τινα, ὁ γενόμενος κύβος ἐστίν· καὶ ὁ Β ἄρα as A (measures) B. And the unit measures A according κύβος ἐστίν. καὶ ἐπεὶ τέσσαρες ἀριθμοὶ οἱ Α, Β, Γ, Δ ἑξῆς to the units in it. Thus, A also measures B according to ἀνάλογόν εἰσιν, καί ἐστιν ὁ Α κύβος, καὶ ὁ Δ ἄρα κύβος the units in (A). A has thus made B (by) multiplying it- ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ε κύβος ἐστίν, καὶ ὁμοίως οἱ self. And A is cube. And if a cube number makes some λοιποὶ πάντες κύβοι εἰσίν· ὅπερ ἔδει δεῖξαι. (number by) multiplying itself then the created (number) is cube [Prop. 9.3]. Thus, B is also cube. And since the four numbers A, B, C, D are continuously proportional, and A is cube, D is thus also cube [Prop. 8.23]. So, for the same (reasons), E is also cube, and, similarly, all the remaining (numbers) are cube. (Which is) the very thing it was required to show.iþ. Proposition 10 ᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ [ἑξῆς] ἀνάλογον If any multitude whatsoever of numbers is [continu- ὦσιν, ὁ δὲ μετὰ τὴν μονάδα μὴ ᾖ τετράγωνος, οὐδ᾿ ἄλλος ously] proportional, (starting) from a unit, and the (num- οὐδεὶς τετράγωνος ἔσται χωρὶς τοῦ τρίτου ἀπὸ τῆς μονάδος ber) after the unit is not square, then no other (number) καὶ τῶν ἕνα διαλειπόντων πάντων. καὶ ἐὰν ὁ μετὰ τὴν will be square either, apart from the third from the unit, μονάδα κύβος μὴ ᾖ, οὐδὲ ἄλλος οὐδεὶς κύβος ἔσται χωρὶς and all those (numbers after that) which leave an inter- τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ τῶν δύο διαλειπόντων val of one (number). And if the (number) after the unit πάντων. is not cube, then no other (number) will be cube either, apart from the fourth from the unit, and all those (num- bers after that) which leave an interval of two (numbers). Ζ Α Β Γ ∆ Ε F A B C D E ῎Εστωσαν ἀπὸ μονάδος ἑξῆς ἀνάλογον ὁσοιδηποτοῦν Let any multitude whatsoever of numbers, A, B, C, ἀριθμοὶ οἱ Α, Β, Γ, Δ, Ε, Ζ, ὁ μετὰ τὴν μονάδα ὁ Α μὴ D, E, F , be continuously proportional, (starting) from ἔστω τετράγωνος· λέγω, ὅτι οὐδὲ ἄλλος οὐδεὶς τετράγωνος a unit. And let the (number) after the unit, A, not be ἔσται χωρὶς τοῦ τρίτου ἀπὸ τὴς μονάδος [καὶ τῶν ἕνα δια- square. I say that no other (number) will be square ei- λειπόντων]. ther, apart from the third from the unit [and (all) those Εἰ γὰρ δυνατόν, ἔστω ὁ Γ τετράγωνος. ἔστι δὲ καὶ ὁ (numbers after that) which leave an interval of one (num- Β τετράγωνος· οἱ Β, Γ ἄρα πρὸς ἀλλήλους λόγον ἔχου- ber)]. σιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καί For, if possible, let C be square. And B is also ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Α πρὸς τὸν Β· οἱ Α, Β ἄρα square [Prop. 9.8]. Thus, B and C have to one another πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς (the) ratio which (some) square number (has) to (some τετράγωνον ἀριθμόν· ὥστε οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν. other) square number. And as B is to C, (so) A (is) καί ἐστι τετράγωνος ὁ Β· τετράγωνος ἄρα ἐστὶ καὶ ὁ Α· to B. Thus, A and B have to one another (the) ratio ὅπερ οὐχ ὑπέκειτο. οὐκ ἄρα ὁ Γ τετράγωνός ἐστιν. ὁμοίως which (some) square number has to (some other) square δὴ δείξομεν, ὅτι οὐδ᾿ ἄλλος οὐδεὶς τετράγωνός ἐστι χωρὶς number. Hence, A and B are similar plane (numbers) 260 STOIQEIWN jþ. ELEMENTS BOOK 9 τοῦ τρίτου ἀπὸ τῆς μονάδος καὶ τῶν ἕνα διαλειπόντων. [Prop. 8.26]. And B is square. Thus, A is also square. Ἀλλὰ δὴ μὴ ἔστω ὁ Α κύβος. λέγω, ὅτι οὐδ᾿ ἄλλος The very opposite thing was assumed. C is thus not οὐδεὶς κύβος ἔσται χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος square. So, similarly, we can show that no other (number καὶ τῶν δύο διαλειπόντων. is) square either, apart from the third from the unit, and Εἰ γὰρ δυνατόν, ἔστω ὁ Δ κύβος. ἔστι δὲ καὶ ὁ Γ κύβος· (all) those (numbers after that) which leave an interval τέταρτος γάρ ἐστιν ἀπὸ τῆς μονάδος. καί ἐστιν ὡς ὁ Γ πρὸς of one (number). τὸν Δ, ὁ Β πρὸς τὸν Γ· καὶ ὁ Β ἄρα πρὸς τὸν Γ λόγον ἔχει, And so let A not be cube. I say that no other (num- ὃν κύβος πρὸς κύβον. καί ἐστιν ὁ Γ κύβος· καὶ ὁ Β ἄρα ber) will be cube either, apart from the fourth from the κύβος ἐστίν. καὶ ἐπεί ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, ὁ Α πρὸς unit, and (all) those (numbers after that) which leave an τὸν Β, ἡ δὲ μονὰς τὸν Α μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας, interval of two (numbers). καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας· ὁ Α For, if possible, let D be cube. And C is also cube ἄρα ἑαυτὸν πολλαπλασιάσας κύβον τὸν Β πεποίηκεν. ἐὰν [Prop. 9.8]. For it is the fourth (number) from the unit. δὲ ἀριθμὸς ἑαυτὸν πολλαπλασιάσας κύβον ποιῇ, καὶ αὐτὸς And as C is to D, (so) B (is) to C. And B thus has to κύβος ἔσται. κύβος ἄρα καὶ ὁ Α· ὅπερ οὐχ ὑπόκειται. οὐκ C the ratio which (some) cube (number has) to (some ἄρα ὁ Δ κύβος ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾿ ἄλλος other) cube (number). And C is cube. Thus, B is also οὐδεὶς κύβος ἐστὶ χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ cube [Props. 7.13, 8.25]. And since as the unit is to τῶν δύο διαλειπόντων· ὅπερ ἔδει δεῖξαι. A, (so) A (is) to B, and the unit measures A accord- ing to the units in it, A thus also measures B according to the units in (A). Thus, A has made the cube (num- ber) B (by) multiplying itself. And if a number makes a cube (number by) multiplying itself then it itself will be cube [Prop. 9.6]. Thus, A (is) also cube. The very oppo- site thing was assumed. Thus, D is not cube. So, simi- larly, we can show that no other (number) is cube either, apart from the fourth from the unit, and (all) those (num- bers after that) which leave an interval of two (numbers). (Which is) the very thing it was required to show.iaþ. Proposition 11 ᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu- ὦσιν, ὁ ἐλάττων τὸν μείζονα μετρεῖ κατά τινα τῶν ὑπαρχόντ- ously proportional, (starting) from a unit, then a lesser ων ἐν τοῖς ἀνάλογον ἀριθμοῖς. (number) measures a greater according to some existing (number) among the proportional numbers. Ε Α Β Γ ∆ E A B C D ῎Εστωσαν ἀπὸ μονάδος τῆς Α ὁποσοιοῦν ἀριθμοὶ ἑξῆς Let any multitude whatsoever of numbers, B, C, D, ἀνάλογον οἱ Β, Γ, Δ, Ε· λέγω, ὅτι τῶν Β, Γ, Δ, Ε ὁ E, be continuously proportional, (starting) from the unit ἐλάχιστος ὁ Β τὸν Ε μετρεῖ κατά τινα τῶν Γ, Δ. A. I say that, for B, C, D, E, the least (number), B, ᾿Επεὶ γάρ ἐστιν ὡς ἡ Α μονὰς πρὸς τὸν Β, οὕτως ὁ Δ measures E according to some (one) of C, D. πρὸς τὸν Ε, ἰσάκις ἄρα ἡ Α μονὰς τὸν Β ἀριθμὸν μετρεῖ For since as the unit A is to B, so D (is) to E, the καὶ ὁ Δ τὸν Ε· ἐναλλὰξ ἄρα ἰσάκις ἡ Α μονὰς τὸν Δ μετρεῖ unit A thus measures the number B the same number of καὶ ὁ Β τὸν Ε. ἡ δὲ Α μονὰς τὸν Δ μετρεῖ κατὰ τὰς ἐν times as D (measures) E. Thus, alternately, the unit A 261 STOIQEIWN jþ. ELEMENTS BOOK 9 αὐτῷ μονάδας· καὶ ὁ Β ἄρα τὸν Ε μετρεῖ κατὰ τὰς ἐν τῷ measures D the same number of times as B (measures) Δ μονάδας· ὥστε ὁ ἐλάσσων ὁ Β τὸν μείζονα τὸν Ε με- E [Prop. 7.15]. And the unit A measures D according to τρεῖ κατά τινα ἀριθμὸν τῶν ὑπαρχόντων ἐν τοῖς ἀνάλογον the units in it. Thus, B also measures E according to the ἀριθμοῖς. units in D. Hence, the lesser (number) B measures the greater E according to some existing number among the proportional numbers (namely, D).Pìrisma. Corollary Καὶ φανερόν, ὅτι ἣν ἔχει τάξιν ὁ μετρῶν ἀπὸ μονάδος, And (it is) clear that what(ever relative) place the τὴν αὐτὴν ἔχει καὶ ὁ καθ᾿ ὃν μετρεῖ ἀπὸ τοῦ μετρουμένου measuring (number) has from the unit, the (number) ἐπὶ τὸ πρὸ αὐτοῦ. ὅπερ ἔδει δεῖξαι. according to which it measures has the same (relative) place from the measured (number), in (the direction of the number) before it. (Which is) the very thing it was required to show.ibþ. Proposition 12 ᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu- ὦσιν, ὑφ᾿ ὅσων ἂν ὁ ἔσχατος πρώτων ἀριθμῶν μετρῆται, ously proportional, (starting) from a unit, then however ὑπὸ τῶν αὐτῶν καὶ ὁ παρὰ τὴν μονάδα μετρηθήσεται. many prime numbers the last (number) is measured by, the (number) next to the unit will also be measured by the same (prime numbers). Θ Α Β Γ ∆ Ε Ζ Η C A B D E F G H ῎Εστωσαν ἀπὸ μονάδος ὁποσοιδηποτοῦν ἀριθμοὶ ἀνάλογ- Let any multitude whatsoever of numbers, A, B, C, ον οἱ Α, Β, Γ, Δ· λέγω, ὅτι ὑφ᾿ ὅσων ἂν ὁ Δ πρώτων D, be (continuously) proportional, (starting) from a unit. ἀριθμῶν μετρῆται, ὑπὸ τῶν αὐτῶν καὶ ὁ Α μετρηθήσεται. I say that however many prime numbers D is measured Μετρείσθω γὰρ ὁ Δ ὑπό τινος πρώτου ἀριθμοῦ τοῦ Ε· by, A will also be measured by the same (prime numbers). λέγω, ὅτι ὁ Ε τὸν Α μετρεῖ. μὴ γάρ· καί ἐστιν ὁ Ε πρῶτος, For let D be measured by some prime number E. I ἅπας δὲ πρῶτος ἀριθμὸς πρὸς ἅπαντα, ὃν μὴ μετρεῖ, πρῶτός say that E measures A. For (suppose it does) not. E is ἐστιν· οἱ Ε, Α ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ prime, and every prime number is prime to every num- ὁ Ε τὸν Δ μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Ζ· ὁ Ε ἄρα ber which it does not measure [Prop. 7.29]. Thus, E τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν. πάλιν, ἐπεὶ ὁ Α and A are prime to one another. And since E measures τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Γ μονάδας, ὁ Α ἄρα τὸν Γ D, let it measure it according to F . Thus, E has made πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Ε τὸν Ζ D (by) multiplying F . Again, since A measures D ac- πολλαπλασιάσας τὸν Δ πεποίηκεν· ὁ ἄρα ἐκ τῶν Α, Γ ἴσος cording to the units in C [Prop. 9.11 corr.], A has thus ἐστὶ τῷ ἐκ τῶν Ε, Ζ. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Ε, ὁ Ζ made D (by) multiplying C. But, in fact, E has also πρὸς τὸν Γ. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, made D (by) multiplying F . Thus, the (number cre- οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ated) from (multiplying) A, C is equal to the (number ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν created) from (multiplying) E, F . Thus, as A is to E, ἑπόμενον· μετρεῖ ἄρα ὁ Ε τὸν Γ. μετρείτω αὐτὸν κατὰ τὸν (so) F (is) to C [Prop. 7.19]. And A and E (are) prime Η· ὁ Ε ἄρα τὸν Η πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ (to one another), and (numbers) prime (to one another μὴν διὰ τὸ πρὸ τούτου καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν are) also the least (of those numbers having the same ra- Γ πεποίηκεν. ὁ ἄρα ἐκ τῶν Α, Β ἴσος ἐστὶ τῷ ἐκ τῶν Ε, Η. tio as them) [Prop. 7.21], and the least (numbers) mea- ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Ε, ὁ Η πρὸς τὸν Β. οἱ δὲ Α, Ε sure those (numbers) having the same ratio as them an πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ equal number of times, the leading (measuring) the lead- 262 STOIQEIWN jþ. ELEMENTS BOOK 9 μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας αὐτοῖς ἰσάκις ὅ τε ing, and the following the following [Prop. 7.20]. Thus, ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον· E measures C. Let it measure it according to G. Thus, μετρεῖ ἄρα ὁ Ε τὸν Β. μετρείτω αὐτὸν κατὰ τὸν Θ· ὁ Ε ἄρα E has made C (by) multiplying G. But, in fact, via the τὸν Θ πολλαπλασιάσας τὸν Β πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α (proposition) before this, A has also made C (by) multi- ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν· ὁ ἄρα ἐκ τῶν Ε, plying B [Prop. 9.11 corr.]. Thus, the (number created) Θ ἴσος ἐστὶ τῷ ἀπὸ τοῦ Α. ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Α, ὁ Α from (multiplying) A, B is equal to the (number created) πρὸς τὸν Θ. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ from (multiplying) E, G. Thus, as A is to E, (so) G δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις (is) to B [Prop. 7.19]. And A and E (are) prime (to ὅ ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον· one another), and (numbers) prime (to one another are) μετρεῖ ἄρα ὁ Ε τὸν Α ὡς ἡγούμενος ἡγούμενον. ἀλλὰ μὴν also the least (of those numbers having the same ratio καὶ οὐ μετρεῖ· ὅπερ ἀδύνατον. οὐκ ἄρα οἱ Ε, Α πρῶτοι πρὸς as them) [Prop. 7.21], and the least (numbers) measure ἀλλήλους εἰσίν. σύνθετοι ἄρα. οἱ δὲ σύνθετοι ὑπὸ [πρώτου] those (numbers) having the same ratio as them an equal ἀριθμοῦ τινος μετροῦνται. καὶ ἐπεὶ ὁ Ε πρῶτος ὑπόκειται, ὁ number of times, the leading (measuring) the leading, δὲ πρῶτος ὑπὸ ἑτέρου ἀριθμοῦ οὐ μετρεῖται ἢ ὑφ᾿ ἑαυτοῦ, ὁ and the following the following [Prop. 7.20]. Thus, E Ε ἄρα τοὺς Α, Ε μετρεῖ· ὥστε ὁ Ε τὸν Α μετρεῖ. μετρεῖ δὲ measures B. Let it measure it according to H . Thus, καὶ τὸν Δ· ὁ Ε ἄρα τοὺς Α, Δ μετρεῖ. ὁμοίως δὴ δείξομεν, E has made B (by) multiplying H . But, in fact, A has ὅτι ὑφ᾿ ὅσων ἂν ὁ Δ πρώτων ἀριθμῶν μετρῆται, ὑπὸ τῶν also made B (by) multiplying itself [Prop. 9.8]. Thus, αὐτῶν καὶ ὁ Α μετρηθήσεται· ὅπερ ἔδει δεῖξαι. the (number created) from (multiplying) E, H is equal to the (square) on A. Thus, as E is to A, (so) A (is) to H [Prop. 7.19]. And A and E are prime (to one an- other), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures A, as the leading (measuring the) leading. But, in fact, (E) also does not measure (A). The very thing (is) impossible. Thus, E and A are not prime to one another. Thus, (they are) composite (to one another). And (numbers) composite (to one another) are (both) measured by some [prime] number [Def. 7.14]. And since E is assumed (to be) prime, and a prime (number) is not measured by another number (other) than itself [Def. 7.11], E thus measures (both) A and E. Hence, E measures A. And it also measures D. Thus, E measures (both) A and D. So, similarly, we can show that however many prime numbers D is measured by, A will also be measured by the same (prime numbers). (Which is) the very thing it was required to show.igþ. Proposition 13 ᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu- ὦσιν, ὁ δὲ μετὰ τὴν μονάδα πρῶτος ᾖ, ὁ μέγιστος ὑπ᾿ ously proportional, (starting) from a unit, and the (num- οὐδενὸς [ἄλλου] μετρηθήσεται παρὲξ τῶν ὑπαρχόντων ἐν ber) after the unit is prime, then the greatest (number) τοῖς ἀνάλογον ἀριθμοῖς. will be measured by no [other] (numbers) except (num- ῎Εστωσαν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον bers) existing among the proportional numbers. οἱ Α, Β, Γ, Δ, ὁ δὲ μετὰ τὴν μονάδα ὁ Α πρῶτος ἔστω· Let any multitude whatsoever of numbers, A, B, C, λέγω, ὅτι ὁ μέγιστος αὐτῶν ὁ Δ ὑπ᾿ οὐδενὸς ἄλλου με- D, be continuously proportional, (starting) from a unit. τρηθήσεται παρὲξ τῶν Α, Β, Γ. And let the (number) after the unit, A, be prime. I say 263 STOIQEIWN jþ. ELEMENTS BOOK 9 that the greatest of them, D, will be measured by no other (numbers) except A, B, C. Γ Α Β ∆ Ε Ζ Η Θ EA B C D H G F Εἰ γὰρ δυνατόν, μετρείσθω ὑπὸ τοῦ Ε, καὶ ὁ Ε μηδενὶ For, if possible, let it be measured by E, and let E not τῶν Α, Β, Γ ἔστω ὁ αὐτός. φανερὸν δή, ὅτι ὁ Ε πρῶτος be the same as one of A, B, C. So it is clear that E is οὔκ ἐστιν. εἰ γὰρ ὁ Ε πρῶτός ἐστι καὶ μετρεῖ τὸν Δ, καὶ τὸν not prime. For if E is prime, and measures D, then it will Α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν also measure A, (despite A) being prime (and) not being ἀδύνατον. οὐκ ἄρα ὁ Ε πρῶτός ἐστιν. σύνθετος ἄρα. πᾶς the same as it [Prop. 9.12]. The very thing is impossible. δὲ σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται· Thus, E is not prime. Thus, (it is) composite. And every ὁ Ε ἄρα ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. λέγω δή, ὅτι composite number is measured by some prime number ὑπ᾿ οὐδενὸς ἄλλου πρώτου μετρηθήσεται πλὴν τοῦ Α. εἰ γὰρ [Prop. 7.31]. Thus, E is measured by some prime num- ὑφ᾿ ἑτέρου μετρεῖται ὁ Ε, ὁ δὲ Ε τὸν Δ μετρεῖ, κἀκεῖνος ἄρα ber. So I say that it will be measured by no other prime τὸν Δ μετρήσει· ὥστε καὶ τὸν Α μετρήσει πρῶτον ὄντα μὴ number than A. For if E is measured by another (prime ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν ἀδύνατον. ὁ Α ἄρα τὸν Ε number), and E measures D, then this (prime number) μετρεῖ. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν will thus also measure D. Hence, it will also measure A, Ζ. λέγω, ὅτι ὁ Ζ οὐδενὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. εἰ γὰρ ὁ (despite A) being prime (and) not being the same as it Ζ ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτὸς καὶ μετρεῖ τὸν Δ κατὰ τὸν [Prop. 9.12]. The very thing is impossible. Thus, A mea- Ε, καὶ εἷς ἄρα τῶν Α, Β, Γ τὸν Δ μετρεῖ κατά τὸν Ε. ἀλλὰ sures E. And since E measures D, let it measure it ac- εἷς τῶν Α, Β, Γ τὸν Δ μετρεῖ κατά τινα τῶν Α, Β, Γ· καὶ ὁ cording to F . I say that F is not the same as one of A, B, Ε ἄρα ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός· ὅπερ οὐχ ὑπόκειται. C. For if F is the same as one of A, B, C, and measures οὐκ ἄρα ὁ Ζ ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. ὁμοίως δὴ D according to E, then one of A, B, C thus also measures δείξομεν, ὅτι μετρεῖται ὁ Ζ ὑπὸ τοῦ Α, δεικνύντες πάλιν, D according to E. But one of A, B, C (only) measures ὅτι ὁ Ζ οὔκ ἐστι πρῶτος. εἰ γὰρ, καὶ μετρεῖ τὸν Δ, καὶ τὸν D according to some (one) of A, B, C [Prop. 9.11]. And Α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν thus E is the same as one of A, B, C. The very oppo- ἀδύνατον· οὐκ ἄρα πρῶτός ἐστιν ὁ Ζ· σύνθετος ἄρα. ἅπας site thing was assumed. Thus, F is not the same as one δὲ σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται· ὁ of A, B, C. Similarly, we can show that F is measured Ζ ἄρα ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. λέγω δή, ὅτι ὑφ᾿ by A, (by) again showing that F is not prime. For if (F ἑτέρου πρώτου οὐ μετρηθήσεται πλὴν τοῦ Α. εἰ γὰρ ἕτερός is prime), and measures D, then it will also measure A, τις πρῶτος τὸν Ζ μετρεῖ, ὁ δὲ Ζ τὸν Δ μετρεῖ, κἀκεῖνος (despite A) being prime (and) not being the same as it ἄρα τὸν Δ μετρήσει· ὥστε καὶ τὸν Α μετρήσει πρῶτον ὄντα [Prop. 9.12]. The very thing is impossible. Thus, F is μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν ἀδύνατον. ὁ Α ἄρα τὸν Ζ not prime. Thus, (it is) composite. And every composite μετρεῖ. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ κατὰ τὸν Ζ, ὁ Ε ἄρα τὸν number is measured by some prime number [Prop. 7.31]. Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α τὸν Thus, F is measured by some prime number. So I say Γ πολλαπλασιάσας τὸν Δ πεποίηκεν· ὁ ἄρα ἐκ τῶν Α, Γ that it will be measured by no other prime number than ἴσος ἐστὶ τῷ ἐκ τῶν Ε, Ζ. ἀνάλογον ἄρα ἐστὶν ὡς ὁ Α πρὸς A. For if some other prime (number) measures F , and τὸν Ε, οὕτως ὁ Ζ πρὸς τὸν Γ. ὁ δὲ Α τὸν Ε μετρεῖ· καὶ ὁ Ζ F measures D, then this (prime number) will thus also ἄρα τὸν Γ μετρεῖ. μετρείτω αὐτὸν κατὰ τὸν Η. ὁμοίως δὴ measure D. Hence, it will also measure A, (despite A) δείξομεν, ὅτι ὁ Η οὐδενὶ τῶν Α, Β ἐστιν ὁ αὐτός, καὶ ὅτι being prime (and) not being the same as it [Prop. 9.12]. μετρεῖται ὑπὸ τοῦ Α. καὶ ἐπεὶ ὁ Ζ τὸν Γ μετρεῖ κατὰ τὸν Η, The very thing is impossible. Thus, A measures F . And ὁ Ζ ἄρα τὸν Η πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ μὴν since E measures D according to F , E has thus made καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν· ὁ ἄρα ἐκ D (by) multiplying F . But, in fact, A has also made D τῶν Α, Β ἴσος ἐστὶ τῷ ἐκ τῶν Ζ, Η. ἀνάλογον ἄρα ὡς ὁ Α (by) multiplying C [Prop. 9.11 corr.]. Thus, the (number πρὸς τὸν Ζ, ὁ Η πρὸς τὸν Β. μετρεῖ δὲ ὁ Α τὸν Ζ· μετρεῖ created) from (multiplying) A, C is equal to the (number ἄρα καὶ ὁ Η τὸν Β. μετρείτω αὐτὸν κατὰ τὸν Θ. ὁμοίως δὴ created) from (multiplying) E, F . Thus, proportionally, δείξομεν, ὅτι ὁ Θ τῷ Α οὐκ ἔστιν ὁ αὐτός. καὶ ἐπεὶ ὁ Η τὸν as A is to E, so F (is) to C [Prop. 7.19]. And A measures 264 STOIQEIWN jþ. ELEMENTS BOOK 9 Β μετρεῖ κατὰ τὸν Θ, ὁ Η ἄρα τὸν Θ πολλαπλασιάσας τὸν E. Thus, F also measures C. Let it measure it according Β πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας to G. So, similarly, we can show that G is not the same τὸν Β πεποίηκεν· ὁ ἄρα ὑπὸ Θ, Η ἴσος ἐστὶ τῷ ἀπὸ τοῦ Α as one of A, B, and that it is measured by A. And since τετραγώνῳ· ἔστιν ἄρα ὡς ὁ Θ πρὸς τὸν Α, ὁ Α πρὸς τὸν Η. F measures C according to G, F has thus made C (by) μετρεῖ δὲ ὁ Α τὸν Η· μετρεῖ ἄρα καὶ ὁ Θ τὸν Α πρῶτον ὄντα multiplying G. But, in fact, A has also made C (by) mul- μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἄτοπον. οὐκ ἄρα ὁ μέγιστος ὁ tiplying B [Prop. 9.11 corr.]. Thus, the (number created) Δ ὑπὸ ἑτέρου ἀριθμοῦ μετρηθήσεται παρὲξ τῶν Α, Β, Γ· from (multiplying) A, B is equal to the (number created) ὅπερ ἔδει δεῖξαι. from (multiplying) F , G. Thus, proportionally, as A (is) to F , so G (is) to B [Prop. 7.19]. And A measures F . Thus, G also measures B. Let it measure it according to H . So, similarly, we can show that H is not the same as A. And since G measures B according to H , G has thus made B (by) multiplying H . But, in fact, A has also made B (by) multiplying itself [Prop. 9.8]. Thus, the (number created) from (multiplying) H , G is equal to the square on A. Thus, as H is to A, (so) A (is) to G [Prop. 7.19]. And A measures G. Thus, H also measures A, (despite A) being prime (and) not being the same as it. The very thing (is) absurd. Thus, the greatest (number) D cannot be measured by another (number) except (one of) A, B, C. (Which is) the very thing it was required to show.idþ. Proposition 14 ᾿Εὰν ἐλάχιστος ἀριθμὸς ὑπὸ πρώτων ἀριθμῶν μετρῆται, If a least number is measured by (some) prime num- ὑπ᾿ οὑδενὸς ἄλλου πρώτου ἀριθμοῦ μετρηθήσεται παρὲξ bers then it will not be measured by any other prime τῶν ἐξ ἀρχῆς μετρούντων. number except (one of) the original measuring (num- bers). ∆ Α Ε Ζ Β Γ D A E F B C ᾿Ελάχιστος γὰρ ἀριθμὸς ὁ Α ὑπὸ πρώτων ἀριθμῶν τῶν For let A be the least number measured by the prime Β, Γ, Δ μετρείσθω· λέγω, ὅτι ὁ Α ὑπ᾿ οὐδενὸς ἄλλου numbers B, C, D. I say that A will not be measured by πρώτου ἀριθμοῦ μετρηθήσεται παρὲξ τῶν Β, Γ, Δ. any other prime number except (one of) B, C, D. Εἰ γὰρ δυνατόν, μετρείσθω ὑπὸ πρώτου τοῦ Ε, καὶ ὁ For, if possible, let it be measured by the prime (num- Ε μηδενὶ τῶν Β, Γ, Δ ἔστω ὁ αὐτός. καὶ ἐπεὶ ὁ Ε τὸν ber) E. And let E not be the same as one of B, C, D. Α μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Ζ· ὁ Ε ἄρα τὸν Ζ And since E measures A, let it measure it according to F . πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ μετρεῖται ὁ Α ὑπὸ Thus, E has made A (by) multiplying F . And A is mea- πρώτων ἀριθμῶν τῶν Β, Γ, Δ. ἐὰν δὲ δύο ἀριθμοὶ πολ- sured by the prime numbers B, C, D. And if two num- λαπλασιάσαντες ἀλλήλους ποιῶσί τινα, τὸν δὲ γενόμενον bers make some (number by) multiplying one another, ἐξ αὐτῶν μετρῇ τις πρῶτος ἀριθμός, καὶ ἕνα τῶν ἐξ ἀρχῆς and some prime number measures the number created μετρήσει· οἱ Β, Γ, Δ ἄρα ἕνα τῶν Ε, Ζ μετρήσουσιν. τὸν from them, then (the prime number) will also measure μὲν οὖν Ε οὐ μετρήσουσιν· ὁ γὰρ Ε πρῶτός ἐστι καὶ οὐδενὶ one of the original (numbers) [Prop. 7.30]. Thus, B, C, τῶν Β, Γ, Δ ὁ αὐτός. τὸν Ζ ἄρα μετροῦσιν ἐλάσσονα ὄντα D will measure one of E, F . In fact, they do not measure τοῦ Α· ὅπερ ἀδύνατον. ὁ γὰρ Α ὑπόκειται ἐλάχιστος ὑπὸ E. For E is prime, and not the same as one of B, C, D. τῶν Β, Γ, Δ μετρούμενος. οὐκ ἄρα τὸν Α μετρήσει πρῶτος Thus, they (all) measure F , which is less than A. The ἀριθμὸς παρὲξ τῶν Β, Γ, Δ· ὅπερ ἔδει δεῖξαι. very thing (is) impossible. For A was assumed (to be) the least (number) measured by B, C, D. Thus, no prime 265 STOIQEIWN jþ. ELEMENTS BOOK 9 number can measure A except (one of) B, C, D. (Which is) the very thing it was required to show.ieþ. Proposition 15 ᾿Εὰν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν ἐλάχιστοι τῶν If three continuously proportional numbers are the τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, δύο ὁποιοιοῦν συν- least of those (numbers) having the same ratio as them τεθέντες πρὸς τὸν λοιπὸν πρῶτοί εἰσιν. then two (of them) added together in any way are prime to the remaining (one). ∆ Α Β Γ Ε Ζ E A B C FD ῎Εστωσαν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον ἐλάχιστοι τῶν Let A, B, C be three continuously proportional num- τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς οἱ Α, Β, Γ· λέγω, ὅτι τῶν bers (which are) the least of those (numbers) having the Α, Β, Γ δύο ὁποιοιοῦν συντεθέντες πρὸς τὸν λοιπὸν πρῶτοι same ratio as them. I say that two of A, B, C added to- εἰσιν, οἱ μὲν Α, Β πρὸς τὸν Γ, οἱ δὲ Β, Γ πρὸς τὸν Α καὶ gether in any way are prime to the remaining (one), (that ἔτι οἱ Α, Γ πρὸς τὸν Β. is) A and B (prime) to C, B and C to A, and, further, A Εἰλήφθωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν and C to B. λόγον ἐχόντων τοῖς Α, Β, Γ δύο οἱ ΔΕ, ΕΖ. φανερὸν Let the two least numbers, DE and EF , having the δή, ὅτι ὁ μὲν ΔΕ ἑαυτὸν πολλαπλασιάσας τὸν Α πεποίηκεν, same ratio as A, B, C, have been taken [Prop. 8.2]. τὸν δὲ ΕΖ πολλαπλασιάσας τὸν Β πεποίηκεν, καὶ ἔτι ὁ ΕΖ So it is clear that DE has made A (by) multiplying it- ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν. καὶ ἐπεὶ οἱ ΔΕ, self, and has made B (by) multiplying EF , and, fur- ΕΖ ἐλάχιστοί εἰσιν, πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐὰν δὲ ther, EF has made C (by) multiplying itself [Prop. 8.2]. δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ συναμφότερος And since DE, EF are the least (of those numbers hav- πρὸς ἑκάτερον πρῶτός ἐστιν· καὶ ὁ ΔΖ ἄρα πρὸς ἑκάτερον ing the same ratio as them), they are prime to one an- τῶν ΔΕ, ΕΖ πρῶτός ἐστιν. ἀλλὰ μὴν καὶ ὁ ΔΕ πρὸς τὸν other [Prop. 7.22]. And if two numbers are prime to ΕΖ πρῶτός ἐστιν· οἱ ΔΖ, ΔΕ ἄρα πρὸς τὸν ΕΖ πρῶτοί εἰσιν. one another then the sum (of them) is also prime to each ἐὰν δὲ δύο ἀριθμοὶ πρός τινα ἀριθμὸν πρῶτοι ὦσιν, καὶ ὁ [Prop. 7.28]. Thus, DF is also prime to each of DE, EF . ἐξ αὐτῶν γενόμενος πρὸς τὸν λοιπὸν πρῶτός ἐστιν· ὥστε But, in fact, DE is also prime to EF . Thus, DF , DE ὁ ἐκ τῶν ΖΔ, ΔΕ πρὸς τὸν ΕΖ πρῶτός ἐστιν· ὥστε καὶ ὁ are (both) prime to EF . And if two numbers are (both) ἐκ τῶν ΖΔ, ΔΕ πρὸς τὸν ἀπὸ τοῦ ΕΖ πρῶτός ἐστιν. [ἐὰν prime to some number then the (number) created from γὰρ δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, ὁ ἐκ τοῦ ἑνὸς (multiplying) them is also prime to the remaining (num- αὐτῶν γενόμενος πρὸς τὸν λοιπὸν πρῶτός ἐστιν]. ἀλλ᾿ ὁ ber) [Prop. 7.24]. Hence, the (number created) from ἐκ τῶν ΖΔ, ΔΕ ὁ ἀπὸ τοῦ ΔΕ ἐστι μετὰ τοῦ ἐκ τῶν ΔΕ, (multiplying) FD, DE is prime to EF . Hence, the (num- ΕΖ· ὁ ἄρα ἀπὸ τοῦ ΔΕ μετὰ τοῦ ἐκ τῶν ΔΕ, ΕΖ πρὸς τὸν ber created) from (multiplying) FD, DE is also prime ἀπὸ τοῦ ΕΖ πρῶτός ἐστιν. καί ἐστιν ὁ μὲν ἀπὸ τοῦ ΔΕ to the (square) on EF [Prop. 7.25]. [For if two num- ὁ Α, ὁ δὲ ἐκ τῶν ΔΕ, ΕΖ ὁ Β, ὁ δὲ ἀπὸ τοῦ ΕΖ ὁ Γ· οἱ bers are prime to one another then the (number) created Α, Β ἄρα συντεθέντες πρὸς τὸν Γ πρῶτοί εἰσιν. ὁμοίως δὴ from (squaring) one of them is prime to the remaining δείξομεν, ὅτι καὶ οἱ Β, Γ πρὸς τὸν Α πρῶτοί εἰσιν. λέγω (number).] But the (number created) from (multiplying) δή, ὅτι καὶ οἱ Α, Γ πρὸς τὸν Β πρῶτοί εἰσιν. ἐπεὶ γὰρ ὁ ΔΖ FD, DE is the (square) on DE plus the (number cre- πρὸς ἑκάτερον τῶν ΔΕ, ΕΖ πρῶτός ἐστιν, καὶ ὁ ἀπὸ τοῦ ated) from (multiplying) DE, EF [Prop. 2.3]. Thus, the ΔΖ πρὸς τὸν ἐκ τῶν ΔΕ, ΕΖ πρῶτός ἐστιν. ἀλλὰ τῷ ἀπὸ (square) on DE plus the (number created) from (multi- τοῦ ΔΖ ἴσοι εἰσὶν οἱ ἀπὸ τῶν ΔΕ, ΕΖ μετὰ τοῦ δὶς ἐκ τῶν plying) DE, EF is prime to the (square) on EF . And ΔΕ, ΕΖ· καὶ οἱ ἀπὸ τῶν ΔΕ, ΕΖ ἄρα μετὰ τοῦ δὶς ὑπὸ τῶν the (square) on DE is A, and the (number created) from ΔΕ, ΕΖ πρὸς τὸν ὑπὸ τῶν ΔΕ, ΕΖ πρῶτοί [εἰσι]. διελόντι (multiplying) DE, EF (is) B, and the (square) on EF οἱ ἀπὸ τῶν ΔΕ, ΕΖ μετὰ τοῦ ἅπαξ ὑπὸ ΔΕ, ΕΖ πρὸς τὸν (is) C. Thus, A, B summed is prime to C. So, similarly, ὑπὸ ΔΕ, ΕΖ πρῶτοί εἰσιν. ἔτι διελόντι οἱ ἀπὸ τῶν ΔΕ, ΕΖ we can show that B, C (summed) is also prime to A. So ἄρα πρὸς τὸν ὑπὸ ΔΕ, ΕΖ πρῶτοί εἰσιν. καί ἐστιν ὁ μὲν I say that A, C (summed) is also prime to B. For since 266 STOIQEIWN jþ. ELEMENTS BOOK 9 ἀπὸ τοῦ ΔΕ ὁ Α, ὁ δὲ ὑπὸ τῶν ΔΕ, ΕΖ ὁ Β, ὁ δὲ ἀπὸ τοῦ DF is prime to each of DE, EF then the (square) on DF ΕΖ ὁ Γ. οἱ Α, Γ ἄρα συντεθέντες πρὸς τὸν Β πρῶτοί εἰσιν· is also prime to the (number created) from (multiplying) ὅπερ ἔδει δεῖξαι. DE, EF [Prop. 7.25]. But, the (sum of the squares) on DE, EF plus twice the (number created) from (multiply- ing) DE, EF is equal to the (square) on DF [Prop. 2.4]. And thus the (sum of the squares) on DE, EF plus twice the (rectangle contained) by DE, EF [is] prime to the (rectangle contained) by DE, EF . By separation, the (sum of the squares) on DE, EF plus once the (rect- angle contained) by DE, EF is prime to the (rectangle contained) by DE, EF .† Again, by separation, the (sum of the squares) on DE, EF is prime to the (rectangle contained) by DE, EF . And the (square) on DE is A, and the (rectangle contained) by DE, EF (is) B, and the (square) on EF (is) C. Thus, A, C summed is prime to B. (Which is) the very thing it was required to show. † Since if α β measures α2 + β2 + 2 α β then it also measures α2 + β2 + α β, and vice versa.i�þ. Proposition 16 ᾿Εὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, οὐκ ἔσται If two numbers are prime to one another then as the ὡς ὁ πρῶτος πρὸς τὸν δεύτερον, οὕτως ὁ δεύτερος πρὸς first is to the second, so the second (will) not (be) to some ἄλλον τινά. other (number). Γ Α Β C A B Δύο γὰρ ἀριθμοὶ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους ἔστω- For let the two numbers A and B be prime to one σαν· λέγω, ὅτι οὐκ ἔστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Β another. I say that as A is to B, so B is not to some other πρὸς ἄλλον τινά. (number). Εἰ γὰρ δυνατόν, ἔστω ὡς ὁ Α πρὸς τὸν Β, ὁ Β πρὸς For, if possible, let it be that as A (is) to B, (so) τὸν Γ. οἱ δὲ Α, Β πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ B (is) to C. And A and B (are) prime (to one an- ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας other). And (numbers) prime (to one another are) also ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν the least (of those numbers having the same ratio as ἑπόμενον· μετρεῖ ἄρα ὁ Α τὸν Β ὡς ἡγούμενος ἡγούμενον. them) [Prop. 7.21]. And the least numbers measure μετρεῖ δὲ καὶ ἑαυτόν· ὁ Α ἄρα τοὺς Α, Β μετρεῖ πρώτους those (numbers) having the same ratio (as them) an ὄντας πρὸς ἀλλήλους· ὅπερ ἄτοπον. οὐκ ἄρα ἔσται ὡς ὁ Α equal number of times, the leading (measuring) the lead- πρὸς τὸν Β, οὕτως ὁ Β πρὸς τὸν Γ· ὅπερ ἔδει δεῖξαι. ing, and the following the following [Prop. 7.20]. Thus, A measures B, as the leading (measuring) the leading. And (A) also measures itself. Thus, A measures A and B, which are prime to one another. The very thing (is) ab- surd. Thus, as A (is) to B, so B cannot be to C. (Which is) the very thing it was required to show.izþ. Proposition 17 ᾿Εὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, οἱ δὲ If any multitude whatsoever of numbers is continu- ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους ὦσιν, οὐκ ἔσται ὡς ὁ ously proportional, and the outermost of them are prime πρῶτος πρὸς τὸν δεύτερον, οὕτως ὁ ἔσχατος πρὸς ἄλλον to one another, then as the first (is) to the second, so the 267 STOIQEIWN jþ. ELEMENTS BOOK 9 τινά. last will not be to some other (number). ῎Εστωσαν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Let A, B, C, D be any multitude whatsoever of con- Β, Γ, Δ, οἱ δὲ ἄκροι αὐτῶν οἱ Α, Δ πρῶτοι πρὸς ἀλλήλους tinuously proportional numbers. And let the outermost ἔστωσαν· λέγω, ὅτι οὐκ ἔστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ of them, A and D, be prime to one another. I say that as Δ πρὸς ἄλλον τινά. A is to B, so D (is) not to some other (number). Ε Α Β Γ ∆ E A B C D Εἰ γὰρ δυνατόν, ἔστω ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ For, if possible, let it be that as A (is) to B, so D πρὸς τὸν Ε· ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, ὁ Β πρὸς (is) to E. Thus, alternately, as A is to D, (so) B (is) τὸν Ε. οἱ δὲ Α, Δ πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ to E [Prop. 7.13]. And A and D are prime (to one ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας another). And (numbers) prime (to one another are) ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν also the least (of those numbers having the same ra- ἑπόμενον. μετρεῖ ἄρα ὁ Α τὸν Β. καί ἐστιν ὡς ὁ Α πρὸς tio as them) [Prop. 7.21]. And the least numbers mea- τὸν Β, ὁ Β πρὸς τὸν Γ. καὶ ὁ Β ἄρα τὸν Γ μετρεῖ΄· ὥστε sure those (numbers) having the same ratio (as them) an καὶ ὁ Α τὸν Γ μετρεῖ. καὶ ἐπεί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, equal number of times, the leading (measuring) the lead- ὁ Γ πρὸς τὸν Δ, μετρεῖ δὲ ὁ Β τὸν Γ, μετρεῖ ἄρα καὶ ὁ Γ ing, and the following the following [Prop. 7.20]. Thus, τὸν Δ. ἀλλ᾿ ὁ Α τὸν Γ ἐμέτρει· ὥστε ὁ Α καὶ τὸν Δ μετρεῖ. A measures B. And as A is to B, (so) B (is) to C. Thus, B μετρεῖ δὲ καὶ ἑαυτόν. ὁ Α ἄρα τοὺς Α, Δ μετρεῖ πρώτους also measures C. And hence A measures C [Def. 7.20]. ὄντας πρὸς ἀλλήλους· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἔσται And since as B is to C, (so) C (is) to D, and B mea- ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς ἄλλον τινά· ὅπερ ἔδει sures C, C thus also measures D [Def. 7.20]. But, A was δεῖξαι. (found to be) measuring C. And hence A also measures D. And (A) also measures itself. Thus, A measures A and D, which are prime to one another. The very thing is impossible. Thus, as A (is) to B, so D cannot be to some other (number). (Which is) the very thing it was required to show.ihþ. Proposition 18 Δύο ἀριθμῶν δοθέντων ἐπισκέψασθαι, εἰ δυνατόν ἐστιν For two given numbers, to investigate whether it is αὐτοῖς τρίτον ἀνάλογον προσευρεῖν. possible to find a third (number) proportional to them. ∆ Α Β Γ D A B C ῎Εστωσαν οἱ δοθέντες δύο ἀριθμοὶ οἱ Α, Β, καὶ Let A and B be the two given numbers. And let it δέον ἔστω ἐπισκέψασθαι, εἰ δυνατόν ἐστιν αὐτοῖς τρίτον be required to investigate whether it is possible to find a ἀνάλογον προσευρεῖν. third (number) proportional to them. Οἱ δὴ Α, Β ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν ἢ οὔ. καὶ εἰ So A and B are either prime to one another, or not. πρῶτοι πρὸς ἀλλήλους εἰσίν, δέδεικται, ὅτι ἀδύνατόν ἐστιν And if they are prime to one another then it has (already) αὐτοῖς τρίτον ἀνάλογον προσευρεῖν. been show that it is impossible to find a third (number) Ἀλλὰ δὴ μὴ ἔστωσαν οἱ Α, Β πρῶτοι πρὸς ἀλλήλους, proportional to them [Prop. 9.16]. καὶ ὁ Β ἑαυτον πολλαπλασιάσας τὸν Γ ποιείτω. ὁ Α δὴ τὸν And so let A and B not be prime to one another. And Γ ἤτοι μετρεῖ ἢ οὐ μετρεῖ. μετρείτω πρότερον κατὰ τὸν Δ· let B make C (by) multiplying itself. So A either mea- ὁ Α ἄρα τὸν Δ πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλα μὴν sures, or does not measure, C. Let it first of all measure καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν· ὁ ἄρα (C) according to D. Thus, A has made C (by) multiply- 268 STOIQEIWN jþ. ELEMENTS BOOK 9 ἐκ τῶν Α, Δ ἴσος ἐστὶ τῷ ἀπὸ τοῦ Β. ἔστιν ἄρα ὡς ὁ Α ing D. But, in fact, B has also made C (by) multiplying πρὸς τὸν Β, ὁ Β πρὸς τὸν Δ· τοῖς Α, Β ἄρα τρίτος ἀριθμὸς itself. Thus, the (number created) from (multiplying) A, ἀνάλογον προσηύρηται ὁ Δ. D is equal to the (square) on B. Thus, as A is to B, (so) Ἀλλὰ δὴ μὴ μετρείτω ὁ Α τὸν Γ· λέγω, ὅτι τοῖς Α, Β B (is) to D [Prop. 7.19]. Thus, a third number has been ἀδύνατόν ἐστι τρίτον ἀνάλογον προσευρεῖν ἀριθμόν. εἰ γὰρ found proportional to A, B, (namely) D. δυνατόν, προσηυρήσθω ὁ Δ. ὁ ἄρα ἐκ τῶν Α, Δ ἴσος ἐστὶ And so let A not measure C. I say that it is impossi- τῷ ἀπὸ τοῦ Β. ὁ δὲ ἀπὸ τοῦ Β ἐστιν ὁ Γ· ὁ ἄρα ἐκ τῶν Α, ble to find a third number proportional to A, B. For, if Δ ἴσος ἐστὶ τῷ Γ. ὥστε ὁ Α τὸν Δ πολλαπλασιάσας τὸν possible, let it have been found, (and let it be) D. Thus, Γ πεποίηκεν· ὁ Α ἄρα τὸν Γ μετρεῖ κατὰ τὸν Δ. ἀλλα μὴν the (number created) from (multiplying) A, D is equal to ὑπόκειται καὶ μὴ μετρῶν· ὅπερ ἄτοπον. οὐκ ἄρα δυνατόν the (square) on B [Prop. 7.19]. And the (square) on B ἐστι τοῖς Α, Β τρίτον ἀνάλογον προσευρεῖν ἀριθμὸν, ὅταν is C. Thus, the (number created) from (multiplying) A, ὁ Α τὸν Γ μὴ μετρῇ· ὅπερ ἔδει δεῖξαι. D is equal to C. Hence, A has made C (by) multiplying D. Thus, A measures C according to D. But (A) was, in fact, also assumed (to be) not measuring (C). The very thing (is) absurd. Thus, it is not possible to find a third number proportional to A, B when A does not measure C. (Which is) the very thing it was required to show.ijþ. Proposition 19† Τριῶν ἀριθμῶν δοθέντων ἐπισκέψασθαι, πότε δυνατόν For three given numbers, to investigate when it is pos- ἐστιν αὐτοῖς τέταρτον ἀνάλογον προσευρεῖν. sible to find a fourth (number) proportional to them. Ε Β Α Γ ∆ C A B D E ῎Εστωσαν οἱ δοθέντες τρεῖς ἀριθμοὶ οἱ Α, Β, Γ, καὶ δέον Let A, B, C be the three given numbers. And let it be ἔστω επισκέψασθαι, πότε δυνατόν ἐστιν αὐτοῖς τέταρτον required to investigate when it is possible to find a fourth ἀνάλογον προσευρεῖν. (number) proportional to them. ῎Ητοι οὖν οὔκ εἰσιν ἑξῆς ἀνάλογον, καὶ οἱ ἄκροι αὐτῶν In fact, (A, B, C) are either not continuously pro- πρῶτοι πρὸς ἀλλήλους εἰσίν, ἢ ἑξῆς εἰσιν ἀνάλογον, καὶ οἱ portional and the outermost of them are prime to one ἄκροι αὐτῶν οὔκ εἰσι πρῶτοι πρὸς ἀλλήλους, ἢ οὕτε ἑξῆς another, or are continuously proportional and the outer- εἰσιν ἀνάλογον, οὔτε οἱ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους most of them are not prime to one another, or are neither εἰσίν, ἢ καὶ ἑξῆς εἰσιν ἀνάλογον, καὶ οἱ ἄκροι αὐτῶν πρῶτοι continuously proportional nor are the outermost of them πρὸς ἀλλήλους εἰσίν. prime to one another, or are continuously proportional Εἰ μὲν οὖν οἱ Α, Β, Γ ἑξῆς εἰσιν ἀνάλογον, καὶ οἱ and the outermost of them are prime to one another. ἄκροι αὐτῶν οἱ Α, Γ πρῶτοι πρὸς ἀλλήλους εἰσίν, δέδεικται, In fact, if A, B, C are continuously proportional, and ὅτι ἀδύνατόν ἐστιν αὐτοῖς τέταρτον ἀνάλογον προσευρεῖν the outermost of them, A and C, are prime to one an- ἀριθμόν. μὴ ἔστωσαν δὴ οἱ Α, Β, Γ ἑξῆς ἀνάλογον τῶν other, (then) it has (already) been shown that it is im- ἀκρῶν πάλιν ὄντων πρώτων πρὸς ἀλλήλους. λέγω, ὅτι possible to find a fourth number proportional to them καὶ οὕτως ἀδύνατόν ἐστιν αὐτοῖς τέταρτον ἀνάλογον προ- [Prop. 9.17]. So let A, B, C not be continuously propor- σευρεῖν. εἰ γὰρ δυνατόν, προσευρήσθω ὁ Δ, ὥστε εἶναι ὡς tional, (with) the outermost of them again being prime to τὸν Α πρὸς τὸν Β, τὸν Γ πρὸς τὸν Δ, καὶ γεγονέτω ὡς ὁ Β one another. I say that, in this case, it is also impossible πρὸς τὸν Γ, ὁ Δ πρὸς τὸν Ε. καὶ ἐπεί ἐστιν ὡς μὲν ὁ Α πρὸς to find a fourth (number) proportional to them. For, if τὸν Β, ὁ Γ πρὸς τὸν Δ, ὡς δὲ ὁ Β πρὸς τὸν Γ, ὁ Δ πρὸς possible, let it have been found, (and let it be) D. Hence, τὸν Ε, δι᾿ ἴσου ἄρα ὡς ὁ Α πρὸς τὸν Γ, ὁ Γ πρὸς τὸν Ε. οἱ it will be that as A (is) to B, (so) C (is) to D. And let it be δὲ Α, Γ πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι contrived that as B (is) to C, (so) D (is) to E. And since 269 STOIQEIWN jþ. ELEMENTS BOOK 9 μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ὅ τε ἡγούμενος as A is to B, (so) C (is) to D, and as B (is) to C, (so) D τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον. μετρεῖ ἄρα ὁ (is) to E, thus, via equality, as A (is) to C, (so) C (is) to E Α τὸν Γ ὡς ἡγούμενος ἡγούμενον. μετρεῖ δὲ καὶ ἑαυτόν· [Prop. 7.14]. And A and C (are) prime (to one another). ὁ Α ἄρα τοὺς Α, Γ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους· And (numbers) prime (to one another are) also the least ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοῖς Α, Β, Γ δυνατόν ἐστι (numbers having the same ratio as them) [Prop. 7.21]. τέταρτον ἀνάλογον προσευρεῖν. And the least (numbers) measure those numbers having Ἀλλά δὴ πάλιν ἔστωσαν οἱ Α, Β, Γ ἑξῆς ἀνάλογον, οἱ δὲ the same ratio as them (the same number of times), the Α, Γ μὴ ἔστωσαν πρῶτοι πρὸς ἀλλήλους. λέγω, ὅτι δυνατόν leading (measuring) the leading, and the following the ἐστιν αὐτοῖς τέταρτον ἀνάλογον προσευρεῖν. ὁ γὰρ Β τὸν Γ following [Prop. 7.20]. Thus, A measures C, (as) the πολλαπλασιάσας τὸν Δ ποιείτω· ὁ Α ἄρα τὸν Δ ἤτοι μετρεῖ leading (measuring) the leading. And it also measures ἢ οὐ μετρεῖ. μετρείτω αὐτὸν πρότερον κατὰ τὸν Ε· ὁ Α ἄρα itself. Thus, A measures A and C, which are prime to τὸν Ε πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ one another. The very thing is impossible. Thus, it is not Β τὸν Γ πολλαπλασιάσας τὸν Δ πεποίηκεν· ὁ ἄρα ἐκ τῶν possible to find a fourth (number) proportional to A, B, Α, Ε ἴσος ἐστὶ τῷ ἐκ τῶν Β, Γ. ἀνάλογον ἄρα [ἐστὶν] ὡς ὁ C. Α πρὸς τὸν Β, ὁ Γ πρὸς τὸν Ε· τοὶς Α, Β, Γ ἄρα τέταρτος And so let A, B, C again be continuously propor- ἀνάλογον προσηύρηται ὁ Ε. tional, and let A and C not be prime to one another. I Ἀλλὰ δὴ μὴ μετρείτω ὁ Α τὸν Δ· λέγω, ὅτι ἀδύνατόν say that it is possible to find a fourth (number) propor- ἐστι τοῖς Α, Β, Γ τέταρτον ἀνάλογον προσευρεῖν ἀριθμόν. tional to them. For let B make D (by) multiplying C. εἰ γὰρ δυνατόν, προσευρήσθω ὁ Ε· ὁ ἄρα ἐκ τῶν Α, Ε ἴσος Thus, A either measures or does not measure D. Let it, ἐστὶ τῷ ἐκ τῶν Β, Γ. ἀλλὰ ὁ ἑκ τῶν Β, Γ ἐστιν ὁ Δ· καὶ first of all, measure (D) according to E. Thus, A has ὁ ἐκ τῶν Α, Ε ἄρα ἴσος ἐστὶ τῷ Δ. ὁ Α ἄρα τὸν Ε πολλα- made D (by) multiplying E. But, in fact, B has also made πλασιάσας τὸν Δ πεποίηκεν· ὁ Α ἄρα τὸν Δ μετρεῖ κατὰ τὸν D (by) multiplying C. Thus, the (number created) from Ε· ὥστε μετρεῖ ὁ Α τὸν Δ. ἀλλὰ καὶ οὐ μετρεῖ· ὅπερ ἄτοπον. (multiplying) A, E is equal to the (number created) from οὐκ ἄρα δυνάτον ἐστι τοῖς Α, Β, Γ τέταρτον ἀνάλογον προ- (multiplying) B, C. Thus, proportionally, as A [is] to B, σευρεῖν ἀριθμόν, ὅταν ὁ Α τὸν Δ μὴ μετρῇ. ἀλλὰ δὴ οἱ Α, Β, (so) C (is) to E [Prop. 7.19]. Thus, a fourth (number) Γ μήτε ἑξῆς ἔστωσαν ἀνάλογον μήτε οἱ ἄκροι πρῶτοι πρὸς proportional to A, B, C has been found, (namely) E. ἀλλήλους. καὶ ὁ Β τὸν Γ πολλαπλασιάσας τὸν Δ ποιείτω. And so let A not measure D. I say that it is impossible ὁμοίως δὴ δειχθήσεται, ὅτι εἰ μὲν μετρεῖ ὁ Α τὸν Δ, δυ- to find a fourth number proportional to A, B, C. For, if νατόν ἐστιν αὐτοῖς ἀνάλογον προσευρεῖν, εἰ δὲ οὐ μετρεῖ, possible, let it have been found, (and let it be) E. Thus, ἀδύνατον· ὅπερ ἔδει δεῖξαι. the (number created) from (multiplying) A, E is equal to the (number created) from (multiplying) B, C. But, the (number created) from (multiplying) B, C is D. And thus the (number created) from (multiplying) A, E is equal to D. Thus, A has made D (by) multiplying E. Thus, A measures D according to E. Hence, A measures D. But, it also does not measure (D). The very thing (is) absurd. Thus, it is not possible to find a fourth number propor- tional to A, B, C when A does not measure D. And so (let) A, B, C (be) neither continuously proportional, nor (let) the outermost of them (be) prime to one another. And let B make D (by) multiplying C. So, similarly, it can be show that if A measures D then it is possible to find a fourth (number) proportional to (A, B, C), and impossible if (A) does not measure (D). (Which is) the very thing it was required to show. † The proof of this proposition is incorrect. There are, in fact, only two cases. Either A, B, C are continuously proportional, with A and C prime to one another, or not. In the first case, it is impossible to find a fourth proportional number. In the second case, it is possible to find a fourth proportional number provided that A measures B times C. Of the four cases considered by Euclid, the proof given in the second case is incorrect, since it only demonstrates that if A : B :: C : D then a number E cannot be found such that B : C :: D : E. The proofs given in the other three 270 STOIQEIWN jþ. ELEMENTS BOOK 9 cases are correct. kþ. Proposition 20 Οἱ πρῶτοι ἀριθμοὶ πλείους εἰσὶ παντὸς τοῦ προτεθέντος The (set of all) prime numbers is more numerous than πλήθους πρώτων ἀριθμῶν. any assigned multitude of prime numbers. Η Ε ∆ Ζ Α Β Γ FE A B C G D ῎Εστωσαν οἱ προτεθέντες πρῶτοι ἀριθμοὶ οἱ Α, Β, Γ· Let A, B, C be the assigned prime numbers. I say that λέγω, ὅτι τῶν Α, Β, Γ πλείους εἰσὶ πρῶτοι ἀριθμοί. the (set of all) primes numbers is more numerous than A, Εἰλήφθω γὰρ ὁ ὑπὸ τῶν Α, Β, Γ ἐλάχιστος μετρούμενος B, C. καὶ ἔστω ΔΕ, καὶ προσκείσθω τῷ ΔΕ μονὰς ἡ ΔΖ. ὁ δὴ ΕΖ For let the least number measured by A, B, C have ἤτοι πρῶτός ἐστιν ἢ οὔ. ἔστω πρότερον πρῶτος· εὐρημένοι been taken, and let it be DE [Prop. 7.36]. And let the ἄρα εἰσὶ πρῶτοι ἀριθμοὶ οἱ Α, Β, Γ, ΕΖ πλείους τῶν Α, Β, unit DF have been added to DE. So EF is either prime, Γ. or not. Let it, first of all, be prime. Thus, the (set of) Ἀλλὰ δὴ μὴ ἔστω ὁ ΕΖ πρῶτος· ὑπὸ πρώτου ἄρα τινὸς prime numbers A, B, C, EF , (which is) more numerous ἀριθμοῦ μετρεῖται. μετρείσθω ὑπὸ πρώτου τοῦ Η· λέγω, than A, B, C, has been found. ὅτι ὁ Η οὐδενὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. εἰ γὰρ δυνατόν, And so let EF not be prime. Thus, it is measured by ἔστω. οἱ δὲ Α, Β, Γ τὸν ΔΕ μετροῦσιν· καὶ ὁ Η ἄρα τὸν some prime number [Prop. 7.31]. Let it be measured by ΔΕ μετρήσει. μετρεῖ δὲ καὶ τὸν ΕΖ· καὶ λοιπὴν τὴν ΔΖ the prime (number) G. I say that G is not the same as μονάδα μετρήσει ὁ Η ἀριθμὸς ὤν· ὄπερ ἄτοπον. οὐκ ἄρα ὁ any of A, B, C. For, if possible, let it be (the same). And Η ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. καὶ ὑπόκειται πρῶτος. A, B, C (all) measure DE. Thus, G will also measure εὑρημένοι ἄρα εἰσὶ πρῶτοι ἀριθμοὶ πλείους τοῦ προτεθέντος DE. And it also measures EF . (So) G will also mea- πλήθους τῶν Α, Β, Γ οἱ Α, Β, Γ, Η· ὅπερ ἔδει δεῖξαι. sure the remainder, unit DF , (despite) being a number [Prop. 7.28]. The very thing (is) absurd. Thus, G is not the same as one of A, B, C. And it was assumed (to be) prime. Thus, the (set of) prime numbers A, B, C, G, (which is) more numerous than the assigned multitude (of prime numbers), A, B, C, has been found. (Which is) the very thing it was required to show.kaþ. Proposition 21 ᾿Εὰν ἄρτιοι ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, ὁ ὅλος If any multitude whatsoever of even numbers is added ἄρτιός ἐστιν. together then the whole is even. ΕΑ Β Γ ∆ C D EA B Συγκείσθωσαν γὰρ ἄρτιοι ἀριθμοὶ ὁποσοιοῦν οἱ ΑΒ, For let any multitude whatsoever of even numbers, ΒΓ, ΓΔ, ΔΕ· λέγω, ὅτι ὅλος ὁ ΑΕ ἄρτιός ἐστιν. AB, BC, CD, DE, lie together. I say that the whole, ᾿Επεὶ γὰρ ἕκαστος τῶν ΑΒ, ΒΓ, ΓΔ, ΔΕ ἄρτιός ἐστιν, AE, is even. ἔχει μέρος ἥμισυ· ὥστε καὶ ὅλος ὁ ΑΕ ἔχει μέρος ἥμισυ. For since everyone of AB, BC, CD, DE is even, it ἄρτιος δὲ ἀριθμός ἐστιν ὁ δίχα διαιρούμενος· ἄρτιος ἄρα has a half part [Def. 7.6]. And hence the whole AE has ἐστὶν ὁ ΑΕ· ὅπερ ἔδει δεῖξαι. a half part. And an even number is one (which can be) divided in half [Def. 7.6]. Thus, AE is even. (Which is) 271 STOIQEIWN jþ. ELEMENTS BOOK 9 the very thing it was required to show.kbþ. Proposition 22 ᾿Εὰν περισσοὶ ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, τὸ δὲ If any multitude whatsoever of odd numbers is added πλῆθος αὐτῶν ἄρτιον ᾖ, ὁ ὅλος ἄρτιος ἔσται. together, and the multitude of them is even, then the whole will be even. ΕΑ Β Γ ∆ C D EA B Συγκείσθωσαν γὰρ περισσοὶ ἀριθμοὶ ὁσοιδηποτοῦν For let any even multitude whatsoever of odd num- ἄρτιοι τὸ πλῆθος οἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ· λέγω, ὅτι ὅλος bers, AB, BC, CD, DE, lie together. I say that the ὁ ΑΕ ἄρτιός ἐστιν. whole, AE, is even. ᾿Επεὶ γὰρ ἕκαστος τῶν ΑΒ, ΒΓ, ΓΔ, ΔΕ περιττός ἐστιν, For since everyone of AB, BC, CD, DE is odd then, a ἀφαιρεθείσης μονάδος ἀφ᾿ ἑκάστου ἕκαστος τῶν λοιπῶν unit being subtracted from each, everyone of the remain- ἄρτιος ἔσται· ὥστε καὶ ὁ συγκείμενος ἐξ αὐτῶν ἄρτιος ders will be (made) even [Def. 7.7]. And hence the sum ἔσται. ἔστι δὲ καὶ τὸ πλῆθος τῶν μονάδων ἄρτιον. καὶ of them will be even [Prop. 9.21]. And the multitude ὅλος ἄρα ὁ ΑΕ ἄρτιός ἐστιν· ὅπερ ἔδει δεῖξαι. of the units is even. Thus, the whole AE is also even [Prop. 9.21]. (Which is) the very thing it was required to show.kgþ. Proposition 23 ᾿Εὰν περισσοὶ ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, τὸ δὲ If any multitude whatsoever of odd numbers is added πλῆθος αὐτῶν περισσὸν ᾖ, καὶ ὁ ὅλος περισσὸς ἔσται. together, and the multitude of them is odd, then the whole will also be odd. ΕΑ Β Γ ∆ EA B C D Συγκείσθωσαν γὰρ ὁποσοιοῦν περισσοὶ ἀριθμοί, ὧν τὸ For let any multitude whatsoever of odd numbers, πλῆθος περισσὸν ἔστω, οἱ ΑΒ, ΒΓ, ΓΔ· λέγω, ὅτι καὶ ὅλος AB, BC, CD, lie together, and let the multitude of them ὁ ΑΔ περισσός ἐστιν. be odd. I say that the whole, AD, is also odd. Ἀφῃρήσθω ἀπὸ τοῦ ΓΔ μονὰς ἡ ΔΕ· λοιπὸς ἄρα ὁ ΓΕ For let the unit DE have been subtracted from CD. ἄρτιός ἐστιν. ἔστι δὲ καὶ ὁ ΓΑ ἄρτιος· καὶ ὅλος ἄρα ὁ ΑΕ The remainder CE is thus even [Def. 7.7]. And CA ἄρτιός ἐστιν. καί ἐστι μονὰς ἡ ΔΕ. περισσὸς ἄρα ἐστὶν ὁ is also even [Prop. 9.22]. Thus, the whole AE is also ΑΔ· ὅπερ ἔδει δεῖξαι. even [Prop. 9.21]. And DE is a unit. Thus, AD is odd [Def. 7.7]. (Which is) the very thing it was required to show.kdþ. Proposition 24 ᾿Εὰν ἀπὸ ἀρτίου ἀριθμοῦ ἄρτιος ἀφαιρεθῇ, ὁ λοιπὸς If an even (number) is subtracted from an(other) even ἄρτιος ἔσται. number then the remainder will be even. Α Γ Β A C B Ἀπὸ γὰρ ἀρτίου τοῦ ΑΒ ἄρτιος ἀφῃρήσθω ὁ ΒΓ· λέγω, For let the even (number) BC have been subtracted ὅτι ὁ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν. from the even number AB. I say that the remainder CA ᾿Επεὶ γὰρ ὁ ΑΒ ἄρτιός ἐστιν, ἔχει μέρος ἥμισυ. διὰ τὰ is even. αὐτὰ δὴ καὶ ὁ ΒΓ ἔχει μέρος ἥμισυ· ὥστε καὶ λοιπὸς [ὁ ΓΑ For since AB is even, it has a half part [Def. 7.6]. So, ἔχει μέρος ἥμισυ] ἄρτιος [ἄρα] ἐστὶν ὁ ΑΓ· ὅπερ ἔδει δεῖξαι. for the same (reasons), BC also has a half part. And hence the remainder [CA has a half part]. [Thus,] AC is even. (Which is) the very thing it was required to show. 272 STOIQEIWN jþ. ELEMENTS BOOK 9 keþ. Proposition 25 ᾿Εὰν ἀπὸ ἀρτίου ἀριθμοῦ περισσὸς ἀφαιρεθῇ, ὁ λοιπὸς If an odd (number) is subtracted from an even num- περισσὸς ἔσται. ber then the remainder will be odd. ∆Α ΒΓ BA C D Ἀπὸ γὰρ ἀρτίου τοῦ ΑΒ περισσὸς ἀφῃρήσθω ὁ ΒΓ· For let the odd (number) BC have been subtracted λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ περισσός ἐστιν. from the even number AB. I say that the remainder CA Ἀφῃρήσθω γὰρ ἀπὸ τοῦ ΒΓ μονὰς ἡ ΓΔ· ὁ ΔΒ ἄρα is odd. ἄρτιός ἐστιν. ἔστι δὲ καὶ ὁ ΑΒ ἄρτιος· καὶ λοιπὸς ἄρα ὁ For let the unit CD have been subtracted from BC. ΑΔ ἄρτιός ἐστιν. καί ἐστι μονὰς ἡ ΓΔ· ὁ ΓΑ ἄρα περισσός DB is thus even [Def. 7.7]. And AB is also even. And ἐστιν· ὅπερ ἔδει δεῖξαι. thus the remainder AD is even [Prop. 9.24]. And CD is a unit. Thus, CA is odd [Def. 7.7]. (Which is) the very thing it was required to show.k�þ. Proposition 26 ᾿Εὰν ἀπὸ περισσοῦ ἀριθμοῦ περισσὸς ἀφαιρεθῇ, ὁ λοιπὸς If an odd (number) is subtracted from an odd number ἄρτιος ἔσται. then the remainder will be even. ΓΑ ∆ Β CA D B Ἀπὸ γὰρ περισσοῦ τοῦ ΑΒ περισσὸς ἀφῃρήσθω ὁ ΒΓ· For let the odd (number) BC have been subtracted λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν. from the odd (number) AB. I say that the remainder CA ᾿Επεὶ γὰρ ὁ ΑΒ περισσός ἐστιν, ἀφῃρήσθω μονὰς ἡ ΒΔ· is even. λοιπὸς ἄρα ὁ ΑΔ ἄρτιός ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ὁ ΓΔ For since AB is odd, let the unit BD have been ἄρτιός ἐστιν· ὥστε καὶ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν· ὅπερ ἔδει subtracted (from it). Thus, the remainder AD is even δεῖξαι. [Def. 7.7]. So, for the same (reasons), CD is also even. And hence the remainder CA is even [Prop. 9.24]. (Which is) the very thing it was required to show.kzþ. Proposition 27 ᾿Εὰν ἀπὸ περισσοῦ ἀριθμοῦ ἄρτιος ἀφαιρεθῇ, ὁ λοιπὸς If an even (number) is subtracted from an odd num- περισσὸς ἔσται. ber then the remainder will be odd. ΒΑ ∆ Γ CA D B Ἀπὸ γὰρ περισσοῦ τοῦ ΑΒ ἄρτιος ἀφῃρήσθω ὁ ΒΓ· For let the even (number) BC have been subtracted λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ περισσός ἐστιν. from the odd (number) AB. I say that the remainder CA Ἀφῃρήσθω [γὰρ] μονὰς ἡ ΑΔ· ὁ ΔΒ ἄρα ἄρτιός ἐστιν. is odd. ἔστι δὲ καὶ ὁ ΒΓ ἄρτιος· καὶ λοιπὸς ἄρα ὁ ΓΔ ἄρτιός ἐστιν. [For] let the unit AD have been subtracted (from AB). περισσὸς ἄρα ὁ ΓΑ· ὅπερ ἔδει δεῖξαι. DB is thus even [Def. 7.7]. And BC is also even. Thus, the remainder CD is also even [Prop. 9.24]. CA (is) thus odd [Def. 7.7]. (Which is) the very thing it was required to show.khþ. Proposition 28 ᾿Εὰν περισσὸς ἀριθμὸς ἄρτιον πολλαπλασιάσας ποιῇ If an odd number makes some (number by) multiply- τινα, ὁ γενόμενος ἄρτιος ἔσται. ing an even (number) then the created (number) will be even. 273 STOIQEIWN jþ. ELEMENTS BOOK 9 Α Γ Β C A B Περισσὸς γὰρ ἀριθμὸς ὁ Α ἄρτιον τὸν Β πολλα- For let the odd number A make C (by) multiplying πλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ ἄρτιός ἐστιν. the even (number) B. I say that C is even. ᾿Επεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, For since A has made C (by) multiplying B, C is thus ὁ Γ ἄρα σύγκειται ἐκ τοσούτων ἴσων τῷ Β, ὅσαι εἰσὶν ἐν composed out of so many (magnitudes) equal to B, as τῷ Α μονάδες. καί ἐστιν ὁ Β ἄρτιος· ὁ Γ ἄρα σύγκειται many as (there) are units in A [Def. 7.15]. And B is ἐξ ἀρτίων. ἐὰν δὲ ἄρτιοι ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, ὁ even. Thus, C is composed out of even (numbers). And ὅλος ἄρτιός ἐστιν. ἄρτιος ἄρα ἐστὶν ὁ Γ· ὅπερ ἔδει δεῖξαι. if any multitude whatsoever of even numbers is added together then the whole is even [Prop. 9.21]. Thus, C is even. (Which is) the very thing it was required to show.kjþ. Proposition 29 ᾿Εὰν περισσὸς ἀριθμὸς περισσὸν ἀριθμὸν πολλαπλασιάς- If an odd number makes some (number by) multiply- ας ποιῇ τινα, ὁ γενόμενος περισσὸς ἔσται. ing an odd (number) then the created (number) will be odd. Α Γ Β C A B Περισσὸς γὰρ ἀριθμὸς ὁ Α περισσὸν τὸν Β πολλα- For let the odd number A make C (by) multiplying πλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ περισσός ἐστιν. the odd (number) B. I say that C is odd. ᾿Επεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, For since A has made C (by) multiplying B, C is thus ὁ Γ ἄρα σύγκειται ἐκ τοσούτων ἴσων τῷ Β, ὅσαι εἰσὶν ἐν τῷ composed out of so many (magnitudes) equal to B, as Α μονάδες. καί ἐστιν ἑκάτερος τῶν Α, Β περισσός· ὁ Γ ἄρα many as (there) are units in A [Def. 7.15]. And each σύγκειται ἐκ περισσῶν ἀριθμῶν, ὧν τὸ πλῆθος περισσόν of A, B is odd. Thus, C is composed out of odd (num- ἐστιν. ὥστε ὁ Γ περισσός ἐστιν· ὅπερ ἔδει δεῖξαι. bers), (and) the multitude of them is odd. Hence C is odd [Prop. 9.23]. (Which is) the very thing it was required to show.lþ. Proposition 30 ᾿Εὰν περισσὸς ἀριθμὸς ἄρτιον ἀριθμὸν μετρῇ, καὶ τὸν If an odd number measures an even number then it ἥμισυν αὐτοῦ μετρήσει. will also measure (one) half of it. Γ Α Β A C B Περισσὸς γὰρ ἀριθμὸς ὁ Α ἄρτιον τὸν Β μετρείτω· λέγω, For let the odd number A measure the even (number) ὅτι καὶ τὸν ἥμισυν αὐτοῦ μετρήσει. B. I say that (A) will also measure (one) half of (B). ᾿Επεὶ γὰρ ὁ Α τὸν Β μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν For since A measures B, let it measure it according to Γ· λέγω, ὅτι ὁ Γ οὐκ ἔστι περισσός. εἰ γὰρ δυνατόν, ἔστω. C. I say that C is not odd. For, if possible, let it be (odd). καὶ ἐπεὶ ὁ Α τὸν Β μετρεῖ κατὰ τὸν Γ, ὁ Α ἄρα τὸν Γ And since A measures B according to C, A has thus made πολλαπλασιάσας τὸν Β πεποίηκεν. ὁ Β ἄρα σύγκειται ἐκ B (by) multiplying C. Thus, B is composed out of odd περισσῶν ἀριθμῶν, ὧν τὸ πλῆθος περισσόν ἐστιν. ὁ Β ἄρα numbers, (and) the multitude of them is odd. B is thus 274 STOIQEIWN jþ. ELEMENTS BOOK 9 περισσός ἐστιν· ὅπερ ἄτοπον· ὑπόκειται γὰρ ἄρτιος. οὐκ odd [Prop. 9.23]. The very thing (is) absurd. For (B) ἄρα ὁ Γ περισσός ἐστιν· ἄρτιος ἄρα ἐστὶν ὁ Γ. ὥστε ὁ Α was assumed (to be) even. Thus, C is not odd. Thus, C τὸν Β μετρεῖ ἀρτιάκις. διὰ δὴ τοῦτο καὶ τὸν ἥμισυν αὐτοῦ is even. Hence, A measures B an even number of times. μετρήσει· ὅπερ ἔδει δεῖξαι. So, on account of this, (A) will also measure (one) half of (B). (Which is) the very thing it was required to show.laþ. Proposition 31 ᾿Εὰν περισσὸς ἀριθμὸς πρός τινα ἀριθμὸν πρῶτος ᾖ, καὶ If an odd number is prime to some number then it will πρὸς τὸν διπλασίονα αὐτοῦ πρῶτος ἔσται. also be prime to its double. ∆ Α Β Γ D A B C Περισσὸς γὰρ ἀριθμὸς ὁ Α πρός τινα ἀριθμὸν τὸν Β For let the odd number A be prime to some number πρῶτος ἔστω, τοῦ δὲ Β διπλασίων ἔστω ὁ Γ· λέγω, ὅτι ὁ Α B. And let C be double B. I say that A is [also] prime to [καὶ] πρὸς τὸν Γ πρῶτός ἐστιν. C. Εἰ γὰρ μή εἰσιν [οἱ Α, Γ] πρῶτοι, μετρήσει τις αὐτοὺς For if [A and C] are not prime (to one another) then ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. καί ἐστιν ὁ Α περισσός· some number will measure them. Let it measure (them), περισσὸς ἄρα καὶ ὁ Δ. καὶ ἐπεὶ ὁ Δ περισσὸς ὢν τὸν Γ and let it be D. And A is odd. Thus, D (is) also odd. μετρεῖ, καί ἐστιν ὁ Γ ἄρτιος, καὶ τὸν ἥμισυν ἄρα τοῦ Γ And since D, which is odd, measures C, and C is even, μετρήσει [ὁ Δ]. τοῦ δὲ Γ ἥμισύ ἐστιν ὁ Β· ὁ Δ ἄρα τὸν [D] will thus also measure half of C [Prop. 9.30]. And B Β μετρεῖ. μετρεῖ δὲ καὶ τὸν Α. ὁ Δ ἄρα τοὺς Α, Β μετρεῖ is half of C. Thus, D measures B. And it also measures πρώτους ὄντας πρὸς ἀλλήλους· ὅπερ ἐστὶν ἀδύνατον. οὐκ A. Thus, D measures (both) A and B, (despite) them ἄρα ὁ Α πρὸς τὸν Γ πρῶτος οὔκ ἐστιν. οἱ Α, Γ ἄρα πρῶτοι being prime to one another. The very thing is impossible. πρὸς ἀλλήλους εἰσίν· ὅπερ ἔδει δεῖξαι. Thus, A is not unprime to C. Thus, A and C are prime to one another. (Which is) the very thing it was required to show.lbþ. Proposition 32 Τῶν ἀπὸ δύαδος διπλασιαζομένων ἀριθμων ἕκαστος Each of the numbers (which is continually) doubled, ἀρτιάκις ἄρτιός ἐστι μόνον. (starting) from a dyad, is an even-times-even (number) only. ∆ Α Β Γ D A B C Ἀπὸ γὰρ δύαδος τῆς Α δεδιπλασιάσθωσαν ὁσοιδη- For let any multitude of numbers whatsoever, B, C, ποτοῦν ἀριθμοὶ οἱ Β, Γ, Δ· λέγω, ὅτι οἱ Β, Γ, Δ ἀρτιάκις D, have been (continually) doubled, (starting) from the ἄρτιοί εἰσι μόνον. dyad A. I say that B, C, D are even-times-even (num- ῞Οτι μὲν οὖν ἕκαστος [τῶν Β, Γ, Δ] ἀρτιάκις ἄρτιός bers) only. ἐστιν, φανερόν· ἀπὸ γὰρ δυάδος ἐστὶ διπλασιασθείς. λέγω, In fact, (it is) clear that each [of B, C, D] is an ὅτι καὶ μόνον. ἐκκείσθω γὰρ μονάς. ἐπεὶ οὖν ἀπὸ μονάδος even-times-even (number). For it is doubled from a dyad ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ μετὰ τὴν [Def. 7.8]. I also say that (they are even-times-even num- μονάδα ὁ Α πρῶτός ἐστιν, ὁ μέγιστος τῶν Α, Β, Γ, Δ ὁ bers) only. For let a unit be laid down. Therefore, since 275 STOIQEIWN jþ. ELEMENTS BOOK 9 Δ ὑπ᾿ οὐδενὸς ἄλλου μετρηθήσεται παρὲξ τῶν Α, Β, Γ. καί any multitude of numbers whatsoever are continuously ἐστιν ἕκαστος τῶν Α, Β, Γ ἄρτιος· ὁ Δ ἄρα ἀρτιάκις ἄρτιός proportional, starting from a unit, and the (number) A af- ἐστι μόνον. ὁμοίως δὴ δείξομεν, ὅτι [καὶ] ἑκάτερος τῶν Β, ter the unit is prime, the greatest of A, B, C, D, (namely) Γ ἀρτιάκις ἄρτιός ἐστι μόνον· ὅπερ ἔδει δεῖξαι. D, will not be measured by any other (numbers) except A, B, C [Prop. 9.13]. And each of A, B, C is even. Thus, D is an even-time-even (number) only [Def. 7.8]. So, similarly, we can show that each of B, C is [also] an even- time-even (number) only. (Which is) the very thing it was required to show.lgþ. Proposition 33 ᾿Εὰν ἀριθμὸς τὸν ἥμισυν ἔχῃ περισσόν, ἀρτιάκις πε- If a number has an odd half then it is an even-time- ρισσός ἐστι μόνον. odd (number) only. Α A Ἀριθμὸς γὰρ ὁ Α τὸν ἥμισυν ἐχέτω περισσόν· λέγω, ὅτι For let the number A have an odd half. I say that A is ὁ Α ἀρτιάκις περισσός ἐστι μόνον. an even-times-odd (number) only. ῞Οτι μὲν οὖν ἀρτιάκις περισσός ἐστιν, φανερόν· ὁ γὰρ In fact, (it is) clear that (A) is an even-times-odd ἥμισυς αὐτοῦ περισσὸς ὢν μετρεῖ αὐτὸν ἀρτιάκις, λέγω δή, (number). For its half, being odd, measures it an even ὅτι καὶ μόνον. εἰ γὰρ ἔσται ὁ Α καὶ ἀρτιάκις ἄρτιος, με- number of times [Def. 7.9]. So I also say that (it is τρηθήσεται ὑπὸ ἀρτίου κατὰ ἄρτιον ἀριθμόν· ὥστε καὶ ὁ an even-times-odd number) only. For if A is also an ἥμισυς αὐτοῦ μετρηθήσεται ὑπὸ ἀρτίου ἀριθμοῦ περισσὸς even-times-even (number) then it will be measured by an ὤν· ὅπερ ἐστὶν ἄτοπον. ὁ Α ἄρα ἀρτιάκις περισσός ἐστι even (number) according to an even number [Def. 7.8]. μόνον· ὅπερ ἔδει δεῖξαι. Hence, its half will also be measured by an even number, (despite) being odd. The very thing is absurd. Thus, A is an even-times-odd (number) only. (Which is) the very thing it was required to show.ldþ. Proposition 34 ᾿Εὰν ἀριθμὸς μήτε τῶν ἀπὸ δυάδος διπλασιαζομένων ᾖ, If a number is neither (one) of the (numbers) doubled μήτε τὸν ἥμισυν ἔχῃ περισσόν, ἀρτιάκις τε ἄρτιός ἐστι καὶ from a dyad, nor has an odd half, then it is (both) an ἀρτιάκις περισσός. even-times-even and an even-times-odd (number). Α A Ἀριθμὸς γὰρ ὁ Α μήτε τῶν ἀπὸ δυάδος διπλασιαζομένων For let the number A neither be (one) of the (num- ἔστω μήτε τὸν ἥμισυν ἐχέτω περισσόν· λέγω, ὅτι ὁ Α bers) doubled from a dyad, nor let it have an odd half. ἀρτιάκις τέ ἐστιν ἄρτιος καὶ ἀρτιάκις περισσός. I say that A is (both) an even-times-even and an even- ῞Οτι μὲν οὖν ὁ Α ἀρτιάκις ἐστὶν ἄρτιος, φανερόν· τὸν times-odd (number). γὰρ ἥμισυν οὐκ ἔχει περισσόν. λέγω δή, ὅτι καὶ ἀρτιάκις πε- In fact, (it is) clear that A is an even-times-even (num- ρισσός ἐστιν. ἐὰν γὰρ τὸν Α τέμνωμεν δίχα καὶ τὸν ἥμισυν ber) [Def. 7.8]. For it does not have an odd half. So I αὐτοῦ δίχα καὶ τοῦτο ἀεὶ ποιῶμεν, καταντήσομεν εἴς τινα say that it is also an even-times-odd (number). For if we ἀριθμὸν περισσόν, ὃς μετρήσει τὸν Α κατὰ ἄρτιον ἀριθμόν. cut A in half, and (then cut) its half in half, and we do εἰ γὰρ οὔ, καταντήσομεν εἰς δυάδα, καὶ ἔσται ὁ Α τῶν ἀπὸ this continually, then we will arrive at some odd num- δυάδος διπλασιαζομένων· ὅπερ οὐχ ὑπόκειται. ὥστε ὁ Α ber which will measure A according to an even number. ἀρτιάκις περισσόν ἐστιν. ἐδείχθη δὲ καὶ ἀρτιάκις ἄρτιος. ὁ For if not, we will arrive at a dyad, and A will be (one) Α ἄρα ἀρτιάκις τε ἄρτιός ἐστι καὶ ἀρτιάκις περισσός· ὅπερ of the (numbers) doubled from a dyad. The very oppo- ἔδει δεῖξαι. site thing (was) assumed. Hence, A is an even-times-odd (number) [Def. 7.9]. And it was also shown (to be) an even-times-even (number). Thus, A is (both) an even- times-even and an even-times-odd (number). (Which is) 276 STOIQEIWN jþ. ELEMENTS BOOK 9 the very thing it was required to show.leþ. Proposition 35† ᾿Εὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, ἀφαι- If there is any multitude whatsoever of continually ρεθῶσι δὲ ἀπό τε τοῦ δευτέρου καὶ τοῦ ἐσχάτου ἴσοι τῷ proportional numbers, and (numbers) equal to the first πρώτῳ, ἔσται ὡς ἡ τοῦ δευτέρου ὑπεροχὴ πρὸς τὸν πρῶτον, are subtracted from (both) the second and the last, then οὕτως ἡ τοῦ ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ as the excess of the second (number is) to the first, so the πάντας. excess of the last will be to (the sum of) all those (num- bers) before it. Λ Α ∆ Ε Β Γ ΖΘ Η Κ LE B A D C F G HK ῎Εστωσαν ὁποσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Let A, BC, D, EF be any multitude whatsoever of ΒΓ, Δ, ΕΖ ἀφχόμενοι ἀπὸ ἐλαχίστου τοῦ Α, καὶ ἀφῃρήσθω continuously proportional numbers, beginning from the ἀπὸ τοῦ ΒΓ καὶ τοῦ ΕΖ τῲ Α ἴσος ἑκάτερος τῶν ΒΗ, ΖΘ· least A. And let BG and FH , each equal to A, have been λέγω, ὅτι ἐστὶν ὡς ὁ ΗΓ πρὸς τὸν Α, οὕτως ὁ ΕΘ πρὸς subtracted from BC and EF (respectively). I say that as τοὺς Α, ΒΓ, Δ. GC is to A, so EH is to A, BC, D. Κείσθω γὰρ τῷ μὲν ΒΓ ἴσος ὁ ΖΚ, τῷ δὲ Δ ἴσος ὁ ΖΛ. For let FK be made equal to BC, and FL to D. And καὶ ἐπεὶ ὁ ΖΚ τῷ ΒΓ ἴσος ἐστίν, ὧν ὁ ΖΘ τῷ ΒΗ ἴσος ἐστίν, since FK is equal to BC, of which FH is equal to BG, λοιπὸς ἄρα ὁ ΘΚ λοιπῷ τῷ ΗΓ ἐστιν ἴσος. καὶ ἐπεί ἐστιν ὡς the remainder HK is thus equal to the remainder GC. ὁ ΕΖ πρὸς τὸν Δ, οὕτως ὁ Δ πρὸς τὸν ΒΓ καὶ ὁ ΒΓ πρὸς And since as EF is to D, so D (is) to BC, and BC to τὸν Α, ἴσος δὲ ὁ μὲν Δ τῷ ΖΛ, ὁ δὲ ΒΓ τῷ ΖΚ, ὁ δὲ Α τῷ A [Prop. 7.13], and D (is) equal to FL, and BC to FK, ΖΘ, ἔστιν ἄρα ὡς ὁ ΕΖ πρὸς τὸν ΖΛ, οὕτως ὁ ΛΖ πρὸς τὸν and A to FH , thus as EF is to FL, so LF (is) to FK, and ΖΚ καὶ ὁ ΖΚ πρὸς τὸν ΖΘ. διελόντι, ὡς ὁ ΕΛ πρὸς τὸν ΛΖ, FK to FH . By separation, as EL (is) to LF , so LK (is) οὕτως ὁ ΛΚ πρὸς τὸν ΖΚ καὶ ὁ ΚΘ πρὸς τὸν ΖΘ. ἔστιν ἄρα to FK, and KH to FH [Props. 7.11, 7.13]. And thus as καὶ ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως one of the leading (numbers) is to one of the following, ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους· ἔστιν so (the sum of) all of the leading (numbers is) to (the ἄρα ὡς ὁ ΚΘ πρὸς τὸν ΖΘ, οὕτως οἱ ΕΛ, ΛΚ, ΚΘ πρὸς sum of) all of the following [Prop. 7.12]. Thus, as KH τοὺς ΛΖ, ΖΚ, ΘΖ. ἴσος δὲ ὁ μὲν ΚΘ τῷ ΓΗ, ὁ δὲ ΖΘ τῷ is to FH , so EL, LK, KH (are) to LF , FK, HF . And Α, οἱ δὲ ΛΖ, ΖΚ, ΘΖ τοὶς Δ, ΒΓ, Α· ἔστιν ἄρα ὡς ὁ ΓΗ KH (is) equal to CG, and FH to A, and LF , FK, HF πρὸς τὸν Α, οὕτως ὁ ΕΘ πρὸς τοὺς Δ, ΒΓ, Α. ἔστιν ἄρα to D, BC, A. Thus, as CG is to A, so EH (is) to D, ὡς ἡ τοῦ δευτέρου ὑπεροχὴ πρὸς τὸν πρῶτον, οὕτως ἡ τοῦ BC, A. Thus, as the excess of the second (number) is to ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ πάντας· ὅπερ ἔδει the first, so the excess of the last (is) to (the sum of) all δεῖξαι. those (numbers) before it. (Which is) the very thing it was required to show. † This proposition allows us to sum a geometric series of the form a, a r, a r2, a r3, · · · a rn−1. According to Euclid, the sum Sn satisfies (a r − a)/a = (a rn − a)/Sn . Hence, Sn = a (rn − 1)/(r − 1).l�þ. Proposition 36† ᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἐκτεθῶσιν ἐν If any multitude whatsoever of numbers is set out con- τῇ διπλασίονι ἀναλογίᾳ, ἕως οὗ ὁ σύμπας συντεθεὶς πρῶτος tinuously in a double proportion, (starting) from a unit, γένηται, καὶ ὁ σύμπας ἐπὶ τὸν ἔσχατον πολλαπλασιασθεὶς until the whole sum added together becomes prime, and 277 STOIQEIWN jþ. ELEMENTS BOOK 9 ποιῇ τινα, ὁ γενόμενος τέλειος ἔσται. the sum multiplied into the last (number) makes some Ἀπὸ γὰρ μονάδος ἐκκείσθωσαν ὁσοιδηποτοῦν ἀριθμ- (number), then the (number so) created will be perfect. οὶ ἐν τῇ διπλασίονι ἀναλογίᾳ, ἕως οὗ ὁ σύμπας συντεθεὶς For let any multitude of numbers, A, B, C, D, be set πρῶτος γένηται, οἱ Α, Β, Γ, Δ, καὶ τῷ σύμπαντι ἴσος ἔστω out (continuouly) in a double proportion, until the whole ὁ Ε, καὶ ὁ Ε τὸν Δ πολλαπλασιάσας τὸν ΖΗ ποιείτω. λέγω, sum added together is made prime. And let E be equal to ὅτι ὁ ΖΗ τέλειός ἐστιν. the sum. And let E make FG (by) multiplying D. I say that FG is a perfect (number). Α Β ∆ Γ D A B C Ξ Ε Λ Μ Ζ Η Ν ΚΘ Ο Π G N K E L M P Q F O H ῞Οσοι γάρ εἰσιν οἱ Α, Β, Γ, Δ τῷ πλήθει, τοσοῦτοι ἀπὸ For as many as is the multitude of A, B, C, D, let so τοῦ Ε εἰλήφθωσαν ἐν τῇ διπλασίονι ἀναλογίᾳ οἱ Ε, ΘΚ, Λ, many (numbers), E, HK, L, M , have been taken in a Μ· δι᾿ ἴσου ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς double proportion, (starting) from E. Thus, via equal- τὸν Μ. ὁ ἄρα ἐκ τῶν Ε, Δ ἴσος ἐστὶ τῷ ἐκ τῶν Α, Μ. καί ity, as A is to D, so E (is) to M [Prop. 7.14]. Thus, the ἐστιν ὁ ἐκ τῶν Ε, Δ ὁ ΖΗ· καὶ ὁ ἐκ τῶν Α, Μ ἄρα ἐστὶν ὁ (number created) from (multiplying) E, D is equal to the ΖΗ. ὁ Α ἄρα τὸν Μ πολλαπλασιάσας τὸν ΖΗ πεποίηκεν· ὁ (number created) from (multiplying) A, M . And FG is Μ ἄρα τὸν ΖΗ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. καί ἐστι the (number created) from (multiplying) E, D. Thus, δυὰς ὁ Α· διπλάσιος ἄρα ἐστὶν ὁ ΖΗ τοῦ Μ. εἰσὶ δὲ καὶ οἱ Μ, FG is also the (number created) from (multiplying) A, Λ, ΘΚ, Ε ἑξῆς διπλάσιοι ἀλλήλων· οἱ Ε, ΘΚ, Λ, Μ, ΖΗ ἄρα M [Prop. 7.19]. Thus, A has made FG (by) multiplying ἑξῆς ἀνάλογόν εἰσιν ἐν τῇ διπλασίονι ἀναλογίᾳ. ἀφῃρήσθω M . Thus, M measures FG according to the units in A. δὴ ἀπὸ τοῦ δευτέρου τοῦ ΘΚ καὶ τοῦ ἐσχάτου τοῦ ΖΗ τῷ And A is a dyad. Thus, FG is double M . And M , L, πρώτῳ τῷ Ε ἴσος ἑκάτερος τῶν ΘΝ, ΖΞ· ἔστιν ἄρα ὡς ἡ HK, E are also continuously double one another. Thus, τοῦ δευτέρου ἀριθμοῦ ὑπεροχὴ πρὸς τὸν πρῶτον, οὕτως ἡ E, HK, L, M , FG are continuously proportional in a τοῦ ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ πάντας. ἔστιν double proportion. So let HN and FO, each equal to the ἄρα ὡς ὁ ΝΚ πρὸς τὸν Ε, οὕτως ὁ ΞΗ πρὸς τοὺς Μ, Λ, first (number) E, have been subtracted from the second ΚΘ, Ε. καί ἐστιν ὁ ΝΚ ἴσος τῷ Ε· καὶ ὁ ΞΗ ἄρα ἴσος ἐστὶ (number) HK and the last FG (respectively). Thus, as τοῖς Μ, Λ, ΘΚ, Ε. ἔστι δὲ καὶ ὁ ΖΞ τῷ Ε ἴσος, ὁ δὲ Ε the excess of the second number is to the first, so the ex- τοῖς Α, Β, Γ, Δ καὶ τῇ μονάδι. ὅλος ἄρα ὁ ΖΗ ἴσος ἐστὶ cess of the last (is) to (the sum of) all those (numbers) τοῖς τε Ε, ΘΚ, Λ, Μ καὶ τοῖς Α, Β, Γ, Δ καὶ τῇ μονάδι· before it [Prop. 9.35]. Thus, as NK is to E, so OG (is) καὶ μετρεῖται ὑπ᾿ αὐτῶν. λέγω, ὅτι καὶ ὁ ΖΗ ὐπ᾿ οὐδενὸς to M , L, KH , E. And NK is equal to E. And thus OG ἄλλου μετρηθήσεται παρὲξ τῶν Α, Β, Γ, Δ, Ε, ΘΚ, Λ, Μ is equal to M , L, HK, E. And FO is also equal to E, καὶ τῆς μονάδος. εἰ γὰρ δυνατόν, μετρείτω τις τὸν ΖΗ ὁ and E to A, B, C, D, and a unit. Thus, the whole of FG Ο, καὶ ὁ Ο μηδενὶ τῶν Α, Β, Γ, Δ, Ε, ΘΚ, Λ, Μ ἔστω ὁ is equal to E, HK, L, M , and A, B, C, D, and a unit. αὐτός. καὶ ὁσάκις ὁ Ο τὸν ΖΗ μετρεῖ, τοσαῦται μονάδες And it is measured by them. I also say that FG will be 278 STOIQEIWN jþ. ELEMENTS BOOK 9 ἔστωσαν ἐν τῷ Π· ὁ Π ἄρα τὸν Ο πολλαπλασιάσας τὸν ΖΗ measured by no other (numbers) except A, B, C, D, E, πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Ε τὸν Δ πολλαπλασιάσας τὸν HK, L, M , and a unit. For, if possible, let some (num- ΖΗ πεποίηκεν· ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Π, ὁ Ο πρὸς τὸν ber) P measure FG, and let P not be the same as any Δ. καὶ ἐπεὶ ἀπὸ μονάδος ἑξῆς ἀνάλογόν εἰσιν οἱ Α, Β, Γ, of A, B, C, D, E, HK, L, M . And as many times as P Δ, ὁ Δ ἄρα ὑπ᾿ οὐδενὸς ἄλλου ἀριθμοῦ μετρηθήσεται παρὲξ measures FG, so many units let there be in Q. Thus, Q τῶν Α, Β, Γ. καὶ ὑπόκειται ὁ Ο οὐδενὶ τῶν Α, Β, Γ ὁ αὐτός· has made FG (by) multiplying P . But, in fact, E has also οὐκ ἄρα μετρήσει ὁ Ο τὸν Δ. ἀλλ᾿ ὡς ὁ Ο πρὸς τὸν Δ, ὁ made FG (by) multiplying D. Thus, as E is to Q, so P Ε πρὸς τὸν Π· οὐδὲ ὁ Ε ἄρα τὸν Π μετρεῖ. καί ἐστιν ὁ Ε (is) to D [Prop. 7.19]. And since A, B, C, D are con- πρῶτος· πᾶς δὲ πρῶτος ἀριθμὸς πρὸς ἅπαντα, ὃν μὴ μετρεῖ, tinually proportional, (starting) from a unit, D will thus πρῶτός [ἐστιν]. οἱ Ε, Π ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. οἱ not be measured by any other numbers except A, B, C δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν [Prop. 9.13]. And P was assumed not (to be) the same αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον as any of A, B, C. Thus, P does not measure D. But, καὶ ὁ ἑπόμενος τὸν ἑπόμενον· καί ἐστιν ὡς ὁ Ε πρὸς τὸν Π, as P (is) to D, so E (is) to Q. Thus, E does not mea- ὁ Ο πρὸς τὸν Δ. ἰσάκις ἄρα ὁ Ε τὸν Ο μετρεῖ καὶ ὁ Π τὸν sure Q either [Def. 7.20]. And E is a prime (number). Δ. ὁ δὲ Δ ὑπ᾿ οὐδενὸς ἄλλου μετρεῖται παρὲξ τῶν Α, Β, Γ· And every prime number [is] prime to every (number) ὁ Π ἄρα ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. ἔστω τῷ Β ὁ αὐτός. which it does not measure [Prop. 7.29]. Thus, E and Q καὶ ὅσοι εἰσὶν οἱ Β, Γ, Δ τῷ πλήθει τοσοῦτοι εἰλήφθωσαν are prime to one another. And (numbers) prime (to one ἀπὸ τοῦ Ε οἱ Ε, ΘΚ, Λ. καί εἰσιν οἱ Ε, ΘΚ, Λ τοῖς Β, Γ, Δ another are) also the least (of those numbers having the ἐν τῷ αὐτῷ λόγω· δι᾿ ἴσου ἄρα ἐστὶν ὡς ὁ Β πρὸς τὸν Δ, ὁ same ratio as them) [Prop. 7.21], and the least (num- Ε πρὸς τὸν Λ. ὁ ἄρα ἐκ τῶν Β, Λ ἴσος ἐστὶ τῷ ἐκ τῶν Δ, bers) measure those (numbers) having the same ratio as Ε· ἀλλ᾿ ὁ ἐκ τῶν Δ, Ε ἴσος ἐστὶ τῷ ἐκ τῶν Π, Ο· καὶ ὁ ἐκ them an equal number of times, the leading (measuring) τῶν Π, Ο ἄρα ἴσος ἐστὶ τῷ ἐκ τῶν Β, Λ. ἔστιν ἄρα ὡς ὁ Π the leading, and the following the following [Prop. 7.20]. πρὸς τὸν Β, ὁ Λ πρὸς τὸν Ο. καί ἐστιν ὁ Π τῷ Β ὁ αὐτός· And as E is to Q, (so) P (is) to D. Thus, E measures P καὶ ὁ Λ ἄρα τῳ Ο ἐστιν ὁ αὐτός· ὅπερ ἀδύνατον· ὁ γὰρ Ο the same number of times as Q (measures) D. And D ὑπόκειται μηδενὶ τῶν ἐκκειμένων ὁ αὐτός· οὐκ ἄρα τὸν ΖΗ is not measured by any other (numbers) except A, B, C. μετρήσει τις ἀριθμὸς παρὲξ τῶν Α, Β, Γ, Δ, Ε, ΘΚ, Λ, Μ Thus, Q is the same as one of A, B, C. Let it be the same καὶ τῆς μονάδος. καὶ ἐδείχη ὁ ΖΗ τοῖς Α, Β, Γ, Δ, Ε, ΘΚ, as B. And as many as is the multitude of B, C, D, let so Λ, Μ καὶ τῇ μονάδι ἴσος. τέλειος δὲ ἀριθμός ἐστιν ὁ τοῖς many (of the set out numbers) have been taken, (start- ἑαυτοῦ μέρεσιν ἴσος ὤν· τέλειος ἄρα ἐστὶν ὁ ΖΗ· ὅπερ ἔδει ing) from E, (namely) E, HK, L. And E, HK, L are in δεῖξαι. the same ratio as B, C, D. Thus, via equality, as B (is) to D, (so) E (is) to L [Prop. 7.14]. Thus, the (number created) from (multiplying) B, L is equal to the (num- ber created) from multiplying D, E [Prop. 7.19]. But, the (number created) from (multiplying) D, E is equal to the (number created) from (multiplying) Q, P . Thus, the (number created) from (multiplying) Q, P is equal to the (number created) from (multiplying) B, L. Thus, as Q is to B, (so) L (is) to P [Prop. 7.19]. And Q is the same as B. Thus, L is also the same as P . The very thing (is) impossible. For P was assumed not (to be) the same as any of the (numbers) set out. Thus, FG cannot be measured by any number except A, B, C, D, E, HK, L, M , and a unit. And FG was shown (to be) equal to (the sum of) A, B, C, D, E, HK, L, M , and a unit. And a perfect number is one which is equal to (the sum of) its own parts [Def. 7.22]. Thus, FG is a perfect (number). (Which is) the very thing it was required to show. † This proposition demonstrates that perfect numbers take the form 2n−1 (2n − 1) provided that 2n − 1 is a prime number. The ancient Greeks knew of four perfect numbers: 6, 28, 496, and 8128, which correspond to n = 2, 3, 5, and 7, respectively. 279 280 ELEMENTS BOOK 10 Incommensurable Magnitudes† †The theory of incommensurable magntidues set out in this book is generally attributed to Theaetetus of Athens. In the footnotes throughout this book, k, k′, etc. stand for distinct ratios of positive integers. 281 STOIQEIWN iþ. ELEMENTS BOOK 10VOroi. Definitions αʹ. Σύμμετρα μεγέθη λέγεται τὰ τῷ αὐτῷ μετρῳ με- 1. Those magnitudes measured by the same measure τρούμενα, ἀσύμμετρα δέ, ὧν μηδὲν ἐνδέχεται κοινὸν μέτρον are said (to be) commensurable, but (those) of which no γενέσθαι. (magnitude) admits to be a common measure (are said βʹ. Εὐθεῖαι δυνάμει σύμμετροί εἰσιν, ὅταν τὰ ἀπ᾿ αὐτῶν to be) incommensurable.† τετράγωνα τῷ αὐτῷ χωρίῳ μετρῆται, ἀσύμμετροι δέ, ὅταν 2. (Two) straight-lines are commensurable in square‡ τοῖς ἀπ᾿ αὐτῶν τετραγώνοις μηδὲν ἐνδέχηται χωρίον κοινὸν when the squares on them are measured by the same μέτρον γενέσθαι. area, but (are) incommensurable (in square) when no γʹ. Τούτων ὑποκειμένων δείκνυται, ὅτι τῇ προτεθείσῃ area admits to be a common measure of the squares on εὐθείᾳ ὑπάρχουσιν εὐθεῖαι πλήθει ἄπειροι σύμμετροί τε καὶ them.§ ἀσύμμετροι αἱ μὲν μήκει μόνον, αἱ δὲ καὶ δυνάμει. καλείσθω 3. These things being assumed, it is proved that there οὖν ἡ μὲν προτεθεῖσα εὐθεῖα ῥητή, καὶ αἱ ταύτῃ σύμμετροι exist an infinite multitude of straight-lines commensu- εἴτε μήκει καὶ δυνάμει εἴτε δυνάμει μόνον ῥηταί, αἱ δὲ ταύτῃ rable and incommensurable with an assigned straight- ἀσύμμετροι ἄλογοι καλείσθωσαν. line—those (incommensurable) in length only, and those δʹ. Καὶ τὸ μὲν ἀπὸ τῆς προτεθείσης εὐθείας τετράγω- also (commensurable or incommensurable) in square.¶ νον ῥητόν, καὶ τὰ τούτῳ σύμμετρα ῥητά, τὰ δὲ τούτῳ Therefore, let the assigned straight-line be called ratio- ἀσύμμετρα ἄλογα καλείσθω, καὶ αἱ δυνάμεναι αὐτὰ ἄλογοι, nal. And (let) the (straight-lines) commensurable with it, εἰ μὲν τετράγωνα εἴη, αὐταὶ αἱ πλευραί, εἰ δὲ ἕτερά τινα either in length and square, or in square only, (also be εὐθύγραμμα, αἱ ἴσα αὐτοῖς τετράγωνα ἀναγράφουσαι. called) rational. But let the (straight-lines) incommensu- rable with it be called irrational.∗ 4. And let the square on the assigned straight-line be called rational. And (let areas) commensurable with it (also be called) rational. But (let areas) incommensu- rable with it (be called) irrational, and (let) their square- roots$ (also be called) irrational—the sides themselves, if the (areas) are squares, and the (straight-lines) describ- ing squares equal to them, if the (areas) are some other rectilinear (figure).‖ † In other words, two magnitudes α and β are commensurable if α : β :: 1 : k, and incommensurable otherwise. ‡ Literally, “in power”. § In other words, two straight-lines of length α and β are commensurable in square if α : β :: 1 : k1/2, and incommensurable in square otherwise. Likewise, the straight-lines are commensurable in length if α : β :: 1 : k, and incommensurable in length otherwise. ¶ To be more exact, straight-lines can either be commensurable in square only, incommensurable in length only, or commenusrable/incommensurable in both length and square, with an assigned straight-line. ∗ Let the length of the assigned straight-line be unity. Then rational straight-lines have lengths expressible as k or k1/2, depending on whether the lengths are commensurable in length, or in square only, respectively, with unity. All other straight-lines are irrational. $ The square-root of an area is the length of the side of an equal area square. ‖ The area of the square on the assigned straight-line is unity. Rational areas are expressible as k. All other areas are irrational. Thus, squares whose sides are of rational length have rational areas, and vice versa.aþ. Proposition 1† Δύο μεγεθῶν ἀνίσων ἐκκειμένων, ἐὰν ἀπὸ τοῦ μείζονος If, from the greater of two unequal magnitudes ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου μεῖζον (which are) laid out, (a part) greater than half is sub- ἢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ γίγνηται, λειφθήσεταί τι μέγεθος, tracted, and (if from) the remainder (a part) greater than ὃ ἔσται ἔλασσον τοῦ ἐκκειμένου ἐλάσσονος μεγέθους. half (is subtracted), and (if) this happens continually, ῎Εστω δύο μεγέθη ἄνισα τὰ ΑΒ, Γ, ὧν μεῖζον τὸ ΑΒ· then some magnitude will (eventually) be left which will 282 STOIQEIWN iþ. ELEMENTS BOOK 10 λέγω, ὅτι, ἐαν ἀπὸ τοῦ ΑΒ ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ be less than the lesser laid out magnitude. καὶ τοῦ καταλειπομένου μεῖζον ἢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ Let AB and C be two unequal magnitudes, of which γίγνηται, λειφθήσεταί τι μέγεθος, ὃ ἔσται ἔλασσον τοῦ Γ (let) AB (be) the greater. I say that if (a part) greater μεγέθους. than half is subtracted from AB, and (if a part) greater than half (is subtracted) from the remainder, and (if) this happens continually, then some magnitude will (eventu- ally) be left which will be less than the magnitude C. Θ Γ ΕΗΖ∆ Α Κ Β G ED A B C F K H Τὸ Γ γὰρ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ ΑΒ For C, when multiplied (by some number), will some- μεῖζον. πεπολλαπλασιάσθω, καὶ ἔστω τὸ ΔΕ τοῦ μὲν Γ times be greater than AB [Def. 5.4]. Let it have been πολλαπλάσιον, τοῦ δὲ ΑΒ μεῖζον, καὶ διῃρήσθω τὸ ΔΕ εἰς (so) multiplied. And let DE be (both) a multiple of C, τὰ τῷ Γ ἴσα τὰ ΔΖ, ΖΗ, ΗΕ, καὶ ἀφῃρήσθω ἀπὸ μὲν τοῦ and greater than AB. And let DE have been divided into ΑΒ μεῖζον ἢ τὸ ἥμισυ τὸ ΒΘ, ἀπὸ δὲ τοῦ ΑΘ μεῖζον ἢ τὸ the (divisions) DF , FG, GE, equal to C. And let BH , ἥμισυ τὸ ΘΚ, καὶ τοῦτο ἀεὶ γιγνέσθω, ἕως ἂν αἱ ἐν τῷ ΑΒ (which is) greater than half, have been subtracted from διαιρέσεις ἰσοπληθεῖς γένωνται ταῖς ἐν τῷ ΔΕ διαιρέσεσιν. AB. And (let) HK, (which is) greater than half, (have ῎Εστωσαν οὖν αἱ ΑΚ, ΚΘ, ΘΒ διαιρέσεις ἰσοπληθεῖς been subtracted) from AH . And let this happen continu- οὖσαι ταῖς ΔΖ, ΖΗ, ΗΕ· καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΔΕ τοῦ ally, until the divisions in AB become equal in number to ΑΒ, καὶ ἀφῄρηται ἀπὸ μὲν τοῦ ΔΕ ἔλασσον τοῦ ἡμίσεως τὸ the divisions in DE. ΕΗ, ἀπὸ δὲ τοῦ ΑΒ μεῖζον ἢ τὸ ἥμισυ τὸ ΒΘ, λοιπὸν ἄρα Therefore, let the divisions (in AB) be AK, KH , HB, τὸ ΗΔ λοιποῦ τοῦ ΘΑ μεῖζόν ἐστιν. καὶ ἐπεὶ μεῖζόν ἐστι τὸ being equal in number to DF , FG, GE. And since DE is ΗΔ τοῦ ΘΑ, καὶ ἀφῄρηται τοῦ μὲν ΗΔ ἥμισυ τὸ ΗΖ, τοῦ greater than AB, and EG, (which is) less than half, has δὲ ΘΑ μεῖζον ἢ τὸ ἥμισυ τὸ ΘΚ, λοιπὸν ἄρα τὸ ΔΖ λοιποῦ been subtracted from DE, and BH , (which is) greater τοῦ ΑΚ μεῖζόν ἐστιν. ἴσον δὲ τὸ ΔΖ τῷ Γ· καὶ τὸ Γ ἄρα than half, from AB, the remainder GD is thus greater τοῦ ΑΚ μεῖζόν ἐστιν. ἔλασσον ἄρα τὸ ΑΚ τοῦ Γ. than the remainder HA. And since GD is greater than Καταλείπεται ἄρα ἀπὸ τοῦ ΑΒ μεγέθους τὸ ΑΚ μέγεθος HA, and the half GF has been subtracted from GD, and ἔλασσον ὂν τοῦ ἐκκειμένου ἐλάσσονος μεγέθους τοῦ Γ· HK, (which is) greater than half, from HA, the remain- ὅπερ ἔδει δεῖξαι.— ὁμοίως δὲ δειχθήσεται, κἂν ἡμίση ᾖ τὰ der DF is thus greater than the remainder AK. And DF ἀφαιρούμενα. (is) equal to C. C is thus also greater than AK. Thus, AK (is) less than C. Thus, the magnitude AK, which is less than the lesser laid out magnitude C, is left over from the magnitude AB. (Which is) the very thing it was required to show. — (The theorem) can similarly be proved even if the (parts) subtracted are halves. † This theorem is the basis of the so-called method of exhaustion, and is generally attributed to Eudoxus of Cnidus.bþ. Proposition 2 ᾿Εὰν δύο μεγεθῶν [ἐκκειμένων] ἀνίσων ἀνθυφαιρουμένου If the remainder of two unequal magnitudes (which ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος τὸ καταλειπόμενον are) [laid out] never measures the (magnitude) before it, μηδέποτε καταμετρῇ τὸ πρὸ ἑαυτοῦ, ἀσύμμετρα ἔσται τὰ (when) the lesser (magnitude is) continually subtracted μεγέθη. in turn from the greater, then the (original) magnitudes Δύο γὰρ μεγεθῶν ὄντων ἀνίσων τῶν ΑΒ, ΓΔ καὶ will be incommensurable. ἐλάσσονος τοῦ ΑΒ ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος For, AB and CD being two unequal magnitudes, and ἀπὸ τοῦ μείζονος τὸ περιλειπόμενον μηδέποτε καταμε- AB (being) the lesser, let the remainder never measure 283 STOIQEIWN iþ. ELEMENTS BOOK 10 τρείτω τὸ πρὸ ἑαυτοῦ· λέγω, ὅτι ἀσύμμετρά ἐστι τὰ ΑΒ, the (magnitude) before it, (when) the lesser (magnitude ΓΔ μεγέθη. is) continually subtracted in turn from the greater. I say that the magnitudes AB and CD are incommensurable. Ε Ζ Α Η Β ∆Γ C E A B DF G Εἰ γάρ ἐστι σύμμετρα, μετρήσει τι αὐτὰ μέγεθος. με- For if they are commensurable then some magnitude τρείτω, εἰ δυνατόν, καὶ ἔστω τὸ Ε· καὶ τὸ μὲν ΑΒ τὸ ΖΔ will measure them (both). If possible, let it (so) measure καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΓΖ, τὸ δὲ ΓΖ τὸ (them), and let it be E. And let AB leave CF less than ΒΗ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΑΗ, καὶ τοῦτο itself (in) measuring FD, and let CF leave AG less than ἀεὶ γινέσθω, ἕως οὗ λειφθῇ τι μέγεθος, ὅ ἐστιν ἔλασσον τοῦ itself (in) measuring BG, and let this happen continually, Ε. γεγονέτω, καὶ λελείφθω τὸ ΑΗ ἔλασσον τοῦ Ε. ἐπεὶ οὖν until some magnitude which is less than E is left. Let τὸ Ε τὸ ΑΒ μετρεῖ, ἀλλὰ τὸ ΑΒ τὸ ΔΖ μετρεῖ, καὶ τὸ Ε ἄρα (this) have occurred,† and let AG, (which is) less than τὸ ΖΔ μετρήσει. μετρεῖ δὲ καὶ ὅλον τὸ ΓΔ· καὶ λοιπὸν ἄρα E, have been left. Therefore, since E measures AB, but τὸ ΓΖ μετρήσει. ἀλλὰ τὸ ΓΖ τὸ ΒΗ μετρεῖ· καὶ τὸ Ε ἄρα AB measures DF , E will thus also measure FD. And it τὸ ΒΗ μετρεῖ. μετρεῖ δὲ καὶ ὅλον τὸ ΑΒ· καὶ λοιπὸν ἄρα τὸ also measures the whole (of) CD. Thus, it will also mea- ΑΗ μετρήσει, τὸ μεῖζον τὸ ἔλασσον· ὅπερ ἐστὶν ἀδύνατον. sure the remainder CF . But, CF measures BG. Thus, E οὐκ ἄρα τὰ ΑΒ, ΓΔ μεγέθη μετρήσει τι μέγεθος· ἀσύμμετρα also measures BG. And it also measures the whole (of) ἄρα ἐστὶ τὰ ΑΒ, ΓΔ μεγέθη. AB. Thus, it will also measure the remainder AG, the ᾿Εὰν ἄρα δύο μεγεθῶν ἀνίσων, καὶ τὰ ἑξῆς. greater (measuring) the lesser. The very thing is impos- sible. Thus, some magnitude cannot measure (both) the magnitudes AB and CD. Thus, the magnitudes AB and CD are incommensurable [Def. 10.1]. Thus, if . . . of two unequal magnitudes, and so on . . . . † The fact that this will eventually occur is guaranteed by Prop. 10.1.gþ. Proposition 3 Δύο μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον αὐτῶν To find the greatest common measure of two given κοινὸν μέτρον εὑρεῖν. commensurable magnitudes. Ζ Β Ε ∆ Η Α Γ G B E D A F C ῎Εστω τὰ δοθέντα δύο μεγέθη σύμμετρα τὰ ΑΒ, ΓΔ, Let AB and CD be the two given magnitudes, of ὧν ἔλασσον τὸ ΑΒ· δεῖ δὴ τῶν ΑΒ, ΓΔ τὸ μέγιστον κοινὸν which (let) AB (be) the lesser. So, it is required to find μέτρον εὑρεῖν. the greatest common measure of AB and CD. Τὸ ΑΒ γὰρ μέγεθος ἤτοι μετρεῖ τὸ ΓΔ ἢ οὔ. εἰ μὲν For the magnitude AB either measures, or (does) not οὖν μετρεῖ, μετρεῖ δὲ καὶ ἑαυτό, τὸ ΑΒ ἄρα τῶν ΑΒ, ΓΔ (measure), CD. Therefore, if it measures (CD), and κοινὸν μέτρον ἐστίν· καὶ φανερόν, ὅτι καὶ μέγιστον. μεῖζον (since) it also measures itself, AB is thus a common mea- γὰρ τοῦ ΑΒ μεγέθους τὸ ΑΒ οὐ μετρήσει. sure of AB and CD. And (it is) clear that (it is) also (the) Μὴ μετρείτω δὴ τὸ ΑΒ τὸ ΓΔ. καὶ ἀνθυφαιρουμένου greatest. For a (magnitude) greater than magnitude AB ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος, τὸ περιλειπόμενον cannot measure AB. μετρήσει ποτὲ τὸ πρὸ ἑαυτοῦ διὰ τὸ μὴ εἶναι ἀσύμμετρα τὰ So let AB not measure CD. And continually subtract- ΑΒ, ΓΔ· καὶ τὸ μὲν ΑΒ τὸ ΕΔ καταμετροῦν λειπέτω ἑαυτοῦ ing in turn the lesser (magnitude) from the greater, the 284 STOIQEIWN iþ. ELEMENTS BOOK 10 ἔλασσον τὸ ΕΓ, τὸ δὲ ΕΓ τὸ ΖΒ καταμετροῦν λειπέτω remaining (magnitude) will (at) some time measure the ἑαυτοῦ ἔλασσον τὸ ΑΖ, τὸ δὲ ΑΖ τὸ ΓΕ μετρείτω. (magnitude) before it, on account of AB and CD not be- ᾿Επεὶ οὖν τὸ ΑΖ τὸ ΓΕ μετρεῖ, ἀλλὰ τὸ ΓΕ τὸ ΖΒ μετρεῖ, ing incommensurable [Prop. 10.2]. And let AB leave EC καὶ τὸ ΑΖ ἄρα τὸ ΖΒ μετρήσει. μετρεῖ δὲ καὶ ἑαυτό· καὶ less than itself (in) measuring ED, and let EC leave AF ὅλον ἄρα τὸ ΑΒ μετρήσει τὸ ΑΖ. ἀλλὰ τὸ ΑΒ τὸ ΔΕ μετρεῖ· less than itself (in) measuring FB, and let AF measure καὶ τὸ ΑΖ ἄρα τὸ ΕΔ μετρήσει. μετρεῖ δὲ καὶ τὸ ΓΕ· καί CE. ὅλον ἄρα τὸ ΓΔ μετρεῖ· τὸ ΑΖ ἄρα τῶν ΑΒ, ΓΔ κοινὸν Therefore, since AF measures CE, but CE measures μέτρον ἐστίν. λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ μή, ἔσται FB, AF will thus also measure FB. And it also mea- τι μέγεθος μεῖζον τοῦ ΑΖ, ὃ μετρήσει τὰ ΑΒ, ΓΔ. ἔστω τὸ sures itself. Thus, AF will also measure the whole (of) Η. ἐπεὶ οὖν τὸ Η τὸ ΑΒ μετρεῖ, ἀλλὰ τὸ ΑΒ τὸ ΕΔ μετρεῖ, AB. But, AB measures DE. Thus, AF will also mea- καὶ τὸ Η ἄρα τὸ ΕΔ μετρήσει. μετρεῖ δὲ καὶ ὅλον τὸ ΓΔ· sure ED. And it also measures CE. Thus, it also mea- καὶ λοιπὸν ἄρα τὸ ΓΕ μετρήσει τὸ Η. ἀλλὰ τὸ ΓΕ τὸ ΖΒ sures the whole of CD. Thus, AF is a common measure μετρεῖ· καὶ τὸ Η ἄρα τὸ ΖΒ μετρήσει. μετρεῖ δὲ καὶ ὅλον of AB and CD. So I say that (it is) also (the) greatest τὸ ΑΒ, καὶ λοιπὸν τὸ ΑΖ μετρήσει, τὸ μεῖζον τὸ ἔλασσον· (common measure). For, if not, there will be some mag- ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα μεῖζόν τι μέγεθος τοῦ ΑΖ nitude, greater than AF , which will measure (both) AB τὰ ΑΒ, ΓΔ μετρήσει· τὸ ΑΖ ἄρα τῶν ΑΒ, ΓΔ τὸ μέγιστον and CD. Let it be G. Therefore, since G measures AB, κοινὸν μέτρον ἐστίν. but AB measures ED, G will thus also measure ED. And Δύο ἄρα μεγεθῶν συμμέτρων δοθέντων τῶν ΑΒ, ΓΔ it also measures the whole of CD. Thus, G will also mea- τὸ μέγιστον κοινὸν μέτρον ηὕρηται· ὅπερ ἔδει δεῖξαι. sure the remainder CE. But CE measures FB. Thus, G will also measure FB. And it also measures the whole (of) AB. And (so) it will measure the remainder AF , the greater (measuring) the lesser. The very thing is im- possible. Thus, some magnitude greater than AF cannot measure (both) AB and CD. Thus, AF is the greatest common measure of AB and CD. Thus, the greatest common measure of two given commensurable magnitudes, AB and CD, has been found. (Which is) the very thing it was required to show.Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι, ἐὰν μέγεθος δύο μεγέθη So (it is) clear, from this, that if a magnitude measures μετρῇ, καὶ τὸ μέγιστον αὐτῶν κοινὸν μέτρον μετρήσει. two magnitudes then it will also measure their greatest common measure.dþ. Proposition 4 Τριῶν μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον To find the greatest common measure of three given αὐτῶν κοινὸν μέτρον εὑρεῖν. commensurable magnitudes. Γ ∆ Ε Ζ Α Β C D E F A B ῎Εστω τὰ δοθέντα τρία μεγέθη σύμμετρα τὰ Α, Β, Γ· Let A, B, C be the three given commensurable mag- δεῖ δὴ τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον εὑρεῖν. nitudes. So it is required to find the greatest common Εἰλήφθω γὰρ δύο τῶν Α, Β τὸ μέγιστον κοινὸν μέτρον, measure of A, B, C. καὶ ἔστω τὸ Δ· τὸ δὴ Δ τὸ Γ ἤτοι μετρεῖ ἢ οὔ [μετρεῖ]. For let the greatest common measure of the two (mag- μετρείτω πρότερον. ἐπεὶ οὖν τὸ Δ τὸ Γ μετρεῖ, μετρεῖ δὲ nitudes) A and B have been taken [Prop. 10.3], and let it 285 STOIQEIWN iþ. ELEMENTS BOOK 10 καὶ τὰ Α, Β, τὸ Δ ἄρα τὰ Α, Β, Γ μετρεῖ· τὸ Δ ἄρα τῶν Α, be D. So D either measures, or [does] not [measure], C. Β, Γ κοινὸν μέτρον ἐστίν. καὶ φανερόν, ὅτι καὶ μέγιστον· Let it, first of all, measure (C). Therefore, since D mea- μεῖζον γὰρ τοῦ Δ μεγέθους τὰ Α, Β οὐ μετρεῖ. sures C, and it also measures A and B, D thus measures Μὴ μετρείτω δὴ τὸ Δ τὸ Γ. λέγω πρῶτον, ὅτι σύμμετρά A, B, C. Thus, D is a common measure of A, B, C. And ἐστι τὰ Γ, Δ. ἐπεὶ γὰρ σύμμετρά ἐστι τὰ Α, Β, Γ, μετρήσει (it is) clear that (it is) also (the) greatest (common mea- τι αὐτὰ μέγεθος, ὃ δηλαδὴ καὶ τὰ Α, Β μετρήσει· ὥστε sure). For no magnitude larger than D measures (both) καὶ τὸ τῶν Α, Β μέγιστον κοινὸν μέτρον τὸ Δ μετρήσει. A and B. μετρεῖ δὲ καὶ τὸ Γ· ὥστε τὸ εἰρημένον μέγεθος μετρήσει τὰ So let D not measure C. I say, first, that C and D are Γ, Δ· σύμμετρα ἄρα ἐστὶ τὰ Γ, Δ. εἰλήφθω οὖν αὐτῶν τὸ commensurable. For if A, B, C are commensurable then μέγιστον κοινὸν μέτρον, καὶ ἔστω τὸ Ε. ἐπεὶ οὖν τὸ Ε τὸ some magnitude will measure them which will clearly Δ μετρεῖ, ἀλλὰ τὸ Δ τὰ Α, Β μετρεῖ, καὶ τὸ Ε ἄρα τὰ Α, Β also measure A and B. Hence, it will also measure D, the μετρήσει. μετρεῖ δὲ καὶ τὸ Γ. τὸ Ε ἄρα τὰ Α, Β, Γ μετρεῖ· greatest common measure of A and B [Prop. 10.3 corr.]. τὸ Ε ἄρα τῶν Α, Β, Γ κοινόν ἐστι μέτρον. λέγω δή, ὅτι καὶ And it also measures C. Hence, the aforementioned mag- μέγιστον. εἰ γὰρ δυνατόν, ἔστω τι τοῦ Ε μεῖζον μέγεθος nitude will measure (both) C and D. Thus, C and D are τὸ Ζ, καὶ μετρείτω τὰ Α, Β, Γ. καὶ ἐπεὶ τὸ Ζ τὰ Α, Β, Γ commensurable [Def. 10.1]. Therefore, let their greatest μετρεῖ, καὶ τὰ Α, Β ἄρα μετρήσει καὶ τὸ τῶν Α, Β μέγιστον common measure have been taken [Prop. 10.3], and let κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Α, Β μέγιστον κοινὸν it be E. Therefore, since E measures D, but D measures μέτρον ἐστὶ τὸ Δ· τὸ Ζ ἄρα τὸ Δ μετρεῖ. μετρεῖ δὲ καὶ τὸ (both) A and B, E will thus also measure A and B. And Γ· τὸ Ζ ἄρα τὰ Γ, Δ μετρεῖ· καὶ τὸ τῶν Γ, Δ ἄρα μέγιστον it also measures C. Thus, E measures A, B, C. Thus, E κοινὸν μέτρον μετρήσει τὸ Ζ. ἔστι δὲ τὸ Ε· τὸ Ζ ἄρα τὸ Ε is a common measure of A, B, C. So I say that (it is) also μετρήσει, τὸ μεῖζον τὸ ἔλασσον· ὅπερ ἐστὶν ἀδύνατον. οὐκ (the) greatest (common measure). For, if possible, let F ἄρα μεῖζόν τι τοῦ Ε μεγέθους [μέγεθος] τὰ Α, Β, Γ μετρεῖ· be some magnitude greater than E, and let it measure A, τὸ Ε ἄρα τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον ἐστίν, ἐὰν B, C. And since F measures A, B, C, it will thus also μὴ μετρῇ τὸ Δ τὸ Γ, ἐὰν δὲ μετρῇ, αὐτὸ τὸ Δ. measure A and B, and will (thus) measure the greatest Τριῶν ἄρα μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον common measure of A and B [Prop. 10.3 corr.]. And D κοινὸν μέτρον ηὕρηται [ὅπερ ἔδει δεῖξαι]. is the greatest common measure of A and B. Thus, F measures D. And it also measures C. Thus, F measures (both) C and D. Thus, F will also measure the greatest common measure of C and D [Prop. 10.3 corr.]. And it is E. Thus, F will measure E, the greater (measuring) the lesser. The very thing is impossible. Thus, some [magni- tude] greater than the magnitude E cannot measure A, B, C. Thus, if D does not measure C then E is the great- est common measure of A, B, C. And if it does measure (C) then D itself (is the greatest common measure). Thus, the greatest common measure of three given commensurable magnitudes has been found. [(Which is) the very thing it was required to show.]Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι, ἐὰν μέγεθος τρία μεγέθη So (it is) clear, from this, that if a magnitude measures μετρῇ, καὶ τὸ μέγιστον αὐτῶν κοινὸν μέτρον μετρήσει. three magnitudes then it will also measure their greatest ῾Ομοίως δὴ καὶ ἐπὶ πλειόνων τὸ μέγιστον κοινὸν μέτρον common measure. ληφθήσεται, καὶ τὸ πόρισμα προχωρήσει. ὅπερ ἔδει δεῖξαι. So, similarly, the greatest common measure of more (magnitudes) can also be taken, and the (above) corol- lary will go forward. (Which is) the very thing it was required to show. 286 STOIQEIWN iþ. ELEMENTS BOOK 10eþ. Proposition 5 Τὰ σύμμετρα μεγέθη πρὸς ἄλληλα λόγον ἔχει, ὃν Commensurable magnitudes have to one another the ἀριθμὸς πρὸς ἀριθμόν. ratio which (some) number (has) to (some) number. Γ Ε∆ Α Β C ED A B ῎Εστω σύμμετρα μεγέθη τὰ Α, Β· λέγω, ὅτι τὸ Α πρὸς Let A and B be commensurable magnitudes. I say τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. that A has to B the ratio which (some) number (has) to ᾿Επεὶ γὰρ σύμμετρά ἐστι τὰ Α, Β, μετρήσει τι αὐτὰ (some) number. μέγεθος. μετρείτω, καὶ ἔστω τὸ Γ. καὶ ὁσάκις τὸ Γ τὸ For if A and B are commensurable (magnitudes) then Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Δ, ὁσάκις δὲ some magnitude will measure them. Let it (so) measure τὸ Γ τὸ Β μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. (them), and let it be C. And as many times as C measures ᾿Επεὶ οὖν τὸ Γ τὸ Α μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας, A, so many units let there be in D. And as many times as μετρεῖ δὲ καὶ ἡ μονὰς τὸν Δ κατὰ τὰς ἐν αὐτῷ μονάδας, C measures B, so many units let there be in E. ἰσάκις ἄρα ἡ μονὰς τὸν Δ μετρεῖ ἀριθμὸν καὶ τὸ Γ μέγεθος Therefore, since C measures A according to the units τὸ Α· ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς in D, and a unit also measures D according to the units τὸν Δ· ἀνάπαλιν ἄρα, ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς in it, a unit thus measures the number D as many times τὴν μονάδα. πάλιν ἐπεὶ τὸ Γ τὸ Β μετρεῖ κατὰ τὰς ἐν τῷ as the magnitude C (measures) A. Thus, as C is to A, Ε μονάδας, μετρεῖ δὲ καὶ ἡ μονὰς τὸν Ε κατὰ τὰς ἐν αὐτῷ so a unit (is) to D [Def. 7.20].† Thus, inversely, as A (is) μονάδας, ἰσάκις ἄρα ἡ μονὰς τὸν Ε μετρεῖ καὶ τὸ Γ τὸ Β· to C, so D (is) to a unit [Prop. 5.7 corr.]. Again, since ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Β, οὕτως ἡ μονὰς πρὸς τὸν Ε. C measures B according to the units in E, and a unit ἐδείχθη δὲ καὶ ὡς τὸ Α πρὸς τὸ Γ, ὁ Δ πρὸς τὴν μονάδα· also measures E according to the units in it, a unit thus δι᾿ ἴσου ἄρα ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως ὁ Δ ἀριθμὸς measures E the same number of times that C (measures) πρὸς τὸν Ε. B. Thus, as C is to B, so a unit (is) to E [Def. 7.20]. And Τὰ ἄρα σύμμετρα μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον it was also shown that as A (is) to C, so D (is) to a unit. ἔχει, ὃν ἀριθμὸς ὁ Δ πρὸς ἀριθμὸν τὸν Ε· ὅπερ ἔδει δεῖξαι. Thus, via equality, as A is to B, so the number D (is) to the (number) E [Prop. 5.22]. Thus, the commensurable magnitudes A and B have to one another the ratio which the number D (has) to the number E. (Which is) the very thing it was required to show. † There is a slight logical gap here, since Def. 7.20 applies to four numbers, rather than two number and two magnitudes.�þ. Proposition 6 ᾿Εὰν δύο μεγέθη πρὸς ἄλληλα λόγον ἔχῃ, ὃν ἀριθμὸς If two magnitudes have to one another the ratio which πρὸς ἀριθμόν, σύμμετρα ἔσται τὰ μεγέθη. (some) number (has) to (some) number then the magni- tudes will be commensurable. Ζ Α ∆ Γ Β Ε F A D C B E Δύο γὰρ μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον ἐχέτω, ὃν For let the two magnitudes A and B have to one an- ἀριθμὸς ὁ Δ πρὸς ἀριθμὸν τὸν Ε· λέγω, ὅτι σύμμετρά ἐστι other the ratio which the number D (has) to the number τὰ Α, Β μεγέθη. E. I say that the magnitudes A and B are commensu- ῞Οσαι γάρ εἰσιν ἐν τῷ Δ μονάδες, εἰς τοσαῦτα ἴσα rable. 287 STOIQEIWN iþ. ELEMENTS BOOK 10 διῃρήσθω τὸ Α, καὶ ἑνὶ αὐτῶν ἴσον ἔστω τὸ Γ· ὅσαι δέ For, as many units as there are in D, let A have been εἰσιν ἐν τῷ Ε μονάδες, ἐκ τοσούτων μεγεθῶν ἴσων τῷ Γ divided into so many equal (divisions). And let C be συγκείσθω τὸ Ζ. equal to one of them. And as many units as there are ᾿Επεὶ οὖν, ὅσαι εἰσὶν ἐν τῷ Δ μονάδες, τοσαῦτά εἰσι καὶ in E, let F be the sum of so many magnitudes equal to ἐν τῷ Α μεγέθη ἴσα τῷ Γ, ὃ ἄρα μέρος ἐστὶν ἡ μονὰς τοῦ C. Δ, τὸ αὐτὸ μέρος ἐστὶ καὶ τὸ Γ τοῦ Α· ἔστιν ἄρα ὡς τὸ Γ Therefore, since as many units as there are in D, so πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ. μετρεῖ δὲ ἡ μονὰς many magnitudes equal to C are also in A, therefore τὸν Δ ἀριθμόν· μετρεῖ ἄρα καὶ τὸ Γ τὸ Α. καὶ ἐπεί ἐστιν whichever part a unit is of D, C is also the same part of ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ [ἀριθμόν], A. Thus, as C is to A, so a unit (is) to D [Def. 7.20]. And ἀνάπαλιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ ἀριθμὸς πρὸς a unit measures the number D. Thus, C also measures τὴν μονάδα. πάλιν ἐπεί, ὅσαι εἰσὶν ἐν τῷ Ε μονάδες, τοσαῦτά A. And since as C is to A, so a unit (is) to the [number] εἰσι καὶ ἐν τῷ Ζ ἴσα τῷ Γ, ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ζ, D, thus, inversely, as A (is) to C, so the number D (is) οὕτως ἡ μονὰς πρὸς τὸν Ε [ἀριθμόν]. ἐδείχθη δὲ καὶ ὡς to a unit [Prop. 5.7 corr.]. Again, since as many units as τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς τὴν μονάδα· δι᾿ ἴσου ἄρα there are in E, so many (magnitudes) equal to C are also ἐστὶν ὡς τὸ Α πρὸς τὸ Ζ, οὕτως ὁ Δ πρὸς τὸν Ε. ἀλλ᾿ ὡς ὁ in F , thus as C is to F , so a unit (is) to the [number] E Δ πρὸς τὸν Ε, οὕτως ἐστὶ τὸ Α πρὸς τὸ Β· καὶ ὡς ἄρα τὸ Α [Def. 7.20]. And it was also shown that as A (is) to C, πρὸς τὸ Β, οὕτως καὶ πρὸς τὸ Ζ. τὸ Α ἄρα πρὸς ἑκάτερον so D (is) to a unit. Thus, via equality, as A is to F , so D τῶν Β, Ζ τὸν αὐτὸν ἔχει λόγον· ἴσον ἄρα ἐστὶ τὸ Β τῷ Ζ. (is) to E [Prop. 5.22]. But, as D (is) to E, so A is to B. μετρεῖ δὲ τὸ Γ τὸ Ζ· μετρεῖ ἄρα καὶ τὸ Β. ἀλλὰ μὴν καὶ τὸ And thus as A (is) to B, so (it) also is to F [Prop. 5.11]. Α· τὸ Γ ἄρα τὰ Α, Β μετρεῖ. σύμμετρον ἄρα ἐστὶ τὸ Α τῷ Thus, A has the same ratio to each of B and F . Thus, B is Β. equal to F [Prop. 5.9]. And C measures F . Thus, it also ᾿Εὰν ἄρα δύο μεγέθη πρὸς ἄλληλα, καὶ τὰ ἑξῆς. measures B. But, in fact, (it) also (measures) A. Thus, C measures (both) A and B. Thus, A is commensurable with B [Def. 10.1]. Thus, if two magnitudes . . . to one another, and so on . . . .Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι, ἐὰν ὦσι δύο ἀριθμοί, ὡς So it is clear, from this, that if there are two numbers, οἱ Δ, Ε, καὶ εὐθεῖα, ὡς ἡ Α, δύνατόν ἐστι ποιῆσαι ὡς ὁ like D and E, and a straight-line, like A, then it is possible Δ ἀριθμὸς πρὸς τὸν Ε ἀριθμόν, οὕτως τὴν εὐθεῖαν πρὸς to contrive that as the number D (is) to the number E, εὐθεῖαν. ἐὰν δὲ καὶ τῶν Α, Ζ μέση ἀνάλογον ληφθῇ, ὡς ἡ so the straight-line (is) to (another) straight-line (i.e., F ). Β, ἔσται ὡς ἡ Α πρὸς τὴν Ζ, οὕτως τὸ ἀπὸ τῆς Α πρὸς τὸ And if the mean proportion, (say) B, is taken of A and ἀπὸ τῆς Β, τουτέστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως F , then as A is to F , so the (square) on A (will be) to the τὸ ἀπὸ τῆς πρώτης πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ (square) on B. That is to say, as the first (is) to the third, ὁμοίως ἀναγραφόμενον. ἀλλ᾿ ὡς ἡ Α πρὸς τὴν Ζ, οὕτως so the (figure) on the first (is) to the similar, and similarly ἐστὶν ὁ Δ ἀριθμος πρὸς τὸν Ε ἀριθμόν· γέγονεν ἄρα καὶ described, (figure) on the second [Prop. 6.19 corr.]. But, ὡς ὁ Δ ἀριθμὸς πρὸς τὸν Ε ἀριθμόν, οὕτως τὸ ἀπὸ τῆς Α as A (is) to F , so the number D is to the number E. Thus, εὐθείας πρὸς τὸ ἀπὸ τῆς Β εὐθείας· ὅπερ ἔδει δεῖξαι. it has also been contrived that as the number D (is) to the number E, so the (figure) on the straight-line A (is) to the (similar figure) on the straight-line B. (Which is) the very thing it was required to show.zþ. Proposition 7 Τὰ ἀσύμμετρα μεγέθη πρὸς ἄλληλα λόγον οὐκ ἔχει, ὃν Incommensurable magnitudes do not have to one an- ἀριθμὸς πρὸς ἀριθμόν. other the ratio which (some) number (has) to (some) ῎Εστω ἀσύμμετρα μεγέθη τὰ Α, Β· λέγω, ὅτι τὸ Α πρὸς number. τὸ Β λόγον οὐκ ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. Let A and B be incommensurable magnitudes. I say that A does not have to B the ratio which (some) number (has) to (some) number. 288 STOIQEIWN iþ. ELEMENTS BOOK 10 Β Α B A Εἰ γὰρ ἔχει τὸ Α πρὸς τὸ Β λόγον, ὃν ἀριθμὸς πρὸς For if A has to B the ratio which (some) number (has) ἀριθμόν, σύμμετρον ἔσται τὸ Α τῷ Β. οὐκ ἔστι δέ· οὐκ ἄρα to (some) number then A will be commensurable with B τὸ Α πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. [Prop. 10.6]. But it is not. Thus, A does not have to B Τὰ ἄρα ἀσύμμετρα μεγέθη πρὸς ἄλληλα λόγον οὐκ ἔχει, the ratio which (some) number (has) to (some) number. καὶ τὰ ἑξῆς. Thus, incommensurable numbers do not have to one another, and so on . . . .hþ. Proposition 8 ᾿Εὰν δύο μεγέθη πρὸς ἄλληλα λόγον μὴ ἔχῃ, ὃν ἀριθμὸς If two magnitudes do not have to one another the ra- πρὸς ἀριθμόν, ἀσύμμετρα ἔσται τὰ μεγέθη. tio which (some) number (has) to (some) number then the magnitudes will be incommensurable. Β Α B A Δύο γὰρ μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον μὴ ἐχέτω, For let the two magnitudes A and B not have to one ὃν ἀριθμὸς πρὸς ἀριθμόν· λέγω, ὅτι ἀσύμμετρά ἐστι τὰ Α, another the ratio which (some) number (has) to (some) Β μεγέθη. number. I say that the magnitudes A and B are incom- Εἰ γὰρ ἔσται σύμμετρα, τὸ Α πρὸς τὸ Β λόγον ἕξει, ὃν mensurable. ἀριθμὸς πρὸς ἀριθμόν. οὐκ ἔχει δέ. ἀσύμμετρα ἄρα ἐστὶ τὰ For if they are commensurable, A will have to B Α, Β μεγέθη. the ratio which (some) number (has) to (some) number ᾿Εὰν ἄρα δύο μεγέθη πρὸς ἄλληλα, καὶ τὰ ἑξῆς. [Prop. 10.5]. But it does not have (such a ratio). Thus, the magnitudes A and B are incommensurable. Thus, if two magnitudes . . . to one another, and so on . . . .jþ. Proposition 9 Τὰ ἀπὸ τῶν μήκει συμμέτρων εὐθειῶν τετράγωνα Squares on straight-lines (which are) commensurable πρὸς ἄλληλα λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς in length have to one another the ratio which (some) τετράγωνον ἀριθμόν· καὶ τὰ τετράγωνα τὰ πρὸς ἄλληλα square number (has) to (some) square number. And λόγον ἔχοντα, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον squares having to one another the ratio which (some) ἀριθμόν, καὶ τὰς πλευρὰς ἕξει μήκει συμμέτρους. τὰ square number (has) to (some) square number will also δὲ ἀπὸ τῶν μήκει ἀσυμμέτρων εὐθειῶν τετράγωνα πρὸς have sides (which are) commensurable in length. But ἄλληλα λόγον οὐκ ἔχει, ὅνπερ τετράγωνος ἀριθμὸς πρὸς squares on straight-lines (which are) incommensurable τετράγωνον ἀριθμόν· καὶ τὰ τετράγωνα τὰ πρὸς ἄλληλα in length do not have to one another the ratio which λόγον μὴ ἔχοντα, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον (some) square number (has) to (some) square number. ἀριθμόν, οὐδὲ τὰς πλευρὰς ἕξει μήκει συμμέτρους. And squares not having to one another the ratio which (some) square number (has) to (some) square number will not have sides (which are) commensurable in length either. ∆ Α Γ Β D A C B ῎Εστωσαν γὰρ αἱ Α, Β μήκει σύμμετροι· λέγω, ὅτι τὸ For let A and B be (straight-lines which are) commen- ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β τετράγωνον surable in length. I say that the square on A has to the λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον square on B the ratio which (some) square number (has) ἀριθμόν. to (some) square number. 289 STOIQEIWN iþ. ELEMENTS BOOK 10 ᾿Επεὶ γὰρ σύμμετρός ἐστιν ἡ Α τῇ Β μήκει, ἡ Α ἄρα πρὸς For since A is commensurable in length with B, A τὴν Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐχέτω, ὃν ὁ thus has to B the ratio which (some) number (has) to Γ πρὸς τὸν Δ. ἐπεὶ οὖν ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ὁ (some) number [Prop. 10.5]. Let it have (that) which Γ πρὸς τὸν Δ, ἀλλὰ τοῦ μὲν τῆς Α πρὸς τὴν Β λόγου δι- C (has) to D. Therefore, since as A is to B, so C (is) πλασίων ἐστὶν ὁ τοῦ ἀπὸ τῆς Α τετραγώνου πρὸς τὸ ἀπὸ τῆς to D. But the (ratio) of the square on A to the square Β τετράγωνον· τὰ γὰρ ὅμοια σχήματα ἐν διπλασίονι λόγῳ on B is the square of the ratio of A to B. For similar ἐστὶ τῶν ὁμολόγων πλευρῶν· τοῦ δὲ τοῦ Γ [ἀριθμοῦ] πρὸς figures are in the squared ratio of (their) corresponding τὸν Δ [ἀριθμὸν] λόγου διπλασίων ἐστὶν ὁ τοῦ ἀπὸ τοῦ Γ sides [Prop. 6.20 corr.]. And the (ratio) of the square τετραγώνου πρὸς τὸν ἀπὸ τοῦ Δ τετράγωνον· δύο γὰρ τε- on C to the square on D is the square of the ratio of τραγώνων ἀριθμῶν εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός, καί the [number] C to the [number] D. For there exits one ὁ τετράγωνος πρὸς τὸν τετράγωνον [ἀριθμὸν] διπλασίονα number in mean proportion to two square numbers, and λόγον ἔχει, ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν· ἔστιν ἄρα (one) square (number) has to the (other) square [num- καὶ ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β ber] a squared ratio with respect to (that) the side (of the τετράγωνον, οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος [ἀριθμὸς] πρὸς former has) to the side (of the latter) [Prop. 8.11]. And, τὸν ἀπὸ τοῦ Δ [ἀριθμοῦ] τετράγωνον [ἀριθμόν]. thus, as the square on A is to the square on B, so the Ἀλλὰ δὴ ἔστω ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ square [number] on the (number) C (is) to the square ἀπὸ τῆς Β, οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος πρὸς τὸν ἀπὸ [number] on the [number] D.† τοῦ Δ [τετράγωνον]· λέγω, ὅτι σύμμετρός ἐστιν ἡ Α τῇ Β And so let the square on A be to the (square) on B as μήκει. the square (number) on C (is) to the [square] (number) ᾿Επεὶ γάρ ἐστιν ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ on D. I say that A is commensurable in length with B. ἀπὸ τῆς Β [τετράγωνον], οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος For since as the square on A is to the [square] on B, so πρὸς τὸν ἀπὸ τοῦ Δ [τετράγωνον], ἀλλ᾿ ὁ μὲν τοῦ ἀπὸ τῆς the square (number) on C (is) to the [square] (number) Α τετραγώνου πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον] λόγος δι- on D. But, the ratio of the square on A to the (square) πλασίων ἐστὶ τοῦ τῆς Α πρὸς τὴν Β λόγου, ὁ δὲ τοῦ ἀπὸ on B is the square of the (ratio) of A to B [Prop. 6.20 τοῦ Γ [ἀριθμοῦ] τετραγώνου [ἀριθμοῦ] πρὸς τὸν ἀπὸ τοῦ Δ corr.]. And the (ratio) of the square [number] on the [ἀριθμοῦ] τετράγωνον [ἀριθμὸν] λόγος διπλασίων ἐστὶ τοῦ [number] C to the square [number] on the [number] D is τοῦ Γ [ἀριθμοῦ] πρὸς τὸν Δ [ἀριθμὸν] λόγου, ἔστιν ἄρα the square of the ratio of the [number] C to the [number] καὶ ὡς ἡ Α πρὸς τὴν Β, οὕτως ὁ Γ [ἀριθμὸς] πρὸς τὸν Δ D [Prop. 8.11]. Thus, as A is to B, so the [number] C [ἀριθμόν]. ἡ Α ἄρα πρὸς τὴν Β λόγον ἔχει, ὃν ἀριθμὸς ὁ Γ also (is) to the [number] D. A, thus, has to B the ratio πρὸς ἀριθμὸν τὸν Δ· σύμμετρος ἄρα ἐστὶν ἡ Α τῇ Β μήκει. which the number C has to the number D. Thus, A is Ἀλλὰ δὴ ἀσύμμετρος ἔστω ἡ Α τῇ Β μήκει· λέγω, ὅτι commensurable in length with B [Prop. 10.6].‡ τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον] And so let A be incommensurable in length with B. I λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον say that the square on A does not have to the [square] on ἀριθμόν. B the ratio which (some) square number (has) to (some) Εἰ γὰρ ἔχει τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ square number. τῆς Β [τετράγωνον] λόγον, ὃν τετράγωνος ἀριθμὸς πρὸς For if the square on A has to the [square] on B the ra- τετράγωνον ἀριθμόν, σύμμετρος ἔσται ἡ Α τῇ Β. οὐκ ἔστι tio which (some) square number (has) to (some) square δέ· οὐκ ἄρα τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς number then A will be commensurable (in length) with Β [τετράγωνον] λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς B. But it is not. Thus, the square on A does not have τετράγωνον ἀριθμόν. to the [square] on the B the ratio which (some) square Πάλιν δὴ τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς number (has) to (some) square number. Β [τετράγωνον] λόγον μὴ ἐχέτω, ὃν τετράγωνος ἀριθμὸς So, again, let the square on A not have to the [square] πρὸς τετράγωνον ἀριθμόν· λέγω, ὅτι ἀσύμμετρός ἐστιν ἡ Α on B the ratio which (some) square number (has) to τῇ Β μήκει. (some) square number. I say that A is incommensurable Εἰ γάρ ἐστι σύμμετρος ἡ Α τῇ Β, ἕξει τὸ ἀπὸ τῆς Α in length with B. πρὸς τὸ ἀπὸ τῆς Β λόγον, ὃν τετράγωνος ἀριθμὸς πρὸς For if A is commensurable (in length) with B then τετράγωνον ἀριθμόν. οὐκ ἔχει δέ· οὐκ ἄρα σύμμετρός ἐστιν the (square) on A will have to the (square) on B the ra- ἡ Α τῇ Β μήκει. tio which (some) square number (has) to (some) square Τὰ ἄρα ἀπὸ τῶν μήκει συμμέτρων, καὶ τὰ ἑξῆς. number. But it does not have (such a ratio). Thus, A is not commensurable in length with B. Thus, (squares) on (straight-lines which are) com- 290 STOIQEIWN iþ. ELEMENTS BOOK 10 mensurable in length, and so on . . . .Pìrisma. Corollary Καὶ φανερὸν ἐκ τῶν δεδειγμένων ἔσται, ὅτι αἱ μήκει And it will be clear, from (what) has been demon- σύμμετροι πάντως καὶ δυνάμει, αἱ δὲ δυνάμει οὐ πάντως καὶ strated, that (straight-lines) commensurable in length μήκει. (are) always also (commensurable) in square, but (straight- lines commensurable) in square (are) not always also (commensurable) in length. † There is an unstated assumption here that if α : β :: γ : δ then α2 : β2 :: γ2 : δ2. ‡ There is an unstated assumption here that if α2 : β2 :: γ2 : δ2 then α : β :: γ : δ.iþ. Proposition 10† Τῇ προτεθείσῃ εὐθείᾳ προσευρεῖν δύο εὐθείας ἀσυμμέτ- To find two straight-lines incommensurable with a ρους, τὴν μὲν μήκει μόνον, τὴν δὲ καὶ δυνάμει. given straight-line, the one (incommensurable) in length only, the other also (incommensurable) in square. Γ Α Ε ∆ Β C A E D B ῎Εστω ἡ προτεθεῖσα εὐθεῖα ἡ Α· δεῖ δὴ τῇ Α προσευρεῖν Let A be the given straight-line. So it is required to δύο εὐθείας ἀσυμμέτρους, τὴν μὲν μήκει μόνον, τὴν δὲ καὶ find two straight-lines incommensurable with A, the one δυνάμει. (incommensurable) in length only, the other also (incom- ᾿Εκκείσθωσαν γὰρ δύο αριθμοὶ οἱ Β, Γ πρὸς ἀλλήλους mensurable) in square. λόγον μὴ ἔχοντες, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον For let two numbers, B and C, not having to one ἀριθμόν, τουτέστι μὴ ὅμοιοι ἐπίπεδοι, καὶ γεγονέτω ὡς ὁ another the ratio which (some) square number (has) to Β πρὸς τὸν Γ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ (some) square number—that is to say, not (being) simi- ἀπὸ τῆς Δ τετράγωνον· ἐμάθομεν γάρ· σύμμετρον ἄρα lar plane (numbers)—have been taken. And let it be con- τὸ ἀπὸ τῆς Α τῷ ἀπὸ τῆς Δ. καὶ ἐπεὶ ὁ Β πρὸς τὸν Γ trived that as B (is) to C, so the square on A (is) to the λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον square on D. For we learned (how to do this) [Prop. 10.6 ἀριθμόν, οὐδ᾿ ἄρα τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Δ λόγον corr.]. Thus, the (square) on A (is) commensurable with ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· the (square) on D [Prop. 10.6]. And since B does not ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ Δ μήκει. εἰλήφθω τῶν Α, Δ have to C the ratio which (some) square number (has) to μέση ἀνάλογον ἡ Ε· ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Δ, οὕτως (some) square number, the (square) on A thus does not τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Ε. ἀσύμμετρος have to the (square) on D the ratio which (some) square δέ ἐστιν ἡ Α τῇ Δ μήκει· ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ number (has) to (some) square number either. Thus, A τῆς Α τετράγωνον τῷ ἀπὸ τῆς Ε τετραγώνῳ· ἀσύμμετρος is incommensurable in length with D [Prop. 10.9]. Let ἄρα ἐστὶν ἡ Α τῇ Ε δυνάμει. the (straight-line) E (which is) in mean proportion to A Τῂ ἄρα προτεθείσῃ εὐθείᾳ τῇ Α προσεύρηνται δύο and D have been taken [Prop. 6.13]. Thus, as A is to D, εὐθεῖαι ἀσύμμετροι αἱ Δ, Ε, μήκει μὲν μόνον ἡ Δ, δυνάμει so the square on A (is) to the (square) on E [Def. 5.9]. δὲ καὶ μήκει δηλαδὴ ἡ Ε [ὅπερ ἔδει δεῖξαι]. And A is incommensurable in length with D. Thus, the square on A is also incommensurble with the square on E [Prop. 10.11]. Thus, A is incommensurable in square with E. 291 STOIQEIWN iþ. ELEMENTS BOOK 10 Thus, two straight-lines, D and E, (which are) in- commensurable with the given straight-line A, have been found, the one, D, (incommensurable) in length only, the other, E, (incommensurable) in square, and, clearly, also in length. [(Which is) the very thing it was required to show.] † This whole proposition is regarded by Heiberg as an interpolation into the original text.iaþ. Proposition 11 ᾿Εὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ δὲ πρῶτον τῷ If four magnitudes are proportional, and the first is δευτέρῳ σύμμετρον ᾖ, καὶ τὸ τρίτον τῷ τετάρτῳ σύμμετρον commensurable with the second, then the third will also ἔσται· κἂν τὸ πρῶτον τῷ δευτέρῳ ἀσύμμετρον ᾖ, καὶ τὸ be commensurable with the fourth. And if the first is in- τρίτον τῷ τετάρτῳ ἀσύμμετρον ἔσται. commensurable with the second, then the third will also be incommensurable with the fourth. ∆ Α Γ Β D A C B ῎Εστωσαν τέσσαρα μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, Let A, B, C, D be four proportional magnitudes, ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, τὸ Α δὲ τῷ (such that) as A (is) to B, so C (is) to D. And let A Β σύμμετρον ἔστω· λέγω, ὅτι καὶ τὸ Γ τῷ Δ σύμμετρον be commensurable with B. I say that C will also be com- ἔσται. mensurable with D. ᾿Επεὶ γὰρ σύμμετρόν ἐστι τὸ Α τῷ Β, τὸ Α ἄρα πρὸς τὸ For since A is commensurable with B, A thus has to B Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. καί ἐστιν ὡς τὸ Α the ratio which (some) number (has) to (some) number πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ· καὶ τὸ Γ ἄρα πρὸς τὸ Δ [Prop. 10.5]. And as A is to B, so C (is) to D. Thus, λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν· σύμμετρον ἄρα ἐστὶ C also has to D the ratio which (some) number (has) τὸ Γ τῷ Δ. to (some) number. Thus, C is commensurable with D Ἀλλὰ δὴ τὸ Α τῷ Β ἀσύμμετρον ἔστω· λέγω, ὅτι καὶ τὸ [Prop. 10.6]. Γ τῷ Δ ἀσύμμετρον ἔσται. ἐπεὶ γὰρ ἀσύμμετρόν ἐστι τὸ Α And so let A be incommensurable with B. I say that τῷ Β, τὸ Α ἄρα πρὸς τὸ Β λόγον οὐκ ἔχει, ὃν ἀριθμὸς πρὸς C will also be incommensurable with D. For since A ἀριθμόν. καί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ is incommensurable with B, A thus does not have to B Δ· οὐδὲ τὸ Γ ἄρα πρὸς τὸ Δ λόγον ἔχει, ὃν ἀριθμὸς πρὸς the ratio which (some) number (has) to (some) number ἀριθμόν· ἀσύμμετρον ἄρα ἐστὶ τὸ Γ τῷ Δ. [Prop. 10.7]. And as A is to B, so C (is) to D. Thus, C ᾿Εὰν ἄρα τέσσαρα μεγέθη, καὶ τὰ ἑξῆς. does not have to D the ratio which (some) number (has) to (some) number either. Thus, C is incommensurable with D [Prop. 10.8]. Thus, if four magnitudes, and so on . . . .ibþ. Proposition 12 Τὰ τῷ αὐτῷ μεγέθει σύμμετρα καὶ ἀλλήλοις ἐστὶ (Magnitudes) commensurable with the same magni- σύμμετρα. tude are also commensurable with one another. ῾Εκάτερον γὰρ τῶν Α, Β τῷ Γ ἔστω σύμμετρον. λέγω, For let A and B each be commensurable with C. I say ὅτι καὶ τὸ Α τῷ Β ἐστι σύμμετρον. that A is also commensurable with B. ᾿Επεὶ γὰρ σύμμετρόν ἐστι τὸ Α τῷ Γ, τὸ Α ἄρα πρὸς For since A is commensurable with C, A thus has τὸ Γ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐχέτω, ὃν ὁ Δ to C the ratio which (some) number (has) to (some) πρὸς τὸν Ε. πάλιν, ἐπεὶ σύμμετρόν ἐστι τὸ Γ τῷ Β, τὸ Γ ἄρα number [Prop. 10.5]. Let it have (the ratio) which D πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐχέτω, ὃν (has) to E. Again, since C is commensurable with B, ὁ Ζ πρὸς τὸν Η. καὶ λόγων δοθέντων ὁποσωνοῦν τοῦ τε, C thus has to B the ratio which (some) number (has) ὃν ἔχει ὁ Δ πρὸς τὸν Ε, καὶ ὁ Ζ πρὸς τὸν Η εἰλήφθωσαν to (some) number [Prop. 10.5]. Let it have (the ratio) ἀριθμοὶ ἑξῆς ἐν τοῖς δοθεῖσι λόγοις οἱ Θ, Κ, Λ· ὥστε εἶναι which F (has) to G. And for any multitude whatsoever 292 STOIQEIWN iþ. ELEMENTS BOOK 10 ὡς μὲν τὸν Δ πρὸς τὸν Ε, οὕτως τὸν Θ πρὸς τὸν Κ, ὡς δὲ of given ratios—(namely,) those which D has to E, and τὸν Ζ πρὸς τὸν Η, οὕτως τὸν Κ πρὸς τὸν Λ. F to G—let the numbers H , K, L (which are) contin- uously (proportional) in the(se) given ratios have been taken [Prop. 8.4]. Hence, as D is to E, so H (is) to K, and as F (is) to G, so K (is) to L. Λ Α ∆ Ε Ζ Η Γ Κ Β Θ G BC H K L A D E F ᾿Επεὶ οὖν ἐστιν ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς Therefore, since as A is to C, so D (is) to E, but as τὸν Ε, ἀλλ᾿ ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Θ πρὸς τὸν Κ, D (is) to E, so H (is) to K, thus also as A is to C, so H ἔστιν ἄρα καὶ ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Θ πρὸς τὸν Κ. (is) to K [Prop. 5.11]. Again, since as C is to B, so F πάλιν, ἐπεί ἐστιν ὡς τὸ Γ πρὸς τὸ Β, οὕτως ὁ Ζ πρὸς τὸν (is) to G, but as F (is) to G, [so] K (is) to L, thus also Η, ἀλλ᾿ ὡς ὁ Ζ πρὸς τὸν Η, [οὕτως] ὁ Κ πρὸς τὸν Λ, καὶ as C (is) to B, so K (is) to L [Prop. 5.11]. And also as A ὡς ἄρα τὸ Γ πρὸς τὸ Β, οὕτως ὁ Κ πρὸς τὸν Λ. ἔστι δὲ καὶ is to C, so H (is) to K. Thus, via equality, as A is to B, ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Θ πρὸς τὸν Κ· δι᾿ ἴσου ἄρα so H (is) to L [Prop. 5.22]. Thus, A has to B the ratio ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως ὁ Θ πρὸς τὸν Λ. τὸ Α ἄρα which the number H (has) to the number L. Thus, A is πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς ὁ Θ πρὸς ἀριθμὸν τὸν Λ· commensurable with B [Prop. 10.6]. σύμμετρον ἄρα ἐστὶ τὸ Α τῷ Β. Thus, (magnitudes) commensurable with the same Τὰ ἄρα τῷ αὐτῷ μεγέθει σύμμετρα καὶ ἀλλήλοις ἐστὶ magnitude are also commensurable with one another. σύμμετρα· ὅπερ ἔδει δεῖξαι. (Which is) the very thing it was required to show.igþ. Proposition 13 ᾿Εὰν ᾖ δύο μεγέθη σύμμετρα, τὸ δὲ ἕτερον αὐτῶν If two magnitudes are commensurable, and one of μεγέθει τινὶ ἀσύμμετρον ᾖ, καὶ τὸ λοιπὸν τῷ αὐτῷ ἀσύμμετρ- them is incommensurable with some magnitude, then ον ἔσται. the remaining (magnitude) will also be incommensurable with it. Β Α Γ B A C ῎Εστω δύο μεγέθη σύμμετρα τὰ Α, Β, τὸ δὲ ἕτερον Let A and B be two commensurable magnitudes, and αὐτῶν τὸ Α ἄλλῳ τινὶ τῷ Γ ἀσύμμετρον ἔστω· λέγω, ὅτι let one of them, A, be incommensurable with some other καὶ τὸ λοιπὸν τὸ Β τῷ Γ ἀσύμμετρόν ἐστιν. (magnitude), C. I say that the remaining (magnitude), Εἰ γάρ ἐστι σύμμετρον τὸ Β τῷ Γ, ἀλλὰ καὶ τὸ Α τῷ B, is also incommensurable with C. Β σύμμετρόν ἐστιν, καὶ τὸ Α ἄρα τῷ Γ σύμμετρόν ἐστιν. For if B is commensurable with C, but A is also com- ἀλλὰ καὶ ἀσύμμετρον· ὅπερ ἀδύνατον. οὐκ ἄρα σύμμετρόν mensurable with B, A is thus also commensurable with ἐστι τὸ Β τῷ Γ· ἀσύμμετρον ἄρα. C [Prop. 10.12]. But, (it is) also incommensurable (with ᾿Εὰν ἄρα ᾖ δύο μεγέθη σύμμετρα, καὶ τὰ ἑξῆς. C). The very thing (is) impossible. Thus, B is not com- mensurable with C. Thus, (it is) incommensurable. Thus, if two magnitudes are commensurable, and so on . . . .L¨mma. Lemma Δύο δοθεισῶν εὐθειῶν ἀνίσων εὑρεῖν, τίνι μεῖζον For two given unequal straight-lines, to find by (the δύναται ἡ μείζων τῆς ἐλάσσονος. square on) which (straight-line) the square on the greater 293 STOIQEIWN iþ. ELEMENTS BOOK 10 (straight-line is) larger than (the square on) the lesser.† ∆ Α Β Γ C A B D ῎Εστωσαν αἱ δοθεῖσαι δύο ἄνισοι εὐθεῖαι αἱ ΑΒ, Γ, ὧν Let AB and C be the two given unequal straight-lines, μείζων ἔστω ἡ ΑΒ· δεῖ δὴ εὑρεῖν, τίνι μεῖζον δύναται ἡ ΑΒ and let AB be the greater of them. So it is required to τῆς Γ. find by (the square on) which (straight-line) the square Γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ εἰς αὐτὸ on AB (is) greater than (the square on) C. ἐνηρμόσθω τῇ Γ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΒ. φανερὸν Let the semi-circle ADB have been described on AB. δή, ὅτι ὀρθή ἐστιν ἡ ὑπὸ ΑΔΒ γωνία, καὶ ὅτι ἡ ΑΒ τῆς And let AD, equal to C, have been inserted into it ΑΔ, τουτέστι τῆς Γ, μεῖζον δύναται τῇ ΔΒ. [Prop. 4.1]. And let DB have been joined. So (it is) clear ῾Ομοίως δὲ καὶ δύο δοθεισῶν εὐθειῶν ἡ δυναμένη αὐτὰς that the angle ADB is a right-angle [Prop. 3.31], and εὑρίσκεται οὕτως. that the square on AB (is) greater than (the square on) ῎Εστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΑΔ, ΔΒ, καὶ δέον AD—that is to say, (the square on) C—by (the square ἔστω εὑρεῖν τὴν δυναμένην αὐτάς. κείσθωσαν γάρ, ὥστε on) DB [Prop. 1.47]. ὀρθὴν γωνίαν περιέχειν τὴν ὑπὸ ΑΔ, ΔΒ, καὶ ἐπεζεύχθω And, similarly, the square-root of (the sum of the ἡ ΑΒ· φανερὸν πάλιν, ὅτι ἡ τὰς ΑΔ, ΔΒ δυναμένη ἐστὶν ἡ squares on) two given straight-lines is also found likeso. ΑΒ· ὅπερ ἔδει δεῖξαι. Let AD and DB be the two given straight-lines. And let it be necessary to find the square-root of (the sum of the squares on) them. For let them have been laid down such as to encompass a right-angle—(namely), that (angle encompassed) by AD and DB. And let AB have been joined. (It is) again clear that AB is the square-root of (the sum of the squares on) AD and DB [Prop. 1.47]. (Which is) the very thing it was required to show. † That is, if α and β are the lengths of two given straight-lines, with α being greater than β, to find a straight-line of length γ such that α2 = β2 + γ2. Similarly, we can also find γ such that γ2 = α2 + β2.idþ. Proposition 14 ᾿Εὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, δύνηται δὲ ἡ If four straight-lines are proportional, and the square πρώτη τῆς δευτέρας μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ on the first is greater than (the square on) the sec- [μήκει], καὶ ἡ τρίτη τῆς τετάρτης μεῖζον δυνήσεται τῷ ἀπὸ ond by the (square) on (some straight-line) commen- συμμέτρου ἑαυτῇ [μήκει]. καὶ ἐὰν ἡ πρώτη τῆς δευτέρας surable [in length] with the first, then the square on μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ [μήκει], καὶ ἡ the third will also be greater than (the square on) the τρίτη τῆς τετάρτης μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου fourth by the (square) on (some straight-line) commen- ἑαυτῇ [μήκει]. surable [in length] with the third. And if the square on ῎Εστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, Δ, the first is greater than (the square on) the second by ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ, καὶ ἡ Α μὲν the (square) on (some straight-line) incommensurable τῆς Β μεῖζον δυνάσθω τῷ ἀπὸ τῆς Ε, ἡ δὲ Γ τῆς Δ μεῖζον [in length] with the first, then the square on the third δυνάσθω τῷ ἀπὸ τῆς Ζ· λέγω, ὅτι, εἴτε σύμμετρός ἐστιν will also be greater than (the square on) the fourth by ἡ Α τῇ Ε, σύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ, εἴτε ἀσύμμετρός the (square) on (some straight-line) incommensurable ἐστιν ἡ Α τῇ Ε, ἀσύμμετρός ἐστι καὶ ὁ Γ τῇ Ζ. [in length] with the third. Let A, B, C, D be four proportional straight-lines, (such that) as A (is) to B, so C (is) to D. And let the square on A be greater than (the square on) B by the 294 STOIQEIWN iþ. ELEMENTS BOOK 10 (square) on E, and let the square on C be greater than (the square on) D by the (square) on F . I say that A is either commensurable (in length) with E, and C is also commensurable with F , or A is incommensurable (in length) with E, and C is also incommensurable with F . ΖΓ ∆Α Β Ε FDA B E C ᾿Επεὶ γάρ ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν For since as A is to B, so C (is) to D, thus as the Δ, ἔστιν ἄρα καὶ ὡς τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Β, οὕτως (square) on A is to the (square) on B, so the (square) on τὸ ἀπὸ τῆς Γ πρὸς τὸ ἀπὸ τῆς Δ. ἀλλὰ τῷ μὲν ἀπὸ τῆς Α C (is) to the (square) on D [Prop. 6.22]. But the (sum ἴσα ἐστὶ τὰ ἀπὸ τῶν Ε, Β, τῷ δὲ ἀπὸ τῆς Γ ἴσα ἐστὶ τὰ ἀπὸ of the squares) on E and B is equal to the (square) on τῶν Δ, Ζ. ἔστιν ἄρα ὡς τὰ ἀπὸ τῶν Ε, Β πρὸς τὸ ἀπὸ τῆς A, and the (sum of the squares) on D and F is equal Β, οὕτως τὰ ἀπὸ τῶν Δ, Ζ πρὸς τὸ ἀπὸ τῆς Δ· διελόντι ἄρα to the (square) on C. Thus, as the (sum of the squares) ἐστὶν ὡς τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς Β, οὕτως τὸ ἀπὸ on E and B is to the (square) on B, so the (sum of the τῆς Ζ πρὸς τὸ ἀπὸ τῆς Δ· ἔστιν ἄρα καὶ ὡς ἡ Ε πρὸς τὴν squares) on D and F (is) to the (square) on D. Thus, Β, οὕτως ἡ Ζ πρὸς τὴν Δ· ἀνάπαλιν ἄρα ἐστὶν ὡς ἡ Β πρὸς via separation, as the (square) on E is to the (square) τὴν Ε, οὕτως ἡ Δ πρὸς τὴν Ζ. ἔστι δὲ καὶ ὡς ἡ Α πρὸς τὴν on B, so the (square) on F (is) to the (square) on D Β, οὕτως ἡ Γ πρὸς τὴν Δ· δι᾿ ἴσου ἄρα ἐστὶν ὡς ἡ Α πρὸς [Prop. 5.17]. Thus, also, as E is to B, so F (is) to D τὴν Ε, οὕτως ἡ Γ πρὸς τὴν Ζ. εἴτε οὖν σύμμετρός ἐστιν ἡ Α [Prop. 6.22]. Thus, inversely, as B is to E, so D (is) τῇ Ε, συμμετρός ἐστι καὶ ἡ Γ τῇ Ζ, εἴτε ἀσύμμετρός ἐστιν to F [Prop. 5.7 corr.]. But, as A is to B, so C also (is) ἡ Α τῇ Ε, ἀσύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ. to D. Thus, via equality, as A is to E, so C (is) to F ᾿Εὰν ἄρα, καὶ τὰ ἑξῆς. [Prop. 5.22]. Therefore, A is either commensurable (in length) with E, and C is also commensurable with F , or A is incommensurable (in length) with E, and C is also incommensurable with F [Prop. 10.11]. Thus, if, and so on . . . .ieþ. Proposition 15 ᾿Εὰν δύο μεγέθη σύμμετρα συντεθῇ, καὶ τὸ ὅλον If two commensurable magnitudes are added together ἑκατέρῳ αὐτῶν σύμμετρον ἔσται· κἂν τὸ ὅλον ἑνὶ αὐτῶν then the whole will also be commensurable with each of σύμμετρον ᾖ, καὶ τὰ ἐξ ἀρχῆς μεγέθη σύμμετρα ἔσται. them. And if the whole is commensurable with one of Συγκείσθω γὰρ δύο μεγέθη σύμμετρα τὰ ΑΒ, ΒΓ· λέγω, them then the original magnitudes will also be commen- ὅτι καὶ ὅλον τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ ἐστι σύμμετρον. surable (with one another). For let the two commensurable magnitudes AB and BC be laid down together. I say that the whole AC is also commensurable with each of AB and BC. 295 STOIQEIWN iþ. ELEMENTS BOOK 10 ∆ Α Β Γ D A B C ᾿Επεὶ γὰρ σύμμετρά ἐστι τὰ ΑΒ, ΒΓ, μετρήσει τι αὐτὰ For since AB and BC are commensurable, some mag- μέγεθος. μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΑΒ, nitude will measure them. Let it (so) measure (them), ΒΓ μετρεῖ, καὶ ὅλον τὸ ΑΓ μετρήσει. μετρεῖ δὲ καὶ τὰ ΑΒ, and let it be D. Therefore, since D measures (both) AB ΒΓ. τὸ Δ ἄρα τὰ ΑΒ, ΒΓ, ΑΓ μετρεῖ· σύμμετρον ἄρα ἐστὶ and BC, it will also measure the whole AC. And it also τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ. measures AB and BC. Thus, D measures AB, BC, and Ἀλλὰ δὴ τὸ ΑΓ ἔστω σύμμετρον τῷ ΑΒ· λέγω δή, ὅτι AC. Thus, AC is commensurable with each of AB and καὶ τὰ ΑΒ, ΒΓ σύμμετρά ἐστιν. BC [Def. 10.1]. ᾿Επεὶ γὰρ σύμμετρά ἐστι τὰ ΑΓ, ΑΒ, μετρήσει τι αὐτὰ And so let AC be commensurable with AB. I say that μέγεθος. μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΓΑ, AB and BC are also commensurable. ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸ ΒΓ μετρήσει. μετρεῖ δὲ καὶ For since AC and AB are commensurable, some mag- τὸ ΑΒ· τὸ Δ ἄρα τὰ ΑΒ, ΒΓ μετρήσει· σύμμετρα ἄρα ἐστὶ nitude will measure them. Let it (so) measure (them), τὰ ΑΒ, ΒΓ. and let it be D. Therefore, since D measures (both) CA ᾿Εὰν ἄρα δύο μεγέθη, καὶ τὰ ἑξῆς. and AB, it will thus also measure the remainder BC. And it also measures AB. Thus, D will measure (both) AB and BC. Thus, AB and BC are commensurable [Def. 10.1]. Thus, if two magnitudes, and so on . . . .i�þ. Proposition 16 ᾿Εὰν δύο μεγέθη ἀσύμμετρα συντεθῇ, καὶ τὸ ὅλον If two incommensurable magnitudes are added to- ἑκατέρῳ αὐτῶν ἀσύμμετρον ἔσται· κἂν τὸ ὅλον ἑνὶ αὐτῶν gether then the whole will also be incommensurable with ἀσύμμετρον ᾖ, καὶ τὰ ἐξ ἀρχῆς μεγέθη ἀσύμμετρα ἔσται. each of them. And if the whole is incommensurable with one of them then the original magnitudes will also be in- commensurable (with one another). ∆ Α Β Γ D A B C Συγκείσθω γὰρ δύο μεγέθη ἀσύμμετρα τὰ ΑΒ, ΒΓ· For let the two incommensurable magnitudes AB and λέγω, ὅτι καὶ ὅλον τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ ἀσύμμετρόν BC be laid down together. I say that that the whole AC ἐστιν. is also incommensurable with each of AB and BC. Εἰ γὰρ μή ἐστιν ἀσύμμετρα τὰ ΓΑ, ΑΒ, μετρήσει τι For if CA and AB are not incommensurable then [αὐτὰ] μέγεθος. μετρείτω, εἰ δυνατόν, καὶ ἔστω τὸ Δ. ἐπεὶ some magnitude will measure [them]. If possible, let it οὖν τὸ Δ τὰ ΓΑ, ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸ ΒΓ μετρήσει. (so) measure (them), and let it be D. Therefore, since μετρεῖ δὲ καὶ τὸ ΑΒ· τὸ Δ ἄρα τὰ ΑΒ, ΒΓ μετρεῖ. σύμμετρα D measures (both) CA and AB, it will thus also mea- ἄρα ἐστὶ τὰ ΑΒ, ΒΓ· ὑπέκειντο δὲ καὶ ἀσύμμετρα· ὅπερ sure the remainder BC. And it also measures AB. Thus, ἐστὶν ἀδύνατον. οὐκ ἄρα τὰ ΓΑ, ΑΒ μετρήσει τι μέγεθος· D measures (both) AB and BC. Thus, AB and BC are ἀσύμμετρα ἄρα ἐστὶ τὰ ΓΑ, ΑΒ. ὁμοίως δὴ δείξομεν, ὅτι καὶ commensurable [Def. 10.1]. But they were also assumed τὰ ΑΓ, ΓΒ ἀσύμμετρά ἐστιν. τὸ ΑΓ ἄρα ἑκατέρῳ τῶν ΑΒ, (to be) incommensurable. The very thing is impossible. ΒΓ ἀσύμμετρόν ἐστιν. Thus, some magnitude cannot measure (both) CA and Ἀλλὰ δὴ τὸ ΑΓ ἑνὶ τῶν ΑΒ, ΒΓ ἀσύμμετρον ἔστω. AB. Thus, CA and AB are incommensurable [Def. 10.1]. ἔστω δὴ πρότερον τῷ ΑΒ· λέγω, ὅτι καὶ τὰ ΑΒ, ΒΓ So, similarly, we can show that AC and CB are also ἀσύμμετρά ἐστιν. εἰ γὰρ ἔσται σύμμετρα, μετρήσει τι αὐτὰ incommensurable. Thus, AC is incommensurable with μέγεθος. μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΑΒ, each of AB and BC. ΒΓ μετρεῖ, καὶ ὅλον ἄρα τὸ ΑΓ μετρήσει. μετρεῖ δὲ καὶ τὸ And so let AC be incommensurable with one of AB ΑΒ· τὸ Δ ἄρα τὰ ΓΑ, ΑΒ μετρεῖ. σύμμετρα ἄρα ἐστὶ τὰ and BC. So let it, first of all, be incommensurable with 296 STOIQEIWN iþ. ELEMENTS BOOK 10 ΓΑ, ΑΒ· ὑπέκειτο δὲ καὶ ἀσύμμετρα· ὅπερ ἐστὶν ἀδύνατον. AB. I say that AB and BC are also incommensurable. οὐκ ἄρα τὰ ΑΒ, ΒΓ μετρήσει τι μέγεθος· ἀσύμμετρα ἄρα For if they are commensurable then some magnitude will ἐστὶ τὰ ΑΒ, ΒΓ. measure them. Let it (so) measure (them), and let it be ᾿Εὰν ἄρα δύο μεγέθη, καὶ τὰ ἑξῆς. D. Therefore, since D measures (both) AB and BC, it will thus also measure the whole AC. And it also mea- sures AB. Thus, D measures (both) CA and AB. Thus, CA and AB are commensurable [Def. 10.1]. But they were also assumed (to be) incommensurable. The very thing is impossible. Thus, some magnitude cannot mea- sure (both) AB and BC. Thus, AB and BC are incom- mensurable [Def. 10.1]. Thus, if two. . . magnitudes, and so on . . . .L¨mma. Lemma ᾿Εὰν παρά τινα εὐθεῖαν παραβληθῇ παραλληλόγραμμον If a parallelogram,† falling short by a square figure, is ἐλλεῖπον εἴδει τετραγώνῳ, τὸ παραβληθὲν ἴσον ἐστὶ τῷ ὑπὸ applied to some straight-line then the applied (parallelo- τῶν ἐκ τῆς παραβολής γενομένων τμημάτων τῆς εὐθείας. gram) is equal (in area) to the (rectangle contained) by the pieces of the straight-line created via the application (of the parallelogram). Α ∆ Γ Β D BCA Παρὰ γὰρ εὐθεῖαν τὴν ΑΒ παραβεβλήσθω παραλ- For let the parallelogram AD, falling short by the ληλόγραμμον τὸ ΑΔ ἐλλεῖπον εἴδει τετραγώνῳ τῷ ΔΒ· square figure DB, have been applied to the straight-line λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΔ τῷ ὑπὸ τῶν ΑΓ, ΓΒ. AB. I say that AD is equal to the (rectangle contained) Καί ἐστιν αὐτόθεν φανερόν· ἐπεὶ γὰρ τετράγωνόν ἐστι by AC and CB. τὸ ΔΒ, ἴση ἐστὶν ἡ ΔΓ τῇ ΓΒ, καί ἐστι τὸ ΑΔ τὸ ὑπὸ τῶν And it is immediately obvious. For since DB is a ΑΓ, ΓΔ, τουτέστι τὸ ὑπὸ τῶν ΑΓ, ΓΒ. square, DC is equal to CB. And AD is the (rectangle ᾿Εὰν ἄρα παρά τινα εὐθεῖαν, καὶ τὰ ἑξῆς. contained) by AC and CD—that is to say, by AC and CB. Thus, if . . . to some straight-line, and so on . . . . † Note that this lemma only applies to rectangular parallelograms.izþ. Proposition 17† ᾿Εὰν ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετράτῳ μέρει If there are two unequal straight-lines, and a (rect- τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ angle) equal to the fourth part of the (square) on the ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς σύμμετρα αὐτὴν διαιρῇ lesser, falling short by a square figure, is applied to the μήκει, ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ greater, and divides it into (parts which are) commen- συμμέτου ἑαυτῇ [μήκει]. καὶ ἐὰν ἡ μείζων τῆς ἐλάσσονος surable in length, then the square on the greater will be μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ [μήκει], τῷ δὲ larger than (the square on) the lesser by the (square) τετράρτῳ τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα on (some straight-line) commensurable [in length] with παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν the greater. And if the square on the greater is larger διαιρεῖ μήκει. than (the square on) the lesser by the (square) on ῎Εστωσαν δύο εὐθεῖαι ἄνισοι αἱ Α, ΒΓ, ὧν μείζων ἡ (some straight-line) commensurable [in length] with the 297 STOIQEIWN iþ. ELEMENTS BOOK 10 ΒΓ, τῷ δὲ τετράρτῳ μέρει τοῦ ἀπὸ ἐλάσσονος τῆς Α, greater, and a (rectangle) equal to the fourth (part) of the τουτέστι τῷ ἀπὸ τῆς ἡμισείας τῆς Α, ἴσον παρὰ τὴν ΒΓ (square) on the lesser, falling short by a square figure, is παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ applied to the greater, then it divides it into (parts which τῶν ΒΔ, ΔΓ, σύμμετρος δὲ ἔστω ἡ ΒΔ τῇ ΔΓ μήκει· λέγω, are) commensurable in length. ὃτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. Let A and BC be two unequal straight-lines, of which (let) BC (be) the greater. And let a (rectangle) equal to the fourth part of the (square) on the lesser, A—that is, (equal) to the (square) on half of A—falling short by a square figure, have been applied to BC. And let it be the (rectangle contained) by BD and DC [see previous lemma]. And let BD be commensurable in length with DC. I say that that the square on BC is greater than the (square on) A by (the square on some straight-line) commensurable (in length) with (BC). Α ΕΒ Ζ Γ∆ CEB D A F Τετμήσθω γὰρ ἡ ΒΓ δίχα κατὰ τὸ Ε σημεῖον, καὶ κείσθω For let BC have been cut in half at the point E [Prop. τῇ ΔΕ ἴση ἡ ΕΖ. λοιπὴ ἄρα ἡ ΔΓ ἴση ἐστὶ τῇ ΒΖ. καὶ ἐπεὶ 1.10]. And let EF be made equal to DE [Prop. 1.3]. εὐθεῖα ἡ ΒΓ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Ε, εἰς δὲ ἄνισα Thus, the remainder DC is equal to BF . And since the κατὰ τὸ Δ, τὸ ἄρα ὑπὸ ΒΔ, ΔΓ περειχόμενον ὀρθογώνιον straight-line BC has been cut into equal (pieces) at E, μετὰ τοῦ ἀπὸ τῆς ΕΔ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς and into unequal (pieces) at D, the rectangle contained ΕΓ τετραγώνῳ· καὶ τὰ τετραπλάσια· τὸ ἄρα τετράκις ὑπὸ by BD and DC, plus the square on ED, is thus equal to τῶν ΒΔ, ΔΓ μετὰ τοῦ τετραπλασίου τοῦ ἀπὸ τῆς ΔΕ ἴσον the square on EC [Prop. 2.5]. (The same) also (for) the ἐστὶ τῷ τετράκις ἀπὸ τῆς ΕΓ τετραγώνῳ. ἀλλὰ τῷ μὲν quadruples. Thus, four times the (rectangle contained) τετραπλασίῳ τοῦ ὑπὸ τῶν ΒΔ, ΔΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς by BD and DC, plus the quadruple of the (square) on Α τετράγωνον, τῷ δὲ τετραπλασίῳ τοῦ ἀπὸ τῆς ΔΕ ἴσον DE, is equal to four times the square on EC. But, the ἐστὶ τὸ ἀπὸ τῆς ΔΖ τετράγωνον· διπλασίων γάρ ἐστιν ἡ ΔΖ square on A is equal to the quadruple of the (rectangle τῆς ΔΕ. τῷ δὲ τετραπλασίῳ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τὸ contained) by BD and DC, and the square on DF is ἀπὸ τῆς ΒΓ τετράγωνον· διπλασίων γάρ ἐστι πάλιν ἡ ΒΓ equal to the quadruple of the (square) on DE. For DF τῆς ΓΕ. τὰ ἄρα ἀπὸ τῶν Α, ΔΖ τετράγωνα ἴσα ἐστὶ τῷ ἀπὸ is double DE. And the square on BC is equal to the τῆς ΒΓ τετράγωνῳ· ὥστε τὸ ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς Α quadruple of the (square) on EC. For, again, BC is dou- μεῖζόν ἐστι τῷ ἀπὸ τῆς ΔΖ· ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται ble CE. Thus, the (sum of the) squares on A and DF is τῇ ΔΖ. δεικτέον, ὅτι καὶ σύμμετρός ἐστιν ἡ ΒΓ τῇ ΔΖ. equal to the square on BC. Hence, the (square) on BC ἐπεὶ γὰρ σύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει, σύμμετρος is greater than the (square) on A by the (square) on DF . ἄρα ἐστὶ καὶ ἡ ΒΓ τῇ ΓΔ μήκει. ἀλλὰ ἡ ΓΔ ταῖς ΓΔ, ΒΖ Thus, BC is greater in square than A by DF . It must ἐστι σύμμετρος μήκει· ἴση γάρ ἐστιν ἡ ΓΔ τῇ ΒΖ. καὶ ἡ ΒΓ also be shown that BC is commensurable (in length) ἄρα σύμμετρός ἐστι ταῖς ΒΖ, ΓΔ μήκει· ὥστε καὶ λοιπῇ τῇ with DF . For since BD is commensurable in length ΖΔ σύμμετρός ἐστιν ἡ ΒΓ μήκει· ἡ ΒΓ ἄρα τῆς Α μεῖζον with DC, BC is thus also commensurable in length with δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. CD [Prop. 10.15]. But, CD is commensurable in length Ἀλλὰ δὴ ἡ ΒΓ τῆς Α μεῖζον δυνάσθω τῷ ἀπὸ συμμέτρου with CD plus BF . For CD is equal to BF [Prop. 10.6]. ἑαυτῇ, τῷ δὲ τετράτρῳ τοῦ ἀπὸ τῆς Α ἴσον παρὰ τὴν ΒΓ Thus, BC is also commensurable in length with BF plus παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ CD [Prop. 10.12]. Hence, BC is also commensurable τῶν ΒΔ, ΔΓ. δεικτέον, ὅτι σύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ in length with the remainder FD [Prop. 10.15]. Thus, μήκει. the square on BC is greater than (the square on) A by Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, the (square) on (some straight-line) commensurable (in ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. δύναται δὲ ἡ length) with (BC). 298 STOIQEIWN iþ. ELEMENTS BOOK 10 ΒΓ τῆς Α μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ. σύμμετρος ἄρα And so let the square on BC be greater than the ἐστὶν ἡ ΒΓ τῇ ΖΔ μήκει· ὥστε καὶ λοιπῇ συναμφοτέρῳ τῇ (square on) A by the (square) on (some straight-line) ΒΖ, ΔΓ σύμμετρός ἐστιν ἡ ΒΓ μήκει. ἀλλὰ συναμφότερος commensurable (in length) with (BC). And let a (rect- ἡ ΒΖ, ΔΓ σύμμετρός ἐστι τῇ ΔΓ [μήκει]. ὥστε καὶ ἡ ΒΓ angle) equal to the fourth (part) of the (square) on A, τῇ ΓΔ σύμμετρός ἐστι μήκει· καὶ διελόντι ἄρα ἡ ΒΔ τῇ ΔΓ falling short by a square figure, have been applied to BC. ἐστι σύμμετρος μήκει. And let it be the (rectangle contained) by BD and DC. It ᾿Εὰν ἄρα ὦσι δύο εὐθεῖαι ἄνισοι, καὶ τὰ ἑξῆς. must be shown that BD is commensurable in length with DC. For, similarly, by the same construction, we can show that the square on BC is greater than the (square on) A by the (square) on FD. And the square on BC is greater than the (square on) A by the (square) on (some straight- line) commensurable (in length) with (BC). Thus, BC is commensurable in length with FD. Hence, BC is also commensurable in length with the remaining sum of BF and DC [Prop. 10.15]. But, the sum of BF and DC is commensurable [in length] with DC [Prop. 10.6]. Hence, BC is also commensurable in length with CD [Prop. 10.12]. Thus, via separation, BD is also commen- surable in length with DC [Prop. 10.15]. Thus, if there are two unequal straight-lines, and so on . . . . † This proposition states that if α x − x2 = β2/4 (where α = BC, x = DC, and β = A) then α and p α2 − β2 are commensurable when α − x are x are commensurable, and vice versa.ihþ. Proposition 18† ᾿Εὰν ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει If there are two unequal straight-lines, and a (rect- τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ angle) equal to the fourth part of the (square) on the ἐλλεῖπον εἴδει τετραγώνῳ, καὶ εἰς ἀσυμμετρα αὐτὴν διαιρῇ lesser, falling short by a square figure, is applied to the [μήκει], ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ greater, and divides it into (parts which are) incom- ἀσυμμέτρου ἑαυτῇ. καὶ ἐὰν ἡ μείζων τῆς ἐλάσσονος μεῖζον mensurable [in length], then the square on the greater δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, τῷ δὲ τετράρτῳ τοῦ ἀπὸ will be larger than the (square on the) lesser by the τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον (square) on (some straight-line) incommensurable (in εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν διαιρεῖ [μήκει]. length) with the greater. And if the square on the ῎Εστωσαν δύο εὐθεῖαι ἄνισοι αἱ Α, ΒΓ, ὧν μείζων ἡ ΒΓ, greater is larger than the (square on the) lesser by the τῷ δὲ τετάρτῳ [μέρει] τοῦ ἀπὸ τῆς ἐλάσσονος τῆς Α ἴσον (square) on (some straight-line) incommensurable (in παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ length) with the greater, and a (rectangle) equal to the ἔστω τὸ ὑπὸ τῶν ΒΔΓ, ἀσύμμετρος δὲ ἔστω ἡ ΒΔ τῇ fourth (part) of the (square) on the lesser, falling short by ΔΓ μήκει· λέγω, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ a square figure, is applied to the greater, then it divides it ἀσυμμέτρου ἑαυτῇ. into (parts which are) incommensurable [in length]. Let A and BC be two unequal straight-lines, of which (let) BC (be) the greater. And let a (rectangle) equal to the fourth [part] of the (square) on the lesser, A, falling short by a square figure, have been applied to BC. And let it be the (rectangle contained) by BDC. And let BD be incommensurable in length with DC. I say that that the square on BC is greater than the (square on) A by the (square) on (some straight-line) incommensurable (in length) with (BC). 299 STOIQEIWN iþ. ELEMENTS BOOK 10 Ζ ΕΒ Α Γ∆ F EB A CD Τῶν γὰρ αὐτῶν κατασκευασθέντων τῷ πρότερον ὁμοίως For, similarly, by the same construction as before, we δείξομεν, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. can show that the square on BC is greater than the δεικτέον [οὖν], ὅτι ἀσύμμετρός ἐστιν ἡ ΒΓ τῇ ΔΖ μήκει. (square on) A by the (square) on FD. [Therefore] it ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει, ἀσύμμετρος must be shown that BC is incommensurable in length ἄρα ἐστὶ καὶ ἡ ΒΓ τῇ ΓΔ μήκει. ἀλλὰ ἡ ΔΓ σύμμετρός ἐστι with DF . For since BD is incommensurable in length συναμφοτέραις ταῖς ΒΖ, ΔΓ· καὶ ἡ ΒΓ ἄρα ἀσύμμετρός with DC, BC is thus also incommensurable in length ἐστι συναμφοτέραις ταῖς ΒΖ, ΔΓ. ὥστε καὶ λοιπῇ τῇ ΖΔ with CD [Prop. 10.16]. But, DC is commensurable (in ἀσύμμετρός ἔστιν ἡ ΒΓ μήκει. καὶ ἡ ΒΓ τῆς Α μεῖζον length) with the sum of BF and DC [Prop. 10.6]. And, δύναται τῷ ἀπὸ τῆς ΖΔ· ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται τῷ thus, BC is incommensurable (in length) with the sum of ἀπὸ ἀσυμμέτρου ἑαυτῇ. BF and DC [Prop. 10.13]. Hence, BC is also incommen- Δυνάσθω δὴ πάλιν ἡ ΒΓ τῆς Α μεῖζον τῷ ἀπὸ ἀσυμμέτρ- surable in length with the remainder FD [Prop. 10.16]. ου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς Α ἴσον παρὰ τὴν ΒΓ And the square on BC is greater than the (square on) παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ A by the (square) on FD. Thus, the square on BC is τῶν ΒΔ, ΔΓ. δεικτέον, ὅτι ἀσύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ greater than the (square on) A by the (square) on (some μήκει. straight-line) incommensurable (in length) with (BC). Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, So, again, let the square on BC be greater than the ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. ἀλλὰ (square on) A by the (square) on (some straight-line) in- ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. commensurable (in length) with (BC). And let a (rect- ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΓ τῇ ΖΔ μήκει· ὥστε καὶ λοιπῇ angle) equal to the fourth [part] of the (square) on A, συναμφοτέρῳ τῇ ΒΖ, ΔΓ ἀσύμμετρός ἐστιν ἡ ΒΓ. ἀλλὰ συ- falling short by a square figure, have been applied to BC. ναμφότερος ἡ ΒΖ, ΔΓ τῇ ΔΓ σύμμετρός ἐστι μήκει· καὶ ἡ And let it be the (rectangle contained) by BD and DC. ΒΓ ἄρα τῇ ΔΓ ἀσύμμετρός ἐστι μήκει· ὥστε καὶ διελόντι ἡ It must be shown that BD is incommensurable in length ΒΔ τῇ ΔΓ ἀσύμμετρός ἐστι μήκει. with DC. ᾿Εὰν ἄρα ὦσι δύο εὐθεῖαι, καὶ τὰ ἑξῆς. For, similarly, by the same construction, we can show that the square on BC is greater than the (square) on A by the (square) on FD. But, the square on BC is greater than the (square) on A by the (square) on (some straight-line) incommensurable (in length) with (BC). Thus, BC is incommensurable in length with FD. Hence, BC is also incommensurable (in length) with the re- maining sum of BF and DC [Prop. 10.16]. But, the sum of BF and DC is commensurable in length with DC [Prop. 10.6]. Thus, BC is also incommensurable in length with DC [Prop. 10.13]. Hence, via separa- tion, BD is also incommensurable in length with DC [Prop. 10.16]. Thus, if there are two . . . straight-lines, and so on . . . . † This proposition states that if α x − x2 = β2/4 (where α = BC, x = DC, and β = A) then α and p α2 − β2 are incommensurable when α − x are x are incommensurable, and vice versa.ijþ. Proposition 19 Τὸ ὑπὸ ῥητῶν μήκει συμμέτρων εὐθειῶν περιεχόμενον The rectangle contained by rational straight-lines ὀρθογώνιον ῥητόν ἐστιν. (which are) commensurable in length is rational. ῾Υπὸ γὰρ ῥητῶν μήκει συμμέτρων εὐθειῶν τῶν ΑΒ, ΒΓ For let the rectangle AC have been enclosed by the 300 STOIQEIWN iþ. ELEMENTS BOOK 10 ὀρθογώνιον περιεχέσθω τὸ ΑΓ· λέγω, ὅτι ῥητόν ἐστι τὸ rational straight-lines AB and BC (which are) commen- ΑΓ. surable in length. I say that AC is rational. Α Β Γ ∆ A B D C Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔ· For let the square AD have been described on AB. ῥητὸν ἄρα ἐστὶ τὸ ΑΔ. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΒ τῇ AD is thus rational [Def. 10.4]. And since AB is com- ΒΓ μήκει, ἴση δέ ἐστιν ἡ ΑΒ τῇ ΒΔ, σύμμετρος ἄρα ἐστὶν mensurable in length with BC, and AB is equal to BD, ἡ ΒΔ τῇ ΒΓ μήκει. καί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΒΓ, οὕτως BD is thus commensurable in length with BC. And as τὸ ΔΑ πρὸς τὸ ΑΓ. σύμμετρον ἄρα ἐστὶ τὸ ΔΑ τῷ ΑΓ. BD is to BC, so DA (is) to AC [Prop. 6.1]. Thus, DA ῥητὸν δὲ τὸ ΔΑ· ῥητὸν ἄρα ἐστὶ καὶ τὸ ΑΓ. is commensurable with AC [Prop. 10.11]. And DA (is) Τὸ ἄρα ὑπὸ ῥητῶν μήκει συμμέτρων, καὶ τὰ ἑξῆς. rational. Thus, AC is also rational [Def. 10.4]. Thus, the . . . by rational straight-lines . . . commensurable, and so on . . . .kþ. Proposition 20 ᾿Εὰν ῥητὸν παρὰ ῥητὴν παραβληθῇ, πλάτος ποιεῖ ῥητὴν If a rational (area) is applied to a rational (straight- καὶ σύμμετρον τῇ, παρ᾿ ἣν παράκειται, μήκει. line) then it produces as breadth a (straight-line which is) rational, and commensurable in length with the (straight- line) to which it is applied. ∆ Β Γ Α D B A C ῾Ρητὸν γὰρ τὸ ΑΓ παρὰ ῥητὴν τὴν ΑΒ παραβεβλήσθω For let the rational (area) AC have been applied to the πλάτος ποιοῦν τὴν ΒΓ· λέγω, ὅτι ῥητή ἐστιν ἡ ΒΓ καὶ rational (straight-line) AB, producing the (straight-line) σύμμετρος τῇ ΒΑ μήκει. BC as breadth. I say that BC is rational, and commen- Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔ· surable in length with BA. ῥητὸν ἄρα ἐστὶ τὸ ΑΔ. ῥητὸν δὲ καὶ τὸ ΑΓ· σύμμετρον ἄρα For let the square AD have been described on AB. 301 STOIQEIWN iþ. ELEMENTS BOOK 10 ἐστὶ τὸ ΔΑ τῷ ΑΓ. καί ἐστιν ὡς τὸ ΔΑ πρὸς τὸ ΑΓ, οὕτως AD is thus rational [Def. 10.4]. And AC (is) also ratio- ἡ ΔΒ πρὸς τὴν ΒΓ. σύμμετρος ἄρα ἐστὶ καὶ ἡ ΔΒ τῇ ΒΓ· nal. DA is thus commensurable with AC. And as DA ἴση δὲ ἡ ΔΒ τῇ ΒΑ· σύμμετρος ἄρα καὶ ἡ ΑΒ τῇ ΒΓ. ῥητὴ is to AC, so DB (is) to BC [Prop. 6.1]. Thus, DB is δέ ἐστιν ἡ ΑΒ· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΒΓ καὶ σύμμετρος τῇ also commensurable (in length) with BC [Prop. 10.11]. ΑΒ μήκει. And DB (is) equal to BA. Thus, AB (is) also commen- ᾿Εὰν ἄρα ῥητὸν παρὰ ῥητὴν παραβληθῇ, καὶ τὰ ἑξῆς. surable (in length) with BC. And AB is rational. Thus, BC is also rational, and commensurable in length with AB [Def. 10.3]. Thus, if a rational (area) is applied to a rational (straight-line), and so on . . . .kaþ. Proposition 21 Τὸ ὑπὸ ῥητῶν δυνάμει μόνον συμμέτρων εὐθειῶν πε- The rectangle contained by rational straight-lines ριεχόμενον ὀρθογώνιον ἄλογόν ἐστιν, καὶ ἡ δυναμένη αὐτὸ (which are) commensurable in square only is irrational, ἄλογός ἐστιν, καλείσθω δὲ μέση. and its square-root is irrational—let it be called medial.† ∆ Β Γ Α D B A C ῾Υπὸ γὰρ ῥητῶν δυνάμει μόνον συμμέτρων εὐθειῶν τῶν For let the rectangle AC be contained by the rational ΑΒ, ΒΓ ὀρθογώνιον περιεχέσθω τὸ ΑΓ· λέγω, ὅτι ἄλογόν straight-lines AB and BC (which are) commensurable in ἐστι τὸ ΑΓ, καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν, καλείσθω square only. I say that AC is irrational, and its square- δὲ μέση. root is irrational—let it be called medial. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔ· For let the square AD have been described on AB. ῥητὸν ἄρα ἐστὶ τὸ ΑΔ. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΒ AD is thus rational [Def. 10.4]. And since AB is incom- τῇ ΒΓ μήκει· δυνάμει γὰρ μόνον ὑπόκεινται σύμμετροι· ἴση mensurable in length with BC. For they were assumed δὲ ἡ ΑΒ τῇ ΒΔ, ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΔΒ τῇ ΒΓ to be commensurable in square only. And AB (is) equal μήκει. καί ἐστιν ὡς ἡ ΔΒ πρὸς τὴν ΒΓ, οὕτως τὸ ΑΔ to BD. DB is thus also incommensurable in length with πρὸς τὸ ΑΓ· ἀσύμμετρον ἄρα [ἐστὶ] τὸ ΔΑ τῷ ΑΓ. ῥητὸν BC. And as DB is to BC, so AD (is) to AC [Prop. 6.1]. δὲ τὸ ΔΑ· ἄλογον ἄρα ἐστὶ τὸ ΑΓ· ὥστε καὶ ἡ δυναμένη τὸ Thus, DA [is] incommensurable with AC [Prop. 10.11]. ΑΓ [τουτέστιν ἡ ἴσον αὐτῷ τετράγωνον δυναμένη] ἄλογός And DA (is) rational. Thus, AC is irrational [Def. 10.4]. ἐστιν, καλείσθω δε μέση· ὅπερ ἔδει δεῖξαι. Hence, its square-root [that is to say, the square-root of the square equal to it] is also irrational [Def. 10.4]—let it be called medial. (Which is) the very thing it was re- quired to show. † Thus, a medial straight-line has a length expressible as k1/4. 302 STOIQEIWN iþ. ELEMENTS BOOK 10L¨mma. Lemma ᾿Εὰν ὦσι δύο εὐθεῖαι, ἔστιν ὡς ἡ πρώτη πρὸς τὴν If there are two straight-lines then as the first is to the δευτέραν, οὕτως τὸ ἀπὸ τῆς πρώτης πρὸς τὸ ὑπὸ τῶν δύο second, so the (square) on the first (is) to the (rectangle εὐθειῶν. contained) by the two straight-lines. Ζ Ε Η ∆ E D GF ῎Εστωσαν δύο εὐθεῖαι αἱ ΖΕ, ΕΗ. λέγω, ὅτι ἐστὶν ὡς ἡ Let FE and EG be two straight-lines. I say that as ΖΕ πρὸς τὴν ΕΗ, οὕτως τὸ ἀπὸ τῆς ΖΕ πρὸς τὸ ὑπὸ τῶν FE is to EG, so the (square) on FE (is) to the (rectangle ΖΕ, ΕΗ. contained) by FE and EG. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΖΕ τετράγωνον τὸ ΔΖ, καὶ For let the square DF have been described on FE. συμπεπληρώσθω τὸ ΗΔ. ἐπεὶ οὖν ἐστιν ὡς ἡ ΖΕ πρὸς τὴν And let GD have been completed. Therefore, since as ΕΗ, οὕτως τὸ ΖΔ πρὸς τὸ ΔΗ, καί ἐστι τὸ μὲν ΖΔ τὸ ἀπὸ FE is to EG, so FD (is) to DG [Prop. 6.1], and FD is τῆς ΖΕ, τὸ δὲ ΔΗ τὸ ὑπὸ τῶν ΔΕ, ΕΗ, τουτέστι τὸ ὑπὸ the (square) on FE, and DG the (rectangle contained) τῶν ΖΕ, ΕΗ, ἔστιν ἄρα ὡς ἡ ΖΕ πρὸς τὴν ΕΗ, οὕτως τὸ by DE and EG—that is to say, the (rectangle contained) ἀπὸ τῆς ΖΕ πρὸς τὸ ὑπὸ τῶν ΖΕ, ΕΗ. ὁμοίως δὲ καὶ ὡς τὸ by FE and EG—thus as FE is to EG, so the (square) ὑπὸ τῶν ΗΕ, ΕΖ πρὸς τὸ ἀπὸ τῆς ΕΖ, τουτέστιν ὡς τὸ ΗΔ on FE (is) to the (rectangle contained) by FE and EG. πρὸς τὸ ΖΔ, οὕτως ἡ ΗΕ πρὸς τὴν ΕΖ· ὅπερ ἔδει δεῖξαι. And also, similarly, as the (rectangle contained) by GE and EF is to the (square on) EF—that is to say, as GD (is) to FD—so GE (is) to EF . (Which is) the very thing it was required to show.kbþ. Proposition 22 Τὸ ἀπὸ μέσης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ The square on a medial (straight-line), being ap- ῥητὴν καὶ ἀσύμμετρον τῇ, παρ᾿ ἣν παράκειται, μήκει. plied to a rational (straight-line), produces as breadth a (straight-line which is) rational, and incommensurable in length with the (straight-line) to which it is applied. ΕΑ Γ ∆ Ζ Η Β G A D B EC F ῎Εστω μέση μὲν ἡ Α, ῥητὴ δὲ ἡ ΓΒ, καὶ τῷ ἀπὸ τῆς Let A be a medial (straight-line), and CB a rational Α ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω χωρίον ὀρθογώνιον τὸ (straight-line), and let the rectangular area BD, equal to ΒΔ πλάτος ποιοῦν τὴν ΓΔ· λέγω, ὅτι ῥητή ἐστιν ἡ ΓΔ καὶ the (square) on A, have been applied to BC, producing ἀσύμμετρος τῇ ΓΒ μήκει. CD as breadth. I say that CD is rational, and incommen- ᾿Επεὶ γὰρ μέση ἐστὶν ἡ Α, δύναται χωρίον περιεχόμενον surable in length with CB. ὑπὸ ῥητῶν δυνάμει μόνον συμμέτρων. δυνάσθω τὸ ΗΖ. For since A is medial, the square on it is equal to a 303 STOIQEIWN iþ. ELEMENTS BOOK 10 δύναται δὲ καὶ τὸ ΒΔ· ἴσον ἄρα ἐστὶ τὸ ΒΔ τῷ ΗΖ. ἔστι δὲ (rectangular) area contained by rational (straight-lines αὐτῷ καὶ ἰσογώνιον· τῶν δὲ ἴσων τε καὶ ἰσογωνίων παραλ- which are) commensurable in square only [Prop. 10.21]. ληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας Let the square on (A) be equal to GF . And the square on γωνίας· ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΓ πρὸς τὴν ΕΗ, οὕτως (A) is also equal to BD. Thus, BD is equal to GF . And ἡ ΕΖ πρὸς τὴν ΓΔ. ἔστιν ἄρα καὶ ὡς τὸ ἀπὸ τῆς ΒΓ πρὸς (BD) is also equiangular with (GF ). And for equal and τὸ ἀπὸ τῆς ΕΗ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΓΔ. equiangular parallelograms, the sides about the equal an- σύμμετρον δέ ἐστι τὸ ἀπὸ τῆς ΓΒ τῷ ἀπὸ τῆς ΕΗ· ῥητὴ γάρ gles are reciprocally proportional [Prop. 6.14]. Thus, pro- ἐστιν ἑκατέρα αὐτῶν· σύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς portionally, as BC is to EG, so EF (is) to CD. And, also, ΕΖ τῷ ἀπὸ τῆς ΓΔ. ῥητὸν δέ ἐστι τὸ ἀπὸ τῆς ΕΖ· ῥητὸν as the (square) on BC is to the (square) on EG, so the ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΓΔ· ῥητὴ ἄρα ἐστὶν ἡ ΓΔ. καὶ ἐπεὶ (square) on EF (is) to the (square) on CD [Prop. 6.22]. ἀσύμμετρός ἐστιν ἡ ΕΖ τῇ ΕΗ μήκει· δυνάμει γὰρ μόνον And the (square) on CB is commensurable with the εἰσὶ σύμμετροι· ὡς δὲ ἡ ΕΖ πρὸς τὴν ΕΗ, οὕτως τὸ ἀπὸ (square) on EG. For they are each rational. Thus, the τῆς ΕΖ πρὸς τὸ ὑπὸ τῶν ΖΕ, ΕΗ, ἀσύμμετρον ἄρα [ἐστὶ] (square) on EF is also commensurable with the (square) τὸ ἀπὸ τῆς ΕΖ τῷ ὑπὸ τῶν ΖΕ, ΕΗ. ἀλλὰ τῷ μὲν ἀπὸ τῆς on CD [Prop. 10.11]. And the (square) on EF is ratio- ΕΖ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΓΔ· ῥηταὶ γάρ εἰσι δυνάμει· nal. Thus, the (square) on CD is also rational [Def. 10.4]. τῷ δὲ ὑπὸ τῶν ΖΕ, ΕΗ σύμμετρόν ἐστι τὸ ὑπὸ τῶν ΔΓ, Thus, CD is rational. And since EF is incommensurable ΓΒ· ἴσα γάρ ἐστι τῷ ἀπὸ τῆς Α· ἀσύμμετρον ἄρα ἐστὶ καὶ in length with EG. For they are commensurable in square τὸ ἀπὸ τῆς ΓΔ τῷ ὑπὸ τῶν ΔΓ, ΓΒ. ὡς δὲ τὸ ἀπὸ τῆς ΓΔ only. And as EF (is) to EG, so the (square) on EF (is) πρὸς τὸ ὑπὸ τῶν ΔΓ, ΓΒ, οὕτως ἐστὶν ἡ ΔΓ πρὸς τὴν ΓΒ· to the (rectangle contained) by FE and EG [see previ- ἀσύμμετρος ἄρα ἐστὶν ἡ ΔΓ τῇ ΓΒ μήκει. ῥητὴ ἄρα ἐστὶν ἡ ous lemma]. The (square) on EF [is] thus incommen- ΓΔ καὶ ἀσύμμετρος τῇ ΓΒ μήκει· ὅπερ ἔδει δεῖξαι. surable with the (rectangle contained) by FE and EG [Prop. 10.11]. But, the (square) on CD is commensu- rable with the (square) on EF . For they are rational in square. And the (rectangle contained) by DC and CB is commensurable with the (rectangle contained) by FE and EG. For they are (both) equal to the (square) on A. Thus, the (square) on CD is also incommensurable with the (rectangle contained) by DC and CB [Prop. 10.13]. And as the (square) on CD (is) to the (rectangle con- tained) by DC and CB, so DC is to CB [see previous lemma]. Thus, DC is incommensurable in length with CB [Prop. 10.11]. Thus, CD is rational, and incommen- surable in length with CB. (Which is) the very thing it was required to show. † Literally, “rational”. kgþ. Proposition 23 ῾Η τῇ μέσῃ σύμμετρος μέση ἐστίν. A (straight-line) commensurable with a medial (straight- ῎Εστω μέση ἡ Α, καὶ τῇ Α σύμμετρος ἔστω ἡ Β· λέγω, line) is medial. ὅτι καὶ ἡ Β μέση ἐστίν. Let A be a medial (straight-line), and let B be com- ᾿Εκκείσθω γὰρ ῥητὴ ἡ ΓΔ, καὶ τῷ μὲν ἀπὸ τῆς Α mensurable with A. I say that B is also a medial (staight- ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω χωρίον ὀρθογώνιον τὸ line). ΓΕ πλάτος ποιοῦν τὴν ΕΔ· ῥητὴ ἄρα ἐστὶν ἡ ΕΔ καὶ Let the rational (straight-line) CD be set out, and let ἀσύμμετρος τῂ ΓΔ μήκει. τῷ δὲ ἀπὸ τῆς Β ἴσον παρὰ the rectangular area CE, equal to the (square) on A, τὴν ΓΔ παραβεβλήσθω χωρίον ὀρθογώνιον τὸ ΓΖ πλάτος have been applied to CD, producing ED as width. ED ποιοῦν τὴν ΔΖ. ἐπεὶ οὖν σύμμετρός ἐστιν ἡ Α τῇ Β, is thus rational, and incommensurable in length with CD σύμμετρόν ἐστι καὶ τὸ ἀπὸ τῆς Α τῷ ἀπὸ τῆς Β. ἀλλὰ [Prop. 10.22]. And let the rectangular area CF , equal τῷ μὲν ἀπὸ τῆς Α ἴσον ἐστὶ τὸ ΕΓ, τῷ δὲ ἀπὸ τῆς Β to the (square) on B, have been applied to CD, produc- ἴσον ἐστὶ τὸ ΓΖ· σύμμετρον ἄρα ἐστὶ τὸ ΕΓ τῷ ΓΖ. καί ing DF as width. Therefore, since A is commensurable ἐστιν ὡς τὸ ΕΓ πρὸς τὸ ΓΖ, οὕτως ἡ ΕΔ πρὸς τὴν ΔΖ· with B, the (square) on A is also commensurable with 304 STOIQEIWN iþ. ELEMENTS BOOK 10 σύμμετρος ἄρα ἐστὶν ἡ ΕΔ τῇ ΔΖ μήκει. ῥητὴ δέ ἐστιν the (square) on B. But, EC is equal to the (square) on A, ἡ ΕΔ καὶ ἀσύμμετρος τῇ ΔΓ μήκει· ῥητὴ ἄρα ἐστὶ καὶ ἡ and CF is equal to the (square) on B. Thus, EC is com- ΔΖ καὶ ἀσύμμετρος τῇ ΔΓ μήκει· αἱ ΓΔ, ΔΖ ἄρα ῥηταί mensurable with CF . And as EC is to CF , so ED (is) to εἰσι δυνάμει μόνον σύμμετροι. ἡ δὲ τὸ ὑπὸ ῥητῶν δυνάμει DF [Prop. 6.1]. Thus, ED is commensurable in length μόνον συμμέτρων δυναμένη μέση ἐστίν. ἡ ἄρα τὸ ὑπὸ τῶν with DF [Prop. 10.11]. And ED is rational, and incom- ΓΔ, ΔΖ δυναμένη μέση ἐστίν· καὶ δύναται τὸ ὑπὸ τῶν ΓΔ, mensurable in length with CD. DF is thus also ratio- ΔΖ ἡ Β· μέση ἄρα ἐστὶν ἡ Β. nal [Def. 10.3], and incommensurable in length with DC [Prop. 10.13]. Thus, CD and DF are rational, and com- mensurable in square only. And the square-root of a (rect- angle contained) by rational (straight-lines which are) commensurable in square only is medial [Prop. 10.21]. Thus, the square-root of the (rectangle contained) by CD and DF is medial. And the square on B is equal to the (rectangle contained) by CD and DF . Thus, B is a me- dial (straight-line). ∆ Α Β Γ Ε Ζ F A B E D C Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι τὸ τῷ μέσῳ χωρίῳ σύμμετρ- And (it is) clear, from this, that an (area) commensu- ον μέσον ἐστίν. rable with a medial area† is medial. † A medial area is equal to the square on some medial straight-line. Hence, a medial area is expressible as k1/2.kdþ. Proposition 24 Τὸ ὑπὸ μέσων μήκει συμμέτρων εὐθειῶν περιεχόμενον A rectangle contained by medial straight-lines (which ὀρθογώνιον μέσον ἐστίν. are) commensurable in length is medial. ῾Υπὸ γὰρ μέσων μήκει συμμέτρων εὐθειῶν τῶν ΑΒ, ΒΓ For let the rectangle AC be contained by the medial περιεχέσθω ὀρθογώνιον τὸ ΑΓ· λέγω, ὅτι τὸ ΑΓ μέσον straight-lines AB and BC (which are) commensurable in ἐστίν. length. I say that AC is medial. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔ· For let the square AD have been described on AB. μέσον ἄρα ἐστὶ τὸ ΑΔ. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΒ τῇ AD is thus medial [see previous footnote]. And since ΒΓ μήκει, ἴση δὲ ἡ ΑΒ τῇ ΒΔ, σύμμετρος ἄρα ἐστὶ καὶ ἡ AB is commensurable in length with BC, and AB (is) ΔΒ τῇ ΒΓ μήκει· ὥστε καὶ τὸ ΔΑ τῷ ΑΓ σύμμετρόν ἐστιν. equal to BD, DB is thus also commensurable in length μέσον δὲ τὸ ΔΑ· μέσον ἄρα καὶ τὸ ΑΓ· ὅπερ ἔδει δεῖξαι. with BC. Hence, DA is also commensurable with AC [Props. 6.1, 10.11]. And DA (is) medial. Thus, AC (is) also medial [Prop. 10.23 corr.]. (Which is) the very thing it was required to show. 305 STOIQEIWN iþ. ELEMENTS BOOK 10 Α Β ∆ Γ A B D C keþ. Proposition 25 Τὸ ὑπὸ μέσων δυνάμει μόνον συμμέτρων εὐθειῶν πε- The rectangle contained by medial straight-lines ριεχόμενον ὀρθογώνιον ἤτοι ῥητὸν ἢ μέσον ἐστίν. (which are) commensurable in square only is either ra- tional or medial. Β∆ Α Γ ΕΞ Η Μ Ν Ζ Κ Λ Θ L D A E M N B C G O F H K ῾Υπὸ γὰρ μέσων δυνάμει μόνον συμμέτρων εὐθειῶν τῶν For let the rectangle AC be contained by the medial ΑΒ, ΒΓ ὀρθογώνιον περιεχέσθω τὸ ΑΓ· λέγω, ὅτι τὸ ΑΓ straight-lines AB and BC (which are) commensurable in ἤτοι ῥητὸν ἢ μέσον ἐστίν. square only. I say that AC is either rational or medial. Ἀναγεγράφθω γὰρ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα τὰ ΑΔ, For let the squares AD and BE have been described ΒΕ· μέσον ἄρα ἐστὶν ἑκάτερον τῶν ΑΔ, ΒΕ. καὶ ἐκκείσθω on (the straight-lines) AB and BC (respectively). AD ῥητὴ ἡ ΖΗ, καὶ τῷ μὲν ΑΔ ἴσον παρὰ τὴν ΖΗ παρα- and BE are thus each medial. And let the rational βεβλήσθω ὀρθογώνιον παραλληλόγραμμον τὸ ΗΘ πλάτος (straight-line) FG be laid out. And let the rectangular ποιοῦν τὴν ΖΘ, τῷ δὲ ΑΓ ἴσον παρὰ τὴν ΘΜ παραβεβλήσθω parallelogram GH , equal to AD, have been applied to ὀρθογώνιον παραλληλόγραμμον τὸ ΜΚ πλάτος ποιοῦν τὴν FG, producing FH as breadth. And let the rectangular ΘΚ, καὶ ἔτι τῷ ΒΕ ἴσον ὁμοίως παρὰ τὴν ΚΝ παρα- parallelogram MK, equal to AC, have been applied to βεβλήσθω τὸ ΝΛ πλάτος ποιοῦν τὴν ΚΛ· ἐπ᾿ εὐθείας ἄρα HM , producing HK as breadth. And, finally, let NL, εἰσὶν αἱ ΖΘ, ΘΚ, ΚΛ. ἐπεὶ οὖν μέσον ἐστὶν ἑκάτερον τῶν equal to BE, have similarly been applied to KN , pro- ΑΔ, ΒΕ, καί ἐστιν ἴσον τὸ μὲν ΑΔ τῷ ΗΘ, τὸ δὲ ΒΕ τῷ ducing KL as breadth. Thus, FH , HK, and KL are in ΝΛ, μέσον ἄρα καὶ ἑκάτερον τῶν ΗΘ, ΝΛ. καὶ παρὰ ῥητὴν a straight-line. Therefore, since AD and BE are each τὴν ΖΗ παράκειται· ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΖΘ, ΚΛ καὶ medial, and AD is equal to GH , and BE to NL, GH ἀσύμμετρος τῇ ΖΗ μήκει. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ΑΔ and NL (are) thus each also medial. And they are ap- τῷ ΒΕ, σύμμετρον ἄρα ἐστὶ καὶ τὸ ΗΘ τῷ ΝΛ. καί ἐστιν ὡς plied to the rational (straight-line) FG. FH and KL are τὸ ΗΘ πρὸς τὸ ΝΛ, οὕτως ἡ ΖΘ πρὸς τὴν ΚΛ· σύμμετρος thus each rational, and incommensurable in length with ἄρα ἐστὶν ἡ ΖΘ τῇ ΚΛ μήκει. αἱ ΖΘ, ΚΛ ἄρα ῥηταί εἰσι FG [Prop. 10.22]. And since AD is commensurable with μήκει σύμμετροι· ῥητὸν ἄρα ἐστὶ τὸ ὑπὸ τῶν ΖΘ, ΚΛ. καὶ BE, GH is thus also commensurable with NL. And as 306 STOIQEIWN iþ. ELEMENTS BOOK 10 ἐπεὶ ἴση ἐστὶν ἡ μὲν ΔΒ τῇ ΒΑ, ἡ δὲ ΞΒ τῇ ΒΓ, ἔστιν ἄρα GH is to NL, so FH (is) to KL [Prop. 6.1]. Thus, FH is ὡς ἡ ΔΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΑΒ πρὸς τὴν ΒΞ. ἀλλ᾿ ὡς commensurable in length with KL [Prop. 10.11]. Thus, μὲν ἡ ΔΒ πρὸς τὴν ΒΓ, οὕτως τὸ ΔΑ πρὸς τὸ ΑΓ· ὡς δὲ ἡ FH and KL are rational (straight-lines which are) com- ΑΒ πρὸς τὴν ΒΞ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΞ· ἔστιν ἄρα ὡς mensurable in length. Thus, the (rectangle contained) by τὸ ΔΑ πρὸς τὸ ΑΓ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΞ. ἴσον δέ ἐστι FH and KL is rational [Prop. 10.19]. And since DB is τὸ μὲν ΑΔ τῷ ΗΘ, τὸ δὲ ΑΓ τῷ ΜΚ, τὸ δὲ ΓΞ τῷ ΝΛ· equal to BA, and OB to BC, thus as DB is to BC, so ἔστιν ἄρα ὡς τὸ ΗΘ πρὸς τὸ ΜΚ, οὕτως τὸ ΜΚ πρὸς τὸ AB (is) to BO. But, as DB (is) to BC, so DA (is) to ΝΛ· ἔστιν ἄρα καὶ ὡς ἡ ΖΘ πρὸς τὴν ΘΚ, οὕτως ἡ ΘΚ πρὸς AC [Props. 6.1]. And as AB (is) to BO, so AC (is) to τὴν ΚΛ· τὸ ἄρα ὑπὸ τῶν ΖΘ, ΚΛ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΘΚ. CO [Prop. 6.1]. Thus, as DA is to AC, so AC (is) to ῥητὸν δὲ τὸ ὑπὸ τῶν ΖΘ, ΚΛ· ῥητὸν ἄρα ἐστὶ καὶ τὸ ἀπὸ CO. And AD is equal to GH , and AC to MK, and CO τῆς ΘΚ· ῥητὴ ἄρα ἐστὶν ἡ ΘΚ. καὶ εἰ μὲν σύμμετρός ἐστι to NL. Thus, as GH is to MK, so MK (is) to NL. Thus, τῇ ΖΗ μήκει, ῥητόν ἐστι τὸ ΘΝ· εἰ δὲ ἀσύμμετρός ἐστι τῇ also, as FH is to HK, so HK (is) to KL [Props. 6.1, ΖΗ μήκει, αἱ ΚΘ, ΘΜ ῥηταί εἰσι δυνάμει μόνον σύμμετροι· 5.11]. Thus, the (rectangle contained) by FH and KL μέσον ἄρα τὸ ΘΝ. τὸ ΘΝ ἄρα ἤτοι ῥητὸν ἢ μέσον ἐστίν. is equal to the (square) on HK [Prop. 6.17]. And the ἴσον δὲ τὸ ΘΝ τῷ ΑΓ· τὸ ΑΓ ἄρα ἤτοι ῥητὸν ἢ μέσον ἐστίν. (rectangle contained) by FH and KL (is) rational. Thus, Τὸ ἄρα ὑπὸ μέσων δυνάμει μόνον συμμέτρων, καὶ τὰ the (square) on HK is also rational. Thus, HK is ratio- εξῆς. nal. And if it is commensurable in length with FG then HN is rational [Prop. 10.19]. And if it is incommensu- rable in length with FG then KH and HM are rational (straight-lines which are) commensurable in square only: thus, HN is medial [Prop. 10.21]. Thus, HN is either ra- tional or medial. And HN (is) equal to AC. Thus, AC is either rational or medial. Thus, the . . . by medial straight-lines (which are) com- mensurable in square only, and so on . . . .k�þ. Proposition 26 Μέσον μέσου οὐχ ὑπερέχει ῥητῷ. A medial (area) does not exceed a medial (area) by a rational (area).† ΖΑ ∆ Γ Β Η Ε Θ Κ K A D B E C G H F Εἰ γὰρ δυνατόν, μέσον τὸ ΑΒ μέσου τοῦ ΑΓ ὑπερεχέτω For, if possible, let the medial (area) AB exceed the ῥητῷ τῷ ΔΒ, καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ τῷ ΑΒ ἴσον παρὰ medial (area) AC by the rational (area) DB. And let τὴν ΕΖ παραβεβλήσθω παραλληλόγραμμον ὀρθογώνιον τὸ the rational (straight-line) EF be laid down. And let the ΖΘ πλάτος ποιοῦν τὴν ΕΘ, τῷ δὲ ΑΓ ἴσον ἀφῃρήσθω τὸ rectangular parallelogram FH , equal to AB, have been ΖΗ· λοιπὸν ἄρα τὸ ΒΔ λοιπῷ τῷ ΚΘ ἐστιν ἴσον. ῥητὸν δέ applied to to EF , producing EH as breadth. And let FG, ἐστι τὸ ΔΒ· ῥητὸν ἄρα ἐστὶ καὶ τὸ ΚΘ. ἐπεὶ οὖν μέσον ἐστὶν equal to AC, have been cut off (from FH). Thus, the ἑκάτερον τῶν ΑΒ, ΑΓ, καί ἐστι τὸ μὲν ΑΒ τῷ ΖΘ ἴσον, τὸ remainder BD is equal to the remainder KH . And DB δὲ ΑΓ τῷ ΖΗ, μέσον ἄρα καὶ ἑκάτερον τῶν ΖΘ, ΖΗ. καὶ is rational. Thus, KH is also rational. Therefore, since παρὰ ῥητὴν τὴν ΕΖ παράκειται· ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν AB and AC are each medial, and AB is equal to FH , ΘΕ, ΕΗ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. καὶ ἐπεὶ ῥητόν ἐστι and AC to FG, FH and FG are thus each also medial. 307 STOIQEIWN iþ. ELEMENTS BOOK 10 τὸ ΔΒ καί ἐστιν ἴσον τῷ ΚΘ, ῥητὸν ἄρα ἐστὶ καὶ τὸ ΚΘ. And they are applied to the rational (straight-line) EF . καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται· ῥητὴ ἄρα ἐστὶν ἡ ΗΘ Thus, HE and EG are each rational, and incommensu- καὶ σύμμετρος τῇ ΕΖ μήκει. ἀλλά καὶ ἡ ΕΗ ῥητή ἐστι καὶ rable in length with EF [Prop. 10.22]. And since DB ἀσύμμετρος τῇ ΕΖ μήκει· ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΗ τῇ is rational, and is equal to KH , KH is thus also ratio- ΗΘ μήκει. καί ἐστιν ὡς ἡ ΕΗ πρὸς τὴν ΗΘ, οὕτως τὸ ἀπὸ nal. And (KH) is applied to the rational (straight-line) τῆς ΕΗ πρὸς τὸ ὑπὸ τῶν ΕΗ, ΗΘ· ἀσύμμετρον ἄρα ἐστὶ τὸ EF . GH is thus rational, and commensurable in length ἀπὸ τῆς ΕΗ τῷ ὑπὸ τῶν ΕΗ, ΗΘ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΕΗ with EF [Prop. 10.20]. But, EG is also rational, and in- σύμμετρά ἐστι τὰ ἀπὸ τῶν ΕΗ, ΗΘ τετράγωνα· ῥητὰ γὰρ commensurable in length with EF . Thus, EG is incom- ἀμφότερα· τῷ δὲ ὑπὸ τῶν ΕΗ, ΗΘ σύμμετρόν ἐστι τὸ δὶς mensurable in length with GH [Prop. 10.13]. And as ὑπὸ τῶν ΕΗ, ΗΘ· διπλάσιον γάρ ἐστιν αὐτοῦ· ἀσύμμετρα EG is to GH , so the (square) on EG (is) to the (rectan- ἄρα ἐστὶ τὰ ἀπὸ τῶν ΕΗ, ΗΘ τῷ δὶς ὑπὸ τῶν ΕΗ, ΗΘ· καὶ gle contained) by EG and GH [Prop. 10.13 lem.]. Thus, συναμφότερα ἄρα τά τε ἀπὸ τῶν ΕΗ, ΗΘ καὶ τὸ δὶς ὑπὸ the (square) on EG is incommensurable with the (rect- τῶν ΕΗ, ΗΘ, ὅπερ ἐστὶ τὸ ἀπὸ τῆς ΕΘ, ἀσύμμετρόν ἐστι angle contained) by EG and GH [Prop. 10.11]. But, the τοῖς ἀπὸ τῶν ΕΗ, ΗΘ. ῥητὰ δὲ τὰ ἀπὸ τῶν ΕΗ, ΗΘ· ἄλογον (sum of the) squares on EG and GH is commensurable ἄρα τὸ ἀπὸ τῆς ΕΘ. ἄλογος ἄρα ἐστὶν ἡ ΕΘ. ἀλλὰ καὶ ῥηρή· with the (square) on EG. For (EG and GH are) both ὅπερ ἐστὶν ἀδύνατον. rational. And twice the (rectangle contained) by EG and Μέσον ἄρα μέσου οὐχ ὑπερέχει ῥητῷ· ὅπερ ἔδει δεῖξαι. GH is commensurable with the (rectangle contained) by EG and GH [Prop. 10.6]. For (the former) is double the latter. Thus, the (sum of the squares) on EG and GH is incommensurable with twice the (rectangle con- tained) by EG and GH [Prop. 10.13]. And thus the sum of the (squares) on EG and GH plus twice the (rectan- gle contained) by EG and GH , that is the (square) on EH [Prop. 2.4], is incommensurable with the (sum of the squares) on EG and GH [Prop. 10.16]. And the (sum of the squares) on EG and GH (is) rational. Thus, the (square) on EH is irrational [Def. 10.4]. Thus, EH is irrational [Def. 10.4]. But, (it is) also rational. The very thing is impossible. Thus, a medial (area) does not exceed a medial (area) by a rational (area). (Which is) the very thing it was required to show. † In other words, √ k − √ k′ 6= k′′.kzþ. Proposition 27 Μέσας εὑρεῖν δυνάμει μόνον συμμέτρους ῥητὸν πε- To find (two) medial (straight-lines), containing a ra- ριεχούσας. tional (area), (which are) commensurable in square only. ∆Α Β Γ CA B D ᾿Εκκείσθωσαν δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ Α, Let the two rational (straight-lines) A and B, (which Β, καὶ εἰλήφθω τῶν Α, Β μέση ἀνάλογον ἡ Γ, καὶ γεγονέτω are) commensurable in square only, be laid down. And let ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ. C—the mean proportional (straight-line) to A and B— 308 STOIQEIWN iþ. ELEMENTS BOOK 10 Καὶ ἐπεὶ αἱ Α, Β ῥηταί εἰσι δυνάμει μόνον σύμμετροι, have been taken [Prop. 6.13]. And let it be contrived τὸ ἄρα ὑπὸ τῶν Α, Β, τουτέστι τὸ ἀπὸ τῆς Γ, μέσον ἐστίν. that as A (is) to B, so C (is) to D [Prop. 6.12]. μέση ἄρα ἡ Γ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, [οὕτως] ἡ Γ And since the rational (straight-lines) A and B πρὸς τὴν Δ, αἱ δὲ Α, Β δυνάμει μόνον [εἰσὶ] σύμμετροι, καὶ are commensurable in square only, the (rectangle con- αἱ Γ, Δ ἄρα δυνάμει μόνον εἰσὶ σύμμετροι. καί ἐστι μέση ἡ tained) by A and B—that is to say, the (square) on C Γ· μέση ἄρα καὶ ἡ Δ. αἱ Γ, Δ ἄρα μέσαι εἰσὶ δυνάμει μόνον [Prop. 6.17]—is thus medial [Prop 10.21]. Thus, C is σύμμετροι. λέγω, ὅτι καὶ ῥητὸν περιέχουσιν. ἐπεὶ γάρ ἐστιν medial [Prop. 10.21]. And since as A is to B, [so] C (is) ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ, ἐναλλὰξ ἄρα to D, and A and B [are] commensurable in square only, ἐστὶν ὡς ἡ Α πρὸς τὴν Γ, ἡ Β πρὸς τὴν Δ. ἀλλ᾿ ὡς ἡ Α C and D are thus also commensurable in square only πρὸς τὴν Γ, ἡ Γ πρὸς τὴν Β· καὶ ὡς ἄρα ἡ Γ πρὸς τὴν Β, [Prop. 10.11]. And C is medial. Thus, D is also medial οὕτως ἡ Β πρὸς τὴν Δ· τὸ ἄρα ὑπὸ τῶν Γ, Δ ἴσον ἐστὶ τῷ [Prop. 10.23]. Thus, C and D are medial (straight-lines ἀπὸ τῆς Β. ῥητὸν δὲ τὸ ἀπὸ τῆς Β· ῥητὸν ἄρα [ἐστὶ] καὶ τὸ which are) commensurable in square only. I say that they ὑπὸ τῶν Γ, Δ. also contain a rational (area). For since as A is to B, so Εὕρηνται ἄρα μέσαι δυνάμει μόνον σύμμετροι ῥητὸν C (is) to D, thus, alternately, as A is to C, so B (is) to περιέχουσαι· ὅπερ ἔδει δεῖξαι. D [Prop. 5.16]. But, as A (is) to C, (so) C (is) to B. And thus as C (is) to B, so B (is) to D [Prop. 5.11]. Thus, the (rectangle contained) by C and D is equal to the (square) on B [Prop. 6.17]. And the (square) on B (is) rational. Thus, the (rectangle contained) by C and D [is] also rational. Thus, (two) medial (straight-lines, C and D), con- taining a rational (area), (which are) commensurable in square only, have been found.† (Which is) the very thing it was required to show. † C and D have lengths k1/4 and k3/4 times that of A, respectively, where the length of B is k1/2 times that of A.khþ. Proposition 28 Μέσας εὑρεῖν δυνάμει μόνον συμμέτρους μέσον πει- To find (two) medial (straight-lines), containing a me- ριεχούσας. dial (area), (which are) commensurable in square only. Α Β Γ ∆ Ε C B D E A ᾿Εκκείσθωσαν [τρεῖς] ῥηταὶ δυνάμει μόνον σύμμετροι αἱ Let the [three] rational (straight-lines) A, B, and C, Α, Β, Γ, καὶ εἰλήφθω τῶν Α, Β μέση ἀνάλογον ἡ Δ, καὶ (which are) commensurable in square only, be laid down. γεγονέτω ὡς ἡ Β πρὸς τὴν Γ, ἡ Δ πρὸς τὴν Ε. And let, D, the mean proportional (straight-line) to A ᾿Επεὶ αἱ Α, Β ῥηταί εἰσι δυνάμει μόνον σύμμετροι, τὸ ἄρα and B, have been taken [Prop. 6.13]. And let it be con- ὑπὸ τῶν Α, Β, τουτέστι τὸ ἀπὸ τῆς Δ, μέσον ἐστὶν. μέση trived that as B (is) to C, (so) D (is) to E [Prop. 6.12]. ἄρα ἡ Δ. καὶ ἐπεὶ αἱ Β, Γ δυνάμει μόνον εἰσὶ σύμμετροι, καί Since the rational (straight-lines) A and B are com- ἐστιν ὡς ἡ Β πρὸς τὴν Γ, ἡ Δ πρὸς τὴν Ε, καὶ αἱ Δ, Ε ἄρα mensurable in square only, the (rectangle contained) by δυνάμει μόνον εἰσὶ σύμμετροι. μέση δὲ ἡ Δ· μέση ἄρα καὶ ἡ A and B—that is to say, the (square) on D [Prop. 6.17]— Ε· αἱ Δ, Ε ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι. λέγω is medial [Prop. 10.21]. Thus, D (is) medial [Prop. 10.21]. δή, ὅτι καὶ μέσον περιέχουσιν. ἐπεὶ γάρ ἐστιν ὡς ἡ Β πρὸς And since B and C are commensurable in square only, τὴν Γ, ἡ Δ πρὸς τὴν Ε, ἐναλλὰξ ἄρα ὡς ἡ Β πρὸς τὴν Δ, ἡ and as B is to C, (so) D (is) to E, D and E are thus com- Γ πρὸς τὴν Ε. ὡς δὲ ἡ Β πρὸς τὴν Δ, ἡ Δ πρὸς τὴν Α· καὶ mensurable in square only [Prop. 10.11]. And D (is) me- ὡς ἅρα ἡ Δ πρὸς τὴν Α, ἡ Γ πρὸς τὴν Ε· τὸ ἄρα ὑπὸ τῶν dial. E (is) thus also medial [Prop. 10.23]. Thus, D and Α, Γ ἴσον ἐστὶ τῷ ὑπὸ τῶν Δ, Ε. μέσον δὲ τὸ ὑπὸ τῶν Α, E are medial (straight-lines which are) commensurable Γ· μέσον ἄρα καὶ τὸ ὑπὸ τῶν Δ, Ε. in square only. So, I say that they also enclose a medial ᾿Εὕρηνται ἄρα μέσαι δυνάμει μόνον σύμμετροι μέσον (area). For since as B is to C, (so) D (is) to E, thus, 309 STOIQEIWN iþ. ELEMENTS BOOK 10 περιέχουσαι· ὅπερ ἔδει δεῖξαι. alternately, as B (is) to D, (so) C (is) to E [Prop. 5.16]. And as B (is) to D, (so) D (is) to A. And thus as D (is) to A, (so) C (is) to E. Thus, the (rectangle contained) by A and C is equal to the (rectangle contained) by D and E [Prop. 6.16]. And the (rectangle contained) by A and C is medial [Prop. 10.21]. Thus, the (rectangle contained) by D and E (is) also medial. Thus, (two) medial (straight-lines, D and E), con- taining a medial (area), (which are) commensurable in square only, have been found. (Which is) the very thing it was required to show. † D and E have lengths k1/4 and k′1/2/k1/4 times that of A, respectively, where the lengths of B and C are k1/2 and k′1/2 times that of A, respectively. L¨mma aþ. Lemma I Εὑρειν δύο τετραγώνους ἀριθμούς, ὥστε καὶ τὸν To find two square numbers such that the sum of them συγκείμενον ἐξ αὐτῶν εἶναι τετράγωνον. is also square. ∆ ΓΑ Β DA BC ᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΒ, ΒΓ, ἔστωσαν δὲ ἤτοι Let the two numbers AB and BC be laid down. And ἄρτιοι ἢ περιττοί. καὶ ἐπεὶ, ἐάν τε ἀπὸ ἀρτίου ἄρτιος ἀφαι- let them be either (both) even or (both) odd. And since, if ρεθῇ, ἐάν τε ἀπὸ περισσοῦ περισσός, ὁ λοιπὸς ἄρτιός ἐστιν, an even (number) is subtracted from an even (number), ὁ λοιπὸς ἄρα ὁ ΑΓ ἄρτιός ἐστιν. τετμήσθω ὁ ΑΓ δίχα κατὰ or if an odd (number is subtracted) from an odd (num- τὸ Δ. ἔστωσαν δὲ καὶ οἱ ΑΒ, ΒΓ ἤτοι ὅμοιοι ἐπίπεδοι ἢ ber), then the remainder is even [Props. 9.24, 9.26], the τετράγωνοι, οἳ καὶ αὐτοὶ ὅμοιοί εἰσιν ἐπίπεδοι· ὁ ἄρα ἐκ remainder AC is thus even. Let AC have been cut in τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ [τοῦ] ΓΔ τετραγώνου ἴσος ἐστὶ half at D. And let AB and BC also be either similar τῷ ἀπὸ τοῦ ΒΔ τετραγώνῳ. καί ἐστι τετράγωνος ὁ ἐκ τῶν plane (numbers), or square (numbers)—which are them- ΑΒ, ΒΓ, ἐπειδήπερ ἐδείχθη, ὅτι, ἐὰν δύο ὅμοιοι ἐπίπεδοι selves also similar plane (numbers). Thus, the (num- πολλαπλασιάσαντες ἀλλήλους ποιῶσι τινα, ὁ γενόμενος ber created) from (multiplying) AB and BC, plus the τετράγωνός ἐστιν. εὕρηνται ἄρα δύο τετράγωνοι ἀριθμοὶ square on CD, is equal to the square on BD [Prop. 2.6]. ὅ τε ἐκ τῶν ΑΒ, ΒΓ καὶ ὁ ἀπὸ τοῦ ΓΔ, οἳ συντεθέντες And the (number created) from (multiplying) AB and ποιοῦσι τὸν ἀπὸ τοῦ ΒΔ τετράγωνον. BC is square—inasmuch as it was shown that if two Καὶ φανερόν, ὅτι εὕρηνται πάλιν δύο τετράγωνοι ὅ τε similar plane (numbers) make some (number) by mul- ἀπὸ τοῦ ΒΔ καὶ ὁ ἀπὸ τοῦ ΓΔ, ὥστε τὴν ὑπεροχὴν αὐτῶν tiplying one another then the (number so) created is τὸν ὑπὸ ΑΒ, ΒΓ εἶναι τετράγωνον, ὅταν οἱ ΑΒ, ΒΓ ὅμοιοι square [Prop. 9.1]. Thus, two square numbers have ὦσιν ἐπίπεδοι. ὅταν δὲ μὴ ὦσιν ὅμοιοι ἐπίπεδοι, εὕρηνται been found—(namely,) the (number created) from (mul- δύο τετράγωνοι ὅ τε ἀπὸ τοῦ ΒΔ καὶ ὁ ἀπὸ τοῦ ΔΓ, ὧν tiplying) AB and BC, and the (square) on CD—which, ἡ ὑπεροχὴ ὁ ὑπὸ τῶν ΑΒ, ΒΓ οὐκ ἔστι τετράγωνος· ὅπερ (when) added (together), make the square on BD. ἔδει δεῖξαι. And (it is) clear that two square (numbers) have again been found—(namely,) the (square) on BD, and the (square) on CD—such that their difference—(namely,) the (rectangle) contained by AB and BC—is square whenever AB and BC are similar plane (numbers). But, when they are not similar plane numbers, two square (numbers) have been found—(namely,) the (square) on BD, and the (square) on DC—between which the difference—(namely,) the (rectangle) contained by AB and BC—is not square. (Which is) the very thing it was required to show. 310 STOIQEIWN iþ. ELEMENTS BOOK 10L¨mma bþ. Lemma II Εὑρεῖν δύο τετραγώνους ἀριθμούς, ὥστε τὸν ἐξ αὐτῶν To find two square numbers such that the sum of them συγκείμενον μὴ εἶναι τετράγωνον. is not square. ΗΑ Β∆ Γ Ε Ζ Θ H A BD CF E G ῎Εστω γὰρ ὁ ἐκ τῶν ΑΒ, ΒΓ, ὡς ἔφαμεν, τετράγωνος, For let the (number created) from (multiplying) AB καὶ ἄρτιος ὁ ΓΑ, καὶ τετμήσθω ὁ ΓΑ δίχα τῷ Δ. φανερὸν and BC, as we said, be square. And (let) CA (be) even. δή, ὅτι ὁ ἐκ τῶν ΑΒ, ΒΓ τετράγωνος μετὰ τοῦ ἀπὸ [τοῦ] And let CA have been cut in half at D. So it is clear ΓΔ τετραγώνου ἴσος ἐστὶ τῷ ἀπὸ [τοῦ] ΒΔ τετραγώνῳ. that the square (number created) from (multiplying) AB ἀφῃρήσθω μονὰς ἡ ΔΕ· ὁ ἄρα ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ and BC, plus the square on CD, is equal to the square ἀπὸ [τοῦ] ΓΕ ἐλάσσων ἐστὶ τοῦ ἀπὸ [τοῦ] ΒΔ τετραγώνου. on BD [see previous lemma]. Let the unit DE have λέγω οὖν, ὅτι ὁ ἐκ τῶν ΑΒ, ΒΓ τετράγωνος μετὰ τοῦ ἀπὸ been subtracted (from BD). Thus, the (number created) [τοῦ] ΓΕ οὐκ ἔσται τετράγωνος. from (multiplying) AB and BC, plus the (square) on Εἰ γὰρ ἔσται τετράγωνος, ἤτοι ἴσος ἐστὶ τῷ ἀπὸ [τοῦ] CE, is less than the square on BD. I say, therefore, that ΒΕ ἢ ἐλάσσων τοῦ ἀπὸ [τοῦ] ΒΕ, οὐκέτι δὲ καὶ μείζων, ἵνα the square (number created) from (multiplying) AB and μὴ τμηθῇ ἡ μονάς. ἔστω, εἰ δυνατόν, πρότερον ὁ ἐκ τῶν BC, plus the (square) on CE, is not square. ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ ἴσος τῷ ἀπὸ ΒΕ, καὶ ἔστω τῆς For if it is square, it is either equal to the (square) ΔΕ μονάδος διπλασίων ὁ ΗΑ. ἐπεὶ οὖν ὅλος ὁ ΑΓ ὅλου on BE, or less than the (square) on BE, but cannot any τοῦ ΓΔ ἐστι διπλασίων, ὧν ὁ ΑΗ τοῦ ΔΕ ἐστι διπλασίων, more be greater (than the square on BE), lest the unit be καὶ λοιπὸς ἄρα ὁ ΗΓ λοιποῦ τοῦ ΕΓ ἐστι διπλασίων· δίχα divided. First of all, if possible, let the (number created) ἄρα τέτμηται ὁ ΗΓ τῷ Ε. ὁ ἄρα ἐκ τῶν ΗΒ, ΒΓ μετὰ τοῦ from (multiplying) AB and BC, plus the (square) on CE, ἀπὸ ΓΕ ἴσος ἐστὶ τῷ ἀπὸ ΒΕ τετραγώνῳ. ἀλλὰ καὶ ὁ ἐκ be equal to the (square) on BE. And let GA be double τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ ἴσος ὑπόκειται τῷ ἀπὸ [τοῦ] the unit DE. Therefore, since the whole of AC is double ΒΕ τετραγώνῳ· ὁ ἄρα ἐκ τῶν ΗΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ the whole of CD, of which AG is double DE, the remain- ἴσος ἐστὶ τῷ ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΕ. καὶ κοινοῦ der GC is thus double the remainder EC. Thus, GC has ἀφαιρεθέντος τοῦ ἀπὸ ΓΕ συνάγεται ὁ ΑΒ ἴσος τῷ ΗΒ· been cut in half at E. Thus, the (number created) from ὅπερ ἄτοπον. οὐκ ἄρα ὁ ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ [τοῦ] (multiplying) GB and BC, plus the (square) on CE, is ΓΕ ἴσος ἐστὶ τῷ ἀπὸ ΒΕ. λέγω δή, ὅτι οὐδὲ ἐλάσσων τοῦ equal to the square on BE [Prop. 2.6]. But, the (num- ἀπὸ ΒΕ. εἰ γὰρ δυνατόν, ἔστω τῷ ἀπὸ ΒΖ ἴσος, καὶ τοῦ ber created) from (multiplying) AB and BC, plus the ΔΖ διπλασίων ὁ ΘΑ. καὶ συναχθήσεται πάλιν διπλασίων ὁ (square) on CE, was also assumed (to be) equal to the ΘΓ τοῦ ΓΖ· ὥστε καὶ τὸν ΓΘ δίχα τετμῆσθαι κατὰ τὸ Ζ, square on BE. Thus, the (number created) from (multi- καὶ διὰ τοῦτο τὸν ἐκ τῶν ΘΒ, ΒΓ μετὰ τοῦ ἀπὸ ΖΓ ἴσον plying) GB and BC, plus the (square) on CE, is equal γίνεσθαι τῷ ἀπὸ ΒΖ. ὑπόκειται δὲ καὶ ὁ ἐκ τῶν ΑΒ, ΒΓ to the (number created) from (multiplying) AB and BC, μετὰ τοῦ ἀπὸ ΓΕ ἴσος τῷ ἀπὸ ΒΖ. ὥστε καὶ ὁ ἐκ τῶν ΘΒ, plus the (square) on CE. And subtracting the (square) on ΒΓ μετὰ τοῦ ἀπὸ ΓΖ ἴσος ἔσται τῷ ἐκ τῶν ΑΒ, ΒΓ μετὰ CE from both, AB is inferred (to be) equal to GB. The τοῦ ἀπὸ ΓΕ· ὅπερ ἄτοπον. οὐκ ἄρα ὁ ἐκ τῶν ΑΒ, ΒΓ μετὰ very thing is absurd. Thus, the (number created) from τοῦ ἀπὸ ΓΕ ἴσος ἐστὶ [τῷ] ἐλάσσονι τοῦ ἀπὸ ΒΕ. ἐδείχθη (multiplying) AB and BC, plus the (square) on CE, is δέ, ὅτι οὐδὲ [αὐτῷ] τῷ ἀπὸ ΒΕ. οὐκ ἄρα ὁ ἐκ τῶν ΑΒ, ΒΓ not equal to the (square) on BE. So I say that (it is) not μετὰ τοῦ ἀπὸ ΓΕ τετράγωνός ἐστιν. ὅπερ ἔδει δεῖξαι. less than the (square) on BE either. For, if possible, let it be equal to the (square) on BF . And (let) HA (be) dou- ble DF . And it can again be inferred that HC (is) double CF . Hence, CH has also been cut in half at F . And, on account of this, the (number created) from (multiplying) HB and BC, plus the (square) on FC, becomes equal to the (square) on BF [Prop. 2.6]. And the (number cre- ated) from (multiplying) AB and BC, plus the (square) on CE, was also assumed (to be) equal to the (square) on BF . Hence, the (number created) from (multiplying) HB and BC, plus the (square) on CF , will also be equal to the (number created) from (multiplying) AB and BC, 311 STOIQEIWN iþ. ELEMENTS BOOK 10 plus the (square) on CE. The very thing is absurd. Thus, the (number created) from (multiplying) AB and BC, plus the (square) on CE, is not equal to less than the (square) on BE. And it was shown that (is it) not equal to the (square) on BE either. Thus, the (number created) from (multiplying) AB and BC, plus the square on CE, is not square. (Which is) the very thing it was required to show.kjþ. Proposition 29 Εὑρεῖν δύο ῥητὰς δυνάμει μόνον συμμέτρους, ὥστε τὴν To find two rational (straight-lines which are) com- μείζονα τῆς ἐλάσσονος μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου mensurable in square only, such that the square on the ἑαυτῇ μήκει. greater is larger than the (square on the) lesser by the (square) on (some straight-line which is) commensurable in length with the greater. Ζ Γ Ε ∆ Α Β F E D A B C ᾿Εκκείσθω γάρ τις ῥητὴ ἡ ΑΒ καὶ δύο τετράγωνοι For let some rational (straight-line) AB be laid down, ἀριθμοὶ οἱ ΓΔ, ΔΕ, ὥστε τὴν ὑπεροχὴν αὐτῶν τὸν ΓΕ μὴ and two square numbers, CD and DE, such that the dif- εἶναι τετράγωνον, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ference between them, CE, is not square [Prop. 10.28 ΑΖΒ, καὶ πεποιήσθω ὡς ὁ ΔΓ πρὸς τὸν ΓΕ, οὕτως τὸ ἀπὸ lem. I]. And let the semi-circle AFB have been drawn on τῆς ΒΑ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΑΖ τετράγωνον, καὶ AB. And let it be contrived that as DC (is) to CE, so the ἐπεζεύχθω ἡ ΖΒ. square on BA (is) to the square on AF [Prop. 10.6 corr.]. ᾿Επεὶ [οὖν] ἐστιν ὡς τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΖ, And let FB have been joined. οὕτως ὁ ΔΓ πρὸς τὸν ΓΕ, τὸ ἀπὸ τῆς ΒΑ ἄρα πρὸς τὸ ἀπὸ [Therefore,] since as the (square) on BA is to the τῆς ΑΖ λόγον ἔχει, ὅν ἀριθμὸς ὁ ΔΓ πρὸς ἀριθμὸν τὸν ΓΕ· (square) on AF , so DC (is) to CE, the (square) on σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΑ τῷ ἀπὸ τῆς ΑΖ. ῥητὸν BA thus has to the (square) on AF the ratio which δὲ τὸ ἀπὸ τῆς ΑΒ· ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΑΖ· ῥητὴ ἄρα the number DC (has) to the number CE. Thus, the καὶ ἡ ΑΖ. καὶ ἐπεὶ ὁ ΔΓ πρὸς τὸν ΓΕ λόγον οὐκ ἔχει, ὃν (square) on BA is commensurable with the (square) on τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ AF [Prop. 10.6]. And the (square) on AB (is) rational τῆς ΒΑ ἄρα πρὸς τὸ ἀπὸ τῆς ΑΖ λόγον ἔχει, ὃν τετράγωνος [Def. 10.4]. Thus, the (square) on AF (is) also ratio- ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ nal. Thus, AF (is) also rational. And since DC does ΑΒ τῇ ΑΖ μήκει· αἱ ΒΑ, ΑΖ ἄρα ῥηταί εἰσι δυνάμει μόνον not have to CE the ratio which (some) square num- σύμμετροι. καὶ ἐπεί [ἐστιν] ὡς ὁ ΔΓ πρὸς τὸν ΓΕ, οὕτως ber (has) to (some) square number, the (square) on τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΖ, ἀναστρέψαντι ἄρα ὡς BA thus does not have to the (square) on AF the ra- ὁ ΓΔ πρὸς τὸν ΔΕ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ tio which (some) square number has to (some) square τῆς ΒΖ. ὁ δὲ ΓΔ πρὸς τὸν ΔΕ λόγον ἔχει, ὃν τετράγωνος number either. Thus, AB is incommensurable in length ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· καὶ τὸ ἀπὸ τῆς ΑΒ ἄρα with AF [Prop. 10.9]. Thus, the rational (straight-lines) πρὸς τὸ ἀπὸ τῆς ΒΖ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς BA and AF are commensurable in square only. And πρὸς τετράγωνον ἀριθμόν· σύμμετρος ἄρα ἐστὶν ἡ ΑΒ τῇ since as DC [is] to CE, so the (square) on BA (is) to ΒΖ μήκει. καί ἐστι τὸ ἀπὸ τῆς ΑΒ ἴσον τοῖς ἀπὸ τῶν ΑΖ, the (square) on AF , thus, via conversion, as CD (is) ΖΒ· ἡ ΑΒ ἄρα τῆς ΑΖ μεῖζον δύναται τῇ ΒΖ συμμέτρῳ to DE, so the (square) on AB (is) to the (square) on 312 STOIQEIWN iþ. ELEMENTS BOOK 10 ἑαυτῇ. BF [Props. 5.19 corr., 3.31, 1.47]. And CD has to DE Εὕρηνται ἄρα δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ ΒΑ, the ratio which (some) square number (has) to (some) ΑΖ, ὥστε τὴν μεῖζονα τὴν ΑΒ τῆς ἐλάσσονος τῆς ΑΖ μεῖζον square number. Thus, the (square) on AB also has to the δύνασθαι τῷ ἀπὸ τῆς ΒΖ συμμέτρου ἑαυτῇ μήκει· ὅπερ ἔδει (square) on BF the ratio which (some) square number δεῖξαι. has to (some) square number. AB is thus commensu- rable in length with BF [Prop. 10.9]. And the (square) on AB is equal to the (sum of the squares) on AF and FB [Prop. 1.47]. Thus, the square on AB is greater than (the square on) AF by (the square on) BF , (which is) commensurable (in length) with (AB). Thus, two rational (straight-lines), BA and AF , com- mensurable in square only, have been found such that the square on the greater, AB, is larger than (the square on) the lesser, AF , by the (square) on BF , (which is) com- mensurable in length with (AB).† (Which is) the very thing it was required to show. † BA and AF have lengths 1 and √ 1 − k2 times that of AB, respectively, where k = p DE/CD.lþ. Proposition 30 Εὑρεῖν δύο ῥητὰς δυνάμει μόνον συμμέτρους, ὥστε τὴν To find two rational (straight-lines which are) com- μείζονα τῆς ἐλάσσονος μεῖζον δύνασθαι τῷ ἀπὸ ἀσυμμέτρου mensurable in square only, such that the square on the ἑαυτῇ μήκει. greater is larger than the (the square on) lesser by the (square) on (some straight-line which is) incommensu- rable in length with the greater. Ζ Ε ∆ Α Β Γ F E D A B C ᾿Εκκείσθω ῥητὴ ἡ ΑΒ καὶ δύο τετράγωνοι ἀριθμοὶ οἱ Let the rational (straight-line) AB be laid out, and the ΓΕ, ΕΔ, ὥστε τὸν συγκείμενον ἐξ αὐτῶν τὸν ΓΔ μὴ εἶναι two square numbers, CE and ED, such that the sum of τετράγωνον, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΖΒ, them, CD, is not square [Prop. 10.28 lem. II]. And let the καὶ πεποιήσθω ὡς ὁ ΔΓ πρὸς τὸν ΓΕ, οὕτως τὸ ἀπὸ τῆς semi-circle AFB have been drawn on AB. And let it be ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΖ, καὶ ἐπεζεύχθω ἡ ΖΒ. contrived that as DC (is) to CE, so the (square) on BA ῾Ομοίως δὴ δείξομεν τῷ πρὸ τούτου, ὅτι αἱ ΒΑ, ΑΖ (is) to the (square) on AF [Prop. 10.6 corr]. And let FB ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ ἐπεί ἐστιν ὡς ὁ have been joined. ΔΓ πρὸς τὸν ΓΕ, οὕτως τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς So, similarly to the (proposition) before this, we can ΑΖ, ἀναστρέψαντι ἄρα ὡς ὁ ΓΔ πρὸς τὸν ΔΕ, οὕτως τὸ show that BA and AF are rational (straight-lines which ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΖ. ὁ δὲ ΓΔ πρὸς τὸν ΔΕ are) commensurable in square only. And since as DC is λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον to CE, so the (square) on BA (is) to the (square) on AF , ἀριθμόν· οὐδ᾿ ἄρα τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΖ λόγον thus, via conversion, as CD (is) to DE, so the (square) on ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· AB (is) to the (square) on BF [Props. 5.19 corr., 3.31, ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΒ τῇ ΒΖ μήκει. καὶ δύναται ἡ 1.47]. And CD does not have to DE the ratio which ΑΒ τῆς ΑΖ μεῖζον τῷ ἀπὸ τῆς ΖΒ ἀσυμμέτρου ἑαυτῇ. (some) square number (has) to (some) square number. 313 STOIQEIWN iþ. ELEMENTS BOOK 10 Αἱ ΑΒ, ΑΖ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ Thus, the (square) on AB does not have to the (square) ἡ ΑΒ τῆς ΑΖ μεῖζον δύναται τῷ ἀπὸ τῆς ΖΒ ἀσυμμέτρου on BF the ratio which (some) square number has to ἑαυτῇ μήκει· ὅπερ ἔδει δεῖξαι. (some) square number either. Thus, AB is incommensu- rable in length with BF [Prop. 10.9]. And the square on AB is greater than the (square on) AF by the (square) on FB [Prop. 1.47], (which is) incommensurable (in length) with (AB). Thus, AB and AF are rational (straight-lines which are) commensurable in square only, and the square on AB is greater than (the square on) AF by the (square) on FB, (which is) incommensurable (in length) with (AB).† (Which is) the very thing it was required to show. † AB and AF have lengths 1 and 1/ √ 1 + k2 times that of AB, respectively, where k = p DE/CE.laþ. Proposition 31 Εὑρεῖν δύο μέσας δυνάμει μόνον συμμέτρους ῥητὸν To find two medial (straight-lines), commensurable in περιεχούσας, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον square only, (and) containing a rational (area), such that δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. the square on the greater is larger than the (square on the) lesser by the (square) on (some straight-line) com- mensurable in length with the greater. Α ∆ΓΒ C DBA ᾿Εκκείσθωσαν δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ Α, Let two rational (straight-lines), A and B, commensu- Β, ὥστε τὴν Α μείζονα οὖσαν τῆς ἐλάσσονος τῆς Β μεῖζον rable in square only, be laid out, such that the square on δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. καὶ τῷ ὑπὸ τῶν the greater A is larger than the (square on the) lesser B Α, Β ἴσον ἔστω τὸ ἀπὸ τῆς Γ. μέσον δὲ τὸ ὑπὸ τῶν Α, Β· by the (square) on (some straight-line) commensurable μέσον ἄρα καὶ τὸ ἀπὸ τῆς Γ· μέση ἄρα καὶ ἡ Γ. τῷ δὲ ἀπὸ in length with (A) [Prop. 10.29]. And let the (square) τῆς Β ἴσον ἔστω τὸ ὑπὸ τῶν Γ, Δ· ῥητὸν δὲ τὸ ἀπὸ τῆς Β· on C be equal to the (rectangle contained) by A and B. ῥητὸν ἄρα καὶ τὸ ὑπὸ τῶν Γ, Δ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς And the (rectangle contained by) A and B (is) medial τὴν Β, οὕτως τὸ ὑπὸ τῶν Α, Β πρὸς τὸ ἀπὸ τῆς Β, ἀλλὰ τῷ [Prop. 10.21]. Thus, the (square) on C (is) also medial. μὲν ὑπὸ τῶν Α, Β ἴσον ἐστὶ τὸ ἀπὸ τῆς Γ, τῷ δὲ ἀπὸ τῆς Thus, C (is) also medial [Prop. 10.21]. And let the (rect- Β ἴσον τὸ ὑπὸ τῶν Γ, Δ, ὡς ἄρα ἡ Α πρὸς τὴν Β, οὕτως angle contained) by C and D be equal to the (square) τὸ ἀπὸ τῆς Γ πρὸς τὸ ὑπὸ τῶν Γ, Δ. ὡς δὲ τὸ ἀπὸ τῆς Γ on B. And the (square) on B (is) rational. Thus, the πρὸς τὸ ὑπὸ τῶν Γ, Δ, οὕτως ἡ Γ πρὸς τὴν Δ· καὶ ὡς ἄρα (rectangle contained) by C and D (is) also rational. And ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ. σύμμετρος δὲ ἡ Α since as A is to B, so the (rectangle contained) by A and τῇ Β δυνάμει μόνον· σύμμετρος ἄρα καὶ ἡ Γ τῇ Δ δυνάμει B (is) to the (square) on B [Prop. 10.21 lem.], but the μόνον. καί ἐστι μέση ἡ Γ· μέση ἄρα καὶ ἡ Δ. καὶ ἐπεί ἐστιν (square) on C is equal to the (rectangle contained) by ὡς ἡ Α πρὸς τὴν Β, ἡ Γ πρὸς τὴν Δ, ἡ δὲ Α τῆς Β μεῖζον A and B, and the (rectangle contained) by C and D to δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ Γ ἄρα τῆς Δ μεῖζον the (square) on B, thus as A (is) to B, so the (square) δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. on C (is) to the (rectangle contained) by C and D. And Εὕρηνται ἄρα δύο μέσαι δυνάμει μόνον σύμμετροι αἱ Γ, as the (square) on C (is) to the (rectangle contained) by 314 STOIQEIWN iþ. ELEMENTS BOOK 10 Δ ῥητὸν περιέχουσαι, καὶ ἡ Γ τῆς Δ μεῖζον δυνάται τῷ ἀπὸ C and D, so C (is) to D [Prop. 10.21 lem.]. And thus συμμέτρου ἑαυτῇ μήκει. as A (is) to B, so C (is) to D. And A is commensurable ῾Ομοίως δὴ δειχθήσεται καὶ τῷ ἀπὸ ἀσυμμέτρου, ὅταν ἡ in square only with B. Thus, C (is) also commensurable Α τῆς Β μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. in square only with D [Prop. 10.11]. And C is medial. Thus, D (is) also medial [Prop. 10.23]. And since as A is to B, (so) C (is) to D, and the square on A is greater than (the square on) B by the (square) on (some straight-line) commensurable (in length) with (A), the square on C is thus also greater than (the square on) D by the (square) on (some straight-line) commensurable (in length) with (C) [Prop. 10.14]. Thus, two medial (straight-lines), C and D, commen- surable in square only, (and) containing a rational (area), have been found. And the square on C is greater than (the square on) D by the (square) on (some straight-line) commensurable in length with (C).† So, similarly, (the proposition) can also be demon- strated for (some straight-line) incommensurable (in length with C), provided that the square on A is greater than (the square on B) by the (square) on (some straight-line) incommensurable (in length) with (A) [Prop. 10.30].‡ † C and D have lengths (1 − k2)1/4 and (1 − k2)3/4 times that of A, respectively, where k is defined in the footnote to Prop. 10.29. ‡ C and D would have lengths 1/(1 + k2)1/4 and 1/(1 + k2)3/4 times that of A, respectively, where k is defined in the footnote to Prop. 10.30.lbþ. Proposition 32 Εὑρεῖν δύο μέσας δυνάμει μόνον συμμέτρους μέσον To find two medial (straight-lines), commensurable in περιεχούσας, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον square only, (and) containing a medial (area), such that δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ. the square on the greater is larger than the (square on the) lesser by the (square) on (some straight-line) com- mensurable (in length) with the greater. Γ Ε ∆Α Β C E DA B ᾿Εκκείσθωσαν τρεῖς ῥηταὶ δυνάμει μόνον σύμμετροι αἱ Let three rational (straight-lines), A, B and C, com- Α, Β, Γ, ὥστε τὴν Α τῆς Γ μεῖζον δύνασθαι τῷ ἀπὸ mensurable in square only, be laid out such that the συμμέτρου ἑαυτῇ, καὶ τῷ μὲν ὑπὸ τῶν Α, Β ἴσον ἔστω τὸ square on A is greater than (the square on C) by the ἀπὸ τὴς Δ. μέσον ἄρα τὸ ἀπὸ τῆς Δ· καὶ ἡ Δ ἄρα μέση (square) on (some straight-line) commensurable (in ἐστίν. τῷ δὲ ὑπὸ τῶν Β, Γ ἴσον ἔστω τὸ ὑπὸ τῶν Δ, Ε. length) with (A) [Prop. 10.29]. And let the (square) καὶ ἐπεί ἐστιν ὡς τὸ ὑπὸ τῶν Α, Β πρὸς τὸ ὑπὸ τῶν Β, Γ, on D be equal to the (rectangle contained) by A and B. οὕτως ἡ Α πρὸς τὴν Γ, ἀλλὰ τῷ μὲν ὑπὸ τῶν Α, Β ἴσον Thus, the (square) on D (is) medial. Thus, D is also me- ἐστὶ τὸ ἀπὸ τῆς Δ, τῷ δὲ ὑπὸ τῶν Β, Γ ἴσον τὸ ὑπὸ τῶν dial [Prop. 10.21]. And let the (rectangle contained) by Δ, Ε, ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Γ, οὕτως τὸ ἀπὸ τῆς Δ D and E be equal to the (rectangle contained) by B and πρὸς τὸ ὑπὸ τῶν Δ, Ε. ὡς δὲ τὸ ἀπὸ τῆς Δ πρὸς τὸ ὑπὸ C. And since as the (rectangle contained) by A and B τῶν Δ, Ε, οὕτως ἡ Δ πρὸς τὴν Ε· καὶ ὡς ἄρα ἡ Α πρὸς τὴν is to the (rectangle contained) by B and C, so A (is) to Γ, οὕτως ἡ Δ πρὸς τὴν Ε. σύμμετρος δὲ ἡ Α τῇ Γ δυνάμει C [Prop. 10.21 lem.], but the (square) on D is equal to [μόνον]. σύμμετρος ἄρα καὶ ἡ Δ τῇ Ε δυνάμει μόνον. μέση the (rectangle contained) by A and B, and the (rectangle 315 STOIQEIWN iþ. ELEMENTS BOOK 10 δὲ ἡ Δ· μέση ἄρα καὶ ἡ Ε. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν contained) by D and E to the (rectangle contained) by Γ, ἡ Δ πρὸς τὴν Ε, ἡ δὲ Α τῆς Γ μεῖζον δύναται τῷ ἀπὸ B and C, thus as A is to C, so the (square) on D (is) συμμέτρου ἑαυτῇ, καὶ ἡ Δ ἄρα τῆς Ε μεῖζον δυνήσεται τῷ to the (rectangle contained) by D and E. And as the ἀπὸ συμμέτρου ἑαυτῇ. λέγω δή, ὅτι καὶ μέσον ἐστὶ τὸ ὑπὸ (square) on D (is) to the (rectangle contained) by D and τῶν Δ, Ε. ἐπεὶ γὰρ ἴσον ἐστὶ τὸ ὑπὸ τῶν Β, Γ τῷ ὑπὸ τῶν E, so D (is) to E [Prop. 10.21 lem.]. And thus as A Δ, Ε, μέσον δὲ τὸ ὑπὸ τῶν Β, Γ [αἱ γὰρ Β, Γ ῥηταί εἰσι (is) to C, so D (is) to E. And A (is) commensurable in δυνάμει μόνον σύμμετροι], μέσον ἄρα καὶ τὸ ὑπὸ τῶν Δ, Ε. square [only] with C. Thus, D (is) also commensurable Εὕρηνται ἄρα δύο μέσαι δυνάμει μόνον σύμμετροι αἱ in square only with E [Prop. 10.11]. And D (is) me- Δ, Ε μέσον περιέχουσαι, ὥστε τὴν μείζονα τῆς ἐλάσσονος dial. Thus, E (is) also medial [Prop. 10.23]. And since μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ. as A is to C, (so) D (is) to E, and the square on A is ῾Ομοίως δὴ πάλιν διεχθήσεται καὶ τῷ ἀπὸ ἀσυμμέτρου, greater than (the square on) C by the (square) on (some ὅταν ἡ Α τῆς Γ μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῃ. straight-line) commensurable (in length) with (A), the square on D will thus also be greater than (the square on) E by the (square) on (some straight-line) commen- surable (in length) with (D) [Prop. 10.14]. So, I also say that the (rectangle contained) by D and E is medial. For since the (rectangle contained) by B and C is equal to the (rectangle contained) by D and E, and the (rect- angle contained) by B and C (is) medial [for B and C are rational (straight-lines which are) commensurable in square only] [Prop. 10.21], the (rectangle contained) by D and E (is) thus also medial. Thus, two medial (straight-lines), D and E, commen- surable in square only, (and) containing a medial (area), have been found such that the square on the greater is larger than the (square on the) lesser by the (square) on (some straight-line) commensurable (in length) with the greater.†. So, similarly, (the proposition) can again also be demonstrated for (some straight-line) incommensurable (in length with the greater), provided that the square on A is greater than (the square on) C by the (square) on (some straight-line) incommensurable (in length) with (A) [Prop. 10.30].‡ † D and E have lengths k′1/4 and k′1/4 √ 1 − k2 times that of A, respectively, where the length of B is k′1/2 times that of A, and k is defined in the footnote to Prop. 10.29. ‡ D and E would have lengths k′1/4 and k′1/4/ √ 1 + k2 times that of A, respectively, where the length of B is k′1/2 times that of A, and k is defined in the footnote to Prop. 10.30.L¨mma. Lemma ῎Εστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν Let ABC be a right-angled triangle having the (an- Α, καὶ ἤχθω κάθετος ἡ ΑΔ· λέγω, ὅτι τὸ μὲν ὑπὸ τῶν ΓΒΔ gle) A a right-angle. And let the perpendicular AD have ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΑ, τὸ δὲ ὑπὸ τῶν ΒΓΑ ἴσον τῷ ἀπὸ been drawn. I say that the (rectangle contained) by CBD τῆς ΓΑ, καὶ τὸ ὑπὸ τῶν ΒΔ, ΔΓ ἴσον τῷ ἀπὸ τῆς ΑΔ, καὶ is equal to the (square) on BA, and the (rectangle con- ἔτι τὸ ὑπὸ τῶν ΒΓ, ΑΔ ἴσον [ἐστὶ] τῷ ὑπὸ τῶν ΒΑ, ΑΓ. tained) by BCD (is) equal to the (square) on CA, and Καὶ πρῶτον, ὅτι τὸ ὑπὸ τῶν ΓΒΔ ἴσον [ἐστὶ] τῷ ἀπὸ the (rectangle contained) by BD and DC (is) equal to the τῆς ΒΑ. (square) on AD, and, further, the (rectangle contained) by BC and AD [is] equal to the (rectangle contained) by BA and AC. 316 STOIQEIWN iþ. ELEMENTS BOOK 10 And, first of all, (let us prove) that the (rectangle con- tained) by CBD [is] equal to the (square) on BA. ∆ Α ΓΒ C A B D ᾿Επεὶ γὰρ ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας For since AD has been drawn from the right-angle in ἐπὶ τὴν βάσιν κάθετος ἦκται ἡ ΑΔ, τὰ ΑΒΔ, ΑΔΓ ἄρα a right-angled triangle, perpendicular to the base, ABD τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ τῷ ΑΒΓ καὶ ἀλλήλοις. καὶ and ADC are thus triangles (which are) similar to the ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΑΒΔ τριγώνῳ, ἔστιν whole, ABC, and to one another [Prop. 6.8]. And since ἄρα ὡς ἡ ΓΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΒΑ πρὸς τὴν ΒΔ· τὸ triangle ABC is similar to triangle ABD, thus as CB is ἄρα ὑπὸ τῶν ΓΒΔ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ. to BA, so BA (is) to BD [Prop. 6.4]. Thus, the (rectan- Διὰ τὰ αὐτὰ δὴ καὶ τὸ ὑπὸ τῶν ΒΓΔ ἴσον ἐστὶ τῷ ἀπὸ gle contained) by CBD is equal to the (square) on AB τῆς ΑΓ. [Prop. 6.17]. Καὶ ἐπεί, ἐὰν ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς So, for the same (reasons), the (rectangle contained) γωνίας ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, ἡ ἀχθεῖσα τῶν τῆς by BCD is also equal to the (square) on AC. βάσεως τμημάτων μέση ἀνάλογόν ἐστιν, ἔστιν ἄρα ὡς ἡ ΒΑ And since if a (straight-line) is drawn from the right- πρὸς τὴν ΔΑ, οὕτως ἡ ΑΔ πρὸς τὴν ΔΓ· τὸ ἄρα ὑπὸ τῶν angle in a right-angled triangle, perpendicular to the ΒΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΑ. base, the (straight-line so) drawn is the mean propor- Λέγω, ὅτι καὶ τὸ ὑπὸ τῶν ΒΓ, ΑΔ ἴσον ἐστὶ τῷ ὑπὸ tional to the pieces of the base [Prop. 6.8 corr.], thus as τῶν ΒΑ, ΑΓ. ἐπεὶ γὰρ, ὡς ἔφαμεν, ὅμοιόν ἐστι τὸ ΑΒΓ τῷ BD is to DA, so AD (is) to DC. Thus, the (rectangle ΑΒΔ, ἔστιν ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΒΑ πρὸς contained) by BD and DC is equal to the (square) on τὴν ΑΔ. τὸ ἄρα ὑπὸ τῶν ΒΓ, ΑΔ ἴσον ἐστὶ τῷ ὑπὸ τῶν DA [Prop. 6.17]. ΒΑ, ΑΓ· ὅπερ ἔδει δεῖξαι. I also say that the (rectangle contained) by BC and AD is equal to the (rectangle contained) by BA and AC. For since, as we said, ABC is similar to ABD, thus as BC is to CA, so BA (is) to AD [Prop. 6.4]. Thus, the (rectan- gle contained) by BC and AD is equal to the (rectangle contained) by BA and AC [Prop. 6.16]. (Which is) the very thing it was required to show.lgþ. Proposition 33 Εὑρεῖν δύο εὐθείας δυνάμει ἀσυμμέτρους ποιούσας τὸ To find two straight-lines (which are) incommensu- μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων ῥητόν, τὸ rable in square, making the sum of the squares on them δ᾿ ὑπ᾿ αὐτῶν μέσον. rational, and the (rectangle contained) by them medial. ᾿Εκκείσθωσαν δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ Let the two rational (straight-lines) AB and BC, ΑΒ, ΒΓ, ὥστε τὴν μείζονα τὴν ΑΒ τῆς ἐλάσσονος τῆς ΒΓ (which are) commensurable in square only, be laid out μείζον δύνασθαι τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ τετμήσθω ἡ such that the square on the greater, AB, is larger than ΒΓ δίχα κατὰ τὸ Δ, καὶ τῷ ἀφ᾿ ὁποτέρας τῶν ΒΔ, ΔΓ ἴσον (the square on) the lesser, BC, by the (square) on (some παρὰ τὴν ΑΒ παραβεβλήσθω παραλληλόγραμμον ἐλλεῖπον straight-line which is) incommensurable (in length) with εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΕΒ, καὶ γεγράφθω (AB) [Prop. 10.30]. And let BC have been cut in half at ἐπὶ τῆς ΑΒ ημικύκλιον τὸ ΑΖΒ, καὶ ἤχθω τῇ ΑΒ πρὸς D. And let a parallelogram equal to the (square) on ei- 317 STOIQEIWN iþ. ELEMENTS BOOK 10 ὀρθὰς ἡ ΕΖ, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΖΒ. ther of BD or DC, (and) falling short by a square figure, have been applied to AB [Prop. 6.28], and let it be the (rectangle contained) by AEB. And let the semi-circle AFB have been drawn on AB. And let EF have been drawn at right-angles to AB. And let AF and FB have been joined. DA B GE Z DA B CE F Καὶ ἐπεὶ [δύο] εὐθεῖαι ἄνισοί εἰσιν αἱ ΑΒ, ΒΓ, καὶ ἡ And since AB and BC are [two] unequal straight- ΑΒ τῆς ΒΓ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, lines, and the square on AB is greater than (the square τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς ΒΓ, τουτέστι τῷ ἀπὸ τῆς on) BC by the (square) on (some straight-line which is) ἡμισείας αὐτῆς, ἴσον παρὰ τὴν ΑΒ παραβέβληται παραλ- incommensurable (in length) with (AB). And a paral- ληλόγραμμον ἐλλεῖπον εἴδει τετραγώνῳ καὶ ποιεῖ τὸ ὑπὸ lelogram, equal to one quarter of the (square) on BC— τῶν ΑΕΒ, ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΕ τῇ ΕΒ. καί ἐστιν ὡς that is to say, (equal) to the (square) on half of it—(and) ἡ ΑΕ πρὸς ΕΒ, οὕτως τὸ ὑπὸ τῶν ΒΑ, ΑΕ πρὸς τὸ ὑπὸ falling short by a square figure, has been applied to AB, τῶν ΑΒ, ΒΕ, ἴσον δὲ τὸ μὲν ὑπὸ τῶν ΒΑ, ΑΕ τῷ ἀπὸ τῆς and makes the (rectangle contained) by AEB. AE is thus ΑΖ, τὸ δὲ ὑπὸ τῶν ΑΒ, ΒΕ τῷ ἁπὸ τῆς ΒΖ· ἀσύμμετρον incommensurable (in length) with EB [Prop. 10.18]. ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΖ τῷ ἀπὸ τῆς ΖΒ· αἱ ΑΖ, ΖΒ ἄρα And as AE is to EB, so the (rectangle contained) by BA δυνάμει εἰσὶν ἀσύμμετροι. καὶ ἐπεὶ ἡ ΑΒ ῥητή ἐστιν, ῥητὸν and AE (is) to the (rectangle contained) by AB and BE. ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΒ· ὥστε καὶ τὸ συγκείμενον ἐκ And the (rectangle contained) by BA and AE (is) equal τῶν ἀπὸ τῶν ΑΖ, ΖΒ ῥητόν ἐστιν. καὶ ἐπεὶ πάλιν τὸ ὑπὸ to the (square) on AF , and the (rectangle contained) by τῶν ΑΕ, ΕΒ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΖ, ὑπόκειται δὲ τὸ ὑπὸ AB and BE to the (square) on BF [Prop. 10.32 lem.]. τῶν ΑΕ, ΕΒ καὶ τῷ ἀπὸ τῆς ΒΔ ἴσον, ἴση ἄρα ἐστὶν ἡ ΖΕ The (square) on AF is thus incommensurable with the τῇ ΒΔ· διπλῆ ἄρα ἡ ΒΓ τὴς ΖΕ· ὥστε καὶ τὸ ὑπὸ τῶν ΑΒ, (square) on FB [Prop. 10.11]. Thus, AF and FB are in- ΒΓ σύμμετρόν ἐστι τῷ ὑπὸ τῶν ΑΒ, ΕΖ. μέσον δὲ τὸ ὑπὸ commensurable in square. And since AB is rational, the τῶν ΑΒ, ΒΓ· μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΑΒ, ΕΖ. ἴσον δὲ (square) on AB is also rational. Hence, the sum of the τὸ ὑπὸ τῶν ΑΒ, ΕΖ τῷ ὑπὸ τῶν ΑΖ, ΖΒ· μέσον ἄρα καὶ τὸ (squares) on AF and FB is also rational [Prop. 1.47]. ὑπὸ τῶν ΑΖ, ΖΒ. ἐδείχθη δὲ καὶ ῥητὸν τὸ συγκείμενον ἐκ And, again, since the (rectangle contained) by AE and τῶν ἀπ᾿ αὐτῶν τετραγώνων. EB is equal to the (square) on EF , and the (rectangle Εὕρηνται ἄρα δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ ΑΖ, ΖΒ contained) by AE and EB was assumed (to be) equal to ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων the (square) on BD, FE is thus equal to BD. Thus, BC ῥητόν, τὸ δὲ ὑπ᾿ αὐτῶν μέσον· ὅπερ ἔδει δεῖξαι. is double FE. And hence the (rectangle contained) by AB and BC is commensurable with the (rectangle con- tained) by AB and EF [Prop. 10.6]. And the (rectan- gle contained) by AB and BC (is) medial [Prop. 10.21]. Thus, the (rectangle contained) by AB and EF (is) also medial [Prop. 10.23 corr.]. And the (rectangle contained) by AB and EF (is) equal to the (rectangle contained) by AF and FB [Prop. 10.32 lem.]. Thus, the (rectangle con- tained) by AF and FB (is) also medial. And the sum of the squares on them was also shown (to be) rational. Thus, the two straight-lines, AF and FB, (which are) incommensurable in square, have been found, making the sum of the squares on them rational, and the (rectan- gle contained) by them medial. (Which is) the very thing it was required to show. 318 STOIQEIWN iþ. ELEMENTS BOOK 10 † AF and FB have lengths q [1 + k/(1 + k2)1/2]/2 and q [1 − k/(1 + k2)1/2]/2 times that of AB, respectively, where k is defined in the footnote to Prop. 10.30. ldþ. Proposition 34 Εὑρεῖν δύο εὐθείας δυνάμει ἀσυμμέτρους ποιούσας τὸ To find two straight-lines (which are) incommensu- μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον, τὸ rable in square, making the sum of the squares on them δ᾿ ὑπ᾿ αὐτῶν ῥητόν. medial, and the (rectangle contained) by them rational. ΓΑ Β ∆ Ζ Ε EA B D F C ᾿Εκκείσθωσαν δύο μέσαι δυνάμει μόνον σύμμετροι αἱ Let the two medial (straight-lines) AB and BC, ΑΒ, ΒΓ ῥητὸν περιέχουσαι τὸ ὑπ᾿ αὐτῶν, ὥστε τὴν ΑΒ (which are) commensurable in square only, be laid out τῆς ΒΓ μεῖζον δύνασθαι τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ having the (rectangle contained) by them rational, (and) γεγράφθω ἐπὶ τῆς ΑΒ τὸ ΑΔΒ ἡμικύκλιον, καὶ τετμήσθω such that the square on AB is greater than (the square ἡ ΒΓ δίχα κατὰ τὸ Ε, καὶ παραβεβλήσθω παρὰ τὴν ΑΒ on) BC by the (square) on (some straight-line) incom- τῷ ἀπὸ τῆς ΒΕ ἴσον παραλληλόγραμμον ἐλλεῖπον εἴδει τε- mensurable (in length) with (AB) [Prop. 10.31]. And τραγώνῳ τὸ ὑπὸ τῶν ΑΖΒ· ἀσύμμετρος ἄρα [ἐστὶν] ἡ ΑΖ let the semi-circle ADB have been drawn on AB. And τῇ ΖΒ μήκει. καὶ ἤχθω ἀπὸ τοῦ Ζ τῇ ΑΒ πρὸς ὀρθὰς ἡ ΖΔ, let BC have been cut in half at E. And let a (rectangu- καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΒ. lar) parallelogram equal to the (square) on BE, (and) ᾿Επεὶ ἀσύμμετρός ἐστιν ἡ ΑΖ τῇ ΖΒ, ἀσύμμετρον ἄρα falling short by a square figure, have been applied to ἐστὶ καὶ τὸ ὑπὸ τῶν ΒΑ, ΑΖ τῷ ὑπὸ τῶν ΑΒ, ΒΖ. ἴσον δὲ AB, (and let it be) the (rectangle contained by) AFB τὸ μὲν ὑπὸ τῶν ΒΑ, ΑΖ τῷ ἀπὸ τῆς ΑΔ, τὸ δὲ ὑπὸ τῶν [Prop. 6.28]. Thus, AF [is] incommensurable in length ΑΒ, ΒΖ τῷ ἀπὸ τῆς ΔΒ· ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ with FB [Prop. 10.18]. And let FD have been drawn τῆς ΑΔ τῷ ἀπὸ τῆς ΔΒ. καὶ ἐπεὶ μέσον ἐστὶ τὸ ἀπὸ τῆς from F at right-angles to AB. And let AD and DB have ΑΒ, μέσον ἄρα καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, been joined. ΔΒ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΒΓ τῆς ΔΖ, διπλάσιον ἄρα καὶ Since AF is incommensurable (in length) with FB, τὸ ὑπὸ τῶν ΑΒ, ΒΓ τοῦ ὑπὸ τῶν ΑΒ, ΖΔ. ῥητὸν δὲ τὸ ὑπὸ the (rectangle contained) by BA and AF is thus also τῶν ΑΒ, ΒΓ· ῥητὸν ἄρα καὶ τὸ ὑπὸ τῶν ΑΒ, ΖΔ. τὸ δὲ ὑπὸ incommensurable with the (rectangle contained) by AB τῶν ΑΒ, ΖΔ ἴσον τῷ ὑπὸ τῶν ΑΔ, ΔΒ· ὥστε καὶ τὸ ὑπὸ and BF [Prop. 10.11]. And the (rectangle contained) by τῶν ΑΔ, ΔΒ ῥητόν ἐστιν. BA and AF (is) equal to the (square) on AD, and the Εὕρηνται ἄρα δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ ΑΔ, (rectangle contained) by AB and BF to the (square) on ΔΒ ποιοῦσαι τὸ [μὲν] συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τε- DB [Prop. 10.32 lem.]. Thus, the (square) on AD is also τραγώνων μέσον, τὸ δ᾿ ὑπ᾿ αὐτῶν ῥητόν· ὅπερ ἔδει δεῖξαι. incommensurable with the (square) on DB. And since the (square) on AB is medial, the sum of the (squares) on AD and DB (is) thus also medial [Props. 3.31, 1.47]. And since BC is double DF [see previous proposition], the (rectangle contained) by AB and BC (is) thus also double the (rectangle contained) by AB and FD. And the (rectangle contained) by AB and BC (is) rational. Thus, the (rectangle contained) by AB and FD (is) also rational [Prop. 10.6, Def. 10.4]. And the (rectangle con- tained) by AB and FD (is) equal to the (rectangle con- tained) by AD and DB [Prop. 10.32 lem.]. And hence the (rectangle contained) by AD and DB is rational. Thus, two straight-lines, AD and DB, (which are) in- commensurable in square, have been found, making the sum of the squares on them medial, and the (rectangle 319 STOIQEIWN iþ. ELEMENTS BOOK 10 contained) by them rational.† (Which is) the very thing it was required to show. † AD and DB have lengths q [(1 + k2)1/2 + k]/[2 (1 + k2)] and q [(1 + k2)1/2 − k]/[2 (1 + k2)] times that of AB, respectively, where k is defined in the footnote to Prop. 10.29.leþ. Proposition 35 Εὑρεῖν δύο εὐθείας δυνάμει ἀσυμμέτρους ποιούσας τό To find two straight-lines (which are) incommensu- τε συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον καὶ rable in square, making the sum of the squares on them τὸ ὑπ᾿ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ medial, and the (rectangle contained) by them medial, τῶν ἀπ᾿ αὐτῶν τετραγώνῳ. and, moreover, incommensurable with the sum of the squares on them. ΓΑ Β ∆ Ζ Ε EA B D F C ᾿Εκκείσθωσαν δύο μέσαι δυνάμει μόνον σύμμετροι αἱ Let the two medial (straight-lines) AB and BC, ΑΒ, ΒΓ μέσον περιέχουσαι, ὥστε τὴν ΑΒ τῆς ΒΓ μεῖζον (which are) commensurable in square only, be laid out δύνασθαι τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ γεγράφθω ἐπὶ τῆς containing a medial (area), such that the square on AB ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ τὰ λοιπὰ γεγονέτω τοῖς ἐπάνω is greater than (the square on) BC by the (square) on ὁμοίως. (some straight-line) incommensurable (in length) with Καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΖ τῇ ΖΒ μήκει, ἀσύμμετρ- (AB) [Prop. 10.32]. And let the semi-circle ADB have ός ἐστι καὶ ἡ ΑΔ τῇ ΔΒ δυνάμει. καὶ ἐπεὶ μέσον ἐστὶ τὸ been drawn on AB. And let the remainder (of the figure) ἀπὸ τῆς ΑΒ, μέσον ἄρα καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ be generated similarly to the above (proposition). τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΖ, ΖΒ ἴσον ἐστὶ τῷ ἀφ᾿ And since AF is incommensurable in length with FB ἑκατέρας τῶν ΒΕ, ΔΖ, ἴση ἄρα ἐστὶν ἡ ΒΕ τῇ ΔΖ· διπλῆ [Prop. 10.18], AD is also incommensurable in square ἄρα ἡ ΒΓ τῆς ΖΔ· ὥστε καὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ διπλάσιόν with DB [Prop. 10.11]. And since the (square) on AB ἐστι τοῦ ὑπὸ τῶν ΑΒ, ΖΔ. μέσον δὲ τὸ ὑπὸ τῶν ΑΒ, ΒΓ· is medial, the sum of the (squares) on AD and DB (is) μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΑΒ, ΖΔ. καί ἐστιν ἴσον τῷ ὑπὸ thus also medial [Props. 3.31, 1.47]. And since the (rect- τῶν ΑΔ, ΔΒ· μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ angle contained) by AF and FB is equal to the (square) ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, σύμμετρος δὲ ἡ ΓΒ on each of BE and DF , BE is thus equal to DF . Thus, τῇ ΒΕ, ἀσύμμετρος ἄρα καὶ ἡ ΑΒ τῇ ΒΕ μήκει· ὥστε καὶ τὸ BC (is) double FD. And hence the (rectangle contained) ἀπὸ τῆς ΑΒ τῷ ὑπὸ τῶν ΑΒ, ΒΕ ἀσύμμετρόν ἐστιν. ἀλλὰ by AB and BC is double the (rectangle) contained by τῷ μὲν ἀπὸ τῆς ΑΒ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ, τῷ δὲ AB and FD. And the (rectangle contained) by AB and ὑπὸ τῶν ΑΒ, ΒΕ ἴσον ἐστὶ τὸ ὑπὸ τῶν ΑΒ, ΖΔ, τουτέστι BC (is) medial. Thus, the (rectangle contained) by AB τὸ ὑπὸ τῶν ΑΔ, ΔΒ· ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον and FD (is) also medial. And it is equal to the (rectan- ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ τῷ ὑπὸ τῶν ΑΔ, ΔΒ. gle contained) by AD and DB [Prop. 10.32 lem.]. Thus, Εὕρηνται ἄρα δύο εὐθεῖαι αἱ ΑΔ, ΔΒ δυνάμει ἀσύμμετροι the (rectangle contained) by AD and DB (is) also me- ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν μέσον καὶ dial. And since AB is incommensurable in length with τὸ ὑπ᾿ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ BC, and CB (is) commensurable (in length) with BE, τῶν ἀπ᾿ αὐτῶν τετραγώνων· ὅπερ ἔδει δεῖξαι. AB (is) thus also incommensurable in length with BE [Prop. 10.13]. And hence the (square) on AB is also incommensurable with the (rectangle contained) by AB and BE [Prop. 10.11]. But the (sum of the squares) on AD and DB is equal to the (square) on AB [Prop. 1.47]. And the (rectangle contained) by AB and FD—that is to say, the (rectangle contained) by AD and DB—is equal to the (rectangle contained) by AB and BE. Thus, the 320 STOIQEIWN iþ. ELEMENTS BOOK 10 sum of the (squares) on AD and DB is incommensurable with the (rectangle contained) by AD and DB. Thus, two straight-lines, AD and DB, (which are) in- commensurable in square, have been found, making the sum of the (squares) on them medial, and the (rectangle contained) by them medial, and, moreover, incommensu- rable with the sum of the squares on them.† (Which is) the very thing it was required to show. † AD and DB have lengths k′1/4 q [1 + k/(1 + k2)1/2]/2 and k′1/4 q [1 − k/(1 + k2)1/2]/2 times that of AB, respectively, where k and k′ are defined in the footnote to Prop. 10.32.l�þ. Proposition 36 ᾿Εὰν δύο ῥηταὶ δυνάμει μόνον σύμμετροι συντεθῶσιν, ἡ If two rational (straight-lines which are) commensu- ὅλη ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο ὀνομάτων. rable in square only are added together then the whole (straight-line) is irrational—let it be called a binomial (straight-line).† Α Β Γ A B C Συγκείσθωσαν γὰρ δύο ῥηταὶ δυνάμει μόνον σύμμετροι For let the two rational (straight-lines), AB and BC, αἱ ΑΒ, ΒΓ· λέγω, ὅτι ὅλη ἡ ΑΓ ἄλογός ἐστιν. (which are) commensurable in square only, be laid down ᾿Επεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει· δυνάμει together. I say that the whole (straight-line), AC, is irra- γὰρ μόνον εἰσὶ σύμμετροι· ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως tional. For since AB is incommensurable in length with τὸ ὑπὸ τῶν ΑΒΓ πρὸς τὸ ἀπὸ τῆς ΒΓ, ἀσύμμετρον ἄρα ἐστὶ BC—for they are commensurable in square only—and as τὸ ὑπὸ τῶν ΑΒ, ΒΓ τῷ ἀπὸ τῆς ΒΓ. ἀλλὰ τῷ μὲν ὑπὸ τῶν AB (is) to BC, so the (rectangle contained) by ABC (is) ΑΒ, ΒΓ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, τῷ δὲ ἀπὸ to the (square) on BC, the (rectangle contained) by AB τῆς ΒΓ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ· αἱ γὰρ ΑΒ, ΒΓ and BC is thus incommensurable with the (square) on ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἀσύμμετρον ἄρα ἐστὶ τὸ BC [Prop. 10.11]. But, twice the (rectangle contained) δὶς ὑπὸ τῶν ΑΒ, ΒΓ τοῖς ἀπὸ τῶν ΑΒ, ΒΓ. καὶ συνθέντι τὸ by AB and BC is commensurable with the (rectangle δὶς ὑπὸ τῶν ΑΒ, ΒΓ μετὰ τῶν ἀπὸ τῶν ΑΒ, ΒΓ, τουτέστι contained) by AB and BC [Prop. 10.6]. And (the sum τὸ ἀπὸ τῆς ΑΓ, ἀσύμμετρόν ἐστι τῷ συγκειμένῳ ἐκ τῶν of) the (squares) on AB and BC is commensurable with ἀπὸ τῶν ΑΒ, ΒΓ· ῥητὸν δὲ τὸ συγκείμενον ἐκ τῶν ἀπὸ the (square) on BC—for the rational (straight-lines) AB τῶν ΑΒ, ΒΓ· ἄλογον ἄρα [ἐστὶ] τὸ ἀπὸ τῆς ΑΓ· ὥστε καὶ ἡ and BC are commensurable in square only [Prop. 10.15]. ΑΓ ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο ὀνομάτων· ὅπερ ἔδει Thus, twice the (rectangle contained) by AB and BC is δεῖξαι. incommensurable with (the sum of) the (squares) on AB and BC [Prop. 10.13]. And, via composition, twice the (rectangle contained) by AB and BC, plus (the sum of) the (squares) on AB and BC—that is to say, the (square) on AC [Prop. 2.4]—is incommensurable with the sum of the (squares) on AB and BC [Prop. 10.16]. And the sum of the (squares) on AB and BC (is) rational. Thus, the (square) on AC [is] irrational [Def. 10.4]. Hence, AC is also irrational [Def. 10.4]—let it be called a binomial (straight-line).‡ (Which is) the very thing it was required to show. † Literally, “from two names”. ‡ Thus, a binomial straight-line has a length expressible as 1 + k1/2 [or, more generally, ρ (1 + k1/2), where ρ is rational—the same proviso applies to the definitions in the following propositions]. The binomial and the corresponding apotome, whose length is expressible as 1 − k1/2 321 STOIQEIWN iþ. ELEMENTS BOOK 10 (see Prop. 10.73), are the positive roots of the quartic x4 − 2 (1 + k) x2 + (1 − k)2 = 0.lzþ. Proposition 37 ᾿Εὰν δύο μέσαι δυνάμει μόνον σύμμετροι συντεθῶσι If two medial (straight-lines), commensurable in ῥητὸν περιέχουσαι, ἡ ὅλη ἄλογός ἐστιν, καλείσθω δὲ ἐκ square only, which contain a rational (area), are added δύο μέσων πρώτη. together then the whole (straight-line) is irrational—let it be called a first bimedial (straight-line).† Α Β Γ A B C Συγκείσθωσαν γὰρ δύο μέσαι δυνάμει μόνον σύμμετροι For let the two medial (straight-lines), AB and BC, αἱ ΑΒ, ΒΓ ῥητὸν περιέχουσαι· λέγω, ὅτι ὅλη ἡ ΑΓ ἄλογός commensurable in square only, (and) containing a ratio- ἐστιν. nal (area), be laid down together. I say that the whole ᾿Επεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, καὶ τὰ (straight-line), AC, is irrational. ἀπὸ τῶν ΑΒ, ΒΓ ἄρα ἀσύμμετρά ἐστι τῷ δὶς ὑπὸ τῶν ΑΒ, For since AB is incommensurable in length with BC, ΒΓ· καὶ συνθέντι τὰ ἀπὸ τῶν ΑΒ, ΒΓ μετὰ τοῦ δὶς ὑπὸ τῶν (the sum of) the (squares) on AB and BC is thus also in- ΑΒ, ΒΓ, ὅπερ ἐστὶ τὸ ἀπὸ τῆς ΑΓ, ἀσύμμετρόν ἐστι τῷ ὑπὸ commensurable with twice the (rectangle contained) by τῶν ΑΒ, ΒΓ. ῥητὸν δὲ τὸ ὑπὸ τῶν ΑΒ, ΒΓ· ὑπόκεινται γὰρ AB and BC [see previous proposition]. And, via com- αἱ ΑΒ, ΒΓ ῥητὸν περιέχουσαι· ἄλογον ἄρα τὸ ἀπὸ τῆς ΑΓ· position, (the sum of) the (squares) on AB and BC, ἄλογος ἄρα ἡ ΑΓ, καλείσθω δὲ ἐκ δύο μέσων πρώτη· ὅπερ plus twice the (rectangle contained) by AB and BC— ἔδει δεῖξαι. that is, the (square) on AC [Prop. 2.4]—is incommen- surable with the (rectangle contained) by AB and BC [Prop. 10.16]. And the (rectangle contained) by AB and BC (is) rational—for AB and BC were assumed to en- close a rational (area). Thus, the (square) on AC (is) irrational. Thus, AC (is) irrational [Def. 10.4]—let it be called a first bimedial (straight-line).‡ (Which is) the very thing it was required to show. † Literally, “first from two medials”. ‡ Thus, a first bimedial straight-line has a length expressible as k1/4 + k3/4. The first bimedial and the corresponding first apotome of a medial, whose length is expressible as k1/4 − k3/4 (see Prop. 10.74), are the positive roots of the quartic x4 − 2 √ k (1 + k) x2 + k (1 − k)2 = 0.lhþ. Proposition 38 ᾿Εὰν δύο μέσαι δυνάμει μόνον σύμμετροι συντεθῶσι If two medial (straight-lines), commensurable in square μέσον περιέχουσαι, ἡ ὅλη ἄλογός ἐστιν, καλείσθω δὲ ἐκ only, which contain a medial (area), are added together δύο μέσων δυετέρα. then the whole (straight-line) is irrational—let it be called a second bimedial (straight-line). Η Α Β Γ ∆ Ε Ζ Θ CA B D E F H G Συγκείσθωσαν γὰρ δύο μέσαι δυνάμει μόνον σύμμετροι For let the two medial (straight-lines), AB and BC, αἱ ΑΒ, ΒΓ μέσον περιέχουσαι· λέγω, ὅτι ἄλογός ἐστιν ἡ commensurable in square only, (and) containing a medial 322 STOIQEIWN iþ. ELEMENTS BOOK 10 ΑΓ. (area), be laid down together [Prop. 10.28]. I say that ᾿Εκκείσθω γὰρ ῥητὴ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΓ ἴσον παρὰ AC is irrational. τὴν ΔΕ παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ. καὶ For let the rational (straight-line) DE be laid down, ἐπεὶ τὸ ἀπὸ τῆς ΑΓ ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΑΒ, ΒΓ καὶ τῷ and let (the rectangle) DF , equal to the (square) on δὶς ὑπὸ τῶν ΑΒ, ΒΓ, παραβεβλήσθω δὴ τοῖς ἀπὸ τῶν ΑΒ, AC, have been applied to DE, making DG as breadth ΒΓ παρὰ τὴν ΔΕ ἴσον τὸ ΕΘ· λοιπὸν ἄρα τὸ ΘΖ ἴσον ἐστὶ [Prop. 1.44]. And since the (square) on AC is equal to τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. καὶ ἐπεὶ μέση ἐστὶν ἑκατέρα τῶν (the sum of) the (squares) on AB and BC, plus twice ΑΒ, ΒΓ, μέσα ἄρα ἐστὶ καὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ. μέσον δὲ the (rectangle contained) by AB and BC [Prop. 2.4], so ὑπόκειται καὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. καί ἐστι τοῖς μὲν ἀπὸ let (the rectangle) EH , equal to (the sum of) the squares τῶν ΑΒ, ΒΓ ἴσον τὸ ΕΘ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ἴσον on AB and BC, have been applied to DE. The remain- τὸ ΖΘ· μέσον ἄρα ἑκάτερον τῶν ΕΘ, ΘΖ. καὶ παρὰ ῥητὴν der HF is thus equal to twice the (rectangle contained) τὴν ΔΕ παράκειται· ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΔΘ, ΘΗ by AB and BC. And since AB and BC are each me- καὶ ἀσύμμετρος τῇ ΔΕ μήκει. ἐπεὶ οὖν ἀσύμμετρός ἐστιν ἡ dial, (the sum of) the squares on AB and BC is thus also ΑΒ τῇ ΒΓ μήκει, καί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ medial.‡ And twice the (rectangle contained) by AB and ἀπὸ τῆς ΑΒ πρὸς τὸ ὑπὸ τῶν ΑΒ, ΒΓ, ἀσύμμετρον ἄρα ἐστὶ BC was also assumed (to be) medial. And EH is equal τὸ ἀπὸ τῆς ΑΒ τῷ ὑπὸ τῶν ΑΒ, ΒΓ. ἀλλὰ τῷ μὲν ἀπὸ τῆς to (the sum of) the squares on AB and BC, and FH (is) ΑΒ σύμμετρόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, equal to twice the (rectangle contained) by AB and BC. ΒΓ τετραγώνων, τῷ δὲ ὑπὸ τῶν ΑΒ, ΒΓ σύμμετρόν ἐστι τὸ Thus, EH and HF (are) each medial. And they were ap- δὶς ὑπὸ τῶν ΑΒ, ΒΓ. ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον plied to the rational (straight-line) DE. Thus, DH and ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. ἀλλὰ HG are each rational, and incommensurable in length τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τὸ ΕΘ, τῷ δὲ δὶς ὑπὸ with DE [Prop. 10.22]. Therefore, since AB is incom- τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τὸ ΘΖ. ἀσύμμετρον ἄρα ἐστὶ τὸ ΕΘ mensurable in length with BC, and as AB is to BC, so τῷ ΘΖ· ὥστε καὶ ἡ ΔΘ τῇ ΘΗ ἐστιν ἀσύμμετρος μήκει. αἱ the (square) on AB (is) to the (rectangle contained) by ΔΘ, ΘΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. ὥστε ἡ AB and BC [Prop. 10.21 lem.], the (square) on AB is ΔΗ ἄλογός ἐστιν. ῥητὴ δὲ ἡ ΔΕ· τὸ δὲ ὑπὸ ἀλόγου καὶ thus incommensurable with the (rectangle contained) by ῥητῆς περιεχόμενον ὀρθογώνιον ἄλογόν ἐστιν· ἄλογον ἄρα AB and BC [Prop. 10.11]. But, the sum of the squares ἐστὶ τὸ ΔΖ χωρίον, καὶ ἡ δυναμένη [αὐτὸ] ἄλογός ἐστιν. on AB and BC is commensurable with the (square) on δύναται δὲ τὸ ΔΖ ἡ ΑΓ· ἄλογος ἄρα ἐστὶν ἡ ΑΓ, καλείσθω AB [Prop. 10.15], and twice the (rectangle contained) by δὲ ἐκ δύο μέσων δευτέρα. ὅπερ ἔδει δεῖξαι. AB and BC is commensurable with the (rectangle con- tained) by AB and BC [Prop. 10.6]. Thus, the sum of the (squares) on AB and BC is incommensurable with twice the (rectangle contained) by AB and BC [Prop. 10.13]. But, EH is equal to (the sum of) the squares on AB and BC, and HF is equal to twice the (rectangle) contained by AB and BC. Thus, EH is incommensurable with HF . Hence, DH is also incommensurable in length with HG [Props. 6.1, 10.11]. Thus, DH and HG are ratio- nal (straight-lines which are) commensurable in square only. Hence, DG is irrational [Prop. 10.36]. And DE (is) rational. And the rectangle contained by irrational and rational (straight-lines) is irrational [Prop. 10.20]. The area DF is thus irrational, and (so) the square-root [of it] is irrational [Def. 10.4]. And AC is the square-root of DF . AC is thus irrational—let it be called a second bimedial (straight-line).§ (Which is) the very thing it was required to show. † Literally, “second from two medials”. ‡ Since, by hypothesis, the squares on AB and BC are commensurable—see Props. 10.15, 10.23. § Thus, a second bimedial straight-line has a length expressible as k1/4+k′1/2/k1/4. The second bimedial and the corresponding second apotome of a medial, whose length is expressible as k1/4 − k′1/2/k1/4 (see Prop. 10.75), are the positive roots of the quartic x4 − 2 [(k + k′)/ √ k ] x2 + 323 STOIQEIWN iþ. ELEMENTS BOOK 10 [(k − k′)2/k] = 0. ljþ. Proposition 39 ᾿Εὰν δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι ποιοῦς- If two straight-lines (which are) incommensurable in αι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων ῥητόν, square, making the sum of the squares on them rational, τὸ δ᾿ ὑπ᾿ αὐτῶν μέσον, ἡ ὅλη εὐθεῖα ἄλογός ἐστιν, καλείσθω and the (rectangle contained) by them medial, are added δὲ μείζων. together then the whole straight-line is irrational—let it be called a major (straight-line). Α Β Γ A B C Συγκείσθωσαν γὰρ δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ For let the two straight-lines, AB and BC, incommen- ΑΒ, ΒΓ ποιοῦσαι τὰ προκείμενα· λέγω, ὅτι ἄλογός ἐστιν ἡ surable in square, and fulfilling the prescribed (condi- ΑΓ. tions), be laid down together [Prop. 10.33]. I say that ᾿Επεὶ γὰρ τὸ ὑπὸ τῶν ΑΒ, ΒΓ μέσον ἐστίν, καὶ τὸ δὶς AC is irrational. [ἄρα] ὑπὸ τῶν ΑΒ, ΒΓ μέσον ἐστίν. τὸ δὲ συγκείμενον ἐκ For since the (rectangle contained) by AB and BC is τῶν ἀπὸ τῶν ΑΒ, ΒΓ ῥητόν· ἀσύμμετρον ἄρα ἐστὶ τὸ δὶς medial, twice the (rectangle contained) by AB and BC ὑπὸ τῶν ΑΒ, ΒΓ τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ· is [thus] also medial [Props. 10.6, 10.23 corr.]. And the ὥστε καὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ μετὰ τοῦ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, sum of the (squares) on AB and BC (is) rational. Thus, ὅπερ ἐστὶ τὸ ἀπὸ τῆς ΑΓ, ἀσύμμετρόν ἐστι τῷ συγκειμένῳ twice the (rectangle contained) by AB and BC is incom- ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ [ῥητὸν δὲ τὸ συγμείμενον ἐκ τῶν mensurable with the sum of the (squares) on AB and ἀπὸ τῶν ΑΒ, ΒΓ]· ἄλογον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΓ. ὥστε BC [Def. 10.4]. Hence, (the sum of) the squares on AB καὶ ἡ ΑΓ ἄλογός ἐστιν, καλείσθω δὲ μείζων. ὅπερ ἔδει and BC, plus twice the (rectangle contained) by AB and δεῖξαι. BC—that is, the (square) on AC [Prop. 2.4]—is also in- commensurable with the sum of the (squares) on AB and BC [Prop. 10.16] [and the sum of the (squares) on AB and BC (is) rational]. Thus, the (square) on AC is irra- tional. Hence, AC is also irrational [Def. 10.4]—let it be called a major (straight-line).† (Which is) the very thing it was required to show. † Thus, a major straight-line has a length expressible as q [1 + k/(1 + k2)1/2]/2 + q [1 − k/(1 + k2)1/2]/2. The major and the corresponding minor, whose length is expressible as q [1 + k/(1 + k2)1/2]/2 − q [1 − k/(1 + k2)1/2]/2 (see Prop. 10.76), are the positive roots of the quartic x4 − 2 x2 + k2/(1 + k2) = 0. mþ. Proposition 40 ᾿Εὰν δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι ποιοῦς- If two straight-lines (which are) incommensurable αι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον, in square, making the sum of the squares on them τὸ δ᾿ ὑπ᾿ αὐτῶν ῥητόν, ἡ ὅλη εὐθεῖα ἄλογός ἐστιν, καλείσθω medial, and the (rectangle contained) by them ratio- δὲ ῥητὸν καὶ μέσον δυναμένη. nal, are added together then the whole straight-line is irrational—let it be called the square-root of a rational plus a medial (area). Α Β Γ A B C Συγκείσθωσαν γὰρ δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ For let the two straight-lines, AB and BC, incommen- ΑΒ, ΒΓ ποιοῦσαι τὰ προκείμενα· λέγω, ὅτι ἄλογός ἐστιν ἡ surable in square, (and) fulfilling the prescribed (condi- ΑΓ. tions), be laid down together [Prop. 10.34]. I say that ᾿Επεὶ γὰρ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ AC is irrational. μέσον ἐστίν, τὸ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ῥητόν, ἀσύμμετρον For since the sum of the (squares) on AB and BC is ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς medial, and twice the (rectangle contained) by AB and 324 STOIQEIWN iþ. ELEMENTS BOOK 10 ὑπὸ τῶν ΑΒ, ΒΓ· ὥστε καὶ τὸ ἁπὸ τῆς ΑΓ ἀσύμμετρόν ἐστι BC (is) rational, the sum of the (squares) on AB and τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. ῥητὸν δὲ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ· BC is thus incommensurable with twice the (rectangle ἄλογον ἄρα τὸ ἀπὸ τῆς ΑΓ. ἄλογος ἄρα ἡ ΑΓ, καλείσθω δὲ contained) by AB and BC. Hence, the (square) on AC ῥητὸν καὶ μέσον δυναμένη. ὅπερ ἔδει δεῖξαι. is also incommensurable with twice the (rectangle con- tained) by AB and BC [Prop. 10.16]. And twice the (rectangle contained) by AB and BC (is) rational. The (square) on AC (is) thus irrational. Thus, AC (is) irra- tional [Def. 10.4]—let it be called the square-root of a rational plus a medial (area).† (Which is) the very thing it was required to show. † Thus, the square-root of a rational plus a medial (area) has a length expressible as q [(1 + k2)1/2 + k]/[2 (1 + k2)]+ q [(1 + k2)1/2 − k]/[2 (1 + k2)]. This and the corresponding irrational with a minus sign, whose length is expressible as q [(1 + k2)1/2 + k]/[2 (1 + k2)]− q [(1 + k2)1/2 − k]/[2 (1 + k2)] (see Prop. 10.77), are the positive roots of the quartic x4 − (2/ √ 1 + k2) x2 + k2/(1 + k2)2 = 0.maþ. Proposition 41 ᾿Εὰν δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι ποιοῦς- If two straight-lines (which are) incommensurable in αι τό τε συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον square, making the sum of the squares on them me- καὶ τὸ ὑπ᾿ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ dial, and the (rectangle contained) by them medial, and, ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων, ἡ ὅλη εὐθεῖα ἄλογός ἐστιν, moreover, incommensurable with the sum of the squares καλείσθω δὲ δύο μέσα δυναμένη. on them, are added together then the whole straight-line is irrational—let it be called the square-root of (the sum of) two medial (areas). ∆ Α Β Γ Ε Ζ ΘΚ Η K A B EDC F H G Συγκείσθωσαν γὰρ δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ For let the two straight-lines, AB and BC, incommen- ΑΒ, ΒΓ ποιοῦσαι τὰ προκείμενα· λέγω, ὅτι ἡ ΑΓ ἄλογός surable in square, (and) fulfilling the prescribed (condi- ἐστιν. tions), be laid down together [Prop. 10.35]. I say that ᾿Εκκείσθω ῥητὴ ἡ ΔΕ, καὶ παραβεβλήσθω παρὰ τὴν ΔΕ AC is irrational. τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ ἴσον τὸ ΔΖ, τῷ δὲ δὶς ὑπὸ τῶν Let the rational (straight-line) DE be laid out, and let ΑΒ, ΒΓ ἴσον τὸ ΗΘ· ὅλον ἄρα τὸ ΔΘ ἴσον ἐστὶ τῷ ἀπὸ (the rectangle) DF , equal to (the sum of) the (squares) τῆς ΑΓ τετραγώνῳ. καὶ ἐπεὶ μέσον ἐστὶ τὸ συγκείμενον on AB and BC, and (the rectangle) GH , equal to twice ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ, καί ἐστιν ἴσον τῷ ΔΖ, μέσον ἄρα the (rectangle contained) by AB and BC, have been ap- ἐστὶ καὶ τὸ ΔΖ. καὶ παρὰ ῥητὴν τὴν ΔΕ παράκειται· ῥητὴ plied to DE. Thus, the whole of DH is equal to the ἄρα ἐστὶν ἡ ΔΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει. διὰ τὰ αὐτὰ square on AC [Prop. 2.4]. And since the sum of the δὴ καὶ ἡ ΗΚ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΗΖ, τουτέστι τῇ (squares) on AB and BC is medial, and is equal to DF , ΔΕ, μήκει. καὶ ἐπεὶ ἀσύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ DF is thus also medial. And it is applied to the rational τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, ἀσύμμετρόν ἐστι τὸ ΔΖ τῷ ΗΘ· (straight-line) DE. Thus, DG is rational, and incommen- 325 STOIQEIWN iþ. ELEMENTS BOOK 10 ὥστε καὶ ἡ ΔΗ τῇ ΗΚ ἀσύμμετρός ἐστιν. καὶ εἰσι ῥηταί· αἱ surable in length with DE [Prop. 10.22]. So, for the same ΔΗ, ΗΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἄλογος (reasons), GK is also rational, and incommensurable in ἄρα ἐστὶν ἡ ΔΚ ἡ καλουμένη ἐκ δύο ὀνομάτων. ῥητὴ δὲ ἡ length with GF—that is to say, DE. And since (the sum ΔΕ· ἄλογον ἄρα ἐστὶ τὸ ΔΘ καὶ ἡ δυναμένη αὐτὸ ἄλογός of) the (squares) on AB and BC is incommensurable ἐστιν. δύναται δὲ τὸ ΘΔ ἡ ΑΓ· ἄλογος ἄρα ἐστὶν ἡ ΑΓ, with twice the (rectangle contained) by AB and BC, καλείσθω δὲ δύο μέσα δυναμένη. ὅπερ ἔδει δεῖξαι. DF is incommensurable with GH . Hence, DG is also in- commensurable (in length) with GK [Props. 6.1, 10.11]. And they are rational. Thus, DG and GK are rational (straight-lines which are) commensurable in square only. Thus, DK is irrational, and that (straight-line which is) called binomial [Prop. 10.36]. And DE (is) rational. Thus, DH is irrational, and its square-root is irrational [Def. 10.4]. And AC (is) the square-root of HD. Thus, AC is irrational—let it be called the square-root of (the sum of) two medial (areas).† (Which is) the very thing it was required to show. † Thus, the square-root of (the sum of) two medial (areas) has a length expressible as k′1/4 „ q [1 + k/(1 + k2)1/2]/2 + q [1 − k/(1 + k2)1/2]/2 « . This and the corresponding irrational with a minus sign, whose length is expressible as k′1/4 „ q [1 + k/(1 + k2)1/2]/2 − q [1 − k/(1 + k2)1/2]/2 « (see Prop. 10.78), are the positive roots of the quartic x4 − 2 k′1/2 x2 + k′ k2/(1 + k2) = 0.L¨mma. Lemma ῞Οτι δὲ αἱ εἰρημέναι ἄλογοι μοναχῶς διαιροῦνται εἰς τὰς We will now demonstrate that the aforementioned εὐθείας, ἐξ ὧν σύγκεινται ποιουσῶν τὰ προκείμενα εἴδη, irrational (straight-lines) are uniquely divided into the δείξομεν ἤδη προεκθέμενοι λημμάτιον τοιοῦτον· straight-lines of which they are the sum, and which pro- duce the prescribed types, (after) setting forth the follow- ing lemma. ΒΑ ∆ Ε Γ CA D E B ᾿Εκκείσθω εὐθεῖα ἡ ΑΒ καὶ τετμήσθω ἡ ὅλη εἰς ἄνισα Let the straight-line AB be laid out, and let the whole καθ᾿ ἑκάτερον τῶν Γ, Δ, ὑποκείσθω δὲ μείζων ἡ ΑΓ τῆς (straight-line) have been cut into unequal parts at each ΔΒ· λέγω, ὅτι τὰ ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τῶν ἀπὸ of the (points) C and D. And let AC be assumed (to be) τῶν ΑΔ, ΔΒ. greater than DB. I say that (the sum of) the (squares) on Τετμήσθω γὰρ ἡ ΑΒ δίχα κατὰ τὸ Ε. καὶ ἐπεὶ μείζων AC and CB is greater than (the sum of) the (squares) on ἐστὶν ἡ ΑΓ τῆς ΔΒ, κοινὴ ἀφῃρήσθω ἡ ΔΓ· λοιπὴ ἄρα ἡ ΑΔ AD and DB. λοιπῆς τῆς ΓΒ μείζων ἐστίν. ἴση δὲ ἡ ΑΕ τῇ ΕΒ· ἐλάττων For let AB have been cut in half at E. And since AC is ἄρα ἡ ΔΕ τῆς ΕΓ· τὰ Γ, Δ ἄρα σημεῖα οὐκ ἴσον ἀπέχουσι greater than DB, let DC have been subtracted from both. τῆς διχοτομίας. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΓ, ΓΒ μετὰ τοῦ ἀπὸ Thus, the remainder AD is greater than the remainder τῆς ΕΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΒ, ἀλλὰ μὴν καὶ τὸ ὑπὸ τῶν CB. And AE (is) equal to EB. Thus, DE (is) less than ΑΔ, ΔΒ μετὰ τοῦ ἀπὸ ΔΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΒ, τὸ EC. Thus, points C and D are not equally far from the ἄρα ὑπὸ τῶν ΑΓ, ΓΒ μετὰ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τῷ point of bisection. And since the (rectangle contained) ὑπὸ τῶν ΑΔ, ΔΒ μετὰ τοῦ ἀπὸ τῆς ΔΕ· ὧν τὸ ἀπὸ τῆς by AC and CB, plus the (square) on EC, is equal to the ΔΕ ἔλασσόν ἐστι τοῦ ἀπὸ τῆς ΕΓ· καὶ λοιπὸν ἄρα τὸ ὑπὸ (square) on EB [Prop. 2.5], but, moreover, the (rectan- τῶν ΑΓ, ΓΒ ἔλασσόν ἐστι τοῦ ὑπὸ τῶν ΑΔ, ΔΒ. ὥστε καὶ gle contained) by AD and DB, plus the (square) on DE, τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἔλασσόν ἐστι τοῦ δὶς ὑπὸ τῶν ΑΔ, is also equal to the (square) on EB [Prop. 2.5], the (rect- ΔΒ. καὶ λοιπὸν ἄρα τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, angle contained) by AC and CB, plus the (square) on ΓΒ μεῖζόν ἐστι τοῦ συγκειμένου ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ. EC, is thus equal to the (rectangle contained) by AD and ὅπερ ἔδει δεῖξαι. DB, plus the (square) on DE. And, of these, the (square) on DE is less than the (square) on EC. And, thus, the 326 STOIQEIWN iþ. ELEMENTS BOOK 10 remaining (rectangle contained) by AC and CB is less than the (rectangle contained) by AD and DB. And, hence, twice the (rectangle contained) by AC and CB is less than twice the (rectangle contained) by AD and DB. And thus the remaining sum of the (squares) on AC and CB is greater than the sum of the (squares) on AD and DB.† (Which is) the very thing it was required to show. † Since, AC 2 + CB 2 + 2 AC CB = AD 2 + DB 2 + 2 AD DB = AB 2.mbþ. Proposition 42 ῾Η ἐκ δύο ὀνομάτων κατὰ ἓν μόνον σημεῖον διαιρεῖται A binomial (straight-line) can be divided into its (compo- εἰς τὰ ὀνόματα. nent) terms at one point only.† ΓΑ ∆ Β CA D B ῎Εστω ἐκ δύο ὀνομάτων ἡ ΑΒ διῃρημένη εἰς τὰ ὀνόματα Let AB be a binomial (straight-line) which has been κατὰ τὸ Γ· αἱ ΑΓ, ΓΒ ἄρα ῥηταί εἰσι δυνάμει μόνον divided into its (component) terms at C. AC and CB are σύμμετροι. λέγω, ὅτι ἡ ΑΒ κατ᾿ ἄλλο σημεῖον οὐ διαιρεῖται thus rational (straight-lines which are) commensurable εἰς δύο ῥητὰς δυνάμει μόνον συμμέτρους. in square only [Prop. 10.36]. I say that AB cannot be Εἰ γὰρ δυνατόν, διῃρήσθω καὶ κατὰ τὸ Δ, ὥστε καὶ τὰς divided at another point into two rational (straight-lines ΑΔ, ΔΒ ῥητὰς εἶναι δυνάμει μόνον συμμέτρους. φανερὸν which are) commensurable in square only. δή, ὅτι ἡ ΑΓ τῇ ΔΒ οὐκ ἔστιν ἡ αὐτή. εἰ γὰρ δυνατόν, For, if possible, let it also have been divided at D, such ἔστω. ἔσται δὴ καὶ ἡ ΑΔ τῇ ΓΒ ἡ αὐτή· καὶ ἔσται ὡς ἡ ΑΓ that AD and DB are also rational (straight-lines which πρὸς τὴν ΓΒ, οὕτως ἡ ΒΔ πρὸς τὴν ΔΑ, καὶ ἔσται ἡ ΑΒ are) commensurable in square only. So, (it is) clear that κατὰ τὸ αὐτὸ τῇ κατὰ τὸ Γ διαιρέσει διαιρεθεῖσα καὶ κατὰ AC is not the same as DB. For, if possible, let it be (the τὸ Δ· ὅπερ οὐχ ὑπόκειται. οὐκ ἄρα ἡ ΑΓ τῇ ΔΒ ἐστιν ἡ same). So, AD will also be the same as CB. And as αὐτή. διὰ δὴ τοῦτο καὶ τὰ Γ, Δ σημεῖα οὐκ ἴσον ἀπέχουσι AC will be to CB, so BD (will be) to DA. And AB will τῆς διχοτομίας. ᾧ ἄρα διαφέρει τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν (thus) also be divided at D in the same (manner) as the ἀπὸ τῶν ΑΔ, ΔΒ, τούτῳ διαφέρει καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, division at C. The very opposite was assumed. Thus, AC ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ διὰ τὸ καὶ τὰ ἀπὸ τῶν ΑΓ, is not the same as DB. So, on account of this, points ΓΒ μετὰ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ καὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ C and D are not equally far from the point of bisection. μετὰ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ἴσα εἶναι τῷ ἀπὸ τῆς ΑΒ. Thus, by whatever (amount the sum of) the (squares) on ἀλλὰ τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν ἀπὸ τῶν ΑΔ, ΔΒ διαφέρει AC and CB differs from (the sum of) the (squares) on ῥητῷ· ῥητὰ γὰρ ἀμφότερα· καὶ τὸ δὶς ἄρα ὑπὸ τῶν ΑΔ, ΔΒ AD and DB, twice the (rectangle contained) by AD and τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ διαφέρει ῥητῷ μέσα ὄντα· ὅπερ DB also differs from twice the (rectangle contained) by ἄτοπον· μέσον γὰρ μέσου οὐχ ὑπερέχει ῥητῷ. AC and CB by this (same amount)—on account of both Οὐχ ἄρα ἡ ἐκ δύο ὀνομάτων κατ᾿ ἄλλο καὶ ἄλλο σημεῖον (the sum of) the (squares) on AC and CB, plus twice the διαιρεῖται· καθ᾿ ἓν ἄρα μόνον· ὅπερ ἔδει δεῖξαι. (rectangle contained) by AC and CB, and (the sum of) the (squares) on AD and DB, plus twice the (rectangle contained) by AD and DB, being equal to the (square) on AB [Prop. 2.4]. But, (the sum of) the (squares) on AC and CB differs from (the sum of) the (squares) on AD and DB by a rational (area). For (they are) both rational (areas). Thus, twice the (rectangle contained) by AD and DB also differs from twice the (rectangle contained) by AC and CB by a rational (area, despite both) being medial (areas) [Prop. 10.21]. The very thing is absurd. For a medial (area) cannot exceed a medial (area) by a rational (area) [Prop. 10.26]. 327 STOIQEIWN iþ. ELEMENTS BOOK 10 Thus, a binomial (straight-line) cannot be divided (into its component terms) at different points. Thus, (it can be so divided) at one point only. (Which is) the very thing it was required to show. † In other words, k + k′1/2 = k′′ + k′′′1/2 has only one solution: i.e., k′′ = k and k′′′ = k′. Likewise, k1/2 + k′1/2 = k′′1/2 + k′′′1/2 has only one solution: i.e., k′′ = k and k′′′ = k′ (or, equivalently, k′′ = k′ and k′′′ = k).mgþ. Proposition 43 ῾Η ἐκ δύο μέσων πρώτη καθ᾿ ἓν μόνον σημεῖον διαιρεῖται. A first bimedial (straight-line) can be divided (into its component terms) at one point only.† ΓΑ ∆ Β CA D B ῎Εστω ἐκ δύο μέσων πρώτη ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, Let AB be a first bimedial (straight-line) which has ὥστε τὰς ΑΓ, ΓΒ μέσας εἶναι δυνάμει μόνον συμμέτρους been divided at C, such that AC and CB are medial ῥητὸν περιεχούσας· λέγω, ὅτι ἡ ΑΒ κατ᾿ ἄλλο σημεῖον οὐ (straight-lines), commensurable in square only, (and) διαιρεῖται. containing a rational (area) [Prop. 10.37]. I say that AB Εἰ γὰρ δυνατόν διῃρήσθω καὶ κατὰ τὸ Δ, ὥστε καὶ τὰς cannot be (so) divided at another point. ΑΔ, ΔΒ μέσας εἶναι δυνάμει μόνον συμμέτρους ῥητὸν πε- For, if possible, let it also have been divided at D, ριεχούσας. ἐπεὶ οὖν, ᾧ διαφέρει τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ such that AD and DB are also medial (straight-lines), τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, τούτῳ διαφέρει τὰ ἀπὸ τῶν ΑΓ, commensurable in square only, (and) containing a ratio- ΓΒ τῶν ἀπὸ τῶν ΑΔ, ΔΒ, ῥητῷ δὲ διαφέρει τὸ δὶς ὑπὸ nal (area). Since, therefore, by whatever (amount) twice τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ· ῥητὰ γὰρ ἀμφότερα· the (rectangle contained) by AD and DB differs from ῥητῷ ἄρα διαφέρει καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν ἀπὸ τῶν twice the (rectangle contained) by AC and CB, (the sum ΑΔ, ΔΒ μέσα ὄντα· ὅπερ ἄτοπον. of) the (squares) on AC and CB differs from (the sum Οὐκ ἄρα ἡ ἐκ δύο μέσων πρώτη κατ᾿ ἄλλο καὶ ἄλλο of) the (squares) on AD and DB by this (same amount) σημεῖον διαιρεῖται εἰς τὰ ὀνόματα· καθ᾿ ἓν ἄρα μόνον· ὅπερ [Prop. 10.41 lem.]. And twice the (rectangle contained) ἔδει δεῖξαι. by AD and DB differs from twice the (rectangle con- tained) by AC and CB by a rational (area). For (they are) both rational (areas). (The sum of) the (squares) on AC and CB thus differs from (the sum of) the (squares) on AD and DB by a rational (area, despite both) being medial (areas). The very thing is absurd [Prop. 10.26]. Thus, a first bimedial (straight-line) cannot be divided into its (component) terms at different points. Thus, (it can be so divided) at one point only. (Which is) the very thing it was required to show. † In other words, k1/4 + k3/4 = k′1/4 + k′3/4 has only one solution: i.e., k′ = k.mdþ. Proposition 44 ῾Η ἐκ δύο μέσων δευτέρα καθ᾿ ἓν μόνον σημεῖον A second bimedial (straight-line) can be divided (into διαιρεῖται. its component terms) at one point only.† ῎Εστω ἐκ δύο μέσων δευτέρα ἡ ΑΒ διῃρημένη κατὰ τὸ Let AB be a second bimedial (straight-line) which Γ, ὥστε τὰς ΑΓ, ΓΒ μέσας εἶναι δυνάμει μόνον συμμέτρους has been divided at C, so that AC and BC are medial μέσον περιεχούσας· φανερὸν δή, ὅτι τὸ Γ οὐκ ἔστι κατὰ (straight-lines), commensurable in square only, (and) τῆς διχοτομίας, ὅτι οὐκ εἰσὶ μήκει σύμμετροι. λέγω, ὅτι ἡ containing a medial (area) [Prop. 10.38]. So, (it is) clear ΑΒ κατ᾿ ἄλλο σημεῖον οὐ διαιρεῖται. that C is not (located) at the point of bisection, since (AC and BC) are not commensurable in length. I say that AB cannot be (so) divided at another point. 328 STOIQEIWN iþ. ELEMENTS BOOK 10 Ζ Α ∆ ΒΓ Ε Μ Ν Λ Η Θ Κ G A D B E N K C F H L M Εἰ γὰρ δυνατόν, διῃρήσθω καὶ κατὰ τὸ Δ, ὥστε τὴν For, if possible, let it also have been (so) divided at ΑΓ τῇ ΔΒ μὴ εἶναι τὴν αὐτήν, ἀλλὰ μείζονα καθ᾿ ὑπόθεσιν D, so that AC is not the same as DB, but AC (is), τὴν ΑΓ· δῆλον δή, ὅτι καὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ, ὡς ἐπάνω by hypothesis, greater. So, (it is) clear that (the sum ἐδείξαμεν, ἐλάσσονα τῶν ἀπὸ τῶν ΑΓ, ΓΒ· καὶ τὰς ΑΔ, ΔΒ of) the (squares) on AD and DB is also less than (the μέσας εἶναι δυνάμει μόνον συμμέτρους μέσον περιεχούσας. sum of) the (squares) on AC and CB, as we showed καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ τῷ μὲν ἀπὸ τῆς ΑΒ ἴσον παρὰ above [Prop. 10.41 lem.]. And AD and DB are medial τὴν ΕΖ παραλληλόγραμμον ὀρθογώνιον παραβεβλήσθω τὸ (straight-lines), commensurable in square only, (and) ΕΚ, τοῖς δὲ ἀπὸ τῶν ΑΓ, ΓΒ ἴσον ἀφῃρήσθω τὸ ΕΗ· λοιπὸν containing a medial (area). And let the rational (straight- ἄρα τὸ ΘΚ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. πάλιν δὴ τοῖς line) EF be laid down. And let the rectangular paral- ἀπὸ τῶν ΑΔ, ΔΒ, ἅπερ ἐλάσσονα ἐδείχθη τῶν ἀπὸ τῶν lelogram EK, equal to the (square) on AB, have been ΑΓ, ΓΒ, ἴσον ἀφῃρήσθω τὸ ΕΛ· καὶ λοιπὸν ἄρα τὸ ΜΚ applied to EF . And let EG, equal to (the sum of) the ἴσον τῷ δὶς ὑπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ μέσα ἐστὶ τὰ ἀπὸ (squares) on AC and CB, have been cut off (from EK). τῶν ΑΓ, ΓΒ, μέσον ἄρα [καὶ] τὸ ΕΗ. καὶ παρὰ ῥητὴν τὴν Thus, the remainder, HK, is equal to twice the (rectan- ΕΖ παράκειται· ῥητὴ ἄρα ἐστὶν ἡ ΕΘ καὶ ἀσύμμετρος τῇ ΕΖ gle contained) by AC and CB [Prop. 2.4]. So, again, μήκει. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΘΝ ῥητή ἐστι καὶ ἀσύμμετρος let EL, equal to (the sum of) the (squares) on AD and τῇ ΕΖ μήκει. καὶ ἐπεὶ αἱ ΑΓ, ΓΒ μέσαι εἰσὶ δυνάμει μόνον DB—which was shown (to be) less than (the sum of) the σύμμετροι, ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΓ τῇ ΓΒ μήκει. ὡς δὲ (squares) on AC and CB—have been cut off (from EK). ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως τὸ ἀπὸ τῆς ΑΓ πρὸς τὸ ὑπὸ τῶν And, thus, the remainder, MK, (is) equal to twice the ΑΓ, ΓΒ· ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΓ τῷ ὑπὸ τῶν (rectangle contained) by AD and DB. And since (the ΑΓ, ΓΒ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΓ σύμμετρά ἐστι τὰ ἀπὸ τῶν sum of) the (squares) on AC and CB is medial, EG ΑΓ, ΓΒ· δυνάμει γάρ εἰσι σύμμετροι αἱ ΑΓ, ΓΒ. τῷ δὲ ὑπὸ (is) thus [also] medial. And it is applied to the ratio- τῶν ΑΓ, ΓΒ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. καὶ τὰ nal (straight-line) EF . Thus, EH is rational, and incom- ἀπὸ τῶν ΑΓ, ΓΒ ἄρα ἀσύμμετρά ἐστι τῷ δὶς ὑπὸ τῶν ΑΓ, mensurable in length with EF [Prop. 10.22]. So, for the ΓΒ. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΑΓ, ΓΒ ἴσον ἐστὶ τὸ ΕΗ, τῷ δὲ same (reasons), HN is also rational, and incommensu- δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ ΘΚ· ἀσύμμετρον ἄρα ἐστὶ τὸ rable in length with EF . And since AC and CB are me- ΕΗ τῷ ΘΚ· ὥστε καὶ ἡ ΕΘ τῇ ΘΝ ἀσύμμετρός ἐστι μήκει. dial (straight-lines which are) commensurable in square καί εἰσι ῥηταί· αἱ ΕΘ, ΘΝ ἄρα ῥηταί εἰσι δυνάμει μόνον only, AC is thus incommensurable in length with CB. σύμμετροι. ἐὰν δὲ δύο ῥηταὶ δυνάμει μόνον σύμμετροι συν- And as AC (is) to CB, so the (square) on AC (is) to the τεθῶσιν, ἡ ὅλη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο ὀνομάτων· (rectangle contained) by AC and CB [Prop. 10.21 lem.]. ἡ ΕΝ ἄρα ἐκ δύο ὀνομάτων ἐστὶ διῃρημένη κατὰ τὸ Θ. κατὰ Thus, the (square) on AC is incommensurable with the τὰ αὐτὰ δὴ δειχθήσονται καὶ αἱ ΕΜ, ΜΝ ῥηταὶ δυνάμει (rectangle contained) by AC and CB [Prop. 10.11]. But, μόνον σύμμετροι· καὶ ἔσται ἡ ΕΝ ἐκ δύο ὀνομάτων κατ᾿ (the sum of) the (squares) on AC and CB is commensu- ἄλλο καὶ ἄλλο διῃρημένη τό τε Θ καὶ τὸ Μ, καὶ οὐκ ἔστιν rable with the (square) on AC. For, AC and CB are com- ἡ ΕΘ τῇ ΜΝ ἡ αὐτή, ὅτι τὰ ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι mensurable in square [Prop. 10.15]. And twice the (rect- τῶν ἀπὸ τῶν ΑΔ, ΔΒ. ἀλλὰ τὰ ἀπὸ τῶν ΑΔ, ΔΒ μείζονά angle contained) by AC and CB is commensurable with ἐστι τοῦ δὶς ὑπὸ ΑΔ, ΔΒ· πολλῷ ἄρα καὶ τὰ ἀπὸ τῶν ΑΓ, the (rectangle contained) by AC and CB [Prop. 10.6]. ΓΒ, τουτέστι τὸ ΕΗ, μεῖζόν ἐστι τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ, And thus (the sum of) the squares on AC and CB is in- τουτέστι τοῦ ΜΚ· ὥστε καὶ ἡ ΕΘ τῆς ΜΝ μείζων ἐστίν. ἡ commensurable with twice the (rectangle contained) by ἄρα ΕΘ τῇ ΜΝ οὐκ ἔστιν ἡ αὐτή· ὅπερ ἔδει δεῖξαι. AC and CB [Prop. 10.13]. But, EG is equal to (the sum of) the (squares) on AC and CB, and HK equal to twice the (rectangle contained) by AC and CB. Thus, EG is incommensurable with HK. Hence, EH is also incom- 329 STOIQEIWN iþ. ELEMENTS BOOK 10 mensurable in length with HN [Props. 6.1, 10.11]. And (they are) rational (straight-lines). Thus, EH and HN are rational (straight-lines which are) commensurable in square only. And if two rational (straight-lines which are) commensurable in square only are added together then the whole (straight-line) is that irrational called bi- nomial [Prop. 10.36]. Thus, EN is a binomial (straight- line) which has been divided (into its component terms) at H . So, according to the same (reasoning), EM and MN can be shown (to be) rational (straight-lines which are) commensurable in square only. And EN will (thus) be a binomial (straight-line) which has been divided (into its component terms) at the different (points) H and M (which is absurd [Prop. 10.42]). And EH is not the same as MN , since (the sum of) the (squares) on AC and CB is greater than (the sum of) the (squares) on AD and DB. But, (the sum of) the (squares) on AD and DB is greater than twice the (rectangle contained) by AD and DB [Prop. 10.59 lem.]. Thus, (the sum of) the (squares) on AC and CB—that is to say, EG—is also much greater than twice the (rectangle contained) by AD and DB— that is to say, MK. Hence, EH is also greater than MN [Prop. 6.1]. Thus, EH is not the same as MN . (Which is) the very thing it was required to show. † In other words, k1/4 + k′1/2/k1/4 = k′′1/4 + k′′′1/2/k′′1/4 has only one solution: i.e., k′′ = k and k′′′ = k′.meþ. Proposition 45 ῾Η μείζων κατὰ τὸ αὐτὸ μόνον σημεῖον διαιρεῖται. A major (straight-line) can only be divided (into its component terms) at the same point.† ΓΑ ∆ Β CA D B ῎Εστω μείζων ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε τὰς ΑΓ, Let AB be a major (straight-line) which has been di- ΓΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τὸ μὲν συγκείμενον vided at C, so that AC and CB are incommensurable in ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνων ῥητόν, τὸ δ᾿ ὑπὸ square, making the sum of the squares on AC and CB τῶν ΑΓ, ΓΒ μέσον· λέγω, ὅτι ἡ ΑΒ κατ᾿ ἄλλο σημεῖον rational, and the (rectangle contained) by AC and CD οὐ διαιρεῖται. medial [Prop. 10.39]. I say that AB cannot be (so) di- Εἰ γὰρ δυνατόν, διῃρήσθω καὶ κατὰ τὸ Δ, ὥστε καὶ vided at another point. τὰς ΑΔ, ΔΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τὸ μὲν For, if possible, let it also have been divided at D, such συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ ῥητόν, τὸ δ᾿ ὑπ᾿ that AD and DB are also incommensurable in square, αὐτῶν μέσον. καὶ ἐπεί, ᾧ διαφέρει τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν making the sum of the (squares) on AD and DB ratio- ἀπὸ τῶν ΑΔ, ΔΒ, τούτῳ διαφέρει καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, nal, and the (rectangle contained) by them medial. And ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, ἀλλὰ τὰ ἀπὸ τῶν ΑΓ, ΓΒ since, by whatever (amount the sum of) the (squares) on τῶν ἀπὸ τῶν ΑΔ, ΔΒ ὑπερέχει ῥητῷ· ῥητὰ γὰρ ἀμφότερα· AC and CB differs from (the sum of) the (squares) on καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ἄρα τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ AD and DB, twice the (rectangle contained) by AD and ὑπερέχει ῥητῷ μέσα ὄντα· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ DB also differs from twice the (rectangle contained) by μείζων κατ᾿ ἄλλο καὶ ἄλλο σημεῖον διαιρεῖται· κατὰ τὸ αὐτὸ AC and CB by this (same amount). But, (the sum of) ἄρα μόνον διαιρεῖται· ὅπερ ἔδει δεῖξαι. the (squares) on AC and CB exceeds (the sum of) the (squares) on AD and DB by a rational (area). For (they are) both rational (areas). Thus, twice the (rectangle 330 STOIQEIWN iþ. ELEMENTS BOOK 10 contained) by AD and DB also exceeds twice the (rect- angle contained) by AC and CB by a rational (area), (despite both) being medial (areas). The very thing is impossible [Prop. 10.26]. Thus, a major (straight-line) cannot be divided (into its component terms) at differ- ent points. Thus, it can only be (so) divided at the same (point). (Which is) the very thing it was required to show. † In other words, q [1 + k/(1 + k2)1/2]/2 + q [1 − k/(1 + k2)1/2]/2 = q [1 + k′/(1 + k′2)1/2]/2 + q [1 − k′/(1 + k′2)1/2]/2 has only one solution: i.e., k′ = k. m�þ. Proposition 46 ῾Η ῥητὸν καὶ μέσον δυναμένη καθ᾿ ἓν μόνον σημεῖον The square-root of a rational plus a medial (area) can be διαιρεῖται. divided (into its component terms) at one point only.† ΓΑ ∆ Β CA D B ῎Εστω ῥητὸν καὶ μέσον δυναμένη ἡ ΑΒ διῃρημένη κατὰ Let AB be the square-root of a rational plus a medial τὸ Γ, ὥστε τὰς ΑΓ, ΓΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας (area) which has been divided at C, so that AC and CB τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον, τὸ δὲ are incommensurable in square, making the sum of the δὶς ὑπὸ τῶν ΑΓ, ΓΒ ῥητόν· λέγω, ὅτι ἡ ΑΒ κατ᾿ ἄλλο (squares) on AC and CB medial, and twice the (rectan- σημεῖον οὐ διαιρεῖται. gle contained) by AC and CB rational [Prop. 10.40]. I Εἰ γὰρ δυνατόν, διῃρήσθω καὶ κατὰ τὸ Δ, ὥστε καὶ say that AB cannot be (so) divided at another point. τὰς ΑΔ, ΔΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τὸ μὲν For, if possible, let it also have been divided at D, so συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ μέσον, τὸ δὲ δὶς that AD and DB are also incommensurable in square, ὑπὸ τῶν ΑΔ, ΔΒ ῥητόν. ἐπεὶ οὖν, ᾧ διαφέρει τὸ δὶς ὑπὸ making the sum of the (squares) on AD and DB medial, τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ, τούτῳ διαφέρει καὶ and twice the (rectangle contained) by AD and DB ra- τὰ ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, τὸ δὲ δὶς ὑπὸ tional. Therefore, since by whatever (amount) twice the τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ὑπερέχει ῥητῷ, καὶ (rectangle contained) by AC and CB differs from twice τὰ ἀπὸ τῶν ΑΔ, ΔΒ ἄρα τῶν ἀπὸ τῶν ΑΓ, ΓΒ ὑπερέχει the (rectangle contained) by AD and DB, (the sum of) ῥητῷ μέσα ὄντα· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ῥητὸν the (squares) on AD and DB also differs from (the sum καὶ μέσον δυναμένη κατ᾿ ἄλλο καὶ ἄλλο σημεῖον διαιρεῖται. of) the (squares) on AC and CB by this (same amount). κατὰ ἓν ἄρα σημεῖον διαιρεῖται· ὅπερ ἔδει δεῖξαι. And twice the (rectangle contained) by AC and CB ex- ceeds twice the (rectangle contained) by AD and DB by a rational (area). (The sum of) the (squares) on AD and DB thus also exceeds (the sum of) the (squares) on AC and CB by a rational (area), (despite both) being medial (areas). The very thing is impossible [Prop. 10.26]. Thus, the square-root of a rational plus a medial (area) cannot be divided (into its component terms) at different points. Thus, it can be (so) divided at one point (only). (Which is) the very thing it was required to show. † In other words, q [(1 + k2)1/2 + k]/[2 (1 + k2)] + q [(1 + k2)1/2 − k]/[2 (1 + k2)] = q [(1 + k′2)1/2 + k′]/[2 (1 + k′2)] + q [(1 + k′2)1/2 − k′]/[2 (1 + k′2)] has only one solution: i.e., k′ = k.mzþ. Proposition 47 ῾Η δύο μέσα δυναμένη καθ᾿ ἓν μόνον σημεῖον διαιρεῖται. The square-root of (the sum of) two medial (areas) can be divided (into its component terms) at one point only.† 331 STOIQEIWN iþ. ELEMENTS BOOK 10 Θ Α ∆ ΒΓ Ζ Λ Η Κ Μ Ε Ν E A D B L K M C G F N H ῎Εστω [δύο μέσα δυναμένη] ἡ ΑΒ διῃρημένη κατὰ τὸ Let AB be [the square-root of (the sum of) two me- Γ, ὥστε τὰς ΑΓ, ΓΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας dial (areas)] which has been divided at C, such that AC τό τε συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον καὶ τὸ and CB are incommensurable in square, making the sum ὑπὸ τῶν ΑΓ, ΓΒ μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ of the (squares) on AC and CB medial, and the (rect- ἐκ τῶν ἀπ᾿ αὐτῶν. λέγω, ὅτι ἡ ΑΒ κατ᾿ ἄλλο σημεῖον οὐ angle contained) by AC and CB medial, and, moreover, διαιρεῖται ποιοῦσα τὰ προκείμενα. incommensurable with the sum of the (squares) on (AC Εἰ γὰρ δυνατόν, διῃρήσθω κατὰ τὸ Δ, ὥστε πάλιν δη- and CB) [Prop. 10.41]. I say that AB cannot be divided λονότι τὴν ΑΓ τῇ ΔΒ μὴ εἶναι τὴν αὐτήν, ἀλλὰ μείζονα at another point fulfilling the prescribed (conditions). καθ᾿ ὑπόθεσιν τὴν ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ παρα- For, if possible, let it have been divided at D, such that βεβλήσθω παρὰ τὴν ΕΖ τοῖς μὲν ἀπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ AC is again manifestly not the same as DB, but AC (is), ΕΗ, τῷ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ ΘΚ· ὅλον ἄρα τὸ by hypothesis, greater. And let the rational (straight-line) ΕΚ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. πάλιν δὴ παρα- EF be laid down. And let EG, equal to (the sum of) the βεβλήσθω παρὰ τὴν ΕΖ τοῖς ἀπὸ τῶν ΑΔ, ΔΒ ἴσον τὸ ΕΛ· (squares) on AC and CB, and HK, equal to twice the λοιπὸν ἄρα τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ λοιπῷ τῷ ΜΚ ἴσον (rectangle contained) by AC and CB, have been applied ἐστίν. καὶ ἐπεὶ μέσον ὑπόκειται τὸ συγκείμενον ἐκ τῶν ἀπὸ to EF . Thus, the whole of EK is equal to the square on τῶν ΑΓ, ΓΒ, μέσον ἄρα ἐστὶ καὶ τὸ ΕΗ. καὶ παρὰ ῥητὴν τὴν AB [Prop. 2.4]. So, again, let EL, equal to (the sum of) ΕΖ παράκειται· ῥητὴ ἄρα ἐστὶν ἡ ΘΕ καὶ ἀσύμμετρος τῇ ΕΖ the (squares) on AD and DB, have been applied to EF . μήκει. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΘΝ ῥητή ἐστι καὶ ἀσύμμετρος Thus, the remainder—twice the (rectangle contained) by τῇ ΕΖ μήκει. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ συγκείμενον AD and DB—is equal to the remainder, MK. And since ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, καὶ τὸ the sum of the (squares) on AC and CB was assumed ΕΗ ἄρα τῷ ΗΝ ἀσύμμετρόν ἐστιν· ὥστε καὶ ἡ ΕΘ τῇ ΘΝ (to be) medial, EG is also medial. And it is applied to ἀσύμμετρός ἐστιν. καί εἰσι ῥηταί· αἱ ΕΘ, ΘΝ ἄρα ῥηταί εἰσι the rational (straight-line) EF . HE is thus rational, and δυνάμει μόνον σύμμετροι· ἡ ΕΝ ἄρα ἐκ δύο ὀνομάτων ἐστὶ incommensurable in length with EF [Prop. 10.22]. So, διῃρημένη κατὰ τὸ Θ. ὁμοίως δὴ δείξομεν, ὅτι καὶ κατὰ τὸ for the same (reasons), HN is also rational, and incom- Μ διῄρηται. καὶ οὐκ ἔστιν ἡ ΕΘ τῇ ΜΝ ἡ αὐτή· ἡ ἄρα ἐκ δύο mensurable in length with EF . And since the sum of ὀνομάτων κατ᾿ ἄλλο καὶ ἄλλο σημεῖον διῄρηται· ὅπερ ἐστίν the (squares) on AC and CB is incommensurable with ἄτοπον. οὐκ ἄρα ἡ δύο μέσα δυναμένη κατ᾿ ἀλλο καὶ ἄλλο twice the (rectangle contained) by AC and CB, EG is σημεῖον διαιρεῖται· καθ᾿ ἓν ἄρα μόνον [σημεῖον] διαιρεῖται. thus also incommensurable with GN . Hence, EH is also incommensurable with HN [Props. 6.1, 10.11]. And they are (both) rational (straight-lines). Thus, EH and HN are rational (straight-lines which are) commensu- rable in square only. Thus, EN is a binomial (straight- line) which has been divided (into its component terms) at H [Prop. 10.36]. So, similarly, we can show that it has also been (so) divided at M . And EH is not the same as MN . Thus, a binomial (straight-line) has been divided (into its component terms) at different points. The very thing is absurd [Prop. 10.42]. Thus, the square-root of (the sum of) two medial (areas) cannot be divided (into 332 STOIQEIWN iþ. ELEMENTS BOOK 10 its component terms) at different points. Thus, it can be (so) divided at one [point] only. † In other words, k′1/4 q [1 + k/(1 + k2)1/2]/2 + k′1/4 q [1 − k/(1 + k2)1/2]/2 = k′′′1/4 q [1 + k′′/(1 + k′′2)1/2]/2 +k′′′1/4 q [1 − k′′/(1 + k′′2)1/2]/2 has only one solution: i.e., k′′ = k and k′′′ = k′.VOroi deÔteroi. Definitions II εʹ. ῾Υποκειμένης ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων 5. Given a rational (straight-line), and a binomial διῃρημένης εἰς τὰ ὀνόματα, ἧς τὸ μεῖζον ὄνομα τοῦ (straight-line) which has been divided into its (compo- ἐλάσσονος μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει, nent) terms, of which the square on the greater term is ἐὰν μὲν τὸ μεῖζον ὄνομα σύμμετρον ᾖ μήκει τῇ ἐκκειμένῃ larger than (the square on) the lesser by the (square) ῥητῇ, καλείσθω [ἡ ὅλη] ἐκ δύο ὀνομάτων πρώτη. on (some straight-line) commensurable in length with ϛʹ. ᾿Εὰν δὲ τὸ ἐλάσσον ὄνομα σύμμετρον ᾖ μήκει τῇ (the greater) then, if the greater term is commensurable ἐκκειμένῃ ῥητῇ, καλείσθω ἐκ δύο ὀνομάτων δευτέρα. in length with the rational (straight-line previously) laid ζʹ. ᾿Εὰν δὲ μηδέτερον τῶν ὀνομάτων σύμμετρον ᾖ μήκει out, let [the whole] (straight-line) be called a first bino- τῇ ἐκκειμένῃ ῥητῇ, καλείσθω ἐκ δύο ὀνομάτων τρίτη. mial (straight-line). ηʹ.Πάλιν δὴ ἐὰν τὸ μεῖζον ὄνομα [τοῦ ἐλάσσονος] μεῖζον 6. And if the lesser term is commensurable in length δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει, ἐὰν μὲν τὸ μεῖζον with the rational (straight-line previously) laid out then ὄνομα σύμμετρον ᾖ μήκει τῇ ἐκκειμένῃ ῥητῇ, καλείσθω ἐκ let (the whole straight-line) be called a second binomial δύο ὀνομάτων τετάρτη. (straight-line). θʹ. ᾿Εὰν δὲ τὸ ἔλασσον, πέμπτη. 7. And if neither of the terms is commensurable in ιʹ. ᾿Εὰν δὲ μηδέτερον, ἕκτη. length with the rational (straight-line previously) laid out then let (the whole straight-line) be called a third bino- mial (straight-line). 8. So, again, if the square on the greater term is larger than (the square on) [the lesser] by the (square) on (some straight-line) incommensurable in length with (the greater) then, if the greater term is commensurable in length with the rational (straight-line previously) laid out, let (the whole straight-line) be called a fourth bino- mial (straight-line). 9. And if the lesser (term is commensurable), a fifth (binomial straight-line). 10. And if neither (term is commensurable), a sixth (binomial straight-line).mhþ. Proposition 48 Εὑρεῖν τὴν ἐκ δύο ὀνομάτων πρώτην. To find a first binomial (straight-line). ᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν Let two numbers AC and CB be laid down such that συγκείμενον ἐξ αὐτῶν τὸν ΑΒ πρὸς μὲν τὸν ΒΓ λόγον their sum AB has to BC the ratio which (some) square ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, number (has) to (some) square number, and does not πρὸς δὲ τὸν ΓΑ λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς have to CA the ratio which (some) square number (has) πρὸς τετράγωνον ἀριθμόν, καὶ ἐκκείσθω τις ῥητὴ ἡ Δ, καὶ to (some) square number [Prop. 10.28 lem. I]. And let τῇ Δ σύμμετρος ἔστω μήκει ἡ ΕΖ. ῥητὴ ἄρα ἐστὶ καὶ ἡ some rational (straight-line) D be laid down. And let EF ΕΖ. καὶ γεγονέτω ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως be commensurable in length with D. EF is thus also ra- τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ. ὁ δὲ ΑΒ πρὸς τὸν tional [Def. 10.3]. And let it have been contrived that as ΑΓ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν· καὶ τὸ ἀπὸ τῆς the number BA (is) to AC, so the (square) on EF (is) ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν ἀριθμὸς πρὸς to the (square) on FG [Prop. 10.6 corr.]. And AB has to ἀριθμόν· ὥστε σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΕΖ τῷ ἀπὸ τῆς AC the ratio which (some) number (has) to (some) num- 333 STOIQEIWN iþ. ELEMENTS BOOK 10 ΖΗ. καὶ ἐστι ῥητὴ ἡ ΕΖ· ῥητὴ ἄρα καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ ber. Thus, the (square) on EF also has to the (square) ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς on FG the ratio which (some) number (has) to (some) πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς number. Hence, the (square) on EF is commensurable τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς with the (square) on FG [Prop. 10.6]. And EF is ra- τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ ΖΗ tional. Thus, FG (is) also rational. And since BA does μήκει. αἱ ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· not have to AC the ratio which (some) square number ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΕΗ. λέγω, ὅτι καὶ πρώτη. (has) to (some) square number, thus the (square) on EF does not have to the (square) on FG the ratio which (some) square number (has) to (some) square number either. Thus, EF is incommensurable in length with FG [Prop 10.9]. EF and FG are thus rational (straight-lines which are) commensurable in square only. Thus, EG is a binomial (straight-line) [Prop. 10.36]. I say that (it is) also a first (binomial straight-line). ∆ Ε Ζ Α Γ Β Η Θ H E A B F G C D ᾿Επεὶ γάρ ἐστιν ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ For since as the number BA is to AC, so the (square) ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ, μείζων δὲ ὁ ΒΑ τοῦ ΑΓ, on EF (is) to the (square) on FG, and BA (is) greater μεῖζον ἄρα καὶ τὸ ἀπὸ τῆς ΕΖ τοῦ ἀπὸ τῆς ΖΗ. ἔστω οὖν τῷ than AC, the (square) on EF (is) thus also greater than ἀπὸ τῆς ΕΖ ἴσα τὰ ἀπὸ τῶν ΖΗ, Θ. καὶ ἐπεί ἐστιν ὡς ὁ ΒΑ the (square) on FG [Prop. 5.14]. Therefore, let (the sum πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ, of) the (squares) on FG and H be equal to the (square) ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΑΒ πρὸς τὸν ΒΓ, οὕτως τὸ on EF . And since as BA is to AC, so the (square) ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον on EF (is) to the (square) on FG, thus, via conver- ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καὶ sion, as AB is to BC, so the (square) on EF (is) to the τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν (square) on H [Prop. 5.19 corr.]. And AB has to BC τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. σύμμετρος the ratio which (some) square number (has) to (some) ἄρα ἐστὶν ἡ ΕΖ τῇ Θ μήκει· ἡ ΕΖ ἄρα τῆς ΖΗ μεῖζον δύναται square number. Thus, the (square) on EF also has to τῷ ἀπὸ συμμέτρου ἑαυτῇ. καί εἰσι ῥηταὶ αἱ ΕΖ, ΖΗ, καὶ the (square) on H the ratio which (some) square number σύμμετρος ἡ ΕΖ τῇ Δ μήκει. (has) to (some) square number. Thus, EF is commensu- ᾿Η ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πρώτη· ὅπερ ἔδει rable in length with H [Prop. 10.9]. Thus, the square on δεῖξαι. EF is greater than (the square on) FG by the (square) on (some straight-line) commensurable (in length) with (EF ). And EF and FG are rational (straight-lines). And EF (is) commensurable in length with D. Thus, EG is a first binomial (straight-line) [Def. 10.5].† (Which is) the very thing it was required to show. †If the rational straight-line has unit length then the length of a first binomial straight-line is k + k √ 1 − k′ 2. This, and the first apotome, whose length is k − k √ 1 − k′ 2 [Prop. 10.85], are the roots of x2 − 2 k x + k2 k′ 2 = 0.mjþ. Proposition 49 Εὑρεῖν τὴν ἐκ δύο ὀνομάτων δευτέραν. To find a second binomial (straight-line). 334 STOIQEIWN iþ. ELEMENTS BOOK 10 ∆ Ε Α Β Η Γ Ζ Θ H E A B D F G C ᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν Let the two numbers AC and CB be laid down such συγκείμενον ἐξ αὐτῶν τὸν ΑΒ πρὸς μὲν τὸν ΒΓ λόγον that their sum AB has to BC the ratio which (some) ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, square number (has) to (some) square number, and does πρὸς δὲ τὸν ΑΓ λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς not have to AC the ratio which (some) square number πρὸς τετράγωνον ἀριθμόν, καὶ ἐκκείσθω ῥητὴ ἡ Δ, καὶ τῇ Δ (has) to (some) square number [Prop. 10.28 lem. I]. And σύμμετρος ἔστω ἡ ΕΖ μήκει· ῥητὴ ἄρα ἐστὶν ἡ ΕΖ. γεγονέτω let the rational (straight-line) D be laid down. And let δὴ καὶ ὡς ὁ ΓΑ ἀριθμὸς πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς ΕΖ EF be commensurable in length with D. EF is thus a πρὸς τὸ ἀπὸ τῆς ΖΗ· σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΕΖ τῷ rational (straight-line). So, let it also have been contrived ἀπὸ τῆς ΖΗ. ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ ΓΑ ἀριθμὸς that as the number CA (is) to AB, so the (square) on EF πρὸς τὸν ΑΒ λὸγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς (is) to the (square) on FG [Prop. 10.6 corr.]. Thus, the τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς (square) on EF is commensurable with the (square) on ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον FG [Prop. 10.6]. Thus, FG is also a rational (straight- ἀριθμόν. ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ ΖΗ μήκει· αἱ ΕΖ, line). And since the number CA does not have to AB ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα the ratio which (some) square number (has) to (some) ὀνομάτων ἐστὶν ἡ ΕΗ. δεικτέον δή, ὅτι καὶ δευτέρα. square number, the (square) on EF does not have to the ᾿Επεὶ γὰρ ἀνάπαλίν ἐστιν ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, (square) on FG the ratio which (some) square number οὕτως τὸ ἀπὸ τῆς ΗΖ πρὸς τὸ ἀπὸ τῆς ΖΕ, μείζων δὲ ὁ ΒΑ (has) to (some) square number either. Thus, EF is in- τοῦ ΑΓ, μεῖζον ἄρα [καὶ] τὸ ἀπὸ τῆς ΗΖ τοῦ ἀπὸ τῆς ΖΕ. commensurable in length with FG [Prop. 10.9]. EF and ἔστω τῷ ἀπὸ τῆς ΗΖ ἴσα τὰ ἀπὸ τῶν ΕΖ, Θ· ἀναστρέψαντι FG are thus rational (straight-lines which are) commen- ἄρα ἐστὶν ὡς ὁ ΑΒ πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ surable in square only. Thus, EG is a binomial (straight- πρὸς τὸ ἀπὸ τῆς Θ. ἀλλ᾿ ὁ ΑΒ πρὸς τὸν ΒΓ λόγον ἔχει, ὃν line) [Prop. 10.36]. So, we must show that (it is) also a τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· καὶ τὸ ἀπὸ second (binomial straight-line). τῆς ΖΗ ἄρα πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος For since, inversely, as the number BA is to AC, so ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. σύμμετρος ἄρα ἐστὶν ἡ the (square) on GF (is) to the (square) on FE [Prop. 5.7 ΖΗ τῇ Θ μήκει· ὥστε ἡ ΖΗ τῆς ΖΕ μεῖζον δύναται τῷ ἀπὸ corr.], and BA (is) greater than AC, the (square) on συμμέτρου ἑαυτῇ. καί εἰσι ῥηταὶ αἱ ΖΗ, ΖΕ δυνάμει μόνον GF (is) thus [also] greater than the (square) on FE σύμμετροι, καὶ τὸ ΕΖ ἔλασσον ὄνομα τῇ ἐκκειμένῃ ῥητῇ [Prop. 5.14]. Let (the sum of) the (squares) on EF and σύμμετρόν ἐστι τῇ Δ μήκει. H be equal to the (square) on GF . Thus, via conver- ῾Η ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ δευτέρα· ὅπερ ἔδει sion, as AB is to BC, so the (square) on FG (is) to the δεῖξαι. (square) on H [Prop. 5.19 corr.]. But, AB has to BC the ratio which (some) square number (has) to (some) square number. Thus, the (square) on FG also has to the (square) on H the ratio which (some) square number (has) to (some) square number. Thus, FG is commensu- rable in length with H [Prop. 10.9]. Hence, the square on FG is greater than (the square on) FE by the (square) on (some straight-line) commensurable in length with (FG). And FG and FE are rational (straight-lines which are) commensurable in square only. And the lesser term EF is commensurable in length with the rational (straight- line) D (previously) laid down. Thus, EG is a second binomial (straight-line) [Def. 10.6].† (Which is) the very thing it was required to show. † If the rational straight-line has unit length then the length of a second binomial straight-line is k/ √ 1 − k′ 2 + k. This, and the second apotome, 335 STOIQEIWN iþ. ELEMENTS BOOK 10 whose length is k/ √ 1 − k′ 2 − k [Prop. 10.86], are the roots of x2 − (2 k/ √ 1 − k′ 2) x + k2 [k′ 2/(1 − k′ 2)] = 0.nþ. Proposition 50 Εὑρεῖν τὴν ἐκ δύο ὀνομάτων τρίτην. To find a third binomial (straight-line). Θ Α Γ Β Ε Κ Ζ Η ∆ K A B E D F G H C ᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν Let the two numbers AC and CB be laid down such συγκείμενον ἐξ αὐτῶν τὸν ΑΒ πρὸς μὲν τὸν ΒΓ λόγον that their sum AB has to BC the ratio which (some) ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, square number (has) to (some) square number, and does πρὸς δὲ τὸν ΑΓ λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς not have to AC the ratio which (some) square number πρὸς τετράγωνον ἀριθμόν. ἐκκείσθω δέ τις καὶ ἄλλος μὴ (has) to (some) square number. And let some other non- τετράγωνος ἀριθμὸς ὁ Δ, καὶ πρὸς ἑκάτερον τῶν ΒΑ, ΑΓ square number D also be laid down, and let it not have λόγον μὴ ἐχέτω, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον to each of BA and AC the ratio which (some) square ἀριθμόν· καὶ ἐκκείσθω τις ῥητὴ εὐθεῖα ἡ Ε, καὶ γεγονέτω number (has) to (some) square number. And let some ra- ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ tional straight-line E be laid down, and let it have been τῆς ΖΗ· σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς Ε τῷ ἀπὸ τῆς ΖΗ. contrived that as D (is) to AB, so the (square) on E καί ἐστι ῥητὴ ἡ Ε· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ Δ (is) to the (square) on FG [Prop. 10.6 corr.]. Thus, the πρὸς τὸν ΑΒ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς (square) on E is commensurable with the (square) on τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς FG [Prop. 10.6]. And E is a rational (straight-line). ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον Thus, FG is also a rational (straight-line). And since ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ Ε τῇ ΖΗ μήκει. γεγονέτω D does not have to AB the ratio which (some) square δὴ πάλιν ὡς ἡ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς number has to (some) square number, the (square) on ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ· σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς E does not have to the (square) on FG the ratio which ΖΗ τῷ ἀπὸ τῆς ΗΘ. ῥητὴ δὲ ἡ ΖΗ· ῥητὴ ἄρα καὶ ἡ ΗΘ. καὶ (some) square number (has) to (some) square number ἐπεὶ ὁ ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος either. E is thus incommensurable in length with FG ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΖΗ [Prop. 10.9]. So, again, let it have been contrived that πρὸς τὸ ἀπὸ τῆς ΘΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς as the number BA (is) to AC, so the (square) on FG πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ (is) to the (square) on GH [Prop. 10.6 corr.]. Thus, the τῇ ΗΘ μήκει. αἱ ΖΗ, ΗΘ ἄρα ῥηταί εἰσι δυνάμει μόνον (square) on FG is commensurable with the (square) on σύμμετροι· ἡ ΖΘ ἄρα ἐκ δύο ὀνομάτων ἐστίν. λέγω δή, ὅτι GH [Prop. 10.6]. And FG (is) a rational (straight-line). καὶ τρίτη. Thus, GH (is) also a rational (straight-line). And since ᾿Επεὶ γάρ ἐστιν ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς BA does not have to AC the ratio which (some) square Ε πρὸς τὸ ἀπὸ τῆς ΖΗ, ὡς δὲ ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως number (has) to (some) square number, the (square) on τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, δι᾿ ἴσου ἄρα ἐστὶν ὡς FG does not have to the (square) on HG the ratio which ὁ Δ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς (some) square number (has) to (some) square number ΗΘ. ὁ δὲ Δ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος either. Thus, FG is incommensurable in length with GH ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· οὐδὲ τὸ ἀπὸ τῆς Ε ἄρα [Prop. 10.9]. FG and GH are thus rational (straight- πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς lines which are) commensurable in square only. Thus, πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ Ε τῇ FH is a binomial (straight-line) [Prop. 10.36]. So, I say ΗΘ μήκει. καὶ ἐπεί ἐστιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως that (it is) also a third (binomial straight-line). τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, μεῖζον ἄρα τὸ ἀπὸ For since as D is to AB, so the (square) on E (is) τῆς ΖΗ τοῦ ἀπὸ τῆς ΗΘ. ἔστω οὖν τῷ ἀπὸ τῆς ΖΗ ἴσα τὰ to the (square) on FG, and as BA (is) to AC, so the ἀπὸ τῶν ΗΘ, Κ· ἀναστρέψαντι ἄρα [ἐστὶν] ὡς ὁ ΑΒ πρὸς (square) on FG (is) to the (square) on GH , thus, via τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Κ. ὁ δὲ equality, as D (is) to AC, so the (square) on E (is) ΑΒ πρὸς τὸν ΒΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς to the (square) on GH [Prop. 5.22]. And D does not 336 STOIQEIWN iþ. ELEMENTS BOOK 10 τετράγωνον ἀριθμόν· καὶ τὸ ἀπὸ τῆς ΖΗ ἄρα πρὸς τὸ ἀπὸ have to AC the ratio which (some) square number (has) τῆς Κ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον to (some) square number. Thus, the (square) on E ἀριθμόν· σύμμετρος ἄρα [ἐστὶν] ἡ ΖΗ τῇ Κ μήκει. ἡ ΖΗ ἄρα does not have to the (square) on GH the ratio which τῆς ΗΘ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καί εἰσιν (some) square number (has) to (some) square number αἱ ΖΗ, ΗΘ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα either. Thus, E is incommensurable in length with GH αὐτῶν σύμμετρός ἐστι τῇ Ε μήκει. [Prop. 10.9]. And since as BA is to AC, so the (square) ῾Η ΖΘ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τρίτη· ὅπερ ἔδει δεῖξαι. on FG (is) to the (square) on GH , the (square) on FG (is) thus greater than the (square) on GH [Prop. 5.14]. Therefore, let (the sum of) the (squares) on GH and K be equal to the (square) on FG. Thus, via conver- sion, as AB [is] to BC, so the (square) on FG (is) to the (square) on K [Prop. 5.19 corr.]. And AB has to BC the ratio which (some) square number (has) to (some) square number. Thus, the (square) on FG also has to the (square) on K the ratio which (some) square number (has) to (some) square number. Thus, FG [is] commensurable in length with K [Prop. 10.9]. Thus, the square on FG is greater than (the square on) GH by the (square) on (some straight-line) commensurable (in length) with (FG). And FG and GH are rational (straight-lines which are) commensurable in square only, and neither of them is commensurable in length with E. Thus, FH is a third binomial (straight-line) [Def. 10.7].† (Which is) the very thing it was required to show. † If the rational straight-line has unit length then the length of a third binomial straight-line is k1/2 (1 + √ 1 − k′ 2). This, and the third apotome, whose length is k1/2 (1 − √ 1 − k′ 2) [Prop. 10.87], are the roots of x2 − 2 k1/2 x + k k′ 2 = 0.naþ. Proposition 51 Εὑρεῖν τὴν ἐκ δύο ὀνομάτων τετάρτην. To find a fourth binomial (straight-line). Β Ε Ζ Η Θ Α ∆ Γ B E A D F G H C ᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν ΑΒ Let the two numbers AC and CB be laid down πρὸς τὸν ΒΓ λόγον μὴ ἔχειν μήτε μὴν πρὸς τὸν ΑΓ, such that AB does not have to BC, or to AC either, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καὶ the ratio which (some) square number (has) to (some) ἐκκείσθω ῥητὴ ἡ Δ, καὶ τῇ Δ σύμμετρος ἔστω μήκει ἡ ΕΖ· square number [Prop. 10.28 lem. I]. And let the rational ῥητὴ ἄρα ἐστὶ καὶ ἡ ΕΖ. καὶ γεγονέτω ὡς ὁ ΒΑ ἀριθμὸς (straight-line) D be laid down. And let EF be com- πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ· mensurable in length with D. Thus, EF is also a ratio- σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΕΖ τῷ ἀπὸ τῆς ΖΗ· ῥητὴ ἄρα nal (straight-line). And let it have been contrived that ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, as the number BA (is) to AC, so the (square) on EF ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ (is) to the (square) on FG [Prop. 10.6 corr.]. Thus, the ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος (square) on EF is commensurable with the (square) on ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ FG [Prop. 10.6]. Thus, FG is also a rational (straight- ΕΖ τῇ ΖΗ μήκει. αἱ ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον line). And since BA does not have to AC the ratio which σύμμετροι· ὥστε ἡ ΕΗ ἐκ δύο ὀνομάτων ἐστίν. λέγω δή, (some) square number (has) to (some) square number, 337 STOIQEIWN iþ. ELEMENTS BOOK 10 ὅτι καὶ τετάρτη. the (square) on EF does not have to the (square) on FG ᾿Επεὶ γάρ ἐστιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ the ratio which (some) square number (has) to (some) τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ [μείζων δὲ ὁ ΒΑ τοῦ ΑΓ], square number either. Thus, EF is incommensurable μεῖζον ἄρα τὸ ἀπὸ τῆς ΕΖ τοῦ ἀπὸ τῆς ΖΗ. ἔστω οὖν τῷ in length with FG [Prop. 10.9]. Thus, EF and FG ἀπὸ τῆς ΕΖ ἴσα τὰ ἀπὸ τῶν ΖΗ, Θ· ἀναστρέψαντι ἄρα ὡς are rational (straight-lines which are) commensurable ὁ ΑΒ ἀριθμὸς πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς in square only. Hence, EG is a binomial (straight-line) τὸ ἀπὸ τῆς Θ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον οὐκ ἔχει, ὃν [Prop. 10.36]. So, I say that (it is) also a fourth (binomial τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· οὐδ᾿ ἄρα τὸ straight-line). ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος For since as BA is to AC, so the (square) on EF (is) to ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἀσύμμετρος ἄρα ἐστὶν the (square) on FG [and BA (is) greater than AC], the ἡ ΕΖ τῇ Θ μήκει· ἡ ΕΖ ἄρα τῆς ΗΖ μεῖζον δύναται τῷ ἀπὸ (square) on EF (is) thus greater than the (square) on FG ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΕΖ, ΖΗ ῥηταὶ δυνάμει μόνον [Prop. 5.14]. Therefore, let (the sum of) the squares on σύμμετροι, καὶ ἡ ΕΖ τῇ Δ σύμμετρός ἐστι μήκει. FG and H be equal to the (square) on EF . Thus, via con- ῾Η ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τετάρτη· ὅπερ ἔδει version, as the number AB (is) to BC, so the (square) on δεῖξαι. EF (is) to the (square) on H [Prop. 5.19 corr.]. And AB does not have to BC the ratio which (some) square num- ber (has) to (some) square number. Thus, the (square) on EF does not have to the (square) on H the ratio which (some) square number (has) to (some) square number either. Thus, EF is incommensurable in length with H [Prop. 10.9]. Thus, the square on EF is greater than (the square on) GF by the (square) on (some straight-line) in- commensurable (in length) with (EF ). And EF and FG are rational (straight-lines which are) commensurable in square only. And EF is commensurable in length with D. Thus, EG is a fourth binomial (straight-line) [Def. 10.8].† (Which is) the very thing it was required to show. † If the rational straight-line has unit length then the length of a fourth binomial straight-line is k (1 + 1/ √ 1 + k′). This, and the fourth apotome, whose length is k (1 − 1/ √ 1 + k′) [Prop. 10.88], are the roots of x2 − 2 k x + k2 k′/(1 + k′) = 0.nbþ. Proposition 52 Εὑρεῖν τὴν ἐκ δύο ὀνομάτων πέμπτην. To find a fifth binomial straight-line. Β Ε Η Θ Α ∆ Ζ Γ B E A D F G H C ᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν ΑΒ Let the two numbers AC and CB be laid down such πρὸς ἑκάτερον αὐτῶν λόγον μὴ ἔχειν, ὃν τετράγωνος that AB does not have to either of them the ratio which ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ ἐκκείσθω ῥητή τις (some) square number (has) to (some) square number εὐθεῖα ἡ Δ, καὶ τῇ Δ σύμμετρος ἔστω [μήκει] ἡ ΕΖ· ῥητὴ [Prop. 10.38 lem.]. And let some rational straight-line ἄρα ἡ ΕΖ. καὶ γεγονέτω ὡς ὁ ΓΑ πρὸς τὸν ΑΒ, οὕτως τὸ D be laid down. And let EF be commensurable [in ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ. ὁ δὲ ΓΑ πρὸς τὸν ΑΒ length] with D. Thus, EF (is) a rational (straight- λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον line). And let it have been contrived that as CA (is) to ἀριθμόν· οὑδὲ τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον AB, so the (square) on EF (is) to the (square) on FG ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. αἱ [Prop. 10.6 corr.]. And CA does not have to AB the ra- 338 STOIQEIWN iþ. ELEMENTS BOOK 10 ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο tio which (some) square number (has) to (some) square ἄρα ὀνομάτων ἐστὶν ἡ ΕΗ. λέγω δή, ὅτι καὶ πέμπτη. number. Thus, the (square) on EF does not have to the ᾿Επεὶ γάρ ἐστιν ὡς ὁ ΓΑ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ (square) on FG the ratio which (some) square number τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ, ἀνάπαλιν ὡς ὁ ΒΑ πρὸς τὸν (has) to (some) square number either. Thus, EF and ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΖΕ· μεῖζον FG are rational (straight-lines which are) commensu- ἄρα τὸ ἀπὸ τῆς ΗΖ τοῦ ἀπὸ τῆς ΖΕ. ἔστω οὖν τῷ ἀπὸ rable in square only [Prop. 10.9]. Thus, EG is a binomial τῆς ΗΖ ἴσα τὰ ἀπὸ τῶν ΕΖ, Θ· ἀναστρέψαντι ἄρα ἐστὶν ὡς (straight-line) [Prop. 10.36]. So, I say that (it is) also a ὁ ΑΒ ἀριθμὸς πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΗΖ πρὸς fifth (binomial straight-line). τὸ ἀπὸ τῆς Θ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον οὐκ ἔχει, ὃν For since as CA is to AB, so the (square) on EF τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· οὐδ᾿ ἄρα τὸ (is) to the (square) on FG, inversely, as BA (is) to ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος AC, so the (square) on FG (is) to the (square) on FE ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἀσύμμετρος ἄρα ἐστὶν [Prop. 5.7 corr.]. Thus, the (square) on GF (is) greater ἡ ΖΗ τῇ Θ μήκει· ὥστε ἡ ΖΗ τῆς ΖΕ μεῖζον δύναται τῷ than the (square) on FE [Prop. 5.14]. Therefore, let ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΗΖ, ΖΕ ῥηταὶ δυνάμει (the sum of) the (squares) on EF and H be equal to μόνον σύμμετροι, καὶ τὸ ΕΖ ἔλαττον ὄνομα σύμμετρόν ἐστι the (square) on GF . Thus, via conversion, as the number τῇ ἐκκειμένῃ ῥητῇ τῇ Δ μήκει. AB is to BC, so the (square) on GF (is) to the (square) ῾Η ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πέμπτη· ὅπερ ἔδει on H [Prop. 5.19 corr.]. And AB does not have to BC δεῖξαι. the ratio which (some) square number (has) to (some) square number. Thus, the (square) on FG does not have to the (square) on H the ratio which (some) square num- ber (has) to (some) square number either. Thus, FG is incommensurable in length with H [Prop. 10.9]. Hence, the square on FG is greater than (the square on) FE by the (square) on (some straight-line) incommensurable (in length) with (FG). And GF and FE are rational (straight-lines which are) commensurable in square only. And the lesser term EF is commensurable in length with the rational (straight-line previously) laid down, D. Thus, EG is a fifth binomial (straight-line).† (Which is) the very thing it was required to show. † If the rational straight-line has unit length then the length of a fifth binomial straight-line is k ( √ 1 + k′ +1). This, and the fifth apotome, whose length is k ( √ 1 + k′ − 1) [Prop. 10.89], are the roots of x2 − 2 k √ 1 + k′ x + k2 k′ = 0.ngþ. Proposition 53 Εὑρεῖν τὴν ἐκ δύο ὀνομάτων ἕκτην. To find a sixth binomial (straight-line). Η Α Γ Β Ζ Θ Ε Κ ∆ G A B E D F H C K ᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν ΑΒ Let the two numbers AC and CB be laid down such πρὸς ἑκάτερον αὐτῶν λόγον μὴ ἔχειν, ὃν τετράγωνος that AB does not have to each of them the ratio which ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἔστω δὲ καὶ ἕτερος (some) square number (has) to (some) square number. ἀριθμὸς ὁ Δ μὴ τετράγωνος ὢν μηδὲ πρὸς ἑκάτερον And let D also be another number, which is not square, τῶν ΒΑ, ΑΓ λόγον ἔχων, ὃν τετράγωνος ἀριθμὸς πρὸς and does not have to each of BA and AC the ratio which τετράγωνον ἀριθμόν· καὶ ἐκκείσθω τις ῥητὴ εὐθεῖα ἡ Ε, (some) square number (has) to (some) square number ei- καὶ γεγονέτω ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς Ε ther [Prop. 10.28 lem. I]. And let some rational straight- πρὸς τὸ ἀπὸ τῆς ΖΗ· σύμμετρον ἄρα τὸ ἀπὸ τῆς Ε τῷ ἀπὸ line E be laid down. And let it have been contrived that 339 STOIQEIWN iþ. ELEMENTS BOOK 10 τῆς ΖΗ. καί ἐστι ῥητὴ ἡ Ε· ῥητὴ ἄρα καὶ ἡ ΖΗ. καὶ ἐπεὶ as D (is) to AB, so the (square) on E (is) to the (square) οὐκ ἔχει ὁ Δ πρὸς τὸν ΑΒ λόγον, ὃν τετράγωνος ἀριθμὸς on FG [Prop. 10.6 corr.]. Thus, the (square) on E (is) πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς Ε ἄρα πρὸς commensurable with the (square) on FG [Prop. 10.6]. τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς And E is rational. Thus, FG (is) also rational. And since τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἡ Ε τῇ ΖΗ μήκει. D does not have to AB the ratio which (some) square γεγονέτω δὴ πάλιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ number (has) to (some) square number, the (square) on τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ. σύμμετρον ἄρα τὸ ἀπὸ τῆς ΖΗ E thus does not have to the (square) on FG the ra- τῷ ἀπὸ τῆς ΘΗ. ῥητὸν ἄρα τὸ ἀπὸ τῆς ΘΗ· ῥητὴ ἄρα ἡ ΘΗ. tio which (some) square number (has) to (some) square καὶ ἐπεὶ ὁ ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος number either. Thus, E (is) incommensurable in length ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΖΗ with FG [Prop. 10.9]. So, again, let it have be contrived πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς that as BA (is) to AC, so the (square) on FG (is) to πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ the (square) on GH [Prop. 10.6 corr.]. The (square) on τῇ ΗΘ μήκει. αἱ ΖΗ, ΗΘ ἄρα ῥηταί εἰσι δυνάμει μόνον FG (is) thus commensurable with the (square) on HG σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΖΘ. δεικτέον δή, [Prop. 10.6]. The (square) on HG (is) thus rational. ὅτι καὶ ἕκτη. Thus, HG (is) rational. And since BA does not have ᾿Επεὶ γάρ ἐστιν ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ to AC the ratio which (some) square number (has) to τῆς Ε πρὸς τὸ ἀπὸ τῆς ΖΗ, ἔστι δὲ καὶ ὡς ὁ ΒΑ πρὸς (some) square number, the (square) on FG does not have τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, δι᾿ to the (square) on GH the ratio which (some) square ἴσου ἄρα ἐστὶν ὡς ὁ Δ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς Ε number (has) to (some) square number either. Thus, πρὸς τὸ ἀπὸ τῆς ΗΘ. ὁ δὲ Δ πρὸς τὸν ΑΓ λόγον οὐκ FG is incommensurable in length with GH [Prop. 10.9]. ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· Thus, FG and GH are rational (straight-lines which are) οὐδὲ τὸ ἀπὸ τῆς Ε ἄρα πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον commensurable in square only. Thus, FH is a binomial ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· (straight-line) [Prop. 10.36]. So, we must show that (it ἀσύμμετρος ἄρα ἐστὶν ἡ Ε τῇ ΗΘ μήκει. ἐδείχθη δὲ καὶ is) also a sixth (binomial straight-line). τῇ ΖΗ ἀσύμμετρος· ἑκατέρα ἄρα τῶν ΖΗ, ΗΘ ἀσύμμετρός For since as D is to AB, so the (square) on E (is) ἐστι τῇ Ε μήκει. καὶ ἐπεί ἐστιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, to the (square) on FG, and also as BA is to AC, so οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, μεῖζον ἄρα the (square) on FG (is) to the (square) on GH , thus, τὸ ἀπὸ τῆς ΖΗ τοῦ ἀπὸ τῆς ΗΘ. ἔστω οὖν τῷ ἀπὸ [τῆς] via equality, as D is to AC, so the (square) on E (is) ΖΗ ἴσα τὰ ἀπὸ τῶν ΗΘ, Κ· ἀναστρέψαντι ἄρα ὡς ὁ ΑΒ to the (square) on GH [Prop. 5.22]. And D does not πρὸς ΒΓ, οὕτως τὸ ἀπὸ ΖΗ πρὸς τὸ ἀπὸ τῆς Κ. ὁ δὲ ΑΒ have to AC the ratio which (some) square number (has) πρὸς τὸν ΒΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς to (some) square number. Thus, the (square) on E τετράγωνον ἀριθμόν· ὥστε οὐδὲ τὸ ἀπὸ ΖΗ πρὸς τὸ ἀπὸ does not have to the (square) on GH the ratio which τῆς Κ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον (some) square number (has) to (some) square number ἀριθμόν. ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ Κ μήκει· ἡ ΖΗ ἄρα either. E is thus incommensurable in length with GH τῆς ΗΘ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί εἰσιν [Prop. 10.9]. And (E) was also shown (to be) incom- αἱ ΖΗ, ΗΘ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα mensurable (in length) with FG. Thus, FG and GH αὐτῶν σύμμετρός ἐστι μήκει τῇ ἐκκειμένῃ ῥητῃ τῇ Ε. are each incommensurable in length with E. And since ῾Η ΖΘ ἄρα ἐκ δύο ὀνομάτων ἐστὶν ἕκτη· ὅπερ ἔδει δεῖξαι. as BA is to AC, so the (square) on FG (is) to the (square) on GH , the (square) on FG (is) thus greater than the (square) on GH [Prop. 5.14]. Therefore, let (the sum of) the (squares) on GH and K be equal to the (square) on FG. Thus, via conversion, as AB (is) to BC, so the (square) on FG (is) to the (square) on K [Prop. 5.19 corr.]. And AB does not have to BC the ra- tio which (some) square number (has) to (some) square number. Hence, the (square) on FG does not have to the (square) on K the ratio which (some) square num- ber (has) to (some) square number either. Thus, FG is incommensurable in length with K [Prop. 10.9]. The square on FG is thus greater than (the square on) GH by the (square) on (some straight-line which is) incom- 340 STOIQEIWN iþ. ELEMENTS BOOK 10 mensurable (in length) with (FG). And FG and GH are rational (straight-lines which are) commensurable in square only, and neither of them is commensurable in length with the rational (straight-line) E (previously) laid down. Thus, FH is a sixth binomial (straight-line) [Def. 10.10].† (Which is) the very thing it was required to show. † If the rational straight-line has unit length then the length of a sixth binomial straight-line is √ k + √ k′. This, and the sixth apotome, whose length is √ k − √ k′ [Prop. 10.90], are the roots of x2 − 2 √ k x + (k − k′) = 0.L¨mma. Lemma ῎Εστω δύο τετράγωνα τὰ ΑΒ, ΒΓ καὶ κείσθωσαν ὥστε Let AB and BC be two squares, and let them be laid ἐπ᾿ εὐθείας εἶναι τὴν ΔΒ τῇ ΒΕ· ἐπ᾿ εὐθείας ἄρα ἐστὶ καὶ ἡ down such that DB is straight-on to BE. FB is, thus, ΖΒ τῇ ΒΗ. καὶ συμπεπληρώσθω τὸ ΑΓ παραλληλόγραμμον· also straight-on to BG. And let the parallelogram AC λέγω, ὅτι τετράγωνόν ἐστι τὸ ΑΓ, καὶ ὅτι τῶν ΑΒ, ΒΓ have been completed. I say that AC is a square, and μέσον ἀνάλογόν ἐστι τὸ ΔΗ, καὶ ἔτι τῶν ΑΓ, ΓΒ μέσον that DG is the mean proportional to AB and BC, and, ἀνάλογόν ἐστι τὸ ΔΓ. moreover, DC is the mean proportional to AC and CB. Γ Α Ε∆ Κ Ζ Η Θ Β C A ED K F H B G ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ μὲν ΔΒ τῇ ΒΖ, ἡ δὲ ΒΕ τῇ ΒΗ, For since DB is equal to BF , and BE to BG, the ὅλη ἄρα ἡ ΔΕ ὅλῃ τῇ ΖΗ ἐστιν ἴση. ἀλλ᾿ ἡ μὲν ΔΕ ἑκατέρᾳ whole of DE is thus equal to the whole of FG. But DE τῶν ΑΘ, ΚΓ ἐστιν ἴση, ἡ δὲ ΖΗ ἑκατέρᾳ τῶν ΑΚ, ΘΓ ἐστιν is equal to each of AH and KC, and FG is equal to each ἴση· καὶ ἑκατέρα ἄρα τῶν ΑΘ, ΚΓ ἑκατέρᾳ τῶν ΑΚ, ΘΓ of AK and HC [Prop. 1.34]. Thus, AH and KC are also ἐστιν ἴση. ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΓ παραλληλόγραμμον· equal to AK and HC, respectively. Thus, the parallel- ἔστι δὲ καὶ ὀρθογώνιον· τετράγωνον ἄρα ἐστὶ τὸ ΑΓ. ogram AC is equilateral. And (it is) also right-angled. Καὶ ἐπεὶ ἐστιν ὡς ἡ ΖΒ πρὸς τὴν ΒΗ, οὕτως ἡ ΔΒ πρὸς Thus, AC is a square. τὴν ΒΕ, ἀλλ᾿ ὡς μὲν ἡ ΖΒ πρὸς τὴν ΒΗ, οὕτως τὸ ΑΒ πρὸς And since as FB is to BG, so DB (is) to BE, but τὸ ΔΗ, ὡς δὲ ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως τὸ ΔΗ πρὸς τὸ as FB (is) to BG, so AB (is) to DG, and as DB (is) to ΒΓ, καὶ ὡς ἄρα τὸ ΑΒ πρὸς τὸ ΔΗ, οὕτως τὸ ΔΗ πρὸς τὸ BE, so DG (is) to BC [Prop. 6.1], thus also as AB (is) ΒΓ. τῶν ΑΒ, ΒΓ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΔΗ. to DG, so DG (is) to BC [Prop. 5.11]. Thus, DG is the Λέγω δή, ὅτι καὶ τῶν ΑΓ, ΓΒ μέσον ἀνάλογόν [ἐστι] τὸ mean proportional to AB and BC. ΔΓ. So I also say that DC [is] the mean proportional to ᾿Επεὶ γάρ ἐστιν ὡς ἡ ΑΔ πρὸς τὴν ΔΚ, οὕτως ἡ ΚΗ AC and CB. πρὸς τὴν ΗΓ· ἴση γάρ [ἐστιν] ἑκατέρα ἑκατέρᾳ· καὶ συνθέντι For since as AD is to DK, so KG (is) to GC. For [they ὡς ἡ ΑΚ πρὸς ΚΔ, οὕτως ἡ ΚΓ πρὸς ΓΗ, ἀλλ᾿ ὡς μὲν ἡ ΑΚ are] respectively equal. And, via composition, as AK (is) πρὸς ΚΔ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΔ, ὡς δὲ ἡ ΚΓ πρὸς ΓΗ, to KD, so KC (is) to CG [Prop. 5.18]. But as AK (is) οὕτως τὸ ΔΓ πρὸς ΓΒ, καὶ ὡς ἄρα τὸ ΑΓ πρὸς ΔΓ, οὕτως to KD, so AC (is) to CD, and as KC (is) to CG, so DC τὸ ΔΓ πρὸς τὸ ΒΓ. τῶν ΑΓ, ΓΒ ἄρα μέσον ἀνάλογόν ἐστι (is) to CB [Prop. 6.1]. Thus, also, as AC (is) to DC, τὸ ΔΓ· ἃ προέκειτο δεῖξαι. so DC (is) to BC [Prop. 5.11]. Thus, DC is the mean proportional to AC and CB. Which (is the very thing) it 341 STOIQEIWN iþ. ELEMENTS BOOK 10 was prescribed to show.ndþ. Proposition 54 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line) ὀνομάτων πρώτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν and a first binomial (straight-line) then the square-root ἡ καλουμένη ἐκ δύο ὀνομάτων. of the area is the irrational (straight-line which is) called binomial.† Ν Α Β Θ ∆ Γ Ρ Ο Η Μ Ξ Ε Κ Ζ Λ Σ Π QA B R N O P H G C D M E K F L S Χωρίον γὰρ τὸ ΑΓ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ For let the area AC be contained by the rational τῆς ἐκ δύο ὀνομάτων πρώτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ (straight-line) AB and by the first binomial (straight- ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο line) AD. I say that square-root of area AC is the ir- ὀνομάτων. rational (straight-line which is) called binomial. ᾿Επεὶ γὰρ ἐκ δύο ὀνομάτων ἐστὶ πρώτη ἡ ΑΔ, διῃρήσθω For since AD is a first binomial (straight-line), let it εἰς τὰ ὀνόματα κατὰ τὸ Ε, καὶ ἔστω τὸ μεῖζον ὄνομα have been divided into its (component) terms at E, and τὸ ΑΕ. φανερὸν δή, ὅτι αἱ ΑΕ, ΕΔ ῥηταί εἰσι δυνάμει let AE be the greater term. So, (it is) clear that AE and μόνον σύμμετροι, καὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ ED are rational (straight-lines which are) commensu- συμμέτρου ἑαυτῃ, καὶ ἡ ΑΕ σύμμετρός ἐστι τῇ ἐκκειμένῃ rable in square only, and that the square on AE is greater ῥητῇ τῇ ΑΒ μήκει. τετμήσθω δὴ ἡ ΕΔ δίχα κατὰ τὸ Ζ than (the square on) ED by the (square) on (some σημεῖον. καὶ ἐπεὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ straight-line) commensurable (in length) with (AE), and συμμέτρου ἑαυτῇ, ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς that AE is commensurable (in length) with the rational ἐλάσσονος, τουτέστι τῷ ἀπὸ τῆς ΕΖ, ἴσον παρὰ τὴν μείζονα (straight-line) AB (first) laid out [Def. 10.5]. So, let ED τὴν ΑΕ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα have been cut in half at point F . And since the square on αὐτὴν διαιρεῖ. παραβεβλήσθω οὖν παρὰ τὴν ΑΕ τῷ ἀπὸ τῆς AE is greater than (the square on) ED by the (square) ΕΖ ἴσον τὸ ὑπὸ ΑΗ, ΗΕ· σύμμετρος ἄρα ἐστὶν ἡ ΑΗ τῇ on (some straight-line) commensurable (in length) with ΕΗ μήκει. καὶ ἤχθωσαν ἀπὸ τῶν Η, Ε, Ζ ὁποτέρᾳ τῶν ΑΒ, (AE), thus if a (rectangle) equal to the fourth part of the ΓΔ παράλληλοι αἱ ΗΘ, ΕΚ, ΖΛ· καὶ τῷ μὲν ΑΘ παραλλη- (square) on the lesser (term)—that is to say, the (square) λογράμμῳ ἴσον τετράγωνον συνεστάτω τὸ ΣΝ, τῷ δὲ ΗΚ on EF—falling short by a square figure, is applied to the ἴσον τὸ ΝΠ, καὶ κείσθω ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΜΝ greater (term) AE, then it divides it into (terms which τῇ ΝΞ· ἐπ᾿ εὐθείας ἄρα ἐστὶ καὶ ἡ ΡΝ τῇ ΝΟ. καὶ συμ- are) commensurable (in length) [Prop 10.17]. Therefore, πεπληρώσθω τὸ ΣΠ παραλληλόγραμμον· τετράγωνον ἄρα let the (rectangle contained) by AG and GE, equal to the ἐστὶ τὸ ΣΠ. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΗ, ΗΕ ἴσον ἐστὶ τῷ ἀπὸ (square) on EF , have been applied to AE. AG is thus τῆς ΕΖ, ἔστιν ἄρα ὡς ἡ ΑΗ πρὸς ΕΖ, οὕτως ἡ ΖΕ πρὸς commensurable in length with EG. And let GH , EK, ΕΗ· καὶ ὡς ἄρα τὸ ΑΘ πρὸς ΕΛ, τὸ ΕΛ πρὸς ΚΗ· τῶν and FL have been drawn from (points) G, E, and F (re- ΑΘ, ΗΚ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΕΛ. ἀλλὰ τὸ μὲν ΑΘ spectively), parallel to either of AB or CD. And let the ἴσον ἐστὶ τῷ ΣΝ, τὸ δὲ ΗΚ ἴσον τῷ ΝΠ· τῶν ΣΝ, ΝΠ ἄρα square SN , equal to the parallelogram AH , have been μέσον ἀνάλογόν ἐστι τὸ ΕΛ. ἔστι δὲ τῶν αὐτῶν τῶν ΣΝ, constructed, and (the square) NQ, equal to (the parallel- ΝΠ μέσον ἀνάλογον καὶ τὸ ΜΡ· ἴσον ἄρα ἐστὶ τὸ ΕΛ τῷ ogram) GK [Prop. 2.14]. And let MN be laid down so ΜΡ· ὥστε καὶ τῷ ΟΞ ἴσον ἐστίν. ἔστι δὲ καὶ τὰ ΑΘ, ΗΚ as to be straight-on to NO. RN is thus also straight-on τοῖς ΣΝ, ΝΠ ἴσα· ὅλον ἄρα τὸ ΑΓ ἴσον ἐστὶν ὅλῳ τῷ ΣΠ, to NP . And let the parallelogram SQ have been com- τουτέστι τῷ ἀπὸ τῆς ΜΞ τετραγώνῳ· τὸ ΑΓ ἄρα δύναται ἡ pleted. SQ is thus a square [Prop. 10.53 lem.]. And since ΜΞ. λέγω, ὅτι ἡ ΜΞ ἐκ δύο ὀνομάτων ἐστίν. the (rectangle contained) by AG and GE is equal to the ᾿Επεὶ γὰρ σύμμετρός ἐστιν ἡ ΑΗ τῇ ΗΕ, σύμμετρός ἐστι (square) on EF , thus as AG is to EF , so FE (is) to EG καὶ ἡ ΑΕ ἑκατέρᾳ τῶν ΑΗ, ΗΕ. ὑπόκειται δὲ καὶ ἡ ΑΕ τῇ [Prop. 6.17]. And thus as AH (is) to EL, (so) EL (is) 342 STOIQEIWN iþ. ELEMENTS BOOK 10 ΑΒ σύμμετρος· καὶ αἱ ΑΗ, ΗΕ ἄρα τῇ ΑΒ σύμμετροί εἰσιν. to KG [Prop. 6.1]. Thus, EL is the mean proportional to καί ἐστι ῥητὴ ἡ ΑΒ· ῥητὴ ἄρα ἐστὶ καὶ ἑκατέρα τῶν ΑΗ, ΗΕ· AH and GK. But, AH is equal to SN , and GK (is) equal ῥητὸν ἄρα ἐστὶν ἑκάτερον τῶν ΑΘ, ΗΚ, καί ἐστι σύμμετρον to NQ. EL is thus the mean proportional to SN and NQ. τὸ ΑΘ τῷ ΗΚ. ἀλλὰ τὸ μὲν ΑΘ τῷ ΣΝ ἴσον ἐστίν, τὸ δὲ And MR is also the mean proportional to the same— ΗΚ τῷ ΝΠ· καὶ τὰ ΣΝ, ΝΠ ἄρα, τουτέστι τὰ ἀπὸ τῶν ΜΝ, (namely), SN and NQ [Prop. 10.53 lem.]. EL is thus ΝΞ, ῥητά ἐστι καὶ σύμμετρα. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ equal to MR. Hence, it is also equal to PO [Prop. 1.43]. ΑΕ τῇ ΕΔ μήκει, ἀλλ᾿ ἡ μὲν ΑΕ τῇ ΑΗ ἐστι σύμμετρος, And AH plus GK is equal to SN plus NQ. Thus, the ἡ δὲ ΔΕ τῇ ΕΖ σύμμετρος, ἀσύμμετρος ἄρα καὶ ἡ ΑΗ τῇ whole of AC is equal to the whole of SQ—that is to say, ΕΖ· ὥστε καὶ τὸ ΑΘ τῷ ΕΛ ἀσύμμετρόν ἐστιν. ἀλλὰ τὸ μὲν to the square on MO. Thus, MO (is) the square-root of ΑΘ τῷ ΣΝ ἐστιν ἴσον, τὸ δὲ ΕΛ τῷ ΜΡ· καὶ τὸ ΣΝ ἄρα (area) AC. I say that MO is a binomial (straight-line). τῷ ΜΡ ἀσύμμετρόν ἐστιν. ἀλλ᾿ ὡς τὸ ΣΝ πρὸς ΜΡ, ἡ ΟΝ For since AG is commensurable (in length) with GE, πρὸς τὴν ΝΡ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΟΝ τῇ ΝΡ. ἴση δὲ ἡ AE is also commensurable (in length) with each of AG μὲν ΟΝ τῇ ΜΝ, ἡ δὲ ΝΡ τῇ ΝΞ· ἀσύμμετρος ἄρα ἐστὶν ἡ and GE [Prop. 10.15]. And AE was also assumed (to ΜΝ τῇ ΝΞ. καί ἐστι τὸ ἀπὸ τῆς ΜΝ σύμμετρον τῷ ἀπὸ τῆς be) commensurable (in length) with AB. Thus, AG ΝΞ, καὶ ῥητὸν ἑκάτερον· αἱ ΜΝ, ΝΞ ἄρα ῥηταί εἰσι δυνάμει and GE are also commensurable (in length) with AB μόνον σύμμετροι. [Prop. 10.12]. And AB is rational. AG and GE are ῾Η ΜΞ ἄρα ἐκ δύο ὀνομάτων ἐστὶ καὶ δύναται τὸ ΑΓ· thus each also rational. Thus, AH and GK are each ὅπερ ἔδει δεῖξαι. rational (areas), and AH is commensurable with GK [Prop. 10.19]. But, AH is equal to SN , and GK to NQ. SN and NQ—that is to say, the (squares) on MN and NO (respectively)—are thus also rational and commen- surable. And since AE is incommensurable in length with ED, but AE is commensurable (in length) with AG, and DE (is) commensurable (in length) with EF , AG (is) thus also incommensurable (in length) with EF [Prop. 10.13]. Hence, AH is also incommensurable with EL [Props. 6.1, 10.11]. But, AH is equal to SN , and EL to MR. Thus, SN is also incommensurable with MR. But, as SN (is) to MR, (so) PN (is) to NR [Prop. 6.1]. PN is thus incommensurable (in length) with NR [Prop. 10.11]. And PN (is) equal to MN , and NR to NO. Thus, MN is incommensurable (in length) with NO. And the (square) on MN is commensurable with the (square) on NO, and each (is) rational. MN and NO are thus rational (straight-lines which are) com- mensurable in square only. Thus, MO is (both) a binomial (straight-line) [Prop. 10.36], and the square-root of AC. (Which is) the very thing it was required to show. † If the rational straight-line has unit length then this proposition states that the square-root of a first binomial straight-line is a binomial straight- line: i.e., a first binomial straight-line has a length k + k √ 1 − k′ 2 whose square-root can be written ρ (1 + √ k′′), where ρ = p k (1 + k′)/2 and k′′ = (1 − k′)/(1 + k′). This is the length of a binomial straight-line (see Prop. 10.36), since ρ is rational.neþ. Proposition 55 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line) and ὁνομάτων δευτέρας, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν a second binomial (straight-line) then the square-root of ἡ καλουμένη ἐκ δύο μέσων πρώτη. the area is the irrational (straight-line which is) called first bimedial.† 343 STOIQEIWN iþ. ELEMENTS BOOK 10 Ν Α Β Θ ∆ Γ Ρ Ο Η Μ Ξ Ε Κ Ζ Λ Σ Π QA B R N O P H G C D M E K F L S Περιεχέσθω γὰρ χωρίον τὸ ΑΒΓΔ ὑπὸ ῥητῆς τῆς ΑΒ For let the area ABCD be contained by the rational καὶ τῆς ἐκ δύο ὀνομάτων δυετέρας τῆς ΑΔ· λέγω, ὅτι ἡ τὸ (straight-line) AB and by the second binomial (straight- ΑΓ χωρίον δυναμένη ἐκ δύο μέσων πρώτη ἐστίν. line) AD. I say that the square-root of area AC is a first ᾿Επεὶ γὰρ ἐκ δύο ὀνομάτων δευτέρα ἐστὶν ἡ ΑΔ, bimedial (straight-line). διῃρήσθω εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον For since AD is a second binomial (straight-line), let it ὄνομα εἶναι τὸ ΑΕ· αἱ ΑΕ, ΕΔ ἄρα ῥηταί εἰσι δυνάμει have been divided into its (component) terms at E, such μόνον σύμμετροι, καὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ that AE is the greater term. Thus, AE and ED are ratio- συμμέτρου ἑαυτῇ, καὶ τὸ ἔλαττον ὄνομα ἡ ΕΔ σύμμετρόν nal (straight-lines which are) commensurable in square ἐστι τῇ ΑΒ μήκει. τετμήσθω ἡ ΕΔ δίχα κατὰ τὸ Ζ, καὶ only, and the square on AE is greater than (the square τῷ ἀπὸ τῆς ΕΖ ἴσον παρὰ τὴν ΑΕ παραβεβλήσθω ἐλλεῖπον on) ED by the (square) on (some straight-line) commen- εἴδει τετραγώνῳ τὸ ὑπὸ τῶν ΑΗΕ· σύμμετρος ἄρα ἡ ΑΗ surable (in length) with (AE), and the lesser term ED τῇ ΗΕ μήκει. καὶ διὰ τῶν Η, Ε, Ζ παράλληλοι ἤχθωσαν is commensurable in length with AB [Def. 10.6]. Let ταῖς ΑΒ, ΓΔ αἱ ΗΘ, ΕΚ, ΖΛ, καὶ τῷ μὲν ΑΘ παραλλη- ED have been cut in half at F . And let the (rectan- λογράμμῳ ἴσον τετράγωνον συνεστάτω τὸ ΣΝ, τῷ δὲ ΗΚ gle contained) by AGE, equal to the (square) on EF , ἴσον τετράγωνον τὸ ΝΠ, καὶ κείσθω ὥστε ἐπ᾿ εὐθείας εἶναι have been applied to AE, falling short by a square fig- τὴν ΜΝ τῇ ΝΞ· ἐπ᾿ εὐθείας ἄρα [ἐστὶ] καὶ ἡ ΡΝ τῆ ΝΟ. καὶ ure. AG (is) thus commensurable in length with GE συμπεπληρώσθω τὸ ΣΠ τετράγωνον· φανερὸν δὴ ἐκ τοῦ [Prop. 10.17]. And let GH , EK, and FL have been προδεδειγμένου, ὅτι τὸ ΜΡ μέσον ἀνάλογόν ἐστι τῶν ΣΝ, drawn through (points) G, E, and F (respectively), par- ΝΠ, καὶ ἴσον τῷ ΕΛ, καὶ ὅτι τὸ ΑΓ χωρίον δύναται ἡ ΜΞ. allel to AB and CD. And let the square SN , equal to δεικτέον δή, ὅτι ἡ ΜΞ ἐκ δύο μέσων ἐστὶ πρώτη. the parallelogram AH , have been constructed, and the ᾿Επεὶ ἀσύμμετρός ἐστιν ἡ ΑΕ τῇ ΕΔ μήκει, σύμμετρος square NQ, equal to GK. And let MN be laid down so δὲ ἡ ΕΔ τῇ ΑΒ, ἀσύμμετρος ἄρα ἡ ΑΕ τῇ ΑΒ. καὶ ἐπεὶ as to be straight-on to NO. Thus, RN [is] also straight-on σύμμετρός ἐστιν ἡ ΑΗ τῇ ΕΗ, σύμμετρός ἐστι καὶ ἡ ΑΕ to NP . And let the square SQ have been completed. So, ἑκατέρᾳ τῶν ΑΗ, ΗΕ. ἀλλὰ ἡ ΑΕ ἀσύμμετρος τῇ ΑΒ μήκει· (it is) clear from what has been previously demonstrated καὶ αἱ ΑΗ, ΗΕ ἄρα ἀσύμμετροί εἰσι τῇ ΑΒ. αἱ ΒΑ, ΑΗ, [Prop. 10.53 lem.] that MR is the mean proportional to ΗΕ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ὥστε μέσον SN and NQ, and (is) equal to EL, and that MO is the ἐστὶν ἑκάτερον τῶν ΑΘ, ΗΚ. ὥστε καὶ ἑκάτερον τῶν ΣΝ, square-root of the area AC. So, we must show that MO ΝΠ μέσον ἐστίν. καὶ αἱ ΜΝ, ΝΞ ἄρα μέσαι εἰσίν. καὶ is a first bimedial (straight-line). ἐπεὶ σύμμετρος ἡ ΑΗ τῇ ΗΕ μήκει, σύμμετρόν ἐστι καὶ Since AE is incommensurable in length with ED, τὸ ΑΘ τῷ ΗΚ, τουτέστι τὸ ΣΝ τῷ ΝΠ, τουτέστι τὸ ἀπὸ and ED (is) commensurable (in length) with AB, τῆς ΜΝ τῷ ἀπὸ τῆς ΝΞ [ὥστε δυνάμει εἰσὶ σύμμετροι αἱ AE (is) thus incommensurable (in length) with AB ΜΝ, ΝΞ]. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΕ τῇ ΕΔ μήκει, [Prop. 10.13]. And since AG is commensurable (in ἀλλ᾿ ἡ μὲν ΑΕ σύμμετρός ἐστι τῇ ΑΗ, ἡ δὲ ΕΔ τῇ ΕΖ length) with EG, AE is also commensurable (in length) σύμμετρος, ἀσύμμετρος ἄρα ἡ ΑΗ τῇ ΕΖ· ὥστε καὶ τὸ with each of AG and GE [Prop. 10.15]. But, AE is in- ΑΘ τῷ ΕΛ ἀσύμμετρόν ἐστιν, τουτέστι τὸ ΣΝ τῷ ΜΡ, commensurable in length with AB. Thus, AG and GE τουτέστιν ὁ ΟΝ τῇ ΝΡ, τουτέστιν ἡ ΜΝ τῇ ΝΞ ἀσύμμετρός are also (both) incommensurable (in length) with AB ἐστι μήκει. ἐδείχθησαν δὲ αἱ ΜΝ, ΝΞ καὶ μέσαι οὖσαι καὶ [Prop. 10.13]. Thus, BA, AG, and (BA, and) GE are δυνάμει σύμμετροι· αἱ ΜΝ, ΝΞ ἄρα μέσαι εἰσὶ δυνάμει μόνον (pairs of) rational (straight-lines which are) commensu- σύμμετροι. λέγω δή, ὅτι καὶ ῥητὸν περιέχουσιν. ἐπεὶ γὰρ ἡ rable in square only. And, hence, each of AH and GK ΔΕ ὑπόκειται ἑκατέρᾳ τῶν ΑΒ, ΕΖ σύμμετρος, σύμμετρος is a medial (area) [Prop. 10.21]. Hence, each of SN ἄρα καὶ ἡ ΕΖ τῇ ΕΚ. καὶ ῥητὴ ἑκατέρα αὐτῶν· ῥητὸν ἄρα and NQ is also a medial (area). Thus, MN and NO τὸ ΕΛ, τουτέστι τὸ ΜΡ· τὸ δὲ ΜΡ ἐστι τὸ ὑπὸ τῶν ΜΝΞ. are medial (straight-lines). And since AG (is) commen- ἐὰν δὲ δύο μέσαι δυνάμει μόνον σύμμετροι συντεθῶσι ῥητὸν surable in length with GE, AH is also commensurable 344 STOIQEIWN iþ. ELEMENTS BOOK 10 περιέχουσαι, ἡ ὅλη ἄλογός ἐστιν, καλεῖται δὲ ἐκ δύο μέσων with GK—that is to say, SN with NQ—that is to say, πρώτη. the (square) on MN with the (square) on NO [hence, ῾Η ἄρα ΜΞ ἐκ δύο μέσων ἐστὶ πρώτη· ὅπερ ἔδει δεῖξαι. MN and NO are commensurable in square] [Props. 6.1, 10.11]. And since AE is incommensurable in length with ED, but AE is commensurable (in length) with AG, and ED commensurable (in length) with EF , AG (is) thus incommensurable (in length) with EF [Prop. 10.13]. Hence, AH is also incommensurable with EL—that is to say, SN with MR—that is to say, PN with NR—that is to say, MN is incommensurable in length with NO [Props. 6.1, 10.11]. But MN and NO have also been shown to be medial (straight-lines) which are commensu- rable in square. Thus, MN and NO are medial (straight- lines which are) commensurable in square only. So, I say that they also contain a rational (area). For since DE was assumed (to be) commensurable (in length) with each of AB and EF , EF (is) thus also commensurable with EK [Prop. 10.12]. And they (are) each rational. Thus, EL— that is to say, MR—(is) rational [Prop. 10.19]. And MR is the (rectangle contained) by MNO. And if two medial (straight-lines), commensurable in square only, which contain a rational (area), are added together, then the whole is (that) irrational (straight-line which is) called first bimedial [Prop. 10.37]. Thus, MO is a first bimedial (straight-line). (Which is) the very thing it was required to show. † If the rational straight-line has unit length then this proposition states that the square-root of a second binomial straight-line is a first bimedial straight-line: i.e., a second binomial straight-line has a length k/ √ 1 − k′ 2 + k whose square-root can be written ρ (k′′1/4 + k′′3/4), where ρ = p (k/2) (1 + k′)/(1 − k′) and k′′ = (1− k′)/(1 + k′). This is the length of a first bimedial straight-line (see Prop. 10.37), since ρ is rational.n�þ. Proposition 56 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line) and ὀνομάτων τρίτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ a third binomial (straight-line) then the square-root of καλουμένη ἐκ δύο μέσων δευτέρα. the area is the irrational (straight-line which is) called second bimedial.† Ν Α Β Θ ∆ Γ Ρ Ο Η Μ Ξ Ε Κ Ζ Λ Σ Π QA B R N O P H G C D M E K F L S Χωρίον γὰρ τὸ ΑΒΓΔ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ For let the area ABCD be contained by the rational καὶ τῆς ἐκ δύο ὀνομάτων τρίτης τῆς ΑΔ διῃρημένης εἰς τὰ (straight-line) AB and by the third binomial (straight- ὀνόματα κατὰ τὸ Ε, ὧν μεῖζόν ἐστι τὸ ΑΕ· λέγω, ὅτι ἡ line) AD, which has been divided into its (component) τὸ ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο terms at E, of which AE is the greater. I say that the μέσων δευτέρα. square-root of area AC is the irrational (straight-line Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρότερον. καὶ ἐπεὶ which is) called second bimedial. 345 STOIQEIWN iþ. ELEMENTS BOOK 10 ἐκ δύο ὀνομάτων ἐστὶ τρίτη ἡ ΑΔ, αἱ ΑΕ, ΕΔ ἄρα ῥηταί For let the same construction be made as previously. εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΕ τῆς ΕΔ μεῖζον And since AD is a third binomial (straight-line), AE and δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ οὐδετέρα τῶν ΑΕ, ED are thus rational (straight-lines which are) commen- ΕΔ σύμμετρός [ἐστι] τῇ ΑΒ μήκει. ὁμοίως δὴ τοῖς προδε- surable in square only, and the square on AE is greater δειγμένοις δείξομεν, ὅτι ἡ ΜΞ ἐστιν ἡ τὸ ΑΓ χωρίον δυ- than (the square on) ED by the (square) on (some ναμένη, καὶ αἱ ΜΝ, ΝΞ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι· straight-line) commensurable (in length) with (AE), and ὥστε ἡ ΜΞ ἐκ δύο μέσων ἐστίν. δεικτέον δή, ὅτι καὶ neither of AE and ED [is] commensurable in length with δευτέρα. AB [Def. 10.7]. So, similarly to that which has been [Καὶ] ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΔΕ τῇ ΑΒ μήκει, previously demonstrated, we can show that MO is the τουτέστι τῇ ΕΚ, σύμμετρος δὲ ἡ ΔΕ τῇ ΕΖ, ἀσύμμετρος square-root of area AC, and MN and NO are medial ἄρα ἐστὶν ἡ ΕΖ τῇ ΕΚ μήκει. καί εἰσι ῥηταί· αἱ ΖΕ, ΕΚ ἄρα (straight-lines which are) commensurable in square only. ῥηταί εἰσι δυνάμει μόνον σύμμετροι. μέσον ἄρα [ἐστὶ] τὸ Hence, MO is bimedial. So, we must show that (it is) ΕΛ, τουτέστι τὸ ΜΡ· καὶ περιέχεται ὑπὸ τῶν ΜΝΞ· μέσον also second (bimedial). ἄρα ἐστὶ τὸ ὑπὸ τῶν ΜΝΞ. [And] since DE is incommensurable in length with ῾Η ΜΞ ἄρα ἐκ δύο μέσων ἐστὶ δευτέρα· ὅπερ ἔδει δεῖξαι. AB—that is to say, with EK—and DE (is) commensu- rable (in length) with EF , EF is thus incommensurable in length with EK [Prop. 10.13]. And they are (both) rational (straight-lines). Thus, FE and EK are rational (straight-lines which are) commensurable in square only. EL—that is to say, MR—[is] thus medial [Prop. 10.21]. And it is contained by MNO. Thus, the (rectangle con- tained) by MNO is medial. Thus, MO is a second bimedial (straight-line) [Prop. 10.38]. (Which is) the very thing it was required to show. † If the rational straight-line has unit length then this proposition states that the square-root of a third binomial straight-line is a second bimedial straight-line: i.e., a third binomial straight-line has a length k1/2 (1+ √ 1 − k′ 2) whose square-root can be written ρ (k1/4 + k′′1/2/k1/4), where ρ = p (1 + k′)/2 and k′′ = k (1 − k′)/(1 + k′). This is the length of a second bimedial straight-line (see Prop. 10.38), since ρ is rational.nzþ. Proposition 57 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line) and ὀνομάτων τετάρτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν a fourth binomial (straight-line) then the square-root of ἡ καλουμένη μείζων. the area is the irrational (straight-line which is) called major.† Ν Α Β Θ ∆ Γ Ρ Ο Η Μ Ξ Ε Κ Ζ Λ Σ Π QA B R N O P H G C D M E K F L S Χωρίον γὰρ τὸ ΑΓ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ For let the area AC be contained by the rational τῆς ἐκ δύο ὀνομάτων τετάρτης τῆς ΑΔ διῃρημένης εἰς τὰ (straight-line) AB and the fourth binomial (straight-line) ὀνόματα κατὰ τὸ Ε, ὧν μεῖζον ἔστω τὸ ΑΕ· λέγω, ὅτι ἡ τὸ AD, which has been divided into its (component) terms ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη μείζων. at E, of which let AE be the greater. I say that the square- ᾿Επεὶ γὰρ ἡ ΑΔ ἐκ δύο ὀνομάτων ἐστὶ τετάρτη, αἱ ΑΕ, root of AC is the irrational (straight-line which is) called ΕΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΕ τῆς major. ΕΔ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ἡ ΑΕ For since AD is a fourth binomial (straight-line), AE τῇ ΑΒ σύμμετρός [ἐστι] μήκει. τετμήσθω ἡ ΔΕ δίχα κατὰ and ED are thus rational (straight-lines which are) com- 346 STOIQEIWN iþ. ELEMENTS BOOK 10 τὸ Ζ, καὶ τῷ ἀπὸ τῆς ΕΖ ἴσον παρὰ τὴν ΑΕ παραβεβλήσθω mensurable in square only, and the square on AE is παραλληλόγραμμον τὸ ὑπὸ ΑΗ, ΗΕ· ἀσύμμετρος ἄρα ἐστὶν greater than (the square on) ED by the (square) on ἡ ΑΗ τῇ ΗΕ μήκει. ἤχθωσαν παράλληλοι τῇ ΑΒ αἱ ΗΘ, (some straight-line) incommensurable (in length) with ΕΚ, ΖΛ, καὶ τὰ λοιπὰ τὰ αὐτὰ τοῖς πρὸ τούτου γεγονέτω· (AE), and AE [is] commensurable in length with AB φανερὸν δή, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἐστὶν ἡ ΜΞ. [Def. 10.8]. Let DE have been cut in half at F , and let δεικτέον δή, ὅτι ἡ ΜΞ ἄλογός ἐστιν ἡ καλουμένη μείζων. the parallelogram (contained by) AG and GE, equal to ᾿Επεὶ ἀσύμμετρός ἐστιν ἡ ΑΗ τῇ ΕΗ μήκει, ἀσύμμετρόν the (square) on EF , (and falling short by a square figure) ἐστι καὶ τὸ ΑΘ τῷ ΗΚ, τουτέστι τὸ ΣΝ τῷ ΝΠ· αἱ ΜΝ, have been applied to AE. AG is thus incommensurable ΝΞ ἄρα δυνάμει εἰσὶν ἀσύμμετροι. καὶ ἐπεὶ σύμμετρός ἐστιν in length with GE [Prop. 10.18]. Let GH , EK, and FL ἡ ΑΕ τῇ ΑΒ μήκει, ῥητόν ἐστι τὸ ΑΚ· καί ἐστιν ἴσον τοῖς have been drawn parallel to AB, and let the rest (of the ἀπὸ τῶν ΜΝ, ΝΞ· ῥητὸν ἄρα [ἐστὶ] καὶ τὸ συγκείμενον ἐκ construction) have been made the same as the (proposi- τῶν ἀπὸ τῶν ΜΝ, ΝΞ. καὶ ἐπεὶ ἀσύμμετρός [ἐστιν] ἡ ΔΕ tion) before this. So, it is clear that MO is the square-root τῇ ΑΒ μήκει, τουτέστι τῇ ΕΚ, ἀλλὰ ἡ ΔΕ σύμμετρός ἐστι of area AC. So, we must show that MO is the irrational τῇ ΕΖ, ἀσύμμετρος ἄρα ἡ ΕΖ τῇ ΕΚ μήκει. αἱ ΕΚ, ΕΖ (straight-line which is) called major. ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· μέσον ἄρα τὸ ΛΕ, Since AG is incommensurable in length with EG, AH τουτέστι τὸ ΜΡ. καὶ περιέχεται ὑπὸ τῶν ΜΝ, ΝΞ· μέσον is also incommensurable with GK—that is to say, SN ἄρα ἐστὶ τὸ ὑπὸ τῶν ΜΝ, ΝΞ. καὶ ῥητὸν τὸ [συγκείμενον] with NQ [Props. 6.1, 10.11]. Thus, MN and NO are ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ, καί εἰσιν ἀσύμμετροι αἱ ΜΝ, ΝΞ incommensurable in square. And since AE is commensu- δυνάμει. ἐὰν δὲ δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι rable in length with AB, AK is rational [Prop. 10.19]. ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων And it is equal to the (sum of the squares) on MN ῥητόν, τὸ δ᾿ ὑπ᾿ αὐτῶν μέσον, ἡ ὅλη ἄλογός ἐστιν, καλεῖται and NO. Thus, the sum of the (squares) on MN and δὲ μείζων. NO [is] also rational. And since DE [is] incommensu- ῾Η ΜΞ ἄρα ἄλογός ἐστιν ἡ καλουμένη μείζων, καὶ rable in length with AB [Prop. 10.13]—that is to say, δύναται τὸ ΑΓ χωρίον· ὅπερ ἔδει δεῖξαι. with EK—but DE is commensurable (in length) with EF , EF (is) thus incommensurable in length with EK [Prop. 10.13]. Thus, EK and EF are rational (straight- lines which are) commensurable in square only. LE— that is to say, MR—(is) thus medial [Prop. 10.21]. And it is contained by MN and NO. The (rectangle contained) by MN and NO is thus medial. And the [sum] of the (squares) on MN and NO (is) rational, and MN and NO are incommensurable in square. And if two straight- lines (which are) incommensurable in square, making the sum of the squares on them rational, and the (rectangle contained) by them medial, are added together, then the whole is the irrational (straight-line which is) called ma- jor [Prop. 10.39]. Thus, MO is the irrational (straight-line which is) called major. And (it is) the square-root of area AC. (Which is) the very thing it was required to show. † If the rational straight-line has unit length then this proposition states that the square-root of a fourth binomial straight-line is a major straight- line: i.e., a fourth binomial straight-line has a length k (1 + 1/ √ 1 + k′) whose square-root can be written ρ q [1 + k′′/(1 + k′′ 2)1/2]/2 + ρ q [1 − k′′/(1 + k′′ 2)1/2]/2, where ρ = √ k and k′′ 2 = k′. This is the length of a major straight-line (see Prop. 10.39), since ρ is rational.nhþ. Proposition 58 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line) and ὀνομάτων πέμπτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν a fifth binomial (straight-line) then the square-root of the ἡ καλουμένη ῥητὸν καὶ μέσον δυναμένη. area is the irrational (straight-line which is) called the Χωρίον γὰρ τὸ ΑΓ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ square-root of a rational plus a medial (area).† 347 STOIQEIWN iþ. ELEMENTS BOOK 10 τῆς ἐκ δύο ὀνομάτων πέμπτης τῆς ΑΔ διῃρημένης εἰς τὰ For let the area AC be contained by the rational ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΕ· (straight-line) AB and the fifth binomial (straight-line) λέγω [δή], ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ AD, which has been divided into its (component) terms καλουμένη ῥητὸν καὶ μέσον δυναμένη. at E, such that AE is the greater term. [So] I say that the square-root of area AC is the irrational (straight-line which is) called the square-root of a rational plus a me- dial (area). Ν Α Β Θ ∆ Γ Ρ Ο Η Μ Ξ Ε Κ Ζ Λ Σ Π QA B R N O P H G C D M E K F L S Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρότερον δεδειγμένοις· For let the same construction be made as that shown φανερὸν δή, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἐστὶν ἡ ΜΞ. previously. So, (it is) clear that MO is the square-root of δεικτέον δή, ὅτι ἡ ΜΞ ἐστιν ἡ ῥητὸν καὶ μέσον δυναμένη. area AC. So, we must show that MO is the square-root ᾿Επεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΗ τῇ ΗΕ, ἀσύμμετρον of a rational plus a medial (area). ἄρα ἐστὶ καὶ τὸ ΑΘ τῷ ΘΕ, τουτέστι τὸ ἀπὸ τῆς ΜΝ τῷ ἀπὸ For since AG is incommensurable (in length) with τῆς ΝΞ· αἱ ΜΝ, ΝΞ ἄρα δυνάμει εἰσὶν ἀσύμμετροι. καὶ ἐπεὶ GE [Prop. 10.18], AH is thus also incommensurable ἡ ΑΔ ἐκ δύο ὀνομάτων ἐστὶ πέμπτη, καί [ἐστιν] ἔλασσον with HE—that is to say, the (square) on MN with the αὐτῆς τμῆμα τὸ ΕΔ, σύμμετρος ἄρα ἡ ΕΔ τῇ ΑΒ μήκει. (square) on NO [Props. 6.1, 10.11]. Thus, MN and ἀλλὰ ἡ ΑΕ τῇ ΕΔ ἐστιν ἀσύμμετρος· καὶ ἡ ΑΒ ἄρα τῇ NO are incommensurable in square. And since AD is ΑΕ ἐστιν ἀσύμμετρος μήκει [αἱ ΒΑ, ΑΕ ῥηταί εἰσι δυνάμει a fifth binomial (straight-line), and ED [is] its lesser seg- μόνον σύμμετροι]· μέσον ἄρα ἐστὶ τὸ ΑΚ, τουτέστι τὸ ment, ED (is) thus commensurable in length with AB συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ. καὶ ἐπεὶ σύμμετρός [Def. 10.9]. But, AE is incommensurable (in length) ἐστιν ἡ ΔΕ τῇ ΑΒ μήκει, τουτέστι τῇ ΕΚ, ἀλλὰ ἡ ΔΕ τῇ with ED. Thus, AB is also incommensurable in length ΕΖ σύμμετρός ἐστιν, καὶ ἡ ΕΖ ἄρα τῇ ΕΚ σύμμετρός ἐστιν. with AE [BA and AE are rational (straight-lines which καὶ ῥητὴ ἡ ΕΚ· ῥητὸν ἄρα καὶ τὸ ΕΛ, τουτέστι τὸ ΜΡ, are) commensurable in square only] [Prop. 10.13]. Thus, τουτέστι τὸ ὑπὸ ΜΝΞ· αἱ ΜΝ, ΝΞ ἄρα δυνάμει ἀσύμμετροί AK—that is to say, the sum of the (squares) on MN εἰσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τε- and NO—is medial [Prop. 10.21]. And since DE is τραγώνων μέσον, τὸ δ᾿ ὑπ᾿ αὐτῶν ῥητόν. commensurable in length with AB—that is to say, with ῾Η ΜΞ ἄρα ῥητὸν καὶ μέσον δυναμένη ἐστὶ καὶ δύναται EK—but, DE is commensurable (in length) with EF , τὸ ΑΓ χωρίον· ὅπερ ἔδει δεῖξαι. EF is thus also commensurable (in length) with EK [Prop. 10.12]. And EK (is) rational. Thus, EL—that is to say, MR—that is to say, the (rectangle contained) by MNO—(is) also rational [Prop. 10.19]. MN and NO are thus (straight-lines which are) incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them rational. Thus, MO is the square-root of a rational plus a me- dial (area) [Prop. 10.40]. And (it is) the square-root of area AC. (Which is) the very thing it was required to show. † If the rational straight-line has unit length then this proposition states that the square-root of a fifth binomial straight-line is the square root of a rational plus a medial area: i.e., a fifth binomial straight-line has a length k ( √ 1 + k′ + 1) whose square-root can be written ρ q [(1 + k′′ 2)1/2 + k′′]/[2 (1 + k′′ 2)] + ρ q [(1 + k′′ 2)1/2 − k′′]/[2 (1 + k′′ 2)], where ρ = p k (1 + k′′ 2) and k′′ 2 = k′. This is the length of 348 STOIQEIWN iþ. ELEMENTS BOOK 10 the square root of a rational plus a medial area (see Prop. 10.40), since ρ is rational.njþ. Proposition 59 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line) and ὀνομάτων ἕκτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ a sixth binomial (straight-line) then the square-root of καλουμένη δύο μέσα δυναμένη. the area is the irrational (straight-line which is) called the square-root of (the sum of) two medial (areas).† Ν Α Β Θ ∆ Γ Ρ Ο Η Μ Ξ Ε Κ Ζ Λ Σ Π QA B R N O P H G C D M E K F L S Χωρίον γὰρ τὸ ΑΒΓΔ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ For let the area ABCD be contained by the rational καὶ τῆς ἐκ δύο ὀνομάτων ἕκτης τῆς ΑΔ διῃρημένης εἰς τὰ (straight-line) AB and the sixth binomial (straight-line) ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΕ· AD, which has been divided into its (component) terms λέγω, ὅτι ἡ τὸ ΑΓ δυναμένη ἡ δύο μέσα δυναμένη ἐστίν. at E, such that AE is the greater term. So, I say that the Κατεσκευάσθω [γὰρ] τὰ αὐτὰ τοῖς προδεδειγμένοις. square-root of AC is the square-root of (the sum of) two φανερὸν δή, ὅτι [ἡ] τὸ ΑΓ δυναμένη ἐστὶν ἡ ΜΞ, καὶ medial (areas). ὅτι ἀσύμμετρός ἐστιν ἡ ΜΝ τῇ ΝΞ δυνάμει. καὶ ἐπεὶ [For] let the same construction be made as that shown ἀσύμμετρός ἐστιν ἡ ΕΑ τῇ ΑΒ μήκει, αἱ ΕΑ, ΑΒ ἄρα previously. So, (it is) clear that MO is the square-root of ῥηταί εἰσι δυνάμει μόνον σύμμετροι· μέσον ἄρα ἐστὶ τὸ ΑΚ, AC, and that MN is incommensurable in square with τουτέστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ. πάλιν, NO. And since EA is incommensurable in length with ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΕΔ τῇ ΑΒ μήκει, ἀσύμμετρος ἄρα AB [Def. 10.10], EA and AB are thus rational (straight- ἐστὶ καὶ ἡ ΖΕ τῇ ΕΚ· αἱ ΖΕ, ΕΚ ἄρα ῥηταί εἰσι δυνάμει lines which are) commensurable in square only. Thus, μόνον σύμμετροι· μέσον ἄρα ἐστὶ τὸ ΕΛ, τουτέστι τὸ ΜΡ, AK—that is to say, the sum of the (squares) on MN τουτέστι τὸ ὑπὸ τῶν ΜΝΞ. καὶ ἐπεὶ ἀσύμμετρος ἡ ΑΕ τῇ and NO—is medial [Prop. 10.21]. Again, since ED ΕΖ, καὶ τὸ ΑΚ τῷ ΕΛ ἀσύμμετρόν ἐστιν. ἀλλὰ τὸ μὲν is incommensurable in length with AB [Def. 10.10], ΑΚ ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ, τὸ FE is thus also incommensurable (in length) with EK δὲ ΕΛ ἐστι τὸ ὑπὸ τῶν ΜΝΞ· ἀσύμμετρον ἄρα ἐστὶ τὸ [Prop. 10.13]. Thus, FE and EK are rational (straight- συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝΞ τῷ ὑπὸ τῶν ΜΝΞ. καί lines which are) commensurable in square only. Thus, ἐστι μέσον ἑκάτερον αὐτῶν, καὶ αἱ ΜΝ, ΝΞ δυνάμει εἰσὶν EL—that is to say, MR—that is to say, the (rectangle ἀσύμμετροι. contained) by MNO—is medial [Prop. 10.21]. And since ῾Η ΜΞ ἄρα δύο μέσα δυναμένη ἐστὶ καὶ δύναται τὸ ΑΓ· AE is incommensurable (in length) with EF , AK is also ὅπερ ἔδει δεῖξαι. incommensurable with EL [Props. 6.1, 10.11]. But, AK is the sum of the (squares) on MN and NO, and EL is the (rectangle contained) by MNO. Thus, the sum of the (squares) on MNO is incommensurable with the (rect- angle contained) by MNO. And each of them is medial. And MN and NO are incommensurable in square. Thus, MO is the square-root of (the sum of) two me- dial (areas) [Prop. 10.41]. And (it is) the square-root of AC. (Which is) the very thing it was required to show. † If the rational straight-line has unit length then this proposition states that the square-root of a sixth binomial straight-line is the square root of the sum of two medial areas: i.e., a sixth binomial straight-line has a length √ k + √ k′ whose square-root can be written k1/4 „ q [1 + k′′/(1 + k′′ 2)1/2]/2 + q [1 − k′′/(1 + k′′ 2)1/2]/2 « , where k′′ 2 = (k − k′)/k′. This is the length of the square-root of the sum of 349 STOIQEIWN iþ. ELEMENTS BOOK 10 two medial areas (see Prop. 10.41).L¨mma. Lemma ᾿Εὰν εὐθεῖα γραμμὴ τμηθῇ εἰς ἄνισα, τὰ ἀπὸ τῶν ἀνίσων If a straight-line is cut unequally then (the sum of) the τετράγωνα μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ἀνίσων περιε- squares on the unequal (parts) is greater than twice the χομένου ὀρθογωνίου. rectangle contained by the unequal (parts). ∆Α Γ Β CA BD ῎Εστω εὐθεῖα ἡ ΑΒ καὶ τετμήσθω εἰς ἄνισα κατὰ τὸ Let AB be a straight-line, and let it have been cut Γ, καὶ ἔστω μείζων ἡ ΑΓ· λέγω, ὅτι τὰ ἀπὸ τῶν ΑΓ, ΓΒ unequally at C, and let AC be greater (than CB). I say μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. that (the sum of) the (squares) on AC and CB is greater Τετμήσθω γὰρ ἡ ΑΒ δίχα κατὰ τὸ Δ. ἐπεὶ οὖν εὐθεῖα than twice the (rectangle contained) by AC and CB. γραμμὴ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Δ, εἰς δὲ ἄνισα κατὰ For let AB have been cut in half at D. Therefore, τὸ Γ, τὸ ἄρα ὑπὸ τῶν ΑΓ, ΓΒ μετὰ τοῦ ἀπὸ ΓΔ ἴσον ἐστὶ since a straight-line has been cut into equal (parts) at D, τῷ ἀπὸ ΑΔ· ὥστε τὸ ὑπὸ τῶν ΑΓ, ΓΒ ἔλαττόν ἐστι τοῦ and into unequal (parts) at C, the (rectangle contained) ἀπὸ ΑΔ· τὸ ἄρα δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἔλαττον ἢ διπλάσιόν by AC and CB, plus the (square) on CD, is thus equal ἐστι τοῦ ἀπὸ ΑΔ. ἀλλὰ τὰ ἀπὸ τῶν ΑΓ, ΓΒ διπλάσιά [ἐστι] to the (square) on AD [Prop. 2.5]. Hence, the (rectangle τῶν ἀπὸ τῶν ΑΔ, ΔΓ· τὰ ἄρα ἀπὸ τῶν ΑΓ, ΓΒ μείζονά contained) by AC and CB is less than the (square) on ἐστι τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ· ὅπερ ἔδει δεῖξαι. AD. Thus, twice the (rectangle contained) by AC and CB is less than double the (square) on AD. But, (the sum of) the (squares) on AC and CB [is] double (the sum of) the (squares) on AD and DC [Prop. 2.9]. Thus, (the sum of) the (squares) on AC and CB is greater than twice the (rectangle contained) by AC and CB. (Which is) the very thing it was required to show.xþ. Proposition 60 Τὸ ἀπὸ τῆς ἐκ δύο ὀνομάτων παρὰ ῥητὴν παρα- The square on a binomial (straight-line) applied to a βαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων πρώτην. rational (straight-line) produces as breadth a first bino- mial (straight-line).† Ε Γ ΒΑ ΛΘ Ξ Μ Ν Ζ ΗΚ∆ H BA L M NKD E C G FO ῎Εστω ἐκ δύο ὀνομάτων ἡ ΑΒ διῃρημένη εἰς τὰ ὀνόματα Let AB be a binomial (straight-line), having been di- κατὰ τὸ Γ, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΓ, καὶ ἐκκείσθω vided into its (component) terms at C, such that AC is ῥητὴ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΔΕ παρα- the greater term. And let the rational (straight-line) DE βεβλήσθω τὸ ΔΕΖΗ πλάτος ποιοῦν τὴν ΔΗ· λέγω, ὅτι ἡ be laid down. And let the (rectangle) DEFG, equal to ΔΗ ἐκ δύο ὀνομάτων ἐστὶ πρώτη. the (square) on AB, have been applied to DE, producing Παραβεβλήσθω γὰρ παρὰ τὴν ΔΕ τῷ μὲν ἀπὸ τῆς ΑΓ DG as breadth. I say that DG is a first binomial (straight- ἴσον τὸ ΔΘ, τῷ δὲ ἀπὸ τῆς ΒΓ ἴσον τὸ ΚΛ· λοιπὸν ἄρα line). τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον ἐστὶ τῷ ΜΖ. τετμήσθω ἡ For let DH , equal to the (square) on AC, and KL, ΜΗ δίχα κατὰ τὸ Ν, καὶ παράλληλος ἤχθω ἡ ΝΞ [ἑκατέρᾳ equal to the (square) on BC, have been applied to DE. 350 STOIQEIWN iþ. ELEMENTS BOOK 10 τῶν ΜΛ, ΗΖ]. ἑκάτερον ἄρα τῶν ΜΞ, ΝΖ ἴσον ἐστὶ τῷ Thus, the remaining twice the (rectangle contained) by ἅπαξ ὑπὸ τῶν ΑΓΒ. καὶ ἐπεὶ ἐκ δύο ὀνομάτων ἐστὶν ἡ ΑΒ AC and CB is equal to MF [Prop. 2.4]. Let MG have διῃρημένη εἰς τὰ ὀνόματα κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα ῥηταί been cut in half at N , and let NO have been drawn par- εἰσι δυνάμει μόνον σύμμετροι· τὰ ἄρα ἀπὸ τῶν ΑΓ, ΓΒ ῥητά allel [to each of ML and GF ]. MO and NF are thus ἐστι καὶ σύμμετρα ἀλλήλοις· ὥστε καὶ τὸ συγκείμενον ἐκ each equal to once the (rectangle contained) by ACB. τῶν ἀπὸ τῶν ΑΓ, ΓΒ. καί ἐστιν ἴσον τῷ ΔΛ· ῥητὸν ἄρα And since AB is a binomial (straight-line), having been ἐστὶ τὸ ΔΛ. καὶ παρὰ ῥητὴν τὴν ΔΕ παράκειται· ῥητὴ ἄρα divided into its (component) terms at C, AC and CB are ἐστὶν ἡ ΔΜ καὶ σύμμετρος τῇ ΔΕ μήκει. πάλιν, ἐπεὶ αἱ ΑΓ, thus rational (straight-lines which are) commensurable ΓΒ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, μέσον ἄρα ἐστὶ in square only [Prop. 10.36]. Thus, the (squares) on AC τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, τουτέστι τὸ ΜΖ. καὶ παρὰ ῥητὴν and CB are rational, and commensurable with one an- τὴν ΜΛ παράκειται· ῥητὴ ἄρα καὶ ἡ ΜΗ καὶ ἀσύμμετρος other. And hence the sum of the (squares) on AC and τῇ ΜΛ, τουτέστι τῇ ΔΕ, μήκει. ἔστι δὲ καὶ ἡ ΜΔ ῥητὴ CB (is rational) [Prop. 10.15], and is equal to DL. Thus, καὶ τῇ ΔΕ μήκει σύμμετρος· ἀσύμμετρος ἄρα ἐστὶν ἡ ΔΜ DL is rational. And it is applied to the rational (straight- τῇ ΜΗ μήκει. καί εἰσι ῥηταί· αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι line) DE. DM is thus rational, and commensurable in δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ length with DE [Prop. 10.20]. Again, since AC and CB ΔΗ. δεικτέον δή, ὅτι καὶ πρώτη. are rational (straight-lines which are) commensurable in ᾿Επεὶ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον ἀνάλογόν ἐστι τὸ square only, twice the (rectangle contained) by AC and ὑπὸ τῶν ΑΓΒ, καὶ τῶν ΔΘ, ΚΛ ἄρα μέσον ἀνάλογόν ἐστι CB—that is to say, MF—is thus medial [Prop. 10.21]. τὸ ΜΞ. ἔστιν ἄρα ὡς τὸ ΔΘ πρὸς τὸ ΜΞ, οὕτως τὸ ΜΞ And it is applied to the rational (straight-line) ML. MG πρὸς τὸ ΚΛ, τουτέστιν ὡς ἡ ΔΚ πρὸς τὴν ΜΝ, ἡ ΜΝ πρὸς is thus also rational, and incommensurable in length τὴν ΜΚ· τὸ ἄρα ὑπὸ τῶν ΔΚ, ΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς with ML—that is to say, with DE [Prop. 10.22]. And ΜΝ. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ ἀπὸ τῆς MD is also rational, and commensurable in length with ΓΒ, σύμμετρόν ἐστι καὶ τὸ ΔΘ τῷ ΚΛ· ὥστε καὶ ἡ ΔΚ DE. Thus, DM is incommensurable in length with MG τῇ ΚΜ σύμμετρός ἐστιν. καὶ ἐπεὶ μείζονά ἐστι τὰ ἀπὸ τῶν [Prop. 10.13]. And they are rational. DM and MG are ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, μεῖζον ἄρα καὶ τὸ ΔΛ thus rational (straight-lines which are) commensurable τοῦ ΜΖ· ὥστε καὶ ἡ ΔΜ τῆς ΜΗ μείζων ἐστίν. καί ἐστιν in square only. Thus, DG is a binomial (straight-line) ἴσον τὸ ὑπὸ τῶν ΔΚ, ΚΜ τῷ ἀπὸ τῆς ΜΝ, τουτέστι τῷ [Prop. 10.36]. So, we must show that (it is) also a first τετάρτῳ τοῦ ἀπὸ τῆς ΜΗ, καὶ σύμμετρος ἡ ΔΚ τῇ ΚΜ. (binomial straight-line). ἐὰν δὲ ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ Since the (rectangle contained) by ACB is the τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον mean proportional to the squares on AC and CB εἴδει τετραγώνῳ καὶ εἰς σύμμετρα αὐτὴν διαιρῇ, ἡ μείζων [Prop. 10.53 lem.], MO is thus also the mean propor- τῆς ἐλάσσονος μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ· ἡ tional to DH and KL. Thus, as DH is to MO, so ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. MO (is) to KL—that is to say, as DK (is) to MN , καί εἰσι ῥηταὶ αἱ ΔΜ, ΜΗ, καὶ ἡ ΔΜ μεῖζον ὄνομα οὖσα (so) MN (is) to MK [Prop. 6.1]. Thus, the (rectan- σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΔΕ μήκει. gle contained) by DK and KM is equal to the (square) ῾Η ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πρώτη· ὅπερ ἔδει on MN [Prop. 6.17]. And since the (square) on AC is δεῖξαι. commensurable with the (square) on CB, DH is also commensurable with KL. Hence, DK is also commensu- rable with KM [Props. 6.1, 10.11]. And since (the sum of) the squares on AC and CB is greater than twice the (rectangle contained) by AC and CB [Prop. 10.59 lem.], DL (is) thus also greater than MF . Hence, DM is also greater than MG [Props. 6.1, 5.14]. And the (rectan- gle contained) by DK and KM is equal to the (square) on MN—that is to say, to one quarter the (square) on MG. And DK (is) commensurable (in length) with KM . And if there are two unequal straight-lines, and a (rect- angle) equal to the fourth part of the (square) on the lesser, falling short by a square figure, is applied to the greater, and divides it into (parts which are) commensu- rable (in length), then the square on the greater is larger 351 STOIQEIWN iþ. ELEMENTS BOOK 10 than (the square on) the lesser by the (square) on (some straight-line) commensurable (in length) with the greater [Prop. 10.17]. Thus, the square on DM is greater than (the square on) MG by the (square) on (some straight- line) commensurable (in length) with (DM). And DM and MG are rational. And DM , which is the greater term, is commensurable in length with the (previously) laid down rational (straight-line) DE. Thus, DG is a first binomial (straight-line) [Def. 10.5]. (Which is) the very thing it was required to show. † In other words, the square of a binomial is a first binomial. See Prop. 10.54.xaþ. Proposition 61 Τὸ ἀπὸ τῆς ἐκ δύο μέσων πρώτης παρὰ ῥητὴν παρα- The square on a first bimedial (straight-line) applied βαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων δευτέραν. to a rational (straight-line) produces as breadth a second binomial (straight-line).† Ε Γ ΒΑ ΛΘ Ξ Μ Ν Ζ ΗΚ∆ H BA L M NKD E C G FO ῎Εστω ἐκ δύο μέσων πρώτη ἡ ΑΒ διῃρημένη εἰς τὰς Let AB be a first bimedial (straight-line) having been μέσας κατὰ τὸ Γ, ὧν μείζων ἡ ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ divided into its (component) medial (straight-lines) at ΔΕ, καὶ παραβεβλήσθω παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον C, of which AC (is) the greater. And let the rational παραλληλόγραμμον τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ· λέγω, (straight-line) DE be laid down. And let the parallelo- ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ δευτέρα. gram DF , equal to the (square) on AB, have been ap- Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρὸ τούτου. καὶ plied to DE, producing DG as breadth. I say that DG is ἐπεὶ ἡ ΑΒ ἐκ δύο μέσων ἐστὶ πρώτη διῃρημένη κατὰ τὸ a second binomial (straight-line). Γ, αἱ ΑΓ, ΓΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι For let the same construction have been made as ῥητὸν περιέχουσαι· ὥστε καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ μέσα in the (proposition) before this. And since AB is a ἐστίν. μέσον ἄρα ἐστὶ τὸ ΔΛ. καὶ παρὰ ῥητὴν τὴν ΔΕ πα- first bimedial (straight-line), having been divided at C, ραβέβληται· ῥητὴ ἄρα ἐστίν ἡ ΜΔ καὶ ἀσύμμετρος τῇ ΔΕ AC and CB are thus medial (straight-lines) commen- μήκει. πάλιν, ἐπεὶ ῥητόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, ῥητόν surable in square only, and containing a rational (area) ἐστι καὶ τὸ ΜΖ. καὶ παρὰ ῥητὴν τὴν ΜΛ παράκειται· ῥητὴ [Prop. 10.37]. Hence, the (squares) on AC and CB ἄρα [ἐστὶ] καὶ ἡ ΜΗ καὶ μήκει σύμμετρος τῇ ΜΛ, τουτέστι are also medial [Prop. 10.21]. Thus, DL is medial τῇ ΔΕ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΔΜ τῇ ΜΗ μήκει. καί εἰσι [Props. 10.15, 10.23 corr.]. And it has been applied to the ῥηταί· αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· rational (straight-line) DE. MD is thus rational, and in- ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. δεικτέον δή, ὅτι καὶ commensurable in length with DE [Prop. 10.22]. Again, δευτέρα. since twice the (rectangle contained) by AC and CB is ᾿Επεὶ γὰρ τὰ ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τοῦ δὶς ὑπὸ rational, MF is also rational. And it is applied to the τῶν ΑΓ, ΓΒ, μεῖζον ἄρα καὶ τὸ ΔΛ τοῦ ΜΖ· ὥστε καὶ ἡ rational (straight-line) ML. Thus, MG [is] also ratio- ΔΜ τῆς ΜΗ. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ nal, and commensurable in length with ML—that is to ἀπὸ τῆς ΓΒ, σύμμετρόν ἐστι καὶ τὸ ΔΘ τῷ ΚΛ· ὥστε καὶ say, with DE [Prop. 10.20]. DM is thus incommensu- ἡ ΔΚ τῇ ΚΜ σύμμετρός ἐστιν. καί ἐστι τὸ ὑπὸ τῶν ΔΚΜ rable in length with MG [Prop. 10.13]. And they are ἴσον τῷ ἀπὸ τῆς ΜΝ· ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ rational. DM and MG are thus rational, and commensu- 352 STOIQEIWN iþ. ELEMENTS BOOK 10 ἀπὸ συμμέτρου ἑαυτῇ. καί ἐστιν ἡ ΜΗ σύμμετρος τῇ ΔΕ rable in square only. DG is thus a binomial (straight-line) μήκει. [Prop. 10.36]. So, we must show that (it is) also a second ῾Η ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ δευτέρα. (binomial straight-line). For since (the sum of) the squares on AC and CB is greater than twice the (rectangle contained) by AC and CB [Prop. 10.59], DL (is) thus also greater than MF . Hence, DM (is) also (greater) than MG [Prop. 6.1]. And since the (square) on AC is commensurable with the (square) on CB, DH is also commensurable with KL. Hence, DK is also commensurable (in length) with KM [Props. 6.1, 10.11]. And the (rectangle contained) by DKM is equal to the (square) on MN . Thus, the square on DM is greater than (the square on) MG by the (square) on (some straight-line) commensurable (in length) with (DM) [Prop. 10.17]. And MG is commen- surable in length with DE. Thus, DG is a second binomial (straight-line) [Def. 10.6]. †In other words, the square of a first bimedial is a second binomial. See Prop. 10.55.xbþ. Proposition 62 Τὸ ἀπὸ τῆς ἐκ δύο μέσων δευτέρας παρὰ ῥητὴν παρα- The square on a second bimedial (straight-line) ap- βαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων τρίτην. plied to a rational (straight-line) produces as breadth a third binomial (straight-line).† Ε Γ ΒΑ ΛΘ Ξ Μ Ν Ζ ΗΚ∆ H BA L M NKD E C G FO ῎Εστω ἐκ δύο μέσων δευτέρα ἡ ΑΒ διῃρημένη εἰς τὰς Let AB be a second bimedial (straight-line) having μέσας κατὰ τὸ Γ, ὥστε τὸ μεῖζον τμῆμα εἶναι τὸ ΑΓ, ῥητὴ been divided into its (component) medial (straight-lines) δέ τις ἔστω ἡ ΔΕ, καὶ παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον at C, such that AC is the greater segment. And let DE be παραλληλόγραμμον παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν some rational (straight-line). And let the parallelogram τὴν ΔΗ· λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ τρίτη. DF , equal to the (square) on AB, have been applied to Κατεσκευάσθω τὰ αὐτὰ τοῖς προδεδειγμένοις. καὶ ἐπεὶ DE, producing DG as breadth. I say that DG is a third ἐκ δύο μέσων δευτέρα ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, binomial (straight-line). αἱ ΑΓ, ΓΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον Let the same construction be made as that shown pre- περιέχουσαι· ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, viously. And since AB is a second bimedial (straight- ΓΒ μέσον ἐστίν. καί ἐστιν ἴσον τῷ ΔΛ· μέσον ἄρα καὶ τὸ line), having been divided at C, AC and CB are thus ΔΛ. καὶ παράκειται παρὰ ῥητὴν τὴν ΔΕ· ῥητὴ ἄρα ἐστὶ καὶ ἡ medial (straight-lines) commensurable in square only, ΜΔ καὶ ἀσύμμετρος τῇ ΔΕ μήκει. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΜΗ and containing a medial (area) [Prop. 10.38]. Hence, ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΜΛ, τουτέστι τῇ ΔΕ, μήκει· the sum of the (squares) on AC and CB is also medial ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΔΜ, ΜΗ καὶ ἀσύμμετρος τῇ [Props. 10.15, 10.23 corr.]. And it is equal to DL. Thus, ΔΕ μήκει. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΓ τῇ ΓΒ μήκει, DL (is) also medial. And it is applied to the rational ὡς δὲ ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως τὸ ἀπὸ τῆς ΑΓ πρὸς τὸ (straight-line) DE. MD is thus also rational, and in- 353 STOIQEIWN iþ. ELEMENTS BOOK 10 ὑπὸ τῶν ΑΓΒ, ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΑΓ τῷ ὑπὸ commensurable in length with DE [Prop. 10.22]. So, τῶν ΑΓΒ. ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, for the same (reasons), MG is also rational, and incom- ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓΒ ἀσύμμετρόν ἐστιν, τουτέστι τὸ mensurable in length with ML—that is to say, with DE. ΔΛ τῷ ΜΖ· ὥστε καὶ ἡ ΔΜ τῷ ΜΗ ἀσύμμετρός ἐστιν. καί Thus, DM and MG are each rational, and incommen- εἰσι ῥηταί· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. δεικτέον [δή], surable in length with DE. And since AC is incommen- ὅτι καὶ τρίτη. surable in length with CB, and as AC (is) to CB, so ῾Ομοίως δὴ τοῖς προτέροις ἐπιλογιούμεθα, ὅτι μείζων the (square) on AC (is) to the (rectangle contained) by ἐστὶν ἡ ΔΜ τῆς ΜΗ, καὶ σύμμετρος ἡ ΔΚ τῇ ΚΜ. καί ἐστι ACB [Prop. 10.21 lem.], the (square) on AC (is) also in- τὸ ὑπὸ τῶν ΔΚΜ ἴσον τῷ ἀπὸ τῆς ΜΝ· ἡ ΔΜ ἄρα τῆς ΜΗ commensurable with the (rectangle contained) by ACB μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καὶ οὐδετέρα τῶν [Prop. 10.11]. And hence the sum of the (squares) on ΔΜ, ΜΗ σύμμετρός ἐστι τῇ ΔΕ μήκει. AC and CB is incommensurable with twice the (rect- ῾Η ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τρίτη· ὅπερ ἔδει δεῖξαι. angle contained) by ACB—that is to say, DL with MF [Props. 10.12, 10.13]. Hence, DM is also incommen- surable (in length) with MG [Props. 6.1, 10.11]. And they are rational. DG is thus a binomial (straight-line) [Prop. 10.36]. [So] we must show that (it is) also a third (binomial straight-line). So, similarly to the previous (propositions), we can conclude that DM is greater than MG, and DK (is) com- mensurable (in length) with KM . And the (rectangle contained) by DKM is equal to the (square) on MN . Thus, the square on DM is greater than (the square on) MG by the (square) on (some straight-line) commensu- rable (in length) with (DM) [Prop. 10.17]. And neither of DM and MG is commensurable in length with DE. Thus, DG is a third binomial (straight-line) [Def. 10.7]. (Which is) the very thing it was required to show. † In other words, the square of a second bimedial is a third binomial. See Prop. 10.56.xgþ. Proposition 63 Τὸ ἀπὸ τῆς μείζονος παρὰ ῥητὴν παραβαλλόμενον The square on a major (straight-line) applied to a ra- πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων τετάρτην. tional (straight-line) produces as breadth a fourth bino- mial (straight-line).† Ε Γ ΒΑ ΛΘ Ξ Μ Ν Ζ ΗΚ∆ H BA L M NKD E C G FO ῎Εστω μείζων ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε μείζονα Let AB be a major (straight-line) having been divided εἶναι τὴν ΑΓ τῆς ΓΒ, ῥητὴ δὲ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον at C, such that AC is greater than CB, and (let) DE παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΖ παραλληλόγραμμον (be) a rational (straight-line). And let the parallelogram πλάτος ποιοῦν τὴν ΔΗ· λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων DF , equal to the (square) on AB, have been applied to ἐστὶ τετάρτη. DE, producing DG as breadth. I say that DG is a fourth Κατεσκευάσθω τὰ αὐτὰ τοῖς προδεδειγμένοις. καὶ ἐπεὶ binomial (straight-line). μείζων ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ δυνάμει Let the same construction be made as that shown pre- 354 STOIQEIWN iþ. ELEMENTS BOOK 10 εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾿ viously. And since AB is a major (straight-line), hav- αὐτῶν τετραγώνων ῥητόν, τὸ δὲ ὑπ᾿ αὐτῶν μέσον. ἐπεὶ οὖν ing been divided at C, AC and CB are incommensu- ῥητόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, ῥητὸν rable in square, making the sum of the squares on them ἄρα ἐστὶ τὸ ΔΛ· ῥητὴ ἄρα καὶ ἡ ΔΜ καὶ σύμμετρος τῇ rational, and the (rectangle contained) by them medial ΔΕ μήκει. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, [Prop. 10.39]. Therefore, since the sum of the (squares) τουτέστι τὸ ΜΖ, καὶ παρὰ ῥητήν ἐστι τὴν ΜΛ, ῥητὴ ἄρα on AC and CB is rational, DL is thus rational. Thus, ἐστὶ καὶ ἡ ΜΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει· ἀσύμμετρος DM (is) also rational, and commensurable in length with ἄρα ἐστὶ καὶ ἡ ΔΜ τῇ ΜΗ μήκει. αἱ ΔΜ, ΜΗ ἄρα ῥηταί DE [Prop. 10.20]. Again, since twice the (rectangle con- εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν tained) by AC and CB—that is to say, MF—is medial, ἡ ΔΗ. δεικτέον [δή], ὅτι καὶ τετάρτη. and is (applied to) the rational (straight-line) ML, MG ῾Ομοίως δὴ δείξομεν τοῖς πρότερον, ὅτι μείζων ἐστὶν is thus also rational, and incommensurable in length with ἡ ΔΜ τῆς ΜΗ, καὶ ὅτι τὸ ὑπὸ ΔΚΜ ἴσον ἐστὶ τῷ ἀπὸ DE [Prop. 10.22]. DM is thus also incommensurable τῆς ΜΝ. ἐπεὶ οὖν ἀσύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ in length with MG [Prop. 10.13]. DM and MG are ἀπὸ τῆς ΓΒ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΔΘ τῷ ΚΛ· thus rational (straight-lines which are) commensurable ὥστε ἀσύμμετρος καὶ ἡ ΔΚ τῇ ΚΜ ἐστιν. ἐὰν δὲ ὦσι in square only. Thus, DG is a binomial (straight-line) δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς [Prop. 10.36]. [So] we must show that (it is) also a fourth ἐλάσσονος ἴσον παραλληλόγραμμον παρὰ τὴν μείζονα παρα- (binomial straight-line). βληθῇ ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς ἀσύμμετρα αὐτὴν So, similarly to the previous (propositions), we can διαιρῇ, ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ show that DM is greater than MG, and that the (rectan- ἀσύμμέτρου ἑαυτῇ μήκει· ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται gle contained) by DKM is equal to the (square) on MN . τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΔΜ, ΜΗ ῥηταὶ Therefore, since the (square) on AC is incommensurable δυνάμει μόνον σύμμετροι, καὶ ἡ ΔΜ σύμμετρός ἐστι τῇ with the (square) on CB, DH is also incommensurable ἐκκειμένῃ ῥητῇ τῇ ΔΕ. with KL. Hence, DK is also incommensurable with ῾Η ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τετάρτη· ὅπερ ἔδει KM [Props. 6.1, 10.11]. And if there are two unequal δεῖξαι. straight-lines, and a parallelogram equal to the fourth part of the (square) on the lesser, falling short by a square figure, is applied to the greater, and divides it into (parts which are) incommensurable (in length), then the square on the greater will be larger than (the square on) the lesser by the (square) on (some straight-line) incommen- surable in length with the greater [Prop. 10.18]. Thus, the square on DM is greater than (the square on) MG by the (square) on (some straight-line) incommensurable (in length) with (DM). And DM and MG are rational (straight-lines which are) commensurable in square only. And DM is commensurable (in length) with the (previ- ously) laid down rational (straight-line) DE. Thus, DG is a fourth binomial (straight-line) [Def. 10.8]. (Which is) the very thing it was required to show. † In other words, the square of a major is a fourth binomial. See Prop. 10.57.xdþ. Proposition 64 Τὸ ἀπὸ τῆς ῥητὸν καὶ μέσον δυναμένης παρὰ ῥητὴν πα- The square on the square-root of a rational plus a me- ραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων πέμπτην. dial (area) applied to a rational (straight-line) produces ῎Εστω ῥητὸν καὶ μέσον δυναμένη ἡ ΑΒ διῃρημένη εἰς as breadth a fifth binomial (straight-line).† τὰς εὐθείας κατὰ τὸ Γ, ὥστε μείζονα εἶναι τὴν ΑΓ, καὶ Let AB be the square-root of a rational plus a medial ἐκκείσθω ῥητὴ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν (area) having been divided into its (component) straight- ΔΕ παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ· λέγω, lines at C, such that AC is greater. And let the rational ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ πέμπτη. (straight-line) DE be laid down. And let the (parallelo- gram) DF , equal to the (square) on AB, have been ap- 355 STOIQEIWN iþ. ELEMENTS BOOK 10 plied to DE, producing DG as breadth. I say that DG is a fifth binomial straight-line. Ε Γ ΒΑ ΛΘ Ξ Μ Ν Ζ ΗΚ∆ H BA L M NKD E C G FO Κατεσκευάσθω τὰ αὐτα τοῖς πρὸ τούτου. ἐπεὶ οὖν Let the same construction be made as in the (proposi- ῥητὸν καὶ μέσον δυναμένη ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ tions) before this. Therefore, since AB is the square-root Γ, αἱ ΑΓ, ΓΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν of a rational plus a medial (area), having been divided συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον, τὸ δ᾿ ὑπ᾿ at C, AC and CB are thus incommensurable in square, αὐτῶν ῥητόν. ἐπεὶ οὖν μέσον ἐστὶ τὸ συγκείμενον ἐκ τῶν making the sum of the squares on them medial, and the ἀπὸ τῶν ΑΓ, ΓΒ, μέσον ἄρα ἐστὶ τὸ ΔΛ· ὥστε ῥητή ἐστιν (rectangle contained) by them rational [Prop. 10.40]. ἡ ΔΜ καὶ μήκει ἀσύμμετρος τῇ ΔΕ. πάλιν, ἐπεὶ ῥητόν ἐστι Therefore, since the sum of the (squares) on AC and τὸ δὶς ὑπὸ τῶν ΑΓΒ, τουτέστι τὸ ΜΖ, ῥητὴ ἄρα ἡ ΜΗ καὶ CB is medial, DL is thus medial. Hence, DM is rational σύμμετρος τῇ ΔΕ. ἀσύμμετρος ἄρα ἡ ΔΜ τῇ ΜΗ· αἱ ΔΜ, and incommensurable in length with DE [Prop. 10.22]. ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα Again, since twice the (rectangle contained) by ACB— ὀνομάτων ἐστὶν ἡ ΔΗ. λέγω δή, ὅτι καὶ πέμπτη. that is to say, MF—is rational, MG (is) thus rational ῾Ομοίως γὰρ διεχθήσεται, ὅτι τὸ ὑπὸ τῶν ΔΚΜ ἴσον and commensurable (in length) with DE [Prop. 10.20]. ἐστὶ τῷ ἀπὸ τῆς ΜΝ, καὶ ἀσύμμετρος ἡ ΔΚ τῇ ΚΜ μήκει· ἡ DM (is) thus incommensurable (in length) with MG ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. [Prop. 10.13]. Thus, DM and MG are rational (straight- καί εἰσιν αἱ ΔΜ, ΜΗ [ῥηταὶ] δυνάμει μόνον σύμμετροι, καὶ lines which are) commensurable in square only. Thus, ἡ ἐλάσσων ἡ ΜΗ σύμμετρος τῇ ΔΕ μήκει. DG is a binomial (straight-line) [Prop. 10.36]. So, I say ῾Η ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πέμπτη· ὅπερ ἔδει that (it is) also a fifth (binomial straight-line). δεῖξαι. For, similarly (to the previous propositions), it can be shown that the (rectangle contained) by DKM is equal to the (square) on MN , and DK (is) incommen- surable in length with KM . Thus, the square on DM is greater than (the square on) MG by the (square) on (some straight-line) incommensurable (in length) with (DM) [Prop. 10.18]. And DM and MG are [rational] (straight-lines which are) commensurable in square only, and the lesser MG is commensurable in length with DE. Thus, DG is a fifth binomial (straight-line) [Def. 10.9]. (Which is) the very thing it was required to show. † In other words, the square of the square-root of a rational plus medial is a fifth binomial. See Prop. 10.58.xeþ. Proposition 65 Τὸ ἀπὸ τῆς δύο μέσα δυναμένης παρὰ ῥητὴν παρα- The square on the square-root of (the sum of) two me- βαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων ἕκτην. dial (areas) applied to a rational (straight-line) produces ῎Εστω δύο μέσα δυναμένη ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, as breadth a sixth binomial (straight-line).† ῥητὴ δὲ ἔστω ἡ ΔΕ, καὶ παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον Let AB be the square-root of (the sum of) two me- παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ· λέγω, ὅτι ἡ dial (areas), having been divided at C. And let DE be a ΔΗ ἐκ δύο ὀνομάτων ἐστὶν ἕκτη. rational (straight-line). And let the (parallelogram) DF , equal to the (square) on AB, have been applied to DE, 356 STOIQEIWN iþ. ELEMENTS BOOK 10 producing DG as breadth. I say that DG is a sixth bino- mial (straight-line). Ε Γ ΒΑ ΛΘ Ξ Μ Ν Ζ ΗΚ∆ H BA L M NKD E C G FO Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρότερον. καὶ ἐπεὶ ἡ For let the same construction be made as in the pre- ΑΒ δύο μέσα δυναμένη ἐστὶ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ vious (propositions). And since AB is the square-root ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον of (the sum of) two medial (areas), having been divided ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπ᾿ αὐτῶν at C, AC and CB are thus incommensurable in square, μέσον καὶ ἔτι ἀσύμμετρον τὸ ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων making the sum of the squares on them medial, and the συγκείμενον τῷ ὑπ᾿ αὐτῶν· ὥστε κατὰ τὰ προδεδειγμένα (rectangle contained) by them medial, and, moreover, μέσον ἐστὶν ἑκάτερον τῶν ΔΛ, ΜΖ. καὶ παρὰ ῥητὴν τὴν the sum of the squares on them incommensurable with ΔΕ παράκειται· ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΔΜ, ΜΗ καὶ the (rectangle contained) by them [Prop. 10.41]. Hence, ἀσύμμετρος τῇ ΔΕ μήκει. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ according to what has been previously demonstrated, DL συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, and MF are each medial. And they are applied to the ΓΒ, ἀσύμμετρον ἄρα ἐστὶ τὸ ΔΛ τῷ ΜΖ. ἀσύμμετρος ἄρα rational (straight-line) DE. Thus, DM and MG are καὶ ἡ ΔΜ τῇ ΜΗ· αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον each rational, and incommensurable in length with DE σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. λέγω δή, ὅτι [Prop. 10.22]. And since the sum of the (squares) on καὶ ἕκτη. AC and CB is incommensurable with twice the (rectan- ῾Ομοίως δὴ πάλιν δεῖξομεν, ὅτι τὸ ὑπὸ τῶν ΔΚΜ ἴσον gle contained) by AC and CB, DL is thus incommensu- ἐστὶ τῷ ἀπὸ τῆς ΜΝ, καὶ ὅτι ἡ ΔΚ τῇ ΚΜ μήκει ἐστὶν rable with MF . Thus, DM (is) also incommensurable (in ἀσύμμετρος· καὶ διὰ τὰ αὐτὰ δὴ ἡ ΔΜ τῆς ΜΗ μεῖζον length) with MG [Props. 6.1, 10.11]. DM and MG are δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει. καὶ οὐδετέρα τῶν thus rational (straight-lines which are) commensurable ΔΜ, ΜΗ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΔΕ μήκει. in square only. Thus, DG is a binomial (straight-line) ῾Η ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶν ἕκτη· ὅπερ ἔδει [Prop. 10.36]. So, I say that (it is) also a sixth (binomial δεῖξαι. straight-line). So, similarly (to the previous propositions), we can again show that the (rectangle contained) by DKM is equal to the (square) on MN , and that DK is incom- mensurable in length with KM . And so, for the same (reasons), the square on DM is greater than (the square on) MG by the (square) on (some straight-line) incom- mensurable in length with (DM) [Prop. 10.18]. And nei- ther of DM and MG is commensurable in length with the (previously) laid down rational (straight-line) DE. Thus, DG is a sixth binomial (straight-line) [Def. 10.10]. (Which is) the very thing it was required to show. † In other words, the square of the square-root of two medials is a sixth binomial. See Prop. 10.59.x�þ. Proposition 66 ῾Η τῇ ἐκ δύο ὀνομάτων μήκει σύμμετρος καὶ αὐτὴ ἐκ A (straight-line) commensurable in length with a bi- δύο ὀνομάτων ἐστὶ καὶ τῇ τάξει ἡ αὐτή. nomial (straight-line) is itself also binomial, and the same ῎Εστω ἐκ δύο ὀνομάτων ἡ ΑΒ, καὶ τῇ ΑΒ μήκει in order. 357 STOIQEIWN iþ. ELEMENTS BOOK 10 σύμμετρος ἔστω ἡ ΓΔ· λέγω, ὅτι ἡ ΓΔ ἐκ δύο ὀνομάτων Let AB be a binomial (straight-line), and let CD be ἐστὶ καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ. commensurable in length with AB. I say that CD is a bi- nomial (straight-line), and (is) the same in order as AB. Β Γ Ζ ∆ Α Ε F D A E B C ᾿Επεὶ γὰρ ἐκ δύο ὀνομάτων ἐστὶν ἡ ΑΒ, διῃρήσθω εἰς τὰ For since AB is a binomial (straight-line), let it have ὀνόματα κατὰ τὸ Ε, καὶ ἔστω μεῖζον ὄνομα τὸ ΑΕ· αἱ ΑΕ, been divided into its (component) terms at E, and let ΕΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. γεγονέτω ὡς AE be the greater term. AE and EB are thus rational ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΑΕ πρὸς τὴν ΓΖ· καὶ λοιπὴ (straight-lines which are) commensurable in square only ἄρα ἡ ΕΒ πρὸς λοιπὴν τὴν ΖΔ ἐστιν, ὡς ἡ ΑΒ πρὸς τὴν [Prop. 10.36]. Let it have been contrived that as AB (is) ΓΔ. σύμμετρος δὲ ἡ ΑΒ τῇ ΓΔ μήκει· σύμμετρος ἄρα ἐστὶ to CD, so AE (is) to CF [Prop. 6.12]. Thus, the remain- καὶ ἡ μὲν ΑΕ τῇ ΓΖ, ἡ δὲ ΕΒ τῇ ΖΔ. καί εἰσι ῥηταὶ αἱ ΑΕ, der EB is also to the remainder FD, as AB (is) to CD ΕΒ· ῥηταὶ ἄρα εἰσὶ καὶ αἱ ΓΖ, ΖΔ. καὶ ἐστιν ὡς ἡ ΑΕ πρὸς [Props. 6.16, 5.19 corr.]. And AB (is) commensurable ΓΖ, ἡ ΕΒ πρὸς ΖΔ. ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΕ πρὸς ΕΒ, ἡ in length with CD. Thus, AE is also commensurable ΓΖ πρὸς ΖΔ. αἱ δὲ ΑΕ, ΕΒ δυνάμει μόνον [εἰσὶ] σύμμετροι· (in length) with CF , and EB with FD [Prop. 10.11]. καὶ αἱ ΓΖ, ΖΔ ἄρα δυνάμει μόνον εἰσὶ σύμμετροι. καί εἰσι And AE and EB are rational. Thus, CF and FD are ῥηταί· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΓΔ. λέγω δή, ὅτι τῇ also rational. And as AE is to CF , (so) EB (is) to FD τάξει ἐστὶν ἡ αὐτὴ τῇ ΑΒ. [Prop. 5.11]. Thus, alternately, as AE is to EB, (so) ῾Η γὰρ ΑΕ τῆς ΕΒ μεῖζον δύναται ἤτοι τῷ ἀπὸ CF (is) to FD [Prop. 5.16]. And AE and EB [are] συμμέτρου ἑαυτῇ ἢ τῷ ἀπὸ ἀσυμμέτρου. εἰ μὲν οὖν ἡ commensurable in square only. Thus, CF and FD are ΑΕ τῆς ΕΒ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ also commensurable in square only [Prop. 10.11]. And ἡ ΓΖ τῆς ΖΔ μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ. they are rational. CD is thus a binomial (straight-line) καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΑΕ τῇ ἐκκειμένῃ ῥητῇ, καὶ [Prop. 10.36]. So, I say that it is the same in order as ἡ ΓΖ σύμμετρος αὐτῇ ἔσται, καὶ διὰ τοῦτο ἑκατέρα τῶν AB. ΑΒ, ΓΔ ἐκ δύο ὀνομάτων ἐστὶ πρώτη, τουτέστι τῇ τάξει For the square on AE is greater than (the square on) ἡ αὐτή. εἰ δὲ ἡ ΕΒ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, EB by the (square) on (some straight-line) either com- καὶ ἡ ΖΔ σύμμετρός ἐστιν αὐτῇ, καὶ διὰ τοῦτο πάλιν τῇ mensurable or incommensurable (in length) with (AE). τάξει ἡ αὐτὴ ἔσται τῇ ΑΒ· ἑκατέρα γὰρ αὐτῶν ἔσται ἐκ δύο Therefore, if the square on AE is greater than (the square ὀνομάτων δευτέρα. εἰ δὲ οὐδετέρα τῶν ΑΕ, ΕΒ σύμμετρός on) EB by the (square) on (some straight-line) com- ἐστι τῇ ἐκκειμένῃ ῥητῇ, οὐδετέρα τῶν ΓΖ, ΖΔ σύμμετρος mensurable (in length) with (AE) then the square on αὐτῇ ἔσται, καί ἐστιν ἑκατέρα τρίτη. εἰ δὲ ἡ ΑΕ τῆς ΕΒ CF will also be greater than (the square on) FD by μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ἡ ΓΖ τὴς ΖΔ the (square) on (some straight-line) commensurable (in μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καὶ εἰ μὲν ἡ ΑΕ length) with (CF ) [Prop. 10.14]. And if AE is com- σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῃ, καὶ ἡ ΓΖ σύμμετρός mensurable (in length) with (some previously) laid down ἐστιν αὐτῇ, καὶ ἐστιν ἑκατέρα τετάρτη. εἰ δὲ ἡ ΕΒ, καὶ rational (straight-line) then CF will also be commensu- ἡ ΖΔ, καὶ ἔσται ἑκατέρα πέμπτη. εἰ δὲ οὐδετέρα τῶν ΑΕ, rable (in length) with it [Prop. 10.12]. And, on account ΕΒ, καὶ τῶν ΓΖ, ΖΔ οὐδετέρα σύμμετρός ἐστι τῇ ἐκκειμένῃ of this, AB and CD are each first binomial (straight- ῥητῇ, καὶ ἔσται ἑκατέρα ἕκτη. lines) [Def. 10.5]—that is to say, the same in order. And if ῞Ωστε ἡ τῇ ἐκ δύο ὀνομάτων μήκει σύμμετρος ἐκ δύο EB is commensurable (in length) with the (previously) ὀνομάτων ἐστὶ καὶ τῇ τάξει ἡ αὐτή· ὅπερ ἔδει δεῖξαι. laid down rational (straight-line) then FD is also com- mensurable (in length) with it [Prop. 10.12], and, again, on account of this, (CD) will be the same in order as AB. For each of them will be second binomial (straight- lines) [Def. 10.6]. And if neither of AE and EB is com- mensurable (in length) with the (previously) laid down rational (straight-line) then neither of CF and FD will be commensurable (in length) with it [Prop. 10.13], and each (of AB and CD) is a third (binomial straight-line) 358 STOIQEIWN iþ. ELEMENTS BOOK 10 [Def. 10.7]. And if the square on AE is greater than (the square on) EB by the (square) on (some straight- line) incommensurable (in length) with (AE) then the square on CF is also greater than (the square on) FD by the (square) on (some straight-line) incommensurable (in length) with (CF ) [Prop. 10.14]. And if AE is com- mensurable (in length) with the (previously) laid down rational (straight-line) then CF is also commensurable (in length) with it [Prop. 10.12], and each (of AB and CD) is a fourth (binomial straight-line) [Def. 10.8]. And if EB (is commensurable in length with the previously laid down rational straight-line) then FD (is) also (com- mensurable in length with it), and each (of AB and CD) will be a fifth (binomial straight-line) [Def. 10.9]. And if neither of AE and EB (is commensurable in length with the previously laid down rational straight-line) then also neither of CF and FD is commensurable (in length) with the laid down rational (straight-line), and each (of AB and CD) will be a sixth (binomial straight-line) [Def. 10.10]. Hence, a (straight-line) commensurable in length with a binomial (straight-line) is a binomial (straight- line), and the same in order. (Which is) the very thing it was required to show.xzþ. Proposition 67 ῾Η τῇ ἐκ δύο μέσων μήκει σύμμετρος καὶ αὐτὴ ἐκ δύο A (straight-line) commensurable in length with a bi- μέσων ἐστὶ καὶ τῇ τάξει ἡ αὐτή. medial (straight-line) is itself also bimedial, and the same in order. Β Γ Ζ ∆ Α Ε F D A E B C ῎Εστω ἐκ δύο μέσων ἡ ΑΒ, καὶ τῇ ΑΒ σύμμετρος ἔστω Let AB be a bimedial (straight-line), and let CD be μήκει ἡ ΓΔ· λέγω, ὅτι ἡ ΓΔ ἐκ δύο μέσων ἐστὶ καὶ τῇ τάξει commensurable in length with AB. I say that CD is bi- ἡ αὐτὴ τῇ ΑΒ. medial, and the same in order as AB. ᾿Επεὶ γὰρ ἐκ δύο μέσων ἐστὶν ἡ ΑΒ, διῃρήσθω εἰς τὰς For since AB is a bimedial (straight-line), let it have μέσας κατὰ τὸ Ε· αἱ ΑΕ, ΕΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον been divided into its (component) medial (straight-lines) σύμμετροι. καὶ γεγονέτω ὡς ἡ ΑΒ πρὸς ΓΔ, ἡ ΑΕ πρὸς at E. Thus, AE and EB are medial (straight-lines ΓΖ· καὶ λοιπὴ ἄρα ἡ ΕΒ πρὸς λοιπὴν τὴν ΖΔ ἐστιν, ὡς ἡ which are) commensurable in square only [Props. 10.37, ΑΒ πρὸς ΓΔ. σύμμετρος δὲ ἡ ΑΒ τῇ ΓΔ μήκει· σύμμετρος 10.38]. And let it have been contrived that as AB (is) to ἄρα καὶ ἑκατέρα τῶν ΑΕ, ΕΒ ἑκατέρᾳ τῶν ΓΖ, ΖΔ. μέσαι CD, (so) AE (is) to CF [Prop. 6.12]. And thus as the δὲ αἱ ΑΕ, ΕΒ· μέσαι ἄρα καὶ αἱ ΓΖ, ΖΔ. καὶ ἐπεί ἐστιν remainder EB is to the remainder FD, so AB (is) to CD ὡς ἡ ΑΕ πρὸς ΕΒ, ἡ ΓΖ πρὸς ΖΔ, αἱ δὲ ΑΕ, ΕΒ δυνάμει [Props. 5.19 corr., 6.16]. And AB (is) commensurable μόνον σύμμετροί εἰσιν, καὶ αἱ ΓΖ, ΖΔ [ἄρα] δυνάμει μόνον in length with CD. Thus, AE and EB are also com- σύμμετροί εἰσιν, ἐδείχθησαν δὲ καὶ μέσαι· ἡ ΓΔ ἄρα ἐκ δύο mensurable (in length) with CF and FD, respectively μέσων ἐστίν. λέγω δή, ὅτι καὶ τῇ τάξει ἡ αὐτή ἐστι τῇ ΑΒ. [Prop. 10.11]. And AE and EB (are) medial. Thus, CF ᾿Επεὶ γάρ ἐστιν ὡς ἡ ΑΕ πρὸς ΕΒ, ἡ ΓΖ πρὸς ΖΔ, καὶ and FD (are) also medial [Prop. 10.23]. And since as ὡς ἄρα τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ὑπὸ τῶν ΑΕΒ, οὕτως τὸ AE is to EB, (so) CF (is) to FD, and AE and EB are ἀπὸ τῆς ΓΖ πρὸς τὸ ὑπὸ τῶν ΓΖΔ· ἐναλλὰξ ὡς τὸ ἀπὸ τῆς commensurable in square only, CF and FD are [thus] 359 STOIQEIWN iþ. ELEMENTS BOOK 10 ΑΕ πρὸς τὸ ἀπὸ τῆς ΓΖ, οὕτως τὸ ὑπὸ τῶν ΑΕΒ πρὸς τὸ also commensurable in square only [Prop. 10.11]. And ὑπὸ τῶν ΓΖΔ. σύμμετρον δὲ τὸ ἀπὸ τῆς ΑΕ τῷ ἀπὸ τῆς they were also shown (to be) medial. Thus, CD is a bi- ΓΖ· σύμμετρον ἄρα καὶ τὸ ὑπὸ τῶν ΑΕΒ τῷ ὑπὸ τῶν ΓΖΔ. medial (straight-line). So, I say that it is also the same in εἴτε οὖν ῥητόν ἐστι τὸ ὑπὸ τῶν ΑΕΒ, καὶ τὸ ὑπὸ τῶν ΓΖΔ order as AB. ῥητόν ἐστιν [καὶ διὰ τοῦτό ἐστιν ἐκ δύο μέσων πρώτη]. εἴτε For since as AE is to EB, (so) CF (is) to FD, thus μέσον, μέσον, καί ἐστιν ἑκατέρα δευτέρα. also as the (square) on AE (is) to the (rectangle con- Καὶ διὰ τοῦτο ἔσται ἡ ΓΔ τῇ ΑΒ τῇ τάξει ἡ αὐτή· ὅπερ tained) by AEB, so the (square) on CF (is) to the (rect- ἔδει δεῖξαι. angle contained) by CFD [Prop. 10.21 lem.]. Alter- nately, as the (square) on AE (is) to the (square) on CF , so the (rectangle contained) by AEB (is) to the (rectangle contained) by CFD [Prop. 5.16]. And the (square) on AE (is) commensurable with the (square) on CF . Thus, the (rectangle contained) by AEB (is) also commensurable with the (rectangle contained) by CFD [Prop. 10.11]. Therefore, either the (rectangle contained) by AEB is rational, and the (rectangle con- tained) by CFD is rational [and, on account of this, (AE and CD) are first bimedial (straight-lines)], or (the rectangle contained by AEB is) medial, and (the rect- angle contained by CFD is) medial, and (AB and CD) are each second (bimedial straight-lines) [Props. 10.23, 10.37, 10.38]. And, on account of this, CD will be the same in order as AB. (Which is) the very thing it was required to show.xhþ. Proposition 68 ῾Η τῇ μείζονι σύμμετρος καὶ αὐτὴ μείζων ἐστίν. A (straight-line) commensurable (in length) with a major (straight-line) is itself also major. Β Γ Ζ ∆ Α Ε F D A E B C ῎Εστω μείζων ἡ ΑΒ, καὶ τῇ ΑΒ σύμμετρος ἔστω ἡ ΓΔ· Let AB be a major (straight-line), and let CD be com- λέγω, ὅτι ἡ ΓΔ μείζων ἐστίν. mensurable (in length) with AB. I say that CD is a major Διῃρήσθω ἡ ΑΒ κατὰ τὸ Ε· αἱ ΑΕ, ΕΒ ἄρα δυνάμει εἰσὶν (straight-line). ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν Let AB have been divided (into its component terms) τετραγώνων ῥητόν, τὸ δ᾿ ὑπ᾿ αὐτῶν μέσον· καὶ γεγονέτω at E. AE and EB are thus incommensurable in square, τὰ αὐτὰ τοῖς πρότερον. καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν making the sum of the squares on them rational, and the ΓΔ, οὕτως ἥ τε ΑΕ πρὸς τὴν ΓΖ καὶ ἡ ΕΒ πρὸς τὴν ΖΔ, (rectangle contained) by them medial [Prop. 10.39]. And καὶ ὡς ἄρα ἡ ΑΕ πρὸς τὴν ΓΖ, οὕτως ἡ ΕΒ πρὸς τὴν ΖΔ. let (the) same (things) have been contrived as in the pre- σύμμετρος δὲ ἡ ΑΒ τῇ ΓΔ· σύμμετρος ἄρα καὶ ἑκατέρα τῶν vious (propositions). And since as AB is to CD, so AE ΑΕ, ΕΒ ἑκατέρᾳ τῶν ΓΖ, ΖΔ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΕ πρὸς (is) to CF and EB to FD, thus also as AE (is) to CF , τὴν ΓΖ, οὕτως ἡ ΕΒ πρὸς τὴν ΖΔ, καὶ ἐναλλὰξ ὡς ἡ ΑΕ so EB (is) to FD [Prop. 5.11]. And AB (is) commen- πρὸς ΕΒ, οὕτως ἡ ΓΖ πρὸς ΖΔ, καὶ συνθέντι ἄρα ἑστὶν ὡς surable (in length) with CD. Thus, AE and EB (are) ἡ ΑΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΓΔ πρὸς τὴν ΔΖ· καὶ ὡς ἄρα also commensurable (in length) with CF and FD, re- τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΕ, οὕτως τὸ ἀπὸ τῆς ΓΔ spectively [Prop. 10.11]. And since as AE is to CF , so πρὸς τὸ ἀπὸ τῆς ΔΖ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ὡς τὸ ἀπὸ EB (is) to FD, also, alternately, as AE (is) to EB, so τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΑΕ, οὕτως τὸ ἀπὸ τῆς ΓΔ πρὸς CF (is) to FD [Prop. 5.16], and thus, via composition, τὸ ἀπὸ τῆς ΓΖ. καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΑΒ πρὸς τὰ ἀπὸ τῶν as AB is to BE, so CD (is) to DF [Prop. 5.18]. And thus ΑΕ, ΕΒ, οὕτως τὸ ἀπὸ τῆς ΓΔ πρὸς τὰ ἀπὸ τῶν ΓΖ, ΖΔ· as the (square) on AB (is) to the (square) on BE, so the 360 STOIQEIWN iþ. ELEMENTS BOOK 10 καὶ ἐναλλὰξ ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς (square) on CD (is) to the (square) on DF [Prop. 6.20]. ΓΔ, οὕτως τὰ ἀπὸ τῶν ΑΕ, ΕΒ πρὸς τὰ ἀπὸ τῶν ΓΖ, ΖΔ. So, similarly, we can also show that as the (square) on σύμμετρον δὲ τὸ ἀπὸ τῆς ΑΒ τῷ ἀπὸ τῆς ΓΔ· σύμμετρα AB (is) to the (square) on AE, so the (square) on CD ἄρα καὶ τὰ ἀπὸ τῶν ΑΕ, ΕΒ τοῖς ἀπὸ τῶν ΓΖ, ΖΔ. καί ἐστι (is) to the (square) on CF . And thus as the (square) on τὰ ἀπὸ τῶν ΑΕ, ΕΒ ἅμα ῥητόν, καὶ τὰ ἀπὸ τῶν ΓΖ, ΖΔ AB (is) to (the sum of) the (squares) on AE and EB, so ἅμα ῥητόν ἐστιν. ὁμοίως δὲ καὶ τὸ δὶς ὑπὸ τῶν ΑΕ, ΕΒ the (square) on CD (is) to (the sum of) the (squares) on σύμμετρόν ἐστι τῷ δὶς ὑπὸ τῶν ΓΖ, ΖΔ. καί ἐστι μέσον τὸ CF and FD. And thus, alternately, as the (square) on AB δὶς ὑπὸ τῶν ΑΕ, ΕΒ· μέσον ἄρα καὶ τὸ δὶς ὑπὸ τῶν ΓΖ, is to the (square) on CD, so (the sum of) the (squares) on ΖΔ. αἱ ΓΖ, ΖΔ ἄρα δυνάμει ἀσύμμετροί εἰσι ποιοῦσαι τὸ AE and EB (is) to (the sum of) the (squares) on CF and μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων ἅμα ῥητόν, FD [Prop. 5.16]. And the (square) on AB (is) commen- τὸ δὲ δὶς ὑπ᾿ αὐτῶν μέσον· ὅλη ἄρα ἡ ΓΔ ἄλογός ἐστιν ἡ surable with the (square) on CD. Thus, (the sum of) the καλουμένη μείζων. (squares) on AE and EB (is) also commensurable with ῾Η ἄρα τῇ μείζονι σύμμετρος μείζων ἐστίν· ὅπερ ἔδει (the sum of) the (squares) on CF and FD [Prop. 10.11]. δεῖξαι. And the (squares) on AE and EB (added) together are rational. The (squares) on CF and FD (added) together (are) thus also rational. So, similarly, twice the (rect- angle contained) by AE and EB is also commensurable with twice the (rectangle contained) by CF and FD. And twice the (rectangle contained) by AE and EB is me- dial. Therefore, twice the (rectangle contained) by CF and FD (is) also medial [Prop. 10.23 corr.]. CF and FD are thus (straight-lines which are) incommensurable in square [Prop 10.13], simultaneously making the sum of the squares on them rational, and twice the (rectangle contained) by them medial. The whole, CD, is thus that irrational (straight-line) called major [Prop. 10.39]. Thus, a (straight-line) commensurable (in length) with a major (straight-line) is major. (Which is) the very thing it was required to show.xjþ. Proposition 69 ῾Η τῇ ῥητὸν καὶ μέσον δυναμένῃ σύμμετρος [καὶ αὐτὴ] A (straight-line) commensurable (in length) with the ῥητὸν καὶ μέσον δυναμένη ἐστίν. square-root of a rational plus a medial (area) is [itself also] the square-root of a rational plus a medial (area). Β Γ Ζ ∆ Α Ε F D A E B C ῎Εστω ῥητὸν καὶ μέσον δυναμένη ἡ ΑΒ, καὶ τῇ ΑΒ Let AB be the square-root of a rational plus a medial σύμμετρος ἔστω ἡ ΓΔ· δεικτέον, ὅτι καὶ ἡ ΓΔ ῥητὸν καὶ (area), and let CD be commensurable (in length) with μέσον δυναμένη ἐστίν. AB. We must show that CD is also the square-root of a Διῃρήσθω ἡ ΑΒ εἰς τὰς εὐθείας κατὰ τὸ Ε· αἱ ΑΕ, ΕΒ rational plus a medial (area). ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον Let AB have been divided into its (component) ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον, τὸ δ᾿ ὑπ᾿ αὐτῶν straight-lines at E. AE and EB are thus incommensu- ῥητόν· καὶ τὰ αὐτὰ κατεσκευάσθω τοῖς πρότερον. ὁμοίως rable in square, making the sum of the squares on them δὴ δείξομεν, ὅτι καὶ αἱ ΓΖ, ΖΔ δυνάμει εἰσὶν ἀσύμμετροι, medial, and the (rectangle contained) by them rational καὶ σύμμετρον τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ [Prop. 10.40]. And let the same construction have been τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ, τὸ δὲ ὑπὸ ΑΕ, ΕΒ made as in the previous (propositions). So, similarly, we τῷ ὑπὸ ΓΖ, ΖΔ· ὥστε καὶ τὸ [μὲν] συγκείμενον ἐκ τῶν ἀπὸ can show that CF and FD are also incommensurable τῶν ΓΖ, ΖΔ τετραγώνων ἐστὶ μέσον, τὸ δ᾿ ὑπὸ τῶν ΓΖ, in square, and that the sum of the (squares) on AE and 361 STOIQEIWN iþ. ELEMENTS BOOK 10 ΖΔ ῥητόν. EB (is) commensurable with the sum of the (squares) ῾Ρητὸν ἄρα καὶ μέσον δυναμένη ἐστὶν ἡ ΓΔ· ὅπερ ἔδει on CF and FD, and the (rectangle contained) by AE δεῖξαι. and EB with the (rectangle contained) by CF and FD. And hence the sum of the squares on CF and FD is me- dial, and the (rectangle contained) by CF and FD (is) rational. Thus, CD is the square-root of a rational plus a medial (area) [Prop. 10.40]. (Which is) the very thing it was required to show.oþ. Proposition 70 ῾Η τῇ δύο μέσα δυναμένῃ σύμμετρος δύο μέσα δυναμένη A (straight-line) commensurable (in length) with the ἐστίν. square-root of (the sum of) two medial (areas) is (itself also) the square-root of (the sum of) two medial (areas). Β Γ Ζ ∆ Α Ε F D A E B C ῎Εστω δύο μέσα δυναμένη ἡ ΑΒ, καὶ τῇ ΑΒ σύμμετρος Let AB be the square-root of (the sum of) two medial ἡ ΓΔ· δεικτέον, ὅτι καὶ ἡ ΓΔ δύο μέσα δυναμένη ἐστίν. (areas), and (let) CD (be) commensurable (in length) ᾿Επεὶ γὰρ δύο μέσα δυναμένη ἐστὶν ἡ ΑΒ, διῃρήσθω with AB. We must show that CD is also the square-root εἰς τὰς εὐθείας κατὰ τὸ Ε· αἱ ΑΕ, ΕΒ ἄρα δυνάμει of (the sum of) two medial (areas). εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾿ For since AB is the square-root of (the sum of) two αὐτῶν [τετραγώνων] μέσον καὶ τὸ ὑπ᾿ αὐτῶν μέσον καὶ medial (areas), let it have been divided into its (compo- ἔτι ἀσύμμετρον τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ nent) straight-lines at E. Thus, AE and EB are incom- τετραγώνων τῷ ὑπὸ τῶν ΑΕ, ΕΒ· καὶ κατεσκευάσθω τὰ mensurable in square, making the sum of the [squares] αὐτὰ τοῖς πρότερον. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ΓΖ, ΖΔ on them medial, and the (rectangle contained) by them δυνάμει εἰσὶν ἀσύμμετροι καὶ σύμμετρον τὸ μὲν συγκείμενον medial, and, moreover, the sum of the (squares) on AE ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, and EB incommensurable with the (rectangle) contained ΖΔ, τὸ δὲ ὑπὸ τῶν ΑΕ, ΕΒ τῷ ὑπὸ τῶν ΓΖ, ΖΔ· ὥστε καὶ by AE and EB [Prop. 10.41]. And let the same construc- τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων μέσον tion have been made as in the previous (propositions). ἐστὶ καὶ τὸ ὑπὸ τῶν ΓΖ, ΖΔ μέσον καὶ ἔτι ἀσύμμετρον τὸ So, similarly, we can show that CF and FD are also συγκείμενον ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων τῷ ὑπὸ incommensurable in square, and (that) the sum of the τῶν ΓΖ, ΖΔ. (squares) on AE and EB (is) commensurable with the ῾Η ἄρα ΓΔ δύο μέσα δυναμένη ἐστίν· ὅπερ ἔδει δεῖξαι. sum of the (squares) on CF and FD, and the (rectangle contained) by AE and EB with the (rectangle contained) by CF and FD. Hence, the sum of the squares on CF and FD is also medial, and the (rectangle contained) by CF and FD (is) medial, and, moreover, the sum of the squares on CF and FD (is) incommensurable with the (rectangle contained) by CF and FD. Thus, CD is the square-root of (the sum of) two me- dial (areas) [Prop. 10.41]. (Which is) the very thing it was required to show.oaþ. Proposition 71 ῾Ρητοῦ καὶ μέσου συντιθεμένου τέσσαρες ἄλογοι γίγνον- When a rational and a medial (area) are added to- ται ἤτοι ἐκ δύο ὀνομάτων ἢ ἐκ δύο μέσων πρώτη ἢ μείζων gether, four irrational (straight-lines) arise (as the square- ἢ ῥητὸν καὶ μέσον δυναμένη. roots of the total area)—either a binomial, or a first bi- 362 STOIQEIWN iþ. ELEMENTS BOOK 10 ῎Εστω ῥητὸν μὲν τὸ ΑΒ, μέσον δὲ τὸ ΓΔ· λέγω, ὅτι ἡ medial, or a major, or the square-root of a rational plus a τὸ ΑΔ χωρίον δυναμένη ἤτοι ἐκ δύο ὀνομάτων ἐστὶν ἢ ἐκ medial (area). δύο μέσων πρώτη ἢ μείζων ἢ ῥητὸν καὶ μέσον δυναμένη. Let AB be a rational (area), and CD a medial (area). I say that the square-root of area AD is either binomial, or first bimedial, or major, or the square-root of a rational plus a medial (area). Θ Β ∆ Ζ Η Ι Α Γ Ε Κ H B D I A E KC F G Τὸ γὰρ ΑΒ τοῦ ΓΔ ἤτοι μεῖζόν ἐστιν ἢ ἔλασσον. For AB is either greater or less than CD. Let it, first ἔστω πρότερον μεῖζον· καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ παρα- of all, be greater. And let the rational (straight-line) EF βεβλήσθω παρὰ τὴν ΕΖ τῷ ΑΒ ἴσον τὸ ΕΗ πλάτος ποιοῦν be laid down. And let (the rectangle) EG, equal to AB, τὴν ΕΘ· τῷ δὲ ΔΓ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω τὸ have been applied to EF , producing EH as breadth. And ΘΙ πλάτος ποιοῦν τὴν ΘΚ. καὶ ἐπεὶ ῥητόν ἐστι τὸ ΑΒ καί let (the recatangle) HI, equal to DC, have been ap- ἐστιν ἴσον τῷ ΕΗ, ῥητὸν ἄρα καὶ τὸ ΕΗ. καὶ παρὰ [ῥητὴν] plied to EF , producing HK as breadth. And since AB τὴν ΕΖ παραβέβληται πλάτος ποιοῦν τὴν ΕΘ· ἡ ΕΘ ἄρα is rational, and is equal to EG, EG is thus also rational. ῥητή ἐστι καὶ σύμμετρος τῇ ΕΖ μήκει. πάλιν, ἐπεὶ μέσον And it has been applied to the [rational] (straight-line) ἐστὶ τὸ ΓΔ καί ἐστιν ἴσον τῷ ΘΙ, μέσον ἄρα ἐστὶ καὶ τὸ EF , producing EH as breadth. EH is thus rational, and ΘΙ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος ποιοῦν τὴν commensurable in length with EF [Prop. 10.20]. Again, ΘΚ· ῥητὴ ἄρα ἐστὶν ἡ ΘΚ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. καὶ since CD is medial, and is equal to HI, HI is thus also ἐπεὶ μέσον ἐστὶ τὸ ΓΔ, ῥητὸν δὲ τὸ ΑΒ, ἀσύμμετρον ἄρα medial. And it is applied to the rational (straight-line) ἐστὶ τὸ ΑΒ τῷ ΓΔ· ὥστε καὶ τὸ ΕΗ ἀσύμμετρόν ἐστι τῷ EF , producing HK as breadth. HK is thus rational, ΘΙ. ὡς δὲ τὸ ΕΗ πρὸς τὸ ΘΙ, οὕτως ἐστὶν ἡ ΕΘ πρὸς τὴν and incommensurable in length with EF [Prop. 10.22]. ΘΚ· ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΕΘ τῇ ΘΚ μήκει. καί εἰσιν And since CD is medial, and AB rational, AB is thus ἀμφότεραι ῥηταί· αἱ ΕΘ, ΘΚ ἄρα ῥηταί εἰσι δυνάμει μόνον incommensurable with CD. Hence, EG is also incom- σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΕΚ διῃρημένη mensurable with HI. And as EG (is) to HI, so EH is κατὰ τὸ Θ. καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΑΒ τοῦ ΓΔ, ἴσον δὲ τὸ to HK [Prop. 6.1]. Thus, EH is also incommensurable μὲν ΑΒ τῷ ΕΗ, τὸ δὲ ΓΔ τῷ ΘΙ, μεῖζον ἄρα καὶ τὸ ΕΗ in length with HK [Prop. 10.11]. And they are both ra- τοῦ ΘΙ· καὶ ἡ ΕΘ ἄρα μείζων ἐστὶ τῆς ΘΚ. ἤτοι οὖν ἡ ΕΘ tional. Thus, EH and HK are rational (straight-lines τῆς ΘΚ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει ἢ which are) commensurable in square only. EK is thus τῷ ἀπὸ ἀσυμμέτρου. δυνάσθω πρότερον τῷ ἀπὸ συμμέτρου a binomial (straight-line), having been divided (into its ἑαυτῇ· καί ἐστιν ἡ μείζων ἡ ΘΕ σύμμετρος τῇ ἐκκειμένῃ component terms) at H [Prop. 10.36]. And since AB ῥητῃ τῇ ΕΖ· ἡ ἄρα ΕΚ ἐκ δύο ὀνομάτων ἐστὶ πρώτη. ῥητὴ is greater than CD, and AB (is) equal to EG, and CD δὲ ἡ ΕΖ· ἐὰν δὲ χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο to HI, EG (is) thus also greater than HI. Thus, EH is ὀνομάτων πρώτης, ἡ τὸ χωρίον δυναμένη ἐκ δύο ὀνομάτων also greater than HK [Prop. 5.14]. Therefore, the square ἐστίν. ἡ ἄρα τὸ ΕΙ δυναμένη ἐκ δύο ὀνομάτων ἐστίν· ὥστε on EH is greater than (the square on) HK either by καὶ ἡ τὸ ΑΔ δυναμένη ἐκ δύο ὀνομάτων ἐστίν. ἀλλὰ δὴ the (square) on (some straight-line) commensurable in δυνάσθω ἡ ΕΘ τῆς ΘΚ μεῖζον τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ· length with (EH), or by the (square) on (some straight- καί ἐστιν ἡ μείζων ἡ ΕΘ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ line) incommensurable (in length with EH). Let it, first ΕΖ μήκει· ἡ ἄρα ΕΚ ἐκ δύο ὀνομάτων ἐστὶ τετάρτη. ῥητὴ of all, be greater by the (square) on (some straight-line) δὲ ἡ ΕΖ· ἐὰν δὲ χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο commensurable (in length with EH). And the greater 363 STOIQEIWN iþ. ELEMENTS BOOK 10 ὀνομάτων τετάρτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ (of the two components of EK) HE is commensurable καλουμένη μείζων. ἡ ἄρα τὸ ΕΙ χωρίον δυναμένη μείζων (in length) with the (previously) laid down (straight- ἐστίν· ὥστε καὶ ἡ τὸ ΑΔ δυναμένη μείζων ἐστίν. line) EF . EK is thus a first binomial (straight-line) Ἀλλὰ δὴ ἔστω ἔλασσον τὸ ΑΒ τοῦ ΓΔ· καὶ τὸ ΕΗ [Def. 10.5]. And EF (is) rational. And if an area is con- ἄρα ἔλασσόν ἐστι τοῦ ΘΙ· ὥστε καὶ ἡ ΕΘ ἐλάσσων ἐστὶ tained by a rational (straight-line) and a first binomial τῆς ΘΚ. ἤτοι δὲ ἡ ΘΚ τῆς ΕΘ μεῖζον δύναται τῷ ἀπὸ (straight-line) then the square-root of the area is a bino- συμμέτρου ἑαυτῇ ἢ τῷ ἀπὸ ἀσυμμέτρου. δυνάσθω πρότερον mial (straight-line) [Prop. 10.54]. Thus, the square-root τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει· καί ἐστιν ἡ ἐλάσσων ἡ ΕΘ of EI is a binomial (straight-line). Hence the square- σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ ΕΖ μήκει· ἡ ἄρα ΕΚ ἐκ root of AD is also a binomial (straight-line). And, so, let δύο ὀνομάτων ἐστὶ δευτέρα. ῥητὴ δὲ ἡ ΕΖ· ἐὰν δὲ χωρίον the square on EH be greater than (the square on) HK περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων δευτέρας, ἡ by the (square) on (some straight-line) incommensurable τὸ χωρίον δυναμένη ἐκ δύο μέσων ἐστὶ πρώτη. ἡ ἄρα τὸ ΕΙ (in length) with (EH). And the greater (of the two com- χωρίον δυναμένη ἐκ δύο μέσων ἐστὶ πρώτη· ὥστε καὶ ἡ τὸ ponents of EK) EH is commensurable in length with the ΑΔ δυναμένη ἐκ δύο μέσων ἐστὶ πρώτη. ἀλλὰ δὴ ἡ ΘΚ τῆς (previously) laid down rational (straight-line) EF . Thus, ΘΕ μεῖζον δυνάσθω τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί ἐστιν EK is a fourth binomial (straight-line) [Def. 10.8]. And ἡ ἐλάσσων ἡ ΕΘ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ ΕΖ· ἡ EF (is) rational. And if an area is contained by a rational ἄρα ΕΚ ἐκ δύο ὀνομάτων ἐστὶ πέμπτη. ῥητὴ δὲ ἡ ΕΖ· ἐὰν (straight-line) and a fourth binomial (straight-line) then δὲ χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων the square-root of the area is the irrational (straight-line) πέμπτης, ἡ τὸ χωρίον δυναμένη ῥητὸν καὶ μέσον δυναμένη called major [Prop. 10.57]. Thus, the square-root of area ἐστίν. ἡ ἄρα τὸ ΕΙ χωρίον δυναμένη ῥητὸν καὶ μέσον δυ- EI is a major (straight-line). Hence, the square-root of ναμένη ἐστίν· ὥστε καὶ ἡ τὸ ΑΔ χωρίον δυναμένη ῥητὸν AD is also major. καὶ μέσον δυναμένη ἐστίν. And so, let AB be less than CD. Thus, EG is also less ῾Ρητοῦ ἄρα καὶ μέσου συντιθεμένου τέσσαρες ἄλογοι than HI. Hence, EH is also less than HK [Props. 6.1, γίγνονται ἤτοι ἐκ δύο ὀνομάτων ἢ ἐκ δύο μέσων πρώτη ἢ 5.14]. And the square on HK is greater than (the μείζων ἢ ῥητὸν καὶ μέσον δυναμένη· ὅπερ ἔδει δεῖξαι. square on) EH either by the (square) on (some straight- line) commensurable (in length) with (HK), or by the (square) on (some straight-line) incommensurable (in length) with (HK). Let it, first of all, be greater by the square on (some straight-line) commensurable in length with (HK). And the lesser (of the two components of EK) EH is commensurable in length with the (previ- ously) laid down rational (straight-line) EF . Thus, EK is a second binomial (straight-line) [Def. 10.6]. And EF (is) rational. And if an area is contained by a rational (straight-line) and a second binomial (straight-line) then the square-root of the area is a first bimedial (straight- line) [Prop. 10.55]. Thus, the square-root of area EI is a first bimedial (straight-line). Hence, the square-root of AD is also a first bimedial (straight-line). And so, let the square on HK be greater than (the square on) HE by the (square) on (some straight-line) incommensurable (in length) with (HK). And the lesser (of the two compo- nents of EK) EH is commensurable (in length) with the (previously) laid down rational (straight-line) EF . Thus, EK is a fifth binomial (straight-line) [Def. 10.9]. And EF (is) rational. And if an area is contained by a ratio- nal (straight-line) and a fifth binomial (straight-line) then the square-root of the area is the square-root of a rational plus a medial (area) [Prop. 10.58]. Thus, the square-root of area EI is the square-root of a rational plus a medial (area). Hence, the square-root of area AD is also the 364 STOIQEIWN iþ. ELEMENTS BOOK 10 square-root of a rational plus a medial (area). Thus, when a rational and a medial area are added to- gether, four irrational (straight-lines) arise (as the square- roots of the total area)—either a binomial, or a first bi- medial, or a major, or the square-root of a rational plus a medial (area). (Which is) the very thing it was required to show.xbþ. Proposition 72 Δύο μέσων ἀσυμμέτρων ἀλλήλοις συντιθεμένων αἱ When two medial (areas which are) incommensu- λοιπαὶ δύο ἄλογοι γίγνονται ἤτοι ἐκ δύο μέσων δευτέρα rable with one another are added together, the remaining ἢ [ἡ] δύο μέσα δυναμένη. two irrational (straight-lines) arise (as the square-roots of the total area)—either a second bimedial, or the square- root of (the sum of) two medial (areas). Θ Β ∆ Ζ Η Ι Α Γ Ε Κ H B D I A E KC F G Συγκείσθω γὰρ δύο μέσα ἀσύμμετρα ἀλλήλοις τὰ ΑΒ, For let the two medial (areas) AB and CD, (which ΓΔ· λέγω, ὅτι ἡ τὸ ΑΔ χωρίον δυναμένη ἤτοι ἐκ δύο μέσων are) incommensurable with one another, have been ἐστὶ δευτέρα ἢ δύο μέσα δυναμένη. added together. I say that the square-root of area AD Τὸ γὰρ ΑΒ τοῦ ΓΔ ἤτοι μεῖζόν ἐστιν ἢ ἔλασσον. ἔστω, is either a second bimedial, or the square-root of (the εἰ τύχον, πρότερον μεῖζον τὸ ΑΒ τοῦ ΓΔ· καὶ ἐκκείσθω sum of) two medial (areas). ῥητὴ ἡ ΕΖ, καὶ τῷ μὲν ΑΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω For AB is either greater than or less than CD. By τὸ ΕΗ πλάτος ποιοῦν τὴν ΕΘ, τῷ δὲ ΓΔ ἴσον τὸ ΘΙ πλάτος chance, let AB, first of all, be greater than CD. And ποιοῦν τὴν ΘΚ. καὶ ἐπεὶ μέσον ἐστὶν ἑκάτερον τῶν ΑΒ, ΓΔ, let the rational (straight-line) EF be laid down. And let μέσον ἄρα καὶ ἐκάτερον τῶν ΕΗ, ΘΙ. καὶ παρὰ ῥητὴν τὴν EG, equal to AB, have been applied to EF , producing ΖΕ παράκειται πλάτος ποιοῦν τὰς ΕΘ, ΘΚ· ἑκατέρα ἄρα τῶν EH as breadth, and HI, equal to CD, producing HK ΕΘ, ΘΚ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΕΖ μήκει. καὶ ἐπεὶ as breadth. And since AB and CD are each medial, EG ἀσύμμετρόν ἐστι τὸ ΑΒ τῷ ΓΔ, καί ἐστιν ἴσον τὸ μὲν ΑΒ and HI (are) thus also each medial. And they are ap- τῷ ΕΗ, τὸ δὲ ΓΔ τῷ ΘΙ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΕΗ τῷ plied to the rational straight-line FE, producing EH and ΘΙ. ὡς δὲ τὸ ΕΗ πρὸς τὸ ΘΙ, οὕτως ἐστὶν ἡ ΕΘ πρὸς ΘΚ· HK (respectively) as breadth. Thus, EH and HK are ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΘ τῇ ΘΚ μήκει. αἱ ΕΘ, ΘΚ ἄρα each rational (straight-lines which are) incommensurable ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα ὀνομάτων in length with EF [Prop. 10.22]. And since AB is incom- ἐστὶν ἡ ΕΚ. ἤτοι δὲ ἡ ΕΘ τῆς ΘΚ μεῖζον δύναται τῷ ἀπὸ mensurable with CD, and AB is equal to EG, and CD συμμέτρου ἑαυτῇ ἢ τῷ ἀπὸ ἀσυμμέτρου. δυνάσθω πρότερον to HI, EG is thus also incommensurable with HI. And τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει· καὶ οὐδετέρα τῶν ΕΘ, ΘΚ as EG (is) to HI, so EH is to HK [Prop. 6.1]. EH is σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΕΖ μήκει· ἡ ΕΚ ἄρα thus incommensurable in length with HK [Prop. 10.11]. ἐκ δύο ὀνομάτων ἐστὶ τρίτη. ῥητὴ δὲ ἡ ΕΖ· ἐὰν δὲ χωρίον Thus, EH and HK are rational (straight-lines which are) περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων τρίτης, ἡ commensurable in square only. EK is thus a binomial τὸ χωρίον δυναμένη ἐκ δύο μέσων ἐστὶ δευτέρα· ἡ ἄρα τὸ (straight-line) [Prop. 10.36]. And the square on EH is ΕΙ, τουτέστι τὸ ΑΔ, δυναμένη ἐκ δύο μέσων ἐστὶ δευτέρα. greater than (the square on) HK either by the (square) 365 STOIQEIWN iþ. ELEMENTS BOOK 10 ἀλλα δὴ ἡ ΕΘ τῆς ΘΚ μεῖζον δυνάσθω τῷ ἀπὸ ἀσυμμέτρου on (some straight-line) commensurable (in length) with ἑαυτῇ μήκει· καὶ ἀσύμμετρός ἐστιν ἑκατέρα τῶν ΕΘ, ΘΚ (EH), or by the (square) on (some straight-line) incom- τῇ ΕΖ μήκει· ἡ ἄρα ΕΚ ἐκ δύο ὀνομάτων ἐστὶν ἕκτη. ἐὰν mensurable (in length with EH). Let it, first of all, be δὲ χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων greater by the square on (some straight-line) commensu- ἕκτης, ἡ τὸ χωρίον δυναμένη ἡ δύο μέσα δυναμένη ἐστίν· rable in length with (EH). And neither of EH or HK is ὥστε καὶ ἡ τὸ ΑΔ χωρίον δυναμένη ἡ δύο μέσα δυναμένη commensurable in length with the (previously) laid down ἐστίν. rational (straight-line) EF . Thus, EK is a third binomial [῾Ομοίως δὴ δείξομεν, ὅτι κἂν ἔλαττον ᾖ τὸ ΑΒ τοῦ ΓΔ, (straight-line) [Def. 10.7]. And EF (is) rational. And if ἡ τὸ ΑΔ χωρίον δυναμένη ἢ ἐκ δύο μέσων δευτέρα ἐστὶν an area is contained by a rational (straight-line) and a ἤτοι δύο μέσα δυναμένη]. third binomial (straight-line) then the square-root of the Δύο ἄρα μέσων ἀσυμμέτρων ἀλλήλοις συντιθεμένων αἱ area is a second bimedial (straight-line) [Prop. 10.56]. λοιπαὶ δύο ἄλογοι γίγνονται ἤτοι ἐκ δύο μέσων δευτέρα ἢ Thus, the square-root of EI—that is to say, of AD— δύο μέσα δυναμένη. is a second bimedial. And so, let the square on EH be greater than (the square) on HK by the (square) ᾿Η ἐκ δύο ὀνομάτων καὶ αἱ μετ᾿ αὐτὴν ἄλογοι οὔτε on (some straight-line) incommensurable in length with τῇ μέσῃ οὔτε ἀλλήλαις εἰσὶν αἱ αὐταί. τὸ μὲν γὰρ ἀπὸ (EH). And EH and HK are each incommensurable in μέσης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ῥητὴν καὶ length with EF . Thus, EK is a sixth binomial (straight- ἀσύμμετρον τῇ παρ᾿ ἣν παράκειται μήκει. τὸ δὲ ἀπὸ τῆς line) [Def. 10.10]. And if an area is contained by a ra- ἐκ δύο ὀνομάτων παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ tional (straight-line) and a sixth binomial (straight-line) τὴν ἐκ δύο ὀνομάτων πρώτην. τὸ δὲ ἀπὸ τῆς ἐκ δύο then the square-root of the area is the square-root of (the μέσων πρώτης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ sum of) two medial (areas) [Prop. 10.59]. Hence, the τὴν ἐκ δύο ὀνομάτων δευτέραν. τὸ δὲ ἀπὸ τῆς ἐκ δύο square-root of area AD is also the square-root of (the μέσων δευτέρας παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ sum of) two medial (areas). τὴν ἐκ δύο ὀνομάτων τρίτην. τὸ δὲ ἀπὸ τῆς μείζονος παρὰ [So, similarly, we can show that, even if AB is less ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων than CD, the square-root of area AD is either a second τετάρτην. τὸ δὲ ἀπὸ τῆς ῥητὸν καὶ μέσον δυναμένης παρὰ bimedial or the square-root of (the sum of) two medial ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων (areas).] πέμπτην. τὸ δὲ ἀπὸ τῆς δύο μέσα δυναμένης παρὰ ῥητὴν Thus, when two medial (areas which are) incommen- παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων ἕκτην. surable with one another are added together, the remain- τὰ δ᾿ εἰρημένα πλάτη διαφέρει τοῦ τε πρώτου καὶ ἀλλήλων, ing two irrational (straight-lines) arise (as the square- τοῦ μὲν πρώτου, ὅτι ῥητή ἐστιν, ἀλλήλων δέ, ὅτι τῇ τάξει roots of the total area)—either a second bimedial, or the οὐκ εἰσὶν αἱ αὐταί· ὥστε καὶ αὐταὶ αἱ ἄλογοι διαφέρουσιν square-root of (the sum of) two medial (areas). ἀλλήλων. A binomial (straight-line), and the (other) irrational (straight-lines) after it, are neither the same as a medial (straight-line) nor (the same) as one another. For the (square) on a medial (straight-line), applied to a rational (straight-line), produces as breadth a rational (straight- line which is) also incommensurable in length with (the straight-line) to which it is applied [Prop. 10.22]. And the (square) on a binomial (straight-line), applied to a rational (straight-line), produces as breadth a first bino- mial [Prop. 10.60]. And the (square) on a first bimedial (straight-line), applied to a rational (straight-line), pro- duces as breadth a second binomial [Prop. 10.61]. And the (square) on a second bimedial (straight-line), applied to a rational (straight-line), produces as breadth a third binomial [Prop. 10.62]. And the (square) on a major (straight-line), applied to a rational (straight-line), pro- duces as breadth a fourth binomial [Prop. 10.63]. And the (square) on the square-root of a rational plus a medial 366 STOIQEIWN iþ. ELEMENTS BOOK 10 (area), applied to a rational (straight-line), produces as breadth a fifth binomial [Prop. 10.64]. And the (square) on the square-root of (the sum of) two medial (areas), applied to a rational (straight-line), produces as breadth a sixth binomial [Prop. 10.65]. And the aforementioned breadths differ from the first (breadth), and from one another—from the first, because it is rational—and from one another, because they are not the same in order. Hence, the (previously mentioned) irrational (straight- lines) themselves also differ from one another.ogþ. Proposition 73 ᾿Εὰν ἀπὸ ῥητῆς ῥητὴ ἀφαιρεθῇ δυνάμει μόνον σύμμετρος If a rational (straight-line), which is commensu- οὖσα τῇ ὅλῃ, ἡ λοιπὴ ἄλογός ἐστιν· καλείσθω δὲ ἀποτομή. rable in square only with the whole, is subtracted from a(nother) rational (straight-line) then the remainder is an irrational (straight-line). Let it be called an apotome. Γ ΒΑ BCA Ἀπὸ γὰρ ῥητῆς τῆς ΑΒ ῥητὴ ἀφῃρήσθω ἡ ΒΓ δυνάμει For let the rational (straight-line) BC, which com- μόνον σύμμετρος οὖσα τῇ ὅλῃ· λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ mensurable in square only with the whole, have been ἄλογός ἐστιν ἡ καλουμένη ἀποτομή. subtracted from the rational (straight-line) AB. I say that ᾿Επεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, καί ἐστιν the remainder AC is that irrational (straight-line) called ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ὑπὸ an apotome. τῶν ΑΒ, ΒΓ, ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΒ τῷ ὑπὸ For since AB is incommensurable in length with BC, τῶν ΑΒ, ΒΓ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΒ σύμμετρά ἐστι τὰ ἀπὸ and as AB is to BC, so the (square) on AB (is) to the τῶν ΑΒ, ΒΓ τετράγωνα, τῷ δὲ ὑπὸ τῶν ΑΒ, ΒΓ σύμμετρόν (rectangle contained) by AB and BC [Prop. 10.21 lem.], ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. καὶ ἐπειδήπερ τὰ ἀπὸ τῶν ΑΒ, the (square) on AB is thus incommensurable with the ΒΓ ἴσα ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΑ, καὶ (rectangle contained) by AB and BC [Prop. 10.11]. But, λοιπῷ ἄρα τῷ ἀπὸ τῆς ΑΓ ἀσύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, the (sum of the) squares on AB and BC is commen- ΒΓ. ῥητὰ δὲ τὰ ἀπὸ τῶν ΑΒ, ΒΓ· ἄλογος ἄρα ἐστὶν ἡ ΑΓ· surable with the (square) on AB [Prop. 10.15], and καλείσθω δὲ ἀποτομή. ὅπερ ἔδει δεῖξαι. twice the (rectangle contained) by AB and BC is com- mensurable with the (rectangle contained) by AB and BC [Prop. 10.6]. And, inasmuch as the (sum of the squares) on AB and BC is equal to twice the (rectan- gle contained) by AB and BC plus the (square) on CA [Prop. 2.7], the (sum of the squares) on AB and BC is thus also incommensurable with the remaining (square) on AC [Props. 10.13, 10.16]. And the (sum of the squares) on AB and BC is rational. AC is thus an ir- rational (straight-line) [Def. 10.4]. And let it be called an apotome.† (Which is) the very thing it was required to show. † See footnote to Prop. 10.36. odþ. Proposition 74 ᾿Εὰν ἀπὸ μέσης μέση ἀφαιρεθῇ δυνάμει μόνον σύμμετρος If a medial (straight-line), which is commensurable in οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ῥητὸν περιέχουσα, ἡ λοιπὴ square only with the whole, and which contains a ratio- ἄλογός ἐστιν· καλείσθω δὲ μέσης ἀποτομὴ πρώτη. nal (area) with the whole, is subtracted from a(nother) medial (straight-line) then the remainder is an irrational 367 STOIQEIWN iþ. ELEMENTS BOOK 10 (straight-line). Let it be called a first apotome of a medial (straight-line). Γ ΒΑ BCA Ἀπὸ γὰρ μέσης τῆς ΑΒ μέση ἀφῃρήσθω ἡ ΒΓ δυνάμει For let the medial (straight-line) BC, which is com- μόνον σύμμετρος οὖσα τῇ ΑΒ, μετὰ δὲ τῆς ΑΒ ῥητὸν mensurable in square only with AB, and which makes ποιοῦσα τὸ ὑπὸ τῶν ΑΒ, ΒΓ· λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ with AB the rational (rectangle contained) by AB and ἄλογός ἐστιν· καλείσθω δὲ μέσης ἀποτομὴ πρώτη. BC, have been subtracted from the medial (straight-line) ᾿Επεὶ γὰρ αἱ ΑΒ, ΒΓ μέσαι εἰσίν, μέσα ἐστὶ καὶ τὰ ἀπὸ AB [Prop. 10.27]. I say that the remainder AC is an ir- τῶν ΑΒ, ΒΓ. ῥητὸν δὲ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ· ἀσύμμετρα rational (straight-line). Let it be called the first apotome ἄρα τὰ ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ· καὶ λοιπῷ of a medial (straight-line). ἄρα τῷ ἀπὸ τῆς ΑΓ ἀσύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, For since AB and BC are medial (straight-lines), the ΒΓ, ἐπεὶ κἂν τὸ ὅλον ἑνὶ αὐτῶν ἀσύμμετρον ᾖ, καὶ τὰ ἐξ (sum of the squares) on AB and BC is also medial. And ἀρχῆς μεγέθη ἀσύμμετρα ἔσται. ῥητὸν δὲ τὸ δὶς ὑπὸ τῶν twice the (rectangle contained) by AB and BC (is) ratio- ΑΒ, ΒΓ· ἄλογον ἄρα τὸ ἀπὸ τῆς ΑΓ· ἄλογος ἄρα ἐστὶν ἡ nal. The (sum of the squares) on AB and BC (is) thus in- ΑΓ· καλείσθω δὲ μέσης ἀποτομὴ πρώτη. commensurable with twice the (rectangle contained) by AB and BC. Thus, twice the (rectangle contained) by AB and BC is also incommensurable with the remain- ing (square) on AC [Prop. 2.7], since if the whole is in- commensurable with one of the (constituent magnitudes) then the original magnitudes will also be incommensu- rable (with one another) [Prop. 10.16]. And twice the (rectangle contained) by AB and BC (is) rational. Thus, the (square) on AC is irrational. Thus, AC is an irra- tional (straight-line) [Def. 10.4]. Let it be called a first apotome of a medial (straight-line).† † See footnote to Prop. 10.37. oeþ. Proposition 75 ᾿Εὰν ἀπὸ μέσης μέση ἀφαιρεθῇ δυνάμει μόνον σύμμετρος If a medial (straight-line), which is commensurable in οὖσα τῇ ὅλη, μετὰ δὲ τῆς ὅλης μέσον περιέχουσα, ἡ λοιπὴ square only with the whole, and which contains a me- ἄλογός ἐστιν· καλείσθω δὲ μέσης ἀποτομὴ δευτέρα. dial (area) with the whole, is subtracted from a(nother) Ἀπὸ γὰρ μέσης τῆς ΑΒ μέση ἀφῃρήσθω ἡ ΓΒ δυνάμει medial (straight-line) then the remainder is an irrational μόνον σύμμετρος οὖσα τῇ ὅλῃ τῇ ΑΒ, μετὰ δὲ τῆς ὅλης (straight-line). Let it be called a second apotome of a τῆς ΑΒ μέσον περιέχουσα τὸ ὑπὸ τῶν ΑΒ, ΒΓ· λέγω, ὅτι medial (straight-line). ἡ λοιπὴ ἡ ΑΓ ἄλογός ἐστιν· καλείσθω δὲ μέσης ἀποτομὴ For let the medial (straight-line) CB, which is com- δευτέρα. mensurable in square only with the whole, AB, and which contains with the whole, AB, the medial (rect- angle contained) by AB and BC, have been subtracted from the medial (straight-line) AB [Prop. 10.28]. I say that the remainder AC is an irrational (straight-line). Let it be called a second apotome of a medial (straight-line). 368 STOIQEIWN iþ. ELEMENTS BOOK 10 Ι Β Ε Γ ΗΖ Θ Α ∆ A BC H E F GD I ᾿Εκκείσθω γὰρ ῥητὴ ἡ ΔΙ, καὶ τοῖς μὲν ἀπὸ τῶν ΑΒ, For let the rational (straight-line) DI be laid down. ΒΓ ἴσον παρὰ τὴν ΔΙ παραβεβλήσθω τὸ ΔΕ πλάτος ποιοῦν And let DE, equal to the (sum of the squares) on AB τὴν ΔΗ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ἴσον παρὰ τὴν ΔΙ and BC, have been applied to DI, producing DG as παραβεβλήσθω τὸ ΔΘ πλάτος ποιοῦν τὴν ΔΖ· λοιπὸν ἄρα breadth. And let DH , equal to twice the (rectangle con- τὸ ΖΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ. καὶ ἐπεὶ μέσα καὶ σύμμετρά tained) by AB and BC, have been applied to DI, produc- ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ, μέσον ἄρα καὶ τὸ ΔΕ. καὶ παρὰ ing DF as breadth. The remainder FE is thus equal to ῥητὴν τὴν ΔΙ παράκειται πλάτος ποιοῦν τὴν ΔΗ· ῥητὴ ἄρα the (square) on AC [Prop. 2.7]. And since the (squares) ἐστὶν ἡ ΔΗ καὶ ἀσύμμετρος τῇ ΔΙ μήκει. πάλιν, ἐπεὶ μέσον on AB and BC are medial and commensurable (with ἐστὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ, καὶ τὸ δὶς ἄρα ὑπὸ τῶν ΑΒ, one another), DE (is) thus also medial [Props. 10.15, ΒΓ μέσον ἐστίν. καί ἐστιν ἴσον τῷ ΔΘ· καὶ τὸ ΔΘ ἄρα 10.23 corr.]. And it is applied to the rational (straight- μέσον ἐστίν. καὶ παρὰ ῥητὴν τὴν ΔΙ παραβέβληται πλάτος line) DI, producing DG as breadth. Thus, DG is rational, ποιοῦν τὴν ΔΖ· ῥητὴ ἄρα ἐστὶν ἡ ΔΖ καὶ ἀσύμμετρος τῇ ΔΙ and incommensurable in length with DI [Prop. 10.22]. μήκει. καὶ ἐπεὶ αἱ ΑΒ, ΒΓ δυνάμει μόνον σύμμετροί εἰσιν, Again, since the (rectangle contained) by AB and BC is ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΒ τῇ ΒΓ μήκει· ἀσύμμετρον ἄρα medial, twice the (rectangle contained) by AB and BC καὶ τὸ ἀπὸ τῆς ΑΒ τετράγωνον τῷ ὑπὸ τῶν ΑΒ, ΒΓ. ἀλλὰ is thus also medial [Prop. 10.23 corr.]. And it is equal τῷ μὲν ἀπὸ τῆς ΑΒ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ, τῷ to DH . Thus, DH is also medial. And it has been ap- δὲ ὑπὸ τῶν ΑΒ, ΒΓ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, plied to the rational (straight-line) DI, producing DF as ΒΓ· ἀσύμμετρον ἄρα ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ τοῖς ἀπὸ breadth. DF is thus rational, and incommensurable in τῶν ΑΒ, ΒΓ. ἴσον δὲ τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ τὸ ΔΕ, length with DI [Prop. 10.22]. And since AB and BC are τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ τὸ ΔΘ· ἀσύμμετρον ἄρα [ἐστὶ] commensurable in square only, AB is thus incommensu- τὸ ΔΕ τῷ ΔΘ. ὡς δὲ τὸ ΔΕ πρὸς τὸ ΔΘ, οὕτως ἡ ΗΔ rable in length with BC. Thus, the square on AB (is) πρὸς τὴν ΔΖ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΗΔ τῇ ΔΖ. καί εἰσιν also incommensurable with the (rectangle contained) by ἀμφότεραι ῥηταί· αἱ ἄρα ΗΔ, ΔΖ ῥηταί εἰσι δυνάμει μόνον AB and BC [Props. 10.21 lem., 10.11]. But, the (sum σύμμετροι· ἡ ΖΗ ἄρα ἀποτομή ἐστιν. ῥητὴ δὲ ἡ ΔΙ· τὸ δὲ of the squares) on AB and BC is commensurable with ὑπὸ ῥητῆς καὶ ἀλόγου περιεχόμενον ἄλογόν ἐστιν, καὶ ἡ the (square) on AB [Prop. 10.15], and twice the (rectan- δυναμένη αὐτὸ ἄλογός ἐστιν. καὶ δύναται τὸ ΖΕ ἡ ΑΓ· ἡ gle contained) by AB and BC is commensurable with the ΑΓ ἄρα ἄλογός ἐστιν· καλείσθω δὲ μέσης ἀποτομὴ δευτέρα. (rectangle contained) by AB and BC [Prop. 10.6]. Thus, ὅπερ ἔδει δεῖξαι. twice the (rectangle contained) by AB and BC is incom- mensurable with the (sum of the squares) on AB and BC [Prop. 10.13]. And DE is equal to the (sum of the squares) on AB and BC, and DH to twice the (rectangle contained) by AB and BC. Thus, DE [is] incommensu- rable with DH . And as DE (is) to DH , so GD (is) to DF [Prop. 6.1]. Thus, GD is incommensurable with DF [Prop. 10.11]. And they are both rational (straight-lines). Thus, GD and DF are rational (straight-lines which are) commensurable in square only. Thus, FG is an apotome [Prop. 10.73]. And DI (is) rational. And the (area) con- tained by a rational and an irrational (straight-line) is irrational [Prop. 10.20], and its square-root is irrational. 369 STOIQEIWN iþ. ELEMENTS BOOK 10 And AC is the square-root of FE. Thus, AC is an irra- tional (straight-line) [Def. 10.4]. And let it be called the second apotome of a medial (straight-line).† (Which is) the very thing it was required to show. † See footnote to Prop. 10.38. o�þ. Proposition 76 ᾿Εὰν ἀπὸ εὐθείας εὐθεῖα ἀφαιρεθῂ δυνάμει ἀσύμμετρος If a straight-line, which is incommensurable in square οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τὰ μὲν ἀπ᾿ αὐτῶν with the whole, and with the whole makes the (squares) ἅμα ῥητόν, τὸ δ᾿ ὑπ᾿ αὐτῶν μέσον, ἡ λοιπὴ ἄλογός ἐστιν· on them (added) together rational, and the (rectangle καλείσθω δὲ ἐλάσσων. contained) by them medial, is subtracted from a(nother) straight-line then the remainder is an irrational (straight- line). Let it be called a minor (straight-line). Γ ΒΑ BCA Ἀπὸ γὰρ εὐθείας τῆς ΑΒ εὐθεῖα ἀφῃρήσθω ἡ ΒΓ For let the straight-line BC, which is incommensu- δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ ποιοῦσα τὰ προκείμενα. rable in square with the whole, and fulfils the (other) λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων. prescribed (conditions), have been subtracted from the ᾿Επεὶ γὰρ τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ straight-line AB [Prop. 10.33]. I say that the remainder τετραγώνων ῥητόν ἐστιν, τὸ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ μέσον, AC is that irrational (straight-line) called minor. ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν For since the sum of the squares on AB and BC is ΑΒ, ΒΓ· καὶ ἀναστρέψαντι λοιπῷ τῷ ἀπὸ τῆς ΑΓ ἀσύμμετρά rational, and twice the (rectangle contained) by AB and ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ. ῥητὰ δὲ τὰ ἀπὸ τῶν ΑΒ, ΒΓ· BC (is) medial, the (sum of the squares) on AB and BC ἄλογον ἄρα τὸ ἀπὸ τῆς ΑΓ· ἄλογος ἄρα ἡ ΑΓ· καλείσθω δὲ is thus incommensurable with twice the (rectangle con- ἐλάσσων. ὅπερ ἔδει δεῖξαι. tained) by AB and BC. And, via conversion, the (sum of the squares) on AB and BC is incommensurable with the remaining (square) on AC [Props. 2.7, 10.16]. And the (sum of the squares) on AB and BC (is) rational. The (square) on AC (is) thus irrational. Thus, AC (is) an irrational (straight-line) [Def. 10.4]. Let it be called a minor (straight-line).† (Which is) the very thing it was required to show. † See footnote to Prop. 10.39. ozþ. Proposition 77 ᾿Εὰν ἀπὸ εὐθείας εὐθεῖα ἀφαιρεθῇ δυνάμει ἀσύμμετρος If a straight-line, which is incommensurable in square οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τὸ μὲν συγκείμενον with the whole, and with the whole makes the sum of the ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον, τὸ δὲ δὶς ὑπ᾿ αὐτῶν squares on them medial, and twice the (rectangle con- ῥητόν, ἡ λοιπὴ ἄλογός ἐστιν· καλείσθω δὲ ἡ μετὰ ῥητοῦ tained) by them rational, is subtracted from a(nother) μέσον τὸ ὅλον ποιοῦσα. straight-line then the remainder is an irrational (straight- line). Let it be called that which makes with a rational (area) a medial whole. Γ ΒΑ BCA Ἀπὸ γὰρ εὐθείας τῆς ΑΒ εὐθεῖα ἀφῃρήσθω ἡ ΒΓ For let the straight-line BC, which is incommensu- δυνάμει ἀσύμμετος οὖσα τῇ ΑΒ ποιοῦσα τὰ προκείμενα· rable in square with AB, and fulfils the (other) prescribed λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ ἄλογός ἐστιν ἡ προειρημένη. (conditions), have been subtracted from the straight-line ᾿Επεὶ γὰρ τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ AB [Prop. 10.34]. I say that the remainder AC is the 370 STOIQEIWN iþ. ELEMENTS BOOK 10 τετραγώνων μέσον ἐστίν, τὸ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ῥητόν, aforementioned irrational (straight-line). ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν For since the sum of the squares on AB and BC is ΑΒ, ΒΓ· καὶ λοιπὸν ἄρα τὸ ἀπὸ τῆς ΑΓ ἀσύμμετρόν ἐστι medial, and twice the (rectangle contained) by AB and τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. καί ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ BC rational, the (sum of the squares) on AB and BC ῥητόν· τὸ ἄρα ἀπὸ τῆς ΑΓ ἄλογόν ἐστιν· ἄλογος ἄρα ἐστὶν is thus incommensurable with twice the (rectangle con- ἡ ΑΓ· καλείσθω δὲ ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα. tained) by AB and BC. Thus, the remaining (square) ὅπερ ἔδει δεῖξαι. on AC is also incommensurable with twice the (rectan- gle contained) by AB and BC [Props. 2.7, 10.16]. And twice the (rectangle contained) by AB and BC is ratio- nal. Thus, the (square) on AC is irrational. Thus, AC is an irrational (straight-line) [Def. 10.4]. And let it be called that which makes with a rational (area) a medial whole.† (Which is) the very thing it was required to show. † See footnote to Prop. 10.40. ohþ. Proposition 78 ᾿Εὰν ἀπὸ εὐθείας εὐθεῖα ἀφαιρεθῇ δυνάμει ἀσύμμετρος If a straight-line, which is incommensurable in square οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τό τε συγκείμενον with the whole, and with the whole makes the sum of the ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον τό τε δὶς ὑπ᾿ αὐτῶν squares on them medial, and twice the (rectangle con- μέσον καὶ ἔτι τὰ ἀπ᾿ αὐτῶν τετράγωνα ἀσύμμετρα τῷ δὶς ὑπ᾿ tained) by them medial, and, moreover, the (sum of the) αὐτῶν, ἡ λοιπὴ ἄλογός ἐστιν· καλείσθω δὲ ἡ μετὰ μέσου squares on them incommensurable with twice the (rect- μέσον τὸ ὅλον ποιοῦσα. angle contained) by them, is subtracted from a(nother) straight-line then the remainder is an irrational (straight- line). Let it be called that which makes with a medial (area) a medial whole. Γ ∆ Ζ Η Ι Θ Ε ΒΑ FD I E BA H G C Ἀπὸ γὰρ εὐθείας τῆς ΑΒ εὐθεῖα ἀφῃρήσθω ἡ ΒΓ For let the straight-line BC, which is incommensu- δυνάμει ἀσύμμετρος οὖσα τῇ ΑΒ ποιοῦσα τὰ προκείμενα· rable in square AB, and fulfils the (other) prescribed λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ ἄλογός ἐστιν ἡ καλουμένη ἡ μετὰ (conditions), have been subtracted from the (straight- μέσου μέσον τὸ ὅλον ποιοῦσα. line) AB [Prop. 10.35]. I say that the remainder AC is ᾿Εκκείσθω γὰρ ῥητὴ ἡ ΔΙ, καὶ τοῖς μὲν ἀπὸ τῶν ΑΒ, the irrational (straight-line) called that which makes with ΒΓ ἴσον παρὰ τὴν ΔΙ παραβεβλήσθω τὸ ΔΕ πλάτος ποιοῦν a medial (area) a medial whole. τὴν ΔΗ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἀφῃρήσθω τὸ For let the rational (straight-line) DI be laid down. ΔΘ [πλάτος ποιοῦν τὴν ΔΖ]. λοιπὸν ἄρα τὸ ΖΕ ἴσον ἐστὶ And let DE, equal to the (sum of the squares) on AB and τῷ ἀπὸ τῆς ΑΓ· ὥστε ἡ ΑΓ δύναται τὸ ΖΕ. καὶ ἐπεὶ τὸ BC, have been applied to DI, producing DG as breadth. συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τετραγώνων μέσον And let DH , equal to twice the (rectangle contained) by ἐστὶ καί ἐστιν ἴσον τῷ ΔΕ, μέσον ἄρα [ἐστὶ] τὸ ΔΕ. καὶ παρὰ AB and BC, have been subtracted (from DE) [produc- ῥητὴν τὴν ΔΙ παράκειται πλάτος ποιοῦν τὴν ΔΗ· ῥητὴ ἄρα ing DF as breadth]. Thus, the remainder FE is equal ἐστὶν ἡ ΔΗ καὶ ἀσύμμετρος τῇ ΔΙ μήκει. πάλιν, ἐπεὶ τὸ δὶς to the (square) on AC [Prop. 2.7]. Hence, AC is the ὑπὸ τῶν ΑΒ, ΒΓ μέσον ἐστὶ καί ἐστιν ἴσον τῷ ΔΘ, τὸ ἄρα square-root of FE. And since the sum of the squares on 371 STOIQEIWN iþ. ELEMENTS BOOK 10 ΔΘ μέσον ἐστίν. καὶ παρὰ ῥητὴν τὴν ΔΙ παράκειται πλάτος AB and BC is medial, and is equal to DE, DE [is] thus ποιοῦν τὴν ΔΖ· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΔΖ καὶ ἀσύμμετρος medial. And it is applied to the rational (straight-line) τῇ ΔΙ μήκει. καὶ ἐπεὶ ἀσύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ DI, producing DG as breadth. Thus, DG is rational, and τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, ἀσύμμετρον ἄρα καὶ τὸ ΔΕ τῷ incommensurable in length with DI [Prop 10.22]. Again, ΔΘ. ὡς δὲ τὸ ΔΕ πρὸς τὸ ΔΘ, οὕτως ἐστὶ καὶ ἡ ΔΗ πρὸς since twice the (rectangle contained) by AB and BC is τὴν ΔΖ· ἀσύμμετρος ἄρα ἡ ΔΗ τῇ ΔΖ. καί εἰσιν ἀμφότεραι medial, and is equal to DH , DH is thus medial. And it is ῥηταί· αἱ ΗΔ, ΔΖ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. applied to the rational (straight-line) DI, producing DF ἀποτομὴ ἄρα ἐστίν ἡ ΖΗ· ῥητὴ δὲ ἡ ΖΘ. τὸ δὲ ὑπὸ ῥητῆς as breadth. Thus, DF is also rational, and incommen- καὶ ἀποτομῆς περιεχόμενον [ὀρθογώνιον] ἄλογόν ἐστιν, καὶ surable in length with DI [Prop. 10.22]. And since the ἡ δυναμένη αὐτὸ ἄλογός ἐστιν· καὶ δύναται τὸ ΖΕ ἡ ΑΓ· ἡ (sum of the squares) on AB and BC is incommensurable ΑΓ ἄρα ἄλογός ἐστιν· καλείσθω δὲ ἡ μετὰ μέσου μέσον τὸ with twice the (rectangle contained) by AB and BC, DE ὅλον ποιοῦσα. ὅπερ ἔδει δεῖξαι. (is) also incommensurable with DH . And as DE (is) to DH , so DG also is to DF [Prop. 6.1]. Thus, DG (is) in- commensurable (in length) with DF [Prop. 10.11]. And they are both rational. Thus, GD and DF are ratio- nal (straight-lines which are) commensurable in square only. Thus, FG is an apotome [Prop. 10.73]. And FH (is) rational. And the [rectangle] contained by a rational (straight-line) and an apotome is irrational [Prop. 10.20], and its square-root is irrational. And AC is the square- root of FE. Thus, AC is irrational. Let it be called that which makes with a medial (area) a medial whole.† (Which is) the very thing it was required to show. † See footnote to Prop. 10.41. ojþ. Proposition 79 Τῇ ἀποτομῇ μία [μόνον] προσαρμόζει εὐθεῖα ῥητὴ [Only] one rational straight-line, which is commensu- δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ. rable in square only with the whole, can be attached to an apotome.† ∆Α Β Γ BA C D ῎Εστω ἀποτομὴ ἡ ΑΒ, προσαρμόζουσα δὲ αὐτῇ ἡ ΒΓ· αἱ Let AB be an apotome, with BC (so) attached to it. ΑΓ, ΓΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· λέγω, ὅτι AC and CB are thus rational (straight-lines which are) τῇ ΑΒ ἑτέρα οὐ προσαρμόζει ῥητὴ δυνάμει μόνον σύμμετρος commensurable in square only [Prop. 10.73]. I say that οὖσα τῇ ὅλῇ. another rational (straight-line), which is commensurable Εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ· καὶ αἱ ΑΔ, in square only with the whole, cannot be attached to AB. ΔΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ ἐπεί, ᾧ For, if possible, let BD be (so) attached (to AB). ὑπερέχει τὰ ἀπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ, Thus, AD and DB are also rational (straight-lines which τούτῳ ὑπερέχει καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν are) commensurable in square only [Prop. 10.73]. And ΑΓ, ΓΒ· τῷ γὰρ αὐτῷ τῷ ἀπὸ τῆς ΑΒ ἀμφότερα ὑπερέχει· since by whatever (area) the (sum of the squares) on AD ἐναλλὰξ ἄρα, ᾧ ὑπερέχει τὰ ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν and DB exceeds twice the (rectangle contained) by AD ΑΓ, ΓΒ, τούτῳ ὑπερέχει [καὶ] τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ and DB, the (sum of the squares) on AC and CB also ex- δὶς ὑπὸ τῶν ΑΓ, ΓΒ. τὰ δὲ ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν ceeds twice the (rectangle contained) by AC and CB by ΑΓ, ΓΒ ὑπερέχει ῥητῷ· ῥητὰ γὰρ ἀμφότερα. καὶ τὸ δὶς ἄρα this (same area). For both exceed by the same (area)— ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ· (namely), the (square) on AB [Prop. 2.7]. Thus, alter- ὅπερ ἐστὶν ἀδύνατον· μέσα γὰρ ἀμφότερα, μέσον δὲ μέσου nately, by whatever (area) the (sum of the squares) on οὐχ ὑπερέχει ῥητῷ. τῇ ἄρα ΑΒ ἑτέρα οὐ προσαρμόζει ῥητὴ AD and DB exceeds the (sum of the squares) on AC δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ. and CB, twice the (rectangle contained) by AD and DB Μία ἄρα μόνη τῇ ἀποτομῇ προσαρμόζει ῥητὴ δυνάμει [also] exceeds twice the (rectangle contained) by AC and 372 STOIQEIWN iþ. ELEMENTS BOOK 10 μόνον σύμμετρος οὖσα τῇ ὅλῃ· ὅπερ ἔδει δεῖξαι. CB by this (same area). And the (sum of the squares) on AD and DB exceeds the (sum of the squares) on AC and CB by a rational (area). For both (are) rational (ar- eas). Thus, twice the (rectangle contained) by AD and DB also exceeds twice the (rectangle contained) by AC and CB by a rational (area). The very thing is impos- sible. For both are medial (areas) [Prop. 10.21], and a medial (area) cannot exceed a(nother) medial (area) by a rational (area) [Prop. 10.26]. Thus, another rational (straight-line), which is commensurable in square only with the whole, cannot be attached to AB. Thus, only one rational (straight-line), which is com- mensurable in square only with the whole, can be at- tached to an apotome. (Which is) the very thing it was required to show. † This proposition is equivalent to Prop. 10.42, with minus signs instead of plus signs.pþ. Proposition 80 Τῇ μέσης ἀποτομῇ πρώτῃ μία μόνον προσαρμόζει εὐθεῖα Only one medial straight-line, which is commensu- μέση δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς rable in square only with the whole, and contains a ra- ὅλης ῥητὸν περιέχουσα. tional (area) with the whole, can be attached to a first apotome of a medial (straight-line).† ∆Α Β Γ BA C D ῎Εστω γὰρ μέσης ἀποτομὴ πρώτη ἡ ΑΒ, καὶ τῇ ΑΒ For let AB be a first apotome of a medial (straight- προσαρμοζέτω ἡ ΒΓ· αἱ ΑΓ, ΓΒ ἄρα μέσαι εἰσὶ δυνάμει line), and let BC be (so) attached to AB. Thus, AC μόνον σύμμετροι ῥητὸν περιέχουσαι τὸ ὑπὸ τῶν ΑΓ, ΓΒ· and CB are medial (straight-lines which are) commen- λέγω, ὅτι τῇ ΑΒ ἑτέρα οὐ προσαρμόζει μέση δυνάμει surable in square only, containing a rational (area)— μόνον σύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ῥητὸν (namely, that contained) by AC and CB [Prop. 10.74]. περιέχουσα. I say that a(nother) medial (straight-line), which is com- Εἰ γὰρ δυνατόν, προσαρμοζέτω καὶ ἡ ΔΒ· αἱ ἄρα ΑΔ, mensurable in square only with the whole, and contains ΔΒ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι a rational (area) with the whole, cannot be attached to τὸ ὑπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεί, ᾧ ὑπερέχει τὰ ἀπὸ τῶν ΑΔ, AB. ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ, τούτῳ ὑπερέχει καὶ τὰ ἀπὸ For, if possible, let DB also be (so) attached to τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ· τῷ γὰρ αὐτῷ [πάλιν] AB. Thus, AD and DB are medial (straight-lines which ὑπερέχουσι τῷ ἀπὸ τῆς ΑΒ· ἐναλλὰξ ἄρα, ᾧ ὑπερέχει τὰ are) commensurable in square only, containing a ratio- ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, τούτῳ ὑπερέχει nal (area)—(namely, that) contained by AD and DB καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. τὸ [Prop. 10.74]. And since by whatever (area) the (sum of δὲ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ὑπερέχει the squares) on AD and DB exceeds twice the (rectangle ῥητῷ· ῥητὰ γὰρ ἀμφότερα. καὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ ἄρα contained) by AD and DB, the (sum of the squares) on τῶν ἀπὸ τῶν ΑΓ, ΓΒ [τετραγώνων] ὑπερέχει ῥητῷ· ὅπερ AC and CB also exceeds twice the (rectangle contained) ἐστὶν ἀδύνατον· μέσα γάρ ἐστιν ἀμφότερα, μέσον δὲ μέσου by AC and CB by this (same area). For [again] both ex- οὐχ ὑπερέχει ῥητῷ. ceed by the same (area)—(namely), the (square) on AB Τῇ ἄρα μέσης ἀποτομῇ πρώτῃ μία μόνον προσαρμόζει [Prop. 2.7]. Thus, alternately, by whatever (area) the εὐθεῖα μέση δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ, μετὰ (sum of the squares) on AD and DB exceeds the (sum δὲ τῆς ὅλης ῥητὸν περιέχουσα· ὅπερ ἔδει δεῖξαι. of the squares) on AC and CB, twice the (rectangle con- tained) by AD and DB also exceeds twice the (rectangle contained) by AC and CB by this (same area). And twice the (rectangle contained) by AD and DB exceeds twice 373 STOIQEIWN iþ. ELEMENTS BOOK 10 the (rectangle contained) by AC and CB by a rational (area). For both (are) rational (areas). Thus, the (sum of the squares) on AD and DB also exceeds the (sum of the) [squares] on AC and CB by a rational (area). The very thing is impossible. For both are medial (areas) [Props. 10.15, 10.23 corr.], and a medial (area) can- not exceed a(nother) medial (area) by a rational (area) [Prop. 10.26]. Thus, only one medial (straight-line), which is com- mensurable in square only with the whole, and contains a rational (area) with the whole, can be attached to a first apotome of a medial (straight-line). (Which is) the very thing it was required to show. † This proposition is equivalent to Prop. 10.43, with minus signs instead of plus signs.paþ. Proposition 81 Τῇ μέσης ἀποτομῇ δευτέρᾳ μία μόνον προσαρμόζει Only one medial straight-line, which is commensu- εὐθεῖα μέση δυνάμει μόνον σύμμετρος τῇ ὅλῃ, μετὰ δὲ τῆς rable in square only with the whole, and contains a me- ὅλης μέσον περιέχουσα. dial (area) with the whole, can be attached to a second apotome of a medial (straight-line).† Ζ Α Β Γ ∆ Ε Θ Μ Ν Η ΙΛ CA B D E M N ILF H G ῎Εστω μέσης ἀποτομὴ δευτέρα ἡ ΑΒ καὶ τῇ ΑΒ προ- Let AB be a second apotome of a medial (straight- σαρμόζουσα ἡ ΒΓ· αἱ ἄρα ΑΓ, ΓΒ μέσαι εἰσὶ δυνάμει μόνον line), with BC (so) attached to AB. Thus, AC and CB σύμμετροι μέσον περιέχουσαι τὸ ὑπὸ τῶν ΑΓ, ΓΒ· λέγω, are medial (straight-lines which are) commensurable in ὅτι τῇ ΑΒ ἑτέρα οὐ προσαρμόσει εὐθεῖα μέση δυνάμει square only, containing a medial (area)—(namely, that μόνον σύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης μέσον contained) by AC and CB [Prop. 10.75]. I say that περιέχουσα. a(nother) medial straight-line, which is commensurable Εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ· καὶ αἱ ΑΔ, ΔΒ in square only with the whole, and contains a medial ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον περιέχουσαι (area) with the whole, cannot be attached to AB. τὸ ὑπὸ τῶν ΑΔ, ΔΒ. καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ τοῖς For, if possible, let BD be (so) attached. Thus, AD μὲν ἀπὸ τῶν ΑΓ, ΓΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω τὸ and DB are also medial (straight-lines which are) com- ΕΗ πλάτος ποιοῦν τὴν ΕΜ· τῷ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον mensurable in square only, containing a medial (area)— ἀφῃρήσθω τὸ ΘΗ πλάτος ποιοῦν τὴν ΘΜ· λοιπὸν ἄρα τὸ ΕΛ (namely, that contained) by AD and DB [Prop. 10.75]. ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ· ὥστε ἡ ΑΒ δύναται τὸ ΕΛ. πάλιν And let the rational (straight-line) EF be laid down. And δὴ τοῖς ἀπὸ τῶν ΑΔ, ΔΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω let EG, equal to the (sum of the squares) on AC and τὸ ΕΙ πλάτος ποιοῦν τὴν ΕΝ· ἔστι δὲ καὶ τὸ ΕΛ ἴσον τῷ ἀπὸ CB, have been applied to EF , producing EM as breadth. τῆς ΑΒ τετραγώνῳ· λοιπὸν ἄρα τὸ ΘΙ ἴσον ἐστὶ τῷ δὶς ὑπὸ And let HG, equal to twice the (rectangle contained) by τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ μέσαι εἰσὶν αἱ ΑΓ, ΓΒ, μέσα ἄρα ἐστὶ AC and CB, have been subtracted (from EG), produc- καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ. καί ἐστιν ἴσα τῷ ΕΗ· μέσον ἄρα καὶ ing HM as breadth. The remainder EL is thus equal τὸ ΕΗ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος ποιοῦν to the (square) on AB [Prop. 2.7]. Hence, AB is the 374 STOIQEIWN iþ. ELEMENTS BOOK 10 τὴν ΕΜ· ῥητὴ ἄρα ἐστὶν ἡ ΕΜ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. square-root of EL. So, again, let EI, equal to the (sum πάλιν, ἐπεὶ μέσον ἐστὶ τὸ ὑπὸ τῶν ΑΓ, ΓΒ, καὶ τὸ δὶς ὑπὸ of the squares) on AD and DB have been applied to EF , τῶν ΑΓ, ΓΒ μέσον ἐστίν. καί ἐστιν ἴσον τῷ ΘΗ· καὶ τὸ ΘΗ producing EN as breadth. And EL is also equal to the ἄρα μέσον ἐστίν. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος square on AB. Thus, the remainder HI is equal to twice ποιοῦν τὴν ΘΜ· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΘΜ καὶ ἀσύμμετρος the (rectangle contained) by AD and DB [Prop. 2.7]. τῇ ΕΖ μήκει. καὶ ἐπεὶ αἱ ΑΓ, ΓΒ δυνάμει μόνον σύμμετροί And since AC and CB are (both) medial (straight-lines), εἰσιν, ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΓ τῇ ΓΒ μήκει. ὡς δὲ ἡ ΑΓ the (sum of the squares) on AC and CB is also me- πρὸς τὴν ΓΒ, οὕτως ἐστὶ τὸ ἀπὸ τῆς ΑΓ πρὸς τὸ ὑπὸ τῶν dial. And it is equal to EG. Thus, EG is also medial ΑΓ, ΓΒ· ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΓ τῷ ὑπὸ τῶν [Props. 10.15, 10.23 corr.]. And it is applied to the ratio- ΑΓ, ΓΒ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΓ σύμμετρά ἐστι τὰ ἀπὸ nal (straight-line) EF , producing EM as breadth. Thus, τῶν ΑΓ, ΓΒ, τῷ δὲ ὑπὸ τῶν ΑΓ, ΓΒ σύμμετρόν ἐστι τὸ EM is rational, and incommensurable in length with EF δὶς ὑπὸ τῶν ΑΓ, ΓΒ· ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΓ, [Prop. 10.22]. Again, since the (rectangle contained) by ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. καί ἐστι τοῖς μὲν ἀπὸ τῶν ΑΓ, AC and CB is medial, twice the (rectangle contained) ΓΒ ἴσον τὸ ΕΗ, τῷ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ ΗΘ· by AC and CB is also medial [Prop. 10.23 corr.]. And it ἀσύμμετρον ἄρα ἐστὶ τὸ ΕΗ τῷ ΘΗ. ὡς δὲ τὸ ΕΗ πρὸς τὸ is equal to HG. Thus, HG is also medial. And it is ap- ΘΗ, οὕτως ἐστὶν ἡ ΕΜ πρὸς τὴν ΘΜ· ἀσύμμετρος ἄρα ἐστὶν plied to the rational (straight-line) EF , producing HM ἡ ΕΜ τῇ ΜΘ μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ΕΜ, ΜΘ as breadth. Thus, HM is also rational, and incommen- ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἀποτομὴ ἄρα ἐστὶν surable in length with EF [Prop. 10.22]. And since AC ἡ ΕΘ, προσαρμόζουσα δὲ αὐτῇ ἡ ΘΜ. ὁμοίως δὴ δείξομεν, and CB are commensurable in square only, AC is thus ὅτι καὶ ἡ ΘΝ αὐτῇ προσαρμόζει· τῇ ἄρα ἀποτομῇ ἄλλη καὶ incommensurable in length with CB. And as AC (is) ἄλλη προσαρμόζει εὐθεῖα δυνάμει μόνον σύμμετρος οὖσα to CB, so the (square) on AC is to the (rectangle con- τῇ ὅλῃ· ὅπερ ἐστὶν ἀδύνατον. tained) by AC and CB [Prop. 10.21 corr.]. Thus, the Τῇ ἄρα μέσης ἀποτομῇ δευτέρᾳ μία μόνον προσαρμόζει (square) on AC is incommensurable with the (rectan- εὐθεῖα μέση δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ, μετὰ gle contained) by AC and CB [Prop. 10.11]. But, the δὲ τῆς ὅλης μέσον περιέχουσα· ὅπερ ἔδει δεῖξαι. (sum of the squares) on AC and CB is commensurable with the (square) on AC, and twice the (rectangle con- tained) by AC and CB is commensurable with the (rect- angle contained) by AC and CB [Prop. 10.6]. Thus, the (sum of the squares) on AC and CB is incommen- surable with twice the (rectangle contained) by AC and CB [Prop. 10.13]. And EG is equal to the (sum of the squares) on AC and CB. And GH is equal to twice the (rectangle contained) by AC and CB. Thus, EG is in- commensurable with HG. And as EG (is) to HG, so EM is to HM [Prop. 6.1]. Thus, EM is incommensurable in length with MH [Prop. 10.11]. And they are both rational (straight-lines). Thus, EM and MH are ratio- nal (straight-lines which are) commensurable in square only. Thus, EH is an apotome [Prop. 10.73], and HM (is) attached to it. So, similarly, we can show that HN (is) also (commensurable in square only with EN and is) attached to (EH). Thus, different straight-lines, which are commensurable in square only with the whole, are attached to an apotome. The very thing is impossible [Prop. 10.79]. Thus, only one medial straight-line, which is commen- surable in square only with the whole, and contains a me- dial (area) with the whole, can be attached to a second apotome of a medial (straight-line). (Which is) the very thing it was required to show. 375 STOIQEIWN iþ. ELEMENTS BOOK 10 † This proposition is equivalent to Prop. 10.44, with minus signs instead of plus signs.pbþ. Proposition 82 Τῇ ἐλάσσονι μία μόνον προσαρμόζει εὐθεῖα δυνάμει Only one straight-line, which is incommensurable in ἀσύμμετρος οὖσα τῇ ὅλῃ ποιοῦσα μετὰ τῆς ὅλης τὸ μὲν square with the whole, and (together) with the whole ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων ῥητόν, τὸ δὲ δὶς ὑπ᾿ αὐτῶν makes the (sum of the) squares on them rational, and μέσον. twice the (rectangle contained) by them medial, can be attached to a minor (straight-line). ∆Α Β Γ BA C D ῎Εστω ἡ ἐλάσσων ἡ ΑΒ, καὶ τῇ ΑΒ προσαρμόζουσα Let AB be a minor (straight-line), and let BC be (so) ἔστω ἡ ΒΓ· αἱ ἄρα ΑΓ, ΓΒ δυνάμει εἰσὶν ἀσύμμετροι attached to AB. Thus, AC and CB are (straight-lines ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων which are) incommensurable in square, making the sum ῥητόν, τὸ δὲ δὶς ὑπ᾿ αὐτῶν μέσον· λέγω, ὅτι τῇ ΑΒ ἑτέρα of the squares on them rational, and twice the (rectan- εὐθεῖα οὐ προσαρμόσει τὰ αὐτὰ ποιοῦσα. gle contained) by them medial [Prop. 10.76]. I say that Εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ· καὶ αἱ ΑΔ, ΔΒ another another straight-line fulfilling the same (condi- ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὰ προειρημένα. καὶ tions) cannot be attached to AB. ἐπεί, ᾧ ὑπερέχει τὰ ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, For, if possible, let BD be (so) attached (to AB). τούτῳ ὑπερέχει καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ Thus, AD and DB are also (straight-lines which are) τῶν ΑΓ, ΓΒ, τὰ δὲ ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα τῶν ἀπὸ incommensurable in square, fulfilling the (other) afore- τῶν ΑΓ, ΓΒ τετραγώνων ὑπερέχει ῥητῷ· ῥητὰ γάρ ἐστιν mentioned (conditions) [Prop. 10.76]. And since by ἀμφότερα· καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ἄρα τοῦ δὶς ὑπὸ whatever (area) the (sum of the squares) on AD and DB τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ· ὅπερ ἐστὶν ἀδύνατον· μέσα exceeds the (sum of the squares) on AC and CB, twice γάρ ἐστιν ἀμφότερα. the (rectangle contained) by AD and DB also exceeds Τῇ ἄρα ἐλάσσονι μία μόνον προσαρμόζει εὐθεῖα δυνάμει twice the (rectangle contained) by AC and CB by this ἀσύμμετρος οὖσα τῇ ὅλῃ καὶ ποιοῦσα τὰ μὲν ἀπ᾿ αὐτῶν (same area) [Prop. 2.7]. And the (sum of the) squares τετράγωνα ἅμα ῥητόν, τὸ δὲ δὶς ὑπ᾿ αὐτῶν μέσον· ὅπερ ἔδει on AD and DB exceeds the (sum of the) squares on AC δεῖξαι. and CB by a rational (area). For both are rational (ar- eas). Thus, twice the (rectangle contained) by AD and DB also exceeds twice the (rectangle contained) by AC and CB by a rational (area). The very thing is impossible. For both are medial (areas) [Prop. 10.26]. Thus, only one straight-line, which is incommensu- rable in square with the whole, and (with the whole) makes the squares on them (added) together rational, and twice the (rectangle contained) by them medial, can be attached to a minor (straight-line). (Which is) the very thing it was required to show. † This proposition is equivalent to Prop. 10.45, with minus signs instead of plus signs.pgþ. Proposition 83 Τῇ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιούσῃ μία μόνον προ- Only one straight-line, which is incommensurable in σαρμόζει εὐθεῖα δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ square with the whole, and (together) with the whole τῆς ὅλης ποιοῦσα τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν makes the sum of the squares on them medial, and twice τετραγώνων μέσον, τὸ δὲ δὶς ὑπ᾿ αὐτῶν ῥητόν. the (rectangle contained) by them rational, can be at- tached to that (straight-line) which with a rational (area) makes a medial whole.† ∆Α Β Γ BA C D 376 STOIQEIWN iþ. ELEMENTS BOOK 10 ῎Εστω ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ, καὶ Let AB be a (straight-line) which with a rational τῇ ΑΒ προσαρμοζέτω ἡ ΒΓ· αἱ ἄρα ΑΓ, ΓΒ δυνάμει εἰσὶν (area) makes a medial whole, and let BC be (so) at- ἀσύμμετροι ποιοῦσαι τὰ προκείμενα· λέγω, ὅτι τῇ ΑΒ ἑτέρα tached to AB. Thus, AC and CB are (straight-lines οὐ προσαρμόσει τὰ αὐτὰ ποιοῦσα. which are) incommensurable in square, fulfilling the Εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ· καὶ αἱ ΑΔ, (other) proscribed (conditions) [Prop. 10.77]. I say that ΔΒ ἄρα εὐθεῖαι δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὰ another (straight-line) fulfilling the same (conditions) προκείμενα. ἐπεὶ οὐν, ᾧ ὑπερέχει τὰ ἀπὸ τῶν ΑΔ, ΔΒ cannot be attached to AB. τῶν ἀπὸ τῶν ΑΓ, ΓΒ, τούτῳ ὑπερέχει καὶ τὸ δὶς ὑπὸ τῶν For, if possible, let BD be (so) attached (to AB). ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἀκολούθως τοῖς πρὸ Thus, AD and DB are also straight-lines (which are) αὐτοῦ, τὸ δὲ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ incommensurable in square, fulfilling the (other) pre- ὑπερέχει ῥητῷ· ῥητὰ γάρ ἐστιν ἀμφότερα· καὶ τὰ ἀπὸ τῶν scribed (conditions) [Prop. 10.77]. Therefore, analo- ΑΔ, ΔΒ ἄρα τῶν ἀπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ· ὅπερ gously to the (propositions) before this, since by what- ἐστὶν ἀδύνατον· μέσα γάρ ἐστιν ἀμφότερα. ever (area) the (sum of the squares) on AD and DB ex- Οὐκ ἄρα τῇ ΑΒ ἑτέρα προσαρμόσει εὐθεῖα δυνάμει ceeds the (sum of the squares) on AC and CB, twice the ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τὰ (rectangle contained) by AD and DB also exceeds twice προειρημένα· μία ἄρα μόνον προσαρμόσει· ὅπερ ἔδει δεῖξαι. the (rectangle contained) by AC and CB by this (same area). And twice the (rectangle contained) by AD and DB exceeds twice the (rectangle contained) by AC and CB by a rational (area). For they are (both) rational (ar- eas). Thus, the (sum of the squares) on AD and DB also exceeds the (sum of the squares) on AC and CB by a ra- tional (area). The very thing is impossible. For both are medial (areas) [Prop. 10.26]. Thus, another straight-line cannot be attached to AB, which is incommensurable in square with the whole, and fulfills the (other) aforementioned (conditions) with the whole. Thus, only one (such straight-line) can be (so) attached. (Which is) the very thing it was required to show. † This proposition is equivalent to Prop. 10.46, with minus signs instead of plus signs.pdþ. Proposition 84 Τῇ μετὰ μέσου μέσον τὸ ὅλον ποιούσῃ μία μόνη προ- Only one straight-line, which is incommensurable in σαρμόζει εὐθεῖα δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ square with the whole, and (together) with the whole τῆς ὅλης ποιοῦσα τό τε συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν makes the sum of the squares on them medial, and twice τετραγώνων μέσον τό τε δὶς ὑπ᾿ αὐτῶν μέσον καὶ ἔτι the (rectangle contained) by them medial, and, more- ἀσύμμετρον τῷ συγκειμένῳ ἐκ τῶν ἀπ᾿ αὐτῶν. over, incommensurable with the sum of the (squares) on ῎Εστω ἡ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ, προ- them, can be attached to that (straight-line) which with σαρμόζουσα δὲ αὐτῇ ἡ ΒΓ· αἱ ἄρα ΑΓ, ΓΒ δυνάμει εἰσὶν a medial (area) makes a medial whole.† ἀσύμμετροι ποιοῦσαι τὰ προειρημένα. λέγω, ὅτι τῇ ΑΒ Let AB be a (straight-line) which with a medial ἑτέρα οὐ προσαρμόσει ποιοῦσα προειρημένα. (area) makes a medial whole, BC being (so) attached to it. Thus, AC and CB are incommensurable in square, fulfilling the (other) aforementioned (conditions) [Prop. 10.78]. I say that a(nother) (straight-line) fulfill- ing the aforementioned (conditions) cannot be attached to AB. 377 STOIQEIWN iþ. ELEMENTS BOOK 10 Ζ Α Β Γ ∆ Ε Θ Μ Ν Η ΙΛ CA B D E M N ILF H G Εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ, ὥστε καὶ τὰς For, if possible, let BD be (so) attached. Hence, ΑΔ, ΔΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τά τε ἀπὸ AD and DB are also (straight-lines which are) incom- τῶν ΑΔ, ΔΒ τετράγωνα ἅμα μέσον καὶ τὸ δὶς ὑπὸ τῶν mensurable in square, making the squares on AD and ΑΔ, ΔΒ μέσον καὶ ἔτι τὰ ἀπὸ τῶν ΑΔ, ΔΒ ἀσύμμετρα τῷ DB (added) together medial, and twice the (rectangle δὶς ὑπὸ τῶν ΑΔ, ΔΒ· καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ τοῖς contained) by AD and DB medial, and, moreover, the μὲν ἀπὸ τῶν ΑΓ, ΓΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω τὸ (sum of the squares) on AD and DB incommensurable ΕΗ πλάτος ποιοῦν τὴν ΕΜ, τῷ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ with twice the (rectangle contained) by AD and DB ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω τὸ ΘΗ πλάτος ποιοῦν [Prop. 10.78]. And let the rational (straight-line) EF be τὴν ΘΜ· λοιπὸν ἄρα τὸ ἀπὸ τῆς ΑΒ ἴσον ἐστὶ τῷ ΕΛ· ἡ laid down. And let EG, equal to the (sum of the squares) ἄρα ΑΒ δύναται τὸ ΕΛ. πάλιν τοῖς ἀπὸ τῶν ΑΔ, ΔΒ ἴσον on AC and CB, have been applied to EF , producing EM παρὰ τὴν ΕΖ παραβεβλήσθω τὸ ΕΙ πλάτος ποιοῦν τὴν ΕΝ. as breadth. And let HG, equal to twice the (rectangle ἔστι δὲ καὶ τὸ ἀπὸ τῆς ΑΒ ἴσον τῷ ΕΛ· λοιπὸν ἄρα τὸ δὶς contained) by AC and CB, have been applied to EF , ὑπὸ τῶν ΑΔ, ΔΒ ἴσον [ἐστὶ] τῷ ΘΙ. καὶ ἐπεὶ μέσον ἐστὶ producing HM as breadth. Thus, the remaining (square) τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ καί ἐστιν ἴσον on AB is equal to EL [Prop. 2.7]. Thus, AB is the square- τῷ ΕΗ, μέσον ἄρα ἐστὶ καὶ τὸ ΕΗ. καὶ παρὰ ῥητὴν τὴν ΕΖ root of EL. Again, let EI, equal to the (sum of the παράκειται πλάτος ποιοῦν τὴν ΕΜ· ῥητὴ ἄρα ἐστὶν ἡ ΕΜ καὶ squares) on AD and DB, have been applied to EF , pro- ἀσύμμετρος τῇ ΕΖ μήκει. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ δὶς ὑπὸ ducing EN as breadth. And the (square) on AB is also τῶν ΑΓ, ΓΒ καί ἐστιν ἴσον τῷ ΘΗ, μέσον ἄρα καὶ τὸ ΘΗ. equal to EL. Thus, the remaining twice the (rectangle καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος ποιοῦν τὴν ΘΜ· contained) by AD and DB [is] equal to HI [Prop. 2.7]. ῥητὴ ἄρα ἐστὶν ἡ ΘΜ καὶ ἀσύμμετρος τῇ ΕΖ μήκει. καὶ ἐπεὶ And since the sum of the (squares) on AC and CB is me- ἀσύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, dial, and is equal to EG, EG is thus also medial. And ΓΒ, ἀσύμμετρόν ἐστι καὶ τὸ ΕΗ τῷ ΘΗ· ἀσύμμετρος ἄρα it is applied to the rational (straight-line) EF , producing ἐστὶ καὶ ἡ ΕΜ τῇ ΜΘ μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ EM as breadth. EM is thus rational, and incommen- ἄρα ΕΜ, ΜΘ ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἀποτομὴ surable in length with EF [Prop. 10.22]. Again, since ἄρα ἐστὶν ἡ ΕΘ, προσαρμόζουσα δὲ αὐτῇ ἡ ΘΜ. ὁμοίως δὴ twice the (rectangle contained) by AC and CB is me- δείξομεν, ὅτι ἡ ΕΘ πάλιν ἀποτομή ἐστιν, προσαρμόζουσα dial, and is equal to HG, HG is thus also medial. And δὲ αὐτῇ ἡ ΘΝ. τῇ ἄρα ἀποτομῇ ἄλλη καὶ ἄλλη προσαρμόζει it is applied to the rational (straight-line) EF , produc- ῥητὴ δυνάμει μόνον σύμμετρος οὖσα τῇ ὅλῃ· ὅπερ ἐδείχθη ing HM as breadth. HM is thus rational, and incom- ἀδύνατον. οὐκ ἄρα τῇ ΑΒ ἑτέρα προσαρμόσει εὐθεῖα. mensurable in length with EF [Prop. 10.22]. And since Τῇ ἄρα ΑΒ μία μόνον προσαρμόζει εὐθεῖα δυνάμει the (sum of the squares) on AC and CB is incommen- ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τά surable with twice the (rectangle contained) by AC and τε ἀπ᾿ αὐτῶν τετράγωνα ἅμα μέσον καὶ τὸ δὶς ὑπ᾿ αὐτῶν CB, EG is also incommensurable with HG. Thus, EM μέσον καὶ ἔτι τὰ ἀπ᾿ αὐτῶν τετράγωνα ἀσύμμετρα τῷ δὶς is also incommensurable in length with MH [Props. 6.1, ὑπ᾿ αὐτῶν· ὅπερ ἔδει δεῖξαι. 10.11]. And they are both rational (straight-lines). Thus, EM and MH are rational (straight-lines which are) com- mensurable in square only. Thus, EH is an apotome [Prop. 10.73], with HM attached to it. So, similarly, we can show that EH is again an apotome, with HN attached to it. Thus, different rational (straight-lines), which are commensurable in square only with the whole, are attached to an apotome. The very thing was shown 378 STOIQEIWN iþ. ELEMENTS BOOK 10 (to be) impossible [Prop. 10.79]. Thus, another straight- line cannot be (so) attached to AB. Thus, only one straight-line, which is incommensu- rable in square with the whole, and (together) with the whole makes the squares on them (added) together me- dial, and twice the (rectangle contained) by them medial, and, moreover, the (sum of the) squares on them incom- mensurable with the (rectangle contained) by them, can be attached to AB. (Which is) the very thing it was re- quired to show. † This proposition is equivalent to Prop. 10.47, with minus signs instead of plus signs.VOroi tr�toi. Definitions III ιαʹ. ῾Υποκειμένης ῥητῆς καὶ ἀποτομῆς, ἐὰν μὲν ἡ ὅλη τῆς 11. Given a rational (straight-line) and an apotome, if προσαρμοζούσης μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ the square on the whole is greater than the (square on a μήκει, καὶ ἡ ὅλη σύμμετρος ᾖ τῇ ἐκκειμένῃ ῥητῇ μήκει, straight-line) attached (to the apotome) by the (square) καλείσθω ἀποτομὴ πρώτη. on (some straight-line) commensurable in length with ιβʹ. ᾿Εὰν δὲ ἡ προσαρμόζουσα σύμμετρος ᾖ τῇ ἐκκειμένῃ (the whole), and the whole is commensurable in length ῥητῇ μήκει, καὶ ἡ ὅλη τῆς προσαρμοζούσης μεῖζον δύνηται with the (previously) laid down rational (straight-line), τῷ ἀπὸ συμμέτρου ἑαυτῇ, καλείσθω ἀποτομὴ δευτέρα. then let the (apotome) be called a first apotome. ιγʹ. ᾿Εὰν δὲ μηδετέρα σύμμετρος ᾖ τῇ ἐκκειμένῃ ῥητῇ 12. And if the attached (straight-line) is commen- μήκει, ἡ δὲ ὅλη τῆς προσαρμοζούσης μεῖζον δύνηται τῷ surable in length with the (previously) laid down ra- ἀπὸ συμμέτρου ἑαυτῇ, καλείσθω ἀποτομὴ τρίτη. tional (straight-line), and the square on the whole is ιδʹ. Πάλιν, ἐὰν ἡ ὅλη τῆς προσαρμοζούσης μεῖζον greater than (the square on) the attached (straight-line) δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ [μήκει], ἐὰν μὲν ἡ ὅλη by the (square) on (some straight-line) commensurable σύμμετρος ᾖ τῇ ἐκκειμένῃ ῥητῇ μήκει, καλείσθω ἀποτομὴ (in length) with (the whole), then let the (apotome) be τετάρτη. called a second apotome. ιεʹ. ᾿Εὰν δὲ ἡ προσαρμόζουσα, πέμπτη. 13. And if neither of (the whole or the attached ιϛʹ. ᾿Εὰν δὲ μηδετέρα, ἕκτη. straight-line) is commensurable in length with the (previ- ously) laid down rational (straight-line), and the square on the whole is greater than (the square on) the attached (straight-line) by the (square) on (some straight-line) commensurable (in length) with (the whole), then let the (apotome) be called a third apotome. 14. Again, if the square on the whole is greater than (the square on) the attached (straight-line) by the (square) on (some straight-line) incommensurable [in length] with (the whole), and the whole is commensu- rable in length with the (previously) laid down rational (straight-line), then let the (apotome) be called a fourth apotome. 15. And if the attached (straight-line is commensu- rable), a fifth (apotome). 16. And if neither (the whole nor the attached straight-line is commensurable), a sixth (apotome).peþ. Proposition 85 Εὑρεῖν τὴν πρώτην ἀποτομήν. To find a first apotome. 379 STOIQEIWN iþ. ELEMENTS BOOK 10 Α Β Γ Η Ε Ζ ∆ Θ F B E D A H C G ᾿Εκκείσθω ῥητὴ ἡ Α, καὶ τῇ Α μήκει σύμμετρος ἔστω Let the rational (straight-line) A be laid down. And ἡ ΒΗ· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΒΗ. καὶ ἐκκείσθωσαν δύο let BG be commensurable in length with A. BG is thus τετράγωνοι ἀριθμοὶ οἱ ΔΕ, ΕΖ, ὧν ἡ ὑπεροχὴ ὁ ΖΔ μὴ also a rational (straight-line). And let two square num- ἔστω τετράγωνος· οὐδ᾿ ἄρα ὁ ΕΔ πρὸς τὸν ΔΖ λόγον ἔχει, bers DE and EF be laid down, and let their difference ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καὶ πε- FD be not square [Prop. 10.28 lem. I]. Thus, ED does ποιήσθω ὡς ὁ ΕΔ πρὸς τὸν ΔΖ, οὕτως τὸ ἀπὸ τῆς ΒΗ not have to DF the ratio which (some) square number τετράγωνον πρὸς τὸ ἀπὸ τὴς ΗΓ τετράγωνον· σύμμετρον (has) to (some) square number. And let it have been ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΗ τῷ ἀπὸ τῆς ΗΓ. ῥητὸν δὲ τὸ ἀπὸ contrived that as ED (is) to DF , so the square on BG τῆς ΒΗ· ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΗΓ· ῥητὴ ἄρα ἐστὶ καὶ (is) to the square on GC [Prop. 10.6. corr.]. Thus, the ἡ ΗΓ. καὶ ἐπεὶ ὁ ΕΔ πρὸς τὸν ΔΖ λόγον οὐκ ἔχει, ὃν (square) on BG is commensurable with the (square) on τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾿ ἄρα τὸ GC [Prop. 10.6]. And the (square) on BG (is) ratio- ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ λόγον ἔχει, ὃν τετράγωνος nal. Thus, the (square) on GC (is) also rational. Thus, ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν GC is also rational. And since ED does not have to DF ἡ ΒΗ τῇ ΗΓ μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ΒΗ, ΗΓ the ratio which (some) square number (has) to (some) ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἡ ἄρα ΒΓ ἀποτομή square number, the (square) on BG thus does not have to ἐστιν. λέγω δή, ὅτι καὶ πρώτη. the (square) on GC the ratio which (some) square num- ῟Ωι γὰρ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΒΗ τοῦ ἀπὸ τῆς ΗΓ, ber (has) to (some) square number either. Thus, BG is ἔστω τὸ ἀπὸ τῆς Θ. καὶ ἐπεί ἐστιν ὡς ὁ ΕΔ πρὸς τὸν incommensurable in length with GC [Prop. 10.9]. And ΖΔ, οὕτως τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ, καὶ ἀνα- they are both rational (straight-lines). Thus, BG and GC στρέψαντι ἄρα ἐστὶν ὡς ὁ ΔΕ πρὸς τὸν ΕΖ, οὕτως τὸ ἀπὸ are rational (straight-lines which are) commensurable in τῆς ΗΒ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΔΕ πρὸς τὸν ΕΖ λόγον square only. Thus, BC is an apotome [Prop. 10.73]. So, ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· I say that (it is) also a first (apotome). ἑκάτερος γὰρ τετράγωνός ἐστιν· καὶ τὸ ἀπὸ τῆς ΗΒ ἄρα Let the (square) on H be that (area) by which the πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς (square) on BG is greater than the (square) on GC πρὸς τετράγωνον ἀριθμόν· σύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ [Prop. 10.13 lem.]. And since as ED is to FD, so the Θ μήκει. καὶ δύναται ἡ ΒΗ τῆς ΗΓ μεῖζον τῷ ἀπὸ τῆς Θ· (square) on BG (is) to the (square) on GC, thus, via con- ἡ ΒΗ ἄρα τῆς ΗΓ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ version, as DE is to EF , so the (square) on GB (is) to μήκει. καί ἐστιν ἡ ὅλη ἡ ΒΗ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ the (square) on H [Prop. 5.19 corr.]. And DE has to EF μήκει τῇ Α. ἡ ΒΓ ἄρα ἀποτομή ἐστι πρώτη. the ratio which (some) square-number (has) to (some) Εὕρηται ἄρα ἡ πρώτη ἀποτομὴ ἡ ΒΓ· ὅπερ ἔδει εὑρεῖν. square-number. For each is a square (number). Thus, the (square) on GB also has to the (square) on H the ra- tio which (some) square number (has) to (some) square number. Thus, BG is commensurable in length with H [Prop. 10.9]. And the square on BG is greater than (the square on) GC by the (square) on H . Thus, the square on BG is greater than (the square on) GC by the (square) on (some straight-line) commensurable in length with (BG). And the whole, BG, is commensurable in length with the (previously) laid down rational (straight-line) A. Thus, BC is a first apotome [Def. 10.11]. Thus, the first apotome BC has been found. (Which is) the very thing it was required to find. † See footnote to Prop. 10.48. p�þ. Proposition 86 Εὑρεῖν τὴν δευτέραν ἀποτομήν. To find a second apotome. 380 STOIQEIWN iþ. ELEMENTS BOOK 10 ᾿Εκκείσθω ῥητὴ ἡ Α καὶ τῇ Α σύμμετρος μήκει ἡ ΗΓ. Let the rational (straight-line) A, and GC (which is) ῥητὴ ἄρα ἐστὶν ἡ ΗΓ. καὶ ἐκκείσθωσαν δύο τετράγωνοι commensurable in length with A, be laid down. Thus, ἀριθμοὶ οἱ ΔΕ, ΕΖ, ὧν ἡ ὑπεροχὴ ὁ ΔΖ μὴ ἔστω τετράγωνος. GC is a rational (straight-line). And let the two square καὶ πεποιήσθω ὡς ὁ ΖΔ πρὸς τὸν ΔΕ, οὕτως τὸ ἀπὸ numbers DE and EF be laid down, and let their differ- τῆς ΓΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΒ τετράγωνον. ence DF be not square [Prop. 10.28 lem. I]. And let it σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΓΗ τετράγωνον τῷ ἀπὸ have been contrived that as FD (is) to DE, so the square τῆς ΗΒ τετραγώνῳ. ῥητὸν δὲ τὸ ἀπὸ τῆς ΓΗ. ῥητὸν ἄρα on CG (is) to the square on GB [Prop. 10.6 corr.]. Thus, [ἐστὶ] καὶ τὸ ἀπὸ τῆς ΗΒ· ῥητὴ ἄρα ἐστὶν ἡ ΒΗ. καὶ ἐπεὶ τὸ the square on CG is commensurable with the square on ἀπὸ τῆς ΗΓ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΒ λόγον οὐκ GB [Prop. 10.6]. And the (square) on CG (is) rational. ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, Thus, the (square) on GB [is] also rational. Thus, BG is a ἀσύμμετρός ἐστιν ἡ ΓΗ τῇ ΗΒ μήκει. καί εἰσιν ἀμφότεραι rational (straight-line). And since the square on GC does ῥηταί· αἱ ΓΗ, ΗΒ ἄρα ρηταί εἰσι δυνάμει μόνον σύμμετροι· not have to the (square) on GB the ratio which (some) ἡ ΒΓ ἄρα ἀποτομή ἐστιν. λέγω δή, ὅτι καὶ δευτέρα. square number (has) to (some) square number, CG is incommensurable in length with GB [Prop. 10.9]. And they are both rational (straight-lines). Thus, CG and GB are rational (straight-lines which are) commensurable in square only. Thus, BC is an apotome [Prop. 10.73]. So, I say that it is also a second (apotome). Α Β Γ Η Ε Ζ ∆ Θ GB E D A H C F ῟Ωι γὰρ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΒΗ τοῦ ἀπὸ τῆς ΗΓ, For let the (square) on H be that (area) by which ἔστω τὸ ἀπὸ τῆς Θ. ἐπεὶ οὖν ἐστιν ὡς τὸ ἀπὸ τῆς ΒΗ the (square) on BG is greater than the (square) on GC πρὸς τὸ ἀπὸ τῆς ΗΓ, οὕτως ὁ ΕΔ ἀριθμὸς πρὸς τὸν ΔΖ [Prop. 10.13 lem.]. Therefore, since as the (square) on ἀριθμόν, ἀναστρέψαντι ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΗ πρὸς BG is to the (square) on GC, so the number ED (is) τὸ ἀπὸ τῆς Θ, οὕτως ὁ ΔΕ πρὸς τὸν ΕΖ. καί ἐστιν ἑκάτερος to the number DF , thus, also, via conversion, as the τῶν ΔΕ, ΕΖ τετράγωνος· τὸ ἄρα ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ (square) on BG is to the (square) on H , so DE (is) τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον to EF [Prop. 5.19 corr.]. And DE and EF are each ἀριθμόν· σύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ Θ μήκει. καὶ δύναται square (numbers). Thus, the (square) on BG has to the ἡ ΒΗ τῆς ΗΓ μεῖζον τῷ ἀπὸ τῆς Θ· ἡ ΒΗ ἄρα τῆς ΗΓ (square) on H the ratio which (some) square number μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. καί ἐστιν ἡ (has) to (some) square number. Thus, BG is commen- προσαρμόζουσα ἡ ΓΗ τῇ ἐκκειμένῃ ῥητῇ σύμμετρος τῇ Α. surable in length with H [Prop. 10.9]. And the square on ἡ ΒΓ ἄρα ἀποτομή ἐστι δευτέτα. BG is greater than (the square on) GC by the (square) Εὕρηται ἄρα δευτέρα ἀποτομὴ ἡ ΒΓ· ὅπερ ἔδει δεῖξαι. on H . Thus, the square on BG is greater than (the square on) GC by the (square) on (some straight-line) commen- surable in length with (BG). And the attachment CG is commensurable (in length) with the (prevously) laid down rational (straight-line) A. Thus, BC is a second apotome [Def. 10.12].† Thus, the second apotome BC has been found. (Which is) the very thing it was required to show. † See footnote to Prop. 10.49. pzþ. Proposition 87 Εὑρεῖν τὴν τρίτην ἀποτομήν. To find a third apotome. 381 STOIQEIWN iþ. ELEMENTS BOOK 10 Κ Β Α ∆ Γ Ζ Θ Η Ε CB A D E K F H G ᾿Εκκείσθω ῥητὴ ἡ Α, καὶ ἐκκείσθωσαν τρεῖς ἀριθμοὶ Let the rational (straight-line) A be laid down. And οἱ Ε, ΒΓ, ΓΔ λόγον μὴ ἔχοντες πρὸς ἀλλήλους, ὃν let the three numbers, E, BC, and CD, not having to one τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, ὁ δὲ ΓΒ another the ratio which (some) square number (has) to πρὸς τὸν ΒΔ λόγον ἐχέτω, ὃν τετράγωνος ἀριθμὸς πρὸς (some) square number, be laid down. And let CB have τετράγωνον ἀριθμόν, καὶ πεποιήσθω ὡς μὲν ὁ Ε πρὸς τὸν to BD the ratio which (some) square number (has) to ΒΓ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς (some) square number. And let it have been contrived ΖΗ τετράγωνον, ὡς δὲ ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ that as E (is) to BC, so the square on A (is) to the τῆς ΖΗ τετράγωνον πρὸς τὸ ἀπὸ τὴς ΗΘ. ἐπεὶ οὖν ἐστιν square on FG, and as BC (is) to CD, so the square on ὡς ὁ Ε πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον FG (is) to the (square) on GH [Prop. 10.6 corr.]. There- πρὸς τὸ ἀπὸ τῆς ΖΗ τετράγωνον, σύμμετρον ἄρα ἐστὶ fore, since as E is to BC, so the square on A (is) to the τὸ ἀπὸ τῆς Α τετράγωνον τῷ ἀπὸ τῆς ΖΗ τετραγώνῳ. square on FG, the square on A is thus commensurable ῥητὸν δὲ τὸ ἀπὸ τῆς Α τετράγωνον. ῥητὸν ἄρα καὶ τὸ ἀπὸ with the square on FG [Prop. 10.6]. And the square on τῆς ΖΗ· ῥητὴ ἄρα ἐστὶν ἡ ΖΗ. καὶ ἐπεὶ ὁ Ε πρὸς τὸν ΒΓ A (is) rational. Thus, the (square) on FG (is) also ra- λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον tional. Thus, FG is a rational (straight-line). And since ἀριθμόν, οὐδ᾿ ἄρα τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ E does not have to BC the ratio which (some) square τῆς ΖΗ [τετράγωνον] λόγον ἔχει, ὅν τετράγωνος ἀριθμὸς number (has) to (some) square number, the square on A πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ thus does not have to the [square] on FG the ratio which ΖΗ μήκει. πάλιν, ἐπεί ἐστιν ὡς ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ (some) square number (has) to (some) square number ἀπὸ τῆς ΖΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΘ, σύμμετρον either. Thus, A is incommensurable in length with FG ἄρα ἐστὶ τὸ ἀπὸ τῆς ΖΗ τῷ ἀπὸ τῆς ΗΘ. ῥητὸν δὲ τὸ ἀπὸ [Prop. 10.9]. Again, since as BC is to CD, so the square τῆς ΖΗ· ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΗΘ· ῥητὴ ἄρα ἐστὶν ἡ ΗΘ. on FG is to the (square) on GH , the square on FG is thus καὶ ἐπεὶ ὁ ΒΓ πρὸς τὸν ΓΔ λόγον οὐκ ἔχει, ὃν τετράγωνος commensurable with the (square) on GH [Prop. 10.6]. ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾿ ἄρα τὸ ἀπὸ τῆς ΖΗ And the (square) on FG (is) rational. Thus, the (square) πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς on GH (is) also rational. Thus, GH is a rational (straight- πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ line). And since BC does not have to CD the ratio which ΗΘ μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ΖΗ, ΗΘ ἄρα ῥηταί (some) square number (has) to (some) square number, εἰσι δυνάμει μόνον σύμμετροι· ἀποτομὴ ἄρα ἐστὶν ἡ ΖΘ. the (square) on FG thus does not have to the (square) λέγω δή, ὅτι καὶ τρίτη. on GH the ratio which (some) square number (has) to ᾿Επεὶ γάρ ἐστιν ὡς μὲν ὁ Ε πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ (some) square number either. Thus, FG is incommen- τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΗ, ὡς δὲ ὁ ΒΓ πρὸς surable in length with GH [Prop. 10.9]. And both are τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΘΗ, δι᾿ ἴσου rational (straight-lines). FG and GH are thus rational ἄρα ἐστὶν ὡς ὁ Ε πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς Α πρὸς (straight-lines which are) commensurable in square only. τὸ ἀπὸ τῆς ΘΗ. ὁ δὲ Ε πρὸς τὸν ΓΔ λόγον οὐκ ἔχει, ὃν Thus, FH is an apotome [Prop. 10.73]. So, I say that (it τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· οὐδ᾿ ἄρα τὸ is) also a third (apotome). ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος For since as E is to BC, so the square on A (is) to the ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἡ Α τῇ (square) on FG, and as BC (is) to CD, so the (square) ΗΘ μήκει. οὐδετέρα ἄρα τῶν ΖΗ, ΗΘ σύμμετρός ἐστι τῇ on FG (is) to the (square) on HG, thus, via equality, as ἐκκειμένῃ ῥητῇ τῇ Α μήκει. ᾧ οὖν μεῖζόν ἐστι τὸ ἀπὸ τῆς E is to CD, so the (square) on A (is) to the (square) on ΖΗ τοῦ ἀπὸ τῆς ΗΘ, ἔστω τὸ ἀπὸ τῆς Κ. ἐπεὶ οὖν ἐστιν HG [Prop. 5.22]. And E does not have to CD the ra- ὡς ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ tio which (some) square number (has) to (some) square τῆς ΗΘ, ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΒΓ πρὸς τὸν ΒΔ, number. Thus, the (square) on A does not have to the οὕτως τὸ ἀπὸ τῆς ΖΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς Κ. ὁ δὲ (square) on GH the ratio which (some) square number ΒΓ πρὸς τὸν ΒΔ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς (has) to (some) square number either. A (is) thus incom- τετράγωνον ἀριθμόν. καὶ τὸ ἁπὸ τῆς ΖΗ ἄρα πρὸς τὸ ἀπὸ mensurable in length with GH [Prop. 10.9]. Thus, nei- τῆς Κ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ther of FG and GH is commensurable in length with the 382 STOIQEIWN iþ. ELEMENTS BOOK 10 ἀριθμόν. σύμμετρός ἄρα ἐστὶν ἡ ΖΗ τῇ Κ μήκει, καὶ δύναται (previously) laid down rational (straight-line) A. There- ἡ ΖΗ τῆς ΗΘ μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ. καὶ οὐδετέρα fore, let the (square) on K be that (area) by which the τῶν ΖΗ, ΗΘ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ Α μήκει· (square) on FG is greater than the (square) on GH ἡ ΖΘ ἄρα ἀποτομή ἐστι τρίτη. [Prop. 10.13 lem.]. Therefore, since as BC is to CD, so Εὕρηται ἄρα ἡ τρίτη ἀποτομὴ ἡ ΖΘ· ὅπερ ἔδει δεῖξαι. the (square) on FG (is) to the (square) on GH , thus, via conversion, as BC is to BD, so the square on FG (is) to the square on K [Prop. 5.19 corr.]. And BC has to BD the ratio which (some) square number (has) to (some) square number. Thus, the (square) on FG also has to the (square) on K the ratio which (some) square number (has) to (some) square number. FG is thus commen- surable in length with K [Prop. 10.9]. And the square on FG is (thus) greater than (the square on) GH by the (square) on (some straight-line) commensurable (in length) with (FG). And neither of FG and GH is com- mensurable in length with the (previously) laid down ra- tional (straight-line) A. Thus, FH is a third apotome [Def. 10.13]. Thus, the third apotome FH has been found. (Which is) very thing it was required to show. † See footnote to Prop. 10.50. phþ. Proposition 88 Εὑρεῖν τὴν τετάρτην ἀποτομήν. To find a fourth apotome. Ε Β Α Θ Η Ζ Γ ∆ GB E D A H C F ᾿Εκκείσθω ῥητὴ ἡ Α καὶ τῇ Α μήκει σύμμετρος ἡ ΒΗ· Let the rational (straight-line) A, and BG (which is) ῥητὴ ἄρα ἐστὶ καὶ ἡ ΒΗ. καὶ ἐκκείσθωσαν δύο ἀριθμοὶ οἱ commensurable in length with A, be laid down. Thus, ΔΖ, ΖΕ, ὥστε τὸν ΔΕ ὅλον πρὸς ἑκάτερον τῶν ΔΖ, ΕΖ BG is also a rational (straight-line). And let the two λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον numbers DF and FE be laid down such that the whole, ἀριθμόν. καὶ πεποιήσθω ὡς ὁ ΔΕ πρὸς τὸν ΕΖ, οὕτως τὸ DE, does not have to each of DF and EF the ratio ἀπὸ τῆς ΒΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΓ· σύμμετρον which (some) square number (has) to (some) square ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΗ τῷ ἀπὸ τῆς ΗΓ. ῥητὸν δὲ τὸ ἀπὸ number. And let it have been contrived that as DE (is) to τῆς ΒΗ· ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΗΓ· ῥητὴ ἄρα ἐστὶν ἡ ΗΓ. EF , so the square on BG (is) to the (square) on GC καὶ ἐπεὶ ὁ ΔΕ πρὸς τὸν ΕΖ λόγον οὐκ ἔχει, ὃν τετράγωνος [Prop. 10.6 corr.]. The (square) on BG is thus com- ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾿ ἄρα τὸ ἀπὸ τῆς ΒΗ mensurable with the (square) on GC [Prop. 10.6]. And πρὸς τὸ ἀπὸ τῆς ΗΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς the (square) on BG (is) rational. Thus, the (square) on πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ GC (is) also rational. Thus, GC (is) a rational (straight- ΗΓ μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ΒΗ, ΗΓ ἄρα ῥηταί line). And since DE does not have to EF the ratio which εἰσι δυνάμει μόνον σύμμετροι· ἀποτομὴ ἄρα ἐστὶν ἡ ΒΓ. (some) square number (has) to (some) square number, [λέγω δή, ὅτι καὶ τετάρτη.] the (square) on BG thus does not have to the (square) ῟Ωι οὖν μεῖζόν ἐστι τὸ ἀπὸ τῆς ΒΗ τοῦ ἀπὸ τῆς ΗΓ, on GC the ratio which (some) square number (has) to ἔστω τὸ ἀπὸ τῆς Θ. ἐπεὶ οὖν ἐστιν ὡς ὁ ΔΕ πρὸς τὸν (some) square number either. Thus, BG is incommensu- ΕΖ, οὕτως τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ, καὶ ἀνα- rable in length with GC [Prop. 10.9]. And they are both στρέψαντι ἄρα ἐστὶν ὡς ὁ ΕΔ πρὸς τὸν ΔΖ, οὕτως τὸ rational (straight-lines). Thus, BG and GC are rational ἀπὸ τῆς ΗΒ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΕΔ πρὸς τὸν ΔΖ (straight-lines which are) commensurable in square only. λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον Thus, BC is an apotome [Prop. 10.73]. [So, I say that (it 383 STOIQEIWN iþ. ELEMENTS BOOK 10 ἀριθμόν· οὐδ᾿ ἄρα τὸ ἀπὸ τῆς ΗΒ πρὸς τὸ ἀπὸ τῆς Θ λόγον is) also a fourth (apotome).] ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· Now, let the (square) on H be that (area) by which ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ Θ μήκει. καὶ δύναται ἡ the (square) on BG is greater than the (square) on GC ΒΗ τῆς ΗΓ μεῖζον τῷ ἀπὸ τῆς Θ· ἡ ἄρα ΒΗ τῆς ΗΓ μεῖζον [Prop. 10.13 lem.]. Therefore, since as DE is to EF , δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καὶ ἐστιν ὅλη ἡ ΒΗ so the (square) on BG (is) to the (square) on GC, thus, σύμμετρος τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ Α. ἡ ἄρα ΒΓ ἀπο- also, via conversion, as ED is to DF , so the (square) on τομή ἐστι τετάρτη. GB (is) to the (square) on H [Prop. 5.19 corr.]. And ED Εὕρηται ἄρα ἡ τετάρτη ἀποτομή· ὅπερ ἔδει δεῖξαι. does not have to DF the ratio which (some) square num- ber (has) to (some) square number. Thus, the (square) on GB does not have to the (square) on H the ratio which (some) square number (has) to (some) square number either. Thus, BG is incommensurable in length with H [Prop. 10.9]. And the square on BG is greater than (the square on) GC by the (square) on H . Thus, the square on BG is greater than (the square) on GC by the (square) on (some straight-line) incommensurable (in length) with (BG). And the whole, BG, is commensurable in length with the the (previously) laid down rational (straight- line) A. Thus, BC is a fourth apotome [Def. 10.14].† Thus, a fourth apotome has been found. (Which is) the very thing it was required to show. † See footnote to Prop. 10.51. pjþ. Proposition 89 Εὐρεῖν τὴν πέμπτην ἀποτομήν. To find a fifth apotome. Θ Β ∆ Ε Γ Η Ζ Α GB D E A H F C ᾿Εκκείσθω ῥητὴ ἡ Α, καὶ τῇ Α μήκει σύμμετρος ἔστω ἡ Let the rational (straight-line) A be laid down, and let ΓΗ· ῥητὴ ἄρα [ἐστὶν] ἡ ΓΗ. καὶ ἐκκείσθωσαν δύο ἀριθμοὶ οἱ CG be commensurable in length with A. Thus, CG [is] ΔΖ, ΖΕ, ὥστε τὸν ΔΕ πρὸς ἑκάτερον τῶν ΔΖ, ΖΕ λόγον a rational (straight-line). And let the two numbers DF πάλιν μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον and FE be laid down such that DE again does not have ἀριθμόν· καὶ πεποιήσθω ὡς ὁ ΖΕ πρὸς τὸν ΕΔ, οὕτως τὸ to each of DF and FE the ratio which (some) square ἀπὸ τῆς ΓΗ πρὸς τὸ ἀπὸ τῆς ΗΒ. ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς number (has) to (some) square number. And let it have ΗΒ· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΒΗ. καὶ ἐπεί ἐστιν ὡς ὁ ΔΕ πρὸς been contrived that as FE (is) to ED, so the (square) on τὸν ΕΖ, οὕτως τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ, ὁ δὲ CG (is) to the (square) on GB. Thus, the (square) on GB ΔΕ πρὸς τὸν ΕΖ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς (is) also rational [Prop. 10.6]. Thus, BG is also rational. πρὸς τετράγωνον ἀριθμόν, οὐδ᾿ ἄρα τὸ ἀπὸ τῆς ΒΗ πρὸς And since as DE is to EF , so the (square) on BG (is) to τὸ ἀπὸ τῆς ΗΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς the (square) on GC. And DE does not have to EF the ra- τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ ΗΓ tio which (some) square number (has) to (some) square μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ΒΗ, ΗΓ ἄρα ῥηταί εἰσι number. The (square) on BG thus does not have to the δυνάμει μόνον σύμμετροι· ἡ ΒΓ ἄρα ἀποτομή ἐστιν. λέγω (square) on GC the ratio which (some) square number δή, ὅτι καὶ πέμπτη. (has) to (some) square number either. Thus, BG is in- ῟Ωι γὰρ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΒΗ τοῦ ἀπὸ τῆς ΗΓ, commensurable in length with GC [Prop. 10.9]. And ἔστω τὸ ἀπὸ τῆς Θ. ἐπεὶ οὖν ἐστιν ὡς τὸ ἀπὸ τὴς ΒΗ πρὸς they are both rational (straight-lines). BG and GC are τὸ ἀπὸ τῆς ΗΓ, οὕτως ὁ ΔΕ πρὸς τὸν ΕΖ, ἀναστρέψαντι thus rational (straight-lines which are) commensurable ἄρα ἐστὶν ὡς ὁ ΕΔ πρὸς τὸν ΔΖ, οὕτως τὸ ἀπὸ τῆς ΒΗ πρὸς in square only. Thus, BC is an apotome [Prop. 10.73]. τὸ ἀπὸ τῆς Θ, ὁ δὲ ΕΔ πρὸς τὸν ΔΖ λόγον οὐκ ἔχει, ὃν So, I say that (it is) also a fifth (apotome). 384 STOIQEIWN iþ. ELEMENTS BOOK 10 τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· οὐδ᾿ ἄρα τὸ For, let the (square) on H be that (area) by which ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος the (square) on BG is greater than the (square) on GC ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ [Prop. 10.13 lem.]. Therefore, since as the (square) on ΒΗ τῇ Θ μήκει. καὶ δύναται ἡ ΒΗ τῆς ΗΓ μεῖζον τῷ ἀπὸ BG (is) to the (square) on GC, so DE (is) to EF , thus, τῆς Θ· ἡ ΗΒ ἄρα τῆς ΗΓ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου via conversion, as ED is to DF , so the (square) on BG ἑαυτῇ μήκει. καί ἐστιν ἡ προσαρμόζουσα ἡ ΓΗ σύμμετρος (is) to the (square) on H [Prop. 5.19 corr.]. And ED does τῇ ἐκκειμένῃ ῥητῇ τῇ Α μήκει· ἡ ἄρα ΒΓ ἀποτομή ἐστι not have to DF the ratio which (some) square number πέμπτη. (has) to (some) square number. Thus, the (square) on Εὕρηται ἄρα ἡ πέμπτη ἀποτομὴ ἡ ΒΓ· ὅπερ ἔδει δεῖξαι. BG does not have to the (square) on H the ratio which (some) square number (has) to (some) square number either. Thus, BG is incommensurable in length with H [Prop. 10.9]. And the square on BG is greater than (the square on) GC by the (square) on H . Thus, the square on GB is greater than (the square on) GC by the (square) on (some straight-line) incommensurable in length with (GB). And the attachment CG is commensurable in length with the (previously) laid down rational (straight- line) A. Thus, BC is a fifth apotome [Def. 10.15].† Thus, the fifth apotome BC has been found. (Which is) the very thing it was required to show. † See footnote to Prop. 10.52. �þ. Proposition 90 Εὑρεῖν τὴν ἕκτην ἀποτομήν. To find a sixth apotome. ᾿Εκκείσθω ῥητὴ ἡ Α καὶ τρεῖς ἀριθμοὶ οἱ Ε, ΒΓ, ΓΔ Let the rational (straight-line) A, and the three num- λόγον μὴ ἔχοντες πρὸς ἀλλήλους, ὃν τετράγωνος ἀριθμὸς bers E, BC, and CD, not having to one another the ra- πρὸς τετράγωνον ἀριθμόν· ἔτι δὲ καὶ ὁ ΓΒ πρὸς τὸν ΒΔ tio which (some) square number (has) to (some) square λόγον μὴ ἐχετώ, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον number, be laid down. Furthermore, let CB also not have ἀριθμόν· καὶ πεποιήσθω ὡς μὲν ὁ Ε πρὸς τὸν ΒΓ, οὕτως to BD the ratio which (some) square number (has) to τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΖΗ, ὡς δὲ ὁ ΒΓ πρὸς τὸν (some) square number. And let it have been contrived ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ. that as E (is) to BC, so the (square) on A (is) to the (square) on FG, and as BC (is) to CD, so the (square) on FG (is) to the (square) on GH [Prop. 10.6 corr.]. ∆Β Α Γ Ζ Θ Η Ε Κ G B A E K D C F H ᾿Επεὶ οὖν ἐστιν ὡς ὁ Ε πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς Therefore, since as E is to BC, so the (square) on A Α πρὸς τὸ ἀπὸ τῆς ΖΗ, σύμμετρον ἄρα τὸ ἀπὸ τῆς Α τῷ (is) to the (square) on FG, the (square) on A (is) thus ἀπὸ τῆς ΖΗ. ῥητὸν δὲ τὸ ἀπὸ τῆς Α· ῥητὸν ἄρα καὶ τὸ ἀπὸ commensurable with the (square) on FG [Prop. 10.6]. τῆς ΖΗ· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ Ε πρὸς τὸν ΒΓ And the (square) on A (is) rational. Thus, the (square) λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον on FG (is) also rational. Thus, FG is also a rational ἀριθμόν, οὐδ᾿ ἄρα τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον (straight-line). And since E does not have to BC the ra- ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· tio which (some) square number (has) to (some) square ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῆ ΖΗ μήκει. πάλιν, ἐπεί ἐστιν ὡς number, the (square) on A thus does not have to the ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς (square) on FG the ratio which (some) square number ΗΘ, σύμμετρον ἄρα τὸ ἀπὸ τῆς ΖΗ τῷ ἀπὸ τῆς ΗΘ. ῥητὸν (has) to (some) square number either. Thus, A is in- 385 STOIQEIWN iþ. ELEMENTS BOOK 10 δὲ τὸ ἀπὸ τῆς ΖΗ· ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΗΘ· ῥητὴ ἄρα commensurable in length with FG [Prop. 10.9]. Again, καὶ ἡ ΗΘ. καὶ ἐπεὶ ὁ ΒΓ πρὸς τὸν ΓΔ λόγον οὐκ ἔχει, ὃν since as BC is to CD, so the (square) on FG (is) to the τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ᾿ ἄρα τὸ (square) on GH , the (square) on FG (is) thus commen- ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος surable with the (square) on GH [Prop. 10.6]. And the ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν (square) on FG (is) rational. Thus, the (square) on GH ἡ ΖΗ τῇ ΗΘ μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ΖΗ, ΗΘ (is) also rational. Thus, GH (is) also rational. And since ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἡ ἄρα ΖΘ ἀποτομή BC does not have to CD the ratio which (some) square ἐστιν. λέγω δή, ὅτι καὶ ἕκτη. number (has) to (some) square number, the (square) on ᾿Επεὶ γάρ ἐστιν ὡς μὲν ὁ Ε πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ FG thus does not have to the (square) on GH the ra- τῆς Α πρὸς τὸ ἀπὸ τῆς ΖΗ, ὡς δὲ ὁ ΒΓ πρὸς τὸν ΓΔ, οὕτως tio which (some) square (number) has to (some) square τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, δι᾿ ἴσου ἄρα ἐστὶν ὡς ὁ (number) either. Thus, FG is incommensurable in length Ε πρὸς τὸν ΓΔ, οὕτως τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς ΗΘ. ὁ with GH [Prop. 10.9]. And both are rational (straight- δὲ Ε πρὸς τὸν ΓΔ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς lines). Thus, FG and GH are rational (straight-lines πρὸς τετράγωνον ἀριθμόν· οὐδ᾿ ἄρα τὸ ἀπὸ τῆς Α πρὸς which are) commensurable in square only. Thus, FH is τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς an apotome [Prop. 10.73]. So, I say that (it is) also a τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ ΗΘ sixth (apotome). μήκει· οὐδετέρα ἄρα τῶν ΖΗ, ΗΘ σύμμετρός ἐστι τῇ Α ῥητῇ For since as E is to BC, so the (square) on A (is) μήκει. ᾧ οὖν μεῖζόν ἐστι τὸ ἀπὸ τῆς ΖΗ τοῦ ἀπὸ τῆς ΗΘ, to the (square) on FG, and as BC (is) to CD, so the ἔστω τὸ ἀπὸ τῆς Κ. ἐπεὶ οὖν ἐστιν ὡς ὁ ΒΓ πρὸς τὸν ΓΔ, (square) on FG (is) to the (square) on GH , thus, via οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, ἀναστρέψαντι equality, as E is to CD, so the (square) on A (is) to ἄρα ἐστὶν ὡς ὁ ΓΒ πρὸς τὸν ΒΔ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς the (square) on GH [Prop. 5.22]. And E does not have τὸ ἀπὸ τῆς Κ. ὁ δὲ ΓΒ πρὸς τὸν ΒΔ λόγον οὐκ ἔχει, ὃν to CD the ratio which (some) square number (has) to τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· οὐδ᾿ ἄρα τὸ (some) square number. Thus, the (square) on A does not ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Κ λόγον ἔχει, ὃν τετράγωνος have to the (square) GH the ratio which (some) square ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ number (has) to (some) square number either. A is thus ΖΗ τῇ Κ μήκει. καὶ δύναται ἡ ΖΗ τῆς ΗΘ μεῖζον τῷ ἀπὸ incommensurable in length with GH [Prop. 10.9]. Thus, τῆς Κ· ἡ ΖΗ ἄρα τῆς ΗΘ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου neither of FG and GH is commensurable in length with ἑαυτῇ μήκει. καὶ οὐδετέρα τῶν ΖΗ, ΗΘ σύμμετρός ἐστι τῇ the rational (straight-line) A. Therefore, let the (square) ἐκκειμένῃ ῥητῇ μήκει τῇ Α. ἡ ἄρα ΖΘ ἀποτομή ἐστιν ἕκτη. on K be that (area) by which the (square) on FG is Εὕρηται ἄρα ἡ ἕκτη ἀποτομὴ ἡ ΖΘ· ὅπερ ἔδει δεῖξαι. greater than the (square) on GH [Prop. 10.13 lem.]. Therefore, since as BC is to CD, so the (square) on FG (is) to the (square) on GH , thus, via conversion, as CB is to BD, so the (square) on FG (is) to the (square) on K [Prop. 5.19 corr.]. And CB does not have to BD the ra- tio which (some) square number (has) to (some) square number. Thus, the (square) on FG does not have to the (square) on K the ratio which (some) square number (has) to (some) square number either. FG is thus in- commensurable in length with K [Prop. 10.9]. And the square on FG is greater than (the square on) GH by the (square) on K. Thus, the square on FG is greater than (the square on) GH by the (square) on (some straight- line) incommensurable in length with (FG). And neither of FG and GH is commensurable in length with the (pre- viously) laid down rational (straight-line) A. Thus, FH is a sixth apotome [Def. 10.16]. Thus, the sixth apotome FH has been found. (Which is) the very thing it was required to show. † See footnote to Prop. 10.53. 386 STOIQEIWN iþ. ELEMENTS BOOK 10�aþ. Proposition 91 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς πρώτης, If an area is contained by a rational (straight-line) and ἡ τὸ χωρίον δυναμένη ἀπορομή ἐστιν. a first apotome then the square-root of the area is an apo- Περιεχέσθω γὰρ χωρίον τὸ ΑΒ ὑπὸ ῥητῆς τῆς ΑΓ καὶ tome. ἀποτομῆς πρώτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυ- For let the area AB have been contained by the ratio- ναμένη ἀποτομή ἐστιν. nal (straight-line) AC and the first apotome AD. I say that the square-root of area AB is an apotome. Φ Α ∆ Ζ Η Γ Β Κ Ο Ν Ξ Υ Χ Ε Θ Ι Σ Λ Ρ Μ Τ O A D B K N U C F G V W E H I L S R T M P ᾿Επεὶ γὰρ ἀποτομή ἐστι πρώτη ἡ ΑΔ, ἔστω αὐτῇ προ- For since AD is a first apotome, let DG be its at- σαρμόζουσα ἡ ΔΗ· αἱ ΑΗ, ΗΔ ἄρα ῥηταί εἰσι δυνάμει μόνον tachment. Thus, AG and DG are rational (straight-lines σύμμετροι. καὶ ὅλη ἡ ΑΗ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ which are) commensurable in square only [Prop. 10.73]. τῇ ΑΓ, καὶ ἡ ΑΗ τῆς ΗΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου And the whole, AG, is commensurable (in length) with ἑαυτῇ μήκει· ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ the (previously) laid down rational (straight-line) AC, ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, and the square on AG is greater than (the square on) εἰς σύμμετρα αὐτὴν διαιρεῖ. τετμήσθω ἡ ΔΗ δίχα κατὰ τὸ GD by the (square) on (some straight-line) commensu- Ε, καὶ τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω rable in length with (AG) [Def. 10.11]. Thus, if (an area) ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ· equal to the fourth part of the (square) on DG is applied σύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΖΗ. καὶ διὰ τῶν Ε, Ζ, Η to AG, falling short by a square figure, then it divides σημείων τῇ ΑΓ παράλληλοι ἤχθωσαν αἱ ΕΘ, ΖΙ, ΗΚ. (AG) into (parts which are) commensurable (in length) Καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΖ τῇ ΖΗ μήκει, καὶ ἡ ΑΗ [Prop. 10.17]. Let DG have been cut in half at E. And ἄρα ἑκατέρᾳ τῶν ΑΖ, ΖΗ σύμμετρός ἐστι μήκει. ἀλλὰ ἡ let (an area) equal to the (square) on EG have been ap- ΑΗ σύμμετρός ἐστι τῇ ΑΓ· καὶ ἑκατέρα ἄρα τῶν ΑΖ, ΖΗ plied to AG, falling short by a square figure. And let it σύμμετρός ἐστι τῇ ΑΓ μήκει. καί ἐστι ῥητὴ ἡ ΑΓ· ῥητὴ be the (rectangle contained) by AF and FG. AF is thus ἄρα καὶ ἑκατέρα τῶν ΑΖ, ΖΗ· ὥστε καὶ ἑκάτερον τῶν ΑΙ, commensurable (in length) with FG. And let EH , FI, ΖΚ ῥητόν ἐστιν. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΕΗ and GK have been drawn through points E, F , and G μήκει, καὶ ἡ ΔΗ ἄρα ἑκατέρᾳ τῶν ΔΕ, ΕΗ σύμμετρός ἐστι (respectively), parallel to AC. μήκει. ῥητὴ δὲ ἡ ΔΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει· ῥητὴ And since AF is commensurable in length with FG, ἄρα καὶ ἑκατέρα τῶν ΔΕ, ΕΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει· AG is thus also commensurable in length with each of ἑκάτερον ἄρα τῶν ΔΘ, ΕΚ μέσον ἐστίν. AF and FG [Prop. 10.15]. But AG is commensurable Κείσθω δὴ τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ δὲ (in length) with AC. Thus, each of AF and FG is also ΖΚ ἴσον τετράγωνον ἀφῃρήσθω κοινὴν γωνίαν ἔχον αὐτῷ commensurable in length with AC [Prop. 10.12]. And τὴν ὑπὸ ΛΟΜ τὸ ΝΞ· περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ AC is a rational (straight-line). Thus, AF and FG (are) ΛΜ, ΝΞ τετράγωνα. ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ κα- each also rational (straight-lines). Hence, AI and FK ταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ὑπὸ τῶν ΑΖ, are also each rational (areas) [Prop. 10.19]. And since ΖΗ περιεχόμενον ὀρθογώνιον τῷ ἀπὸ τῆς ΕΗ τετραγώνῳ, DE is commensurable in length with EG, DG is thus ἔστιν ἄρα ὡς ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΗ. also commensurable in length with each of DE and EG ἀλλ᾿ ὡς μὲν ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως τὸ ΑΙ πρὸς τὸ ΕΚ, [Prop. 10.15]. And DG (is) rational, and incommen- ὡς δὲ ἡ ΕΗ πρὸς τὴν ΖΗ, οὕτως ἐστὶ τὸ ΕΚ πρὸς τὸ ΚΖ· surable in length with AC. DE and EG (are) thus τῶν ἄρα ΑΙ, ΚΖ μέσον ἀνάλογόν ἐστι τὸ ΕΚ. ἔστι δὲ καὶ each rational, and incommensurable in length with AC τῶν ΛΜ, ΝΞ μέσον ἀνάλογον τὸ ΜΝ, ὡς ἐν τοῖς ἔμπρο- [Prop. 10.13]. Thus, DH and EK are each medial (ar- σθεν ἐδείχθη, καί ἐστι τὸ [μὲν] ΑΙ τῷ ΛΜ τετραγώνῳ ἴσον, eas) [Prop. 10.21]. τὸ δὲ ΚΖ τῷ ΝΞ· καὶ τὸ ΜΝ ἄρα τῷ ΕΚ ἴσον ἐστίν. ἀλλὰ So let the square LM , equal to AI, be laid down. τὸ μὲν ΕΚ τῷ ΔΘ ἐστιν ἴσον, τὸ δὲ ΜΝ τῷ ΛΞ· τὸ ἄρα And let the square NO, equal to FK, have been sub- 387 STOIQEIWN iþ. ELEMENTS BOOK 10 ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ. ἔστι δὲ καὶ τὸ tracted (from LM), having with it the common angle ΑΚ ἴσον τοῖς ΛΜ, ΝΞ τετραγώνοις· λοιπὸν ἄρα τὸ ΑΒ ἴσον LPM . Thus, the squares LM and NO are about the ἐστὶ τῷ ΣΤ. τὸ δὲ ΣΤ τὸ ἀπὸ τῆς ΛΝ ἐστι τετράγωνον· τὸ same diagonal [Prop. 6.26]. Let PR be their (com- ἄρα ἀπὸ τῆς ΛΝ τετράγωνον ἴσον ἐστὶ τῷ ΑΒ· ἡ ΛΝ ἄρα mon) diagonal, and let the (rest of the) figure have been δύναται τὸ ΑΒ. λέγω δή, ὅτι ἡ ΛΝ ἀποτομή ἐστιν. drawn. Therefore, since the rectangle contained by AF ᾿Επεὶ γὰρ ῥητόν ἐστιν ἑκάτερον τῶν ΑΙ, ΖΚ, καί ἐστιν and FG is equal to the square EG, thus as AF is to ἴσον τοῖς ΛΜ, ΝΞ, καὶ ἑκάτερον ἄρα τῶν ΛΜ, ΝΞ ῥητόν EG, so EG (is) to FG [Prop. 6.17]. But, as AF (is) ἐστιν, τουτέστι τὸ ἀπὸ ἑκατέρας τῶν ΛΟ, ΟΝ· καὶ ἑκατέρα to EG, so AI (is) to EK, and as EG (is) to FG, so ἄρα τῶν ΛΟ, ΟΝ ῥητή ἐστιν. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ ΔΘ EK is to KF [Prop. 6.1]. Thus, EK is the mean pro- καί ἐστιν ἴσον τῷ ΛΞ, μέσον ἄρα ἐστὶ καὶ τὸ ΛΞ. ἐπεὶ οὖν τὸ portional to AI and KF [Prop. 5.11]. And MN is also μὲν ΛΞ μέσον ἐστίν, τὸ δὲ ΝΞ ῥητόν, ἀσύμμετρον ἄρα ἐστὶ the mean proportional to LM and NO, as shown before τὸ ΛΞ τῷ ΝΞ· ὡς δὲ τὸ ΛΞ πρὸς τὸ ΝΞ, οὕτως ἐστὶν ἡ ΛΟ [Prop. 10.53 lem.]. And AI is equal to the square LM , πρὸς τὴν ΟΝ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΛΟ τῇ ΟΝ μήκει. and KF to NO. Thus, MN is also equal to EK. But, EK καί εἰσιν ἀμφότεραι ῥηταί· αἱ ΛΟ, ΟΝ ἄρα ῥηταί εἰσι δυνάμει is equal to DH , and MN to LO [Prop. 1.43]. Thus, DK μόνον σύμμετροι· ἀποτομὴ ἄρα ἐστὶν ἡ ΛΝ. καὶ δύναται τὸ is equal to the gnomon UV W and NO. And AK is also ΑΒ χωρίον· ἡ ἄρα τὸ ΑΒ χωρίον δυναμένη ἀποτομή ἐστιν. equal to (the sum of) the squares LM and NO. Thus, the ᾿Εὰν ἄρα χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τὰ ἑξῆς. remainder AB is equal to ST . And ST is the square on LN . Thus, the square on LN is equal to AB. Thus, LN is the square-root of AB. So, I say that LN is an apotome. For since AI and FK are each rational (areas), and are equal to LM and NO (respectively), thus LM and NO—that is to say, the (squares) on each of LP and PN (respectively)—are also each rational (areas). Thus, LP and PN are also each rational (straight-lines). Again, since DH is a medial (area), and is equal to LO, LO is thus also a medial (area). Therefore, since LO is medial, and NO rational, LO is thus incommensurable with NO. And as LO (is) to NO, so LP is to PN [Prop. 6.1]. LP is thus incommensurable in length with PN [Prop. 10.11]. And they are both rational (straight- lines). Thus, LP and PN are rational (straight-lines which are) commensurable in square only. Thus, LN is an apotome [Prop. 10.73]. And it is the square-root of area AB. Thus, the square-root of area AB is an apo- tome. Thus, if an area is contained by a rational (straight- line), and so on . . . .�bþ. Proposition 92 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς δευτέρας, If an area is contained by a rational (straight-line) and ἡ τὸ χωρίον δυναμένη μέσης ἀποτομή ἐστι πρώτη. a second apotome then the square-root of the area is a first apotome of a medial (straight-line). Φ Α ∆ Ζ Η Γ Β Κ Ο Ν Ξ Ε Θ Ι Σ Λ Ρ Μ Τ Χ Υ Π C A D B K O NE I S L R M T U F G H W V P Q 388 STOIQEIWN iþ. ELEMENTS BOOK 10 Χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ For let the area AB have been contained by the ratio- ἀποτομῆς δευτέρας τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΒ χωρίον nal (straight-line) AC and the second apotome AD. I say δυναμένη μέσης ἀποτομή ἐστι πρώτη. that the square-root of area AB is the first apotome of a ῎Εστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ· αἱ ἄρα medial (straight-line). ΑΗ, ΗΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ προ- For let DG be an attachment to AD. Thus, AG and σαρμόζουσα ἡ ΔΗ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ GD are rational (straight-lines which are) commensu- ΑΓ, ἡ δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΗΔ μεῖζον rable in square only [Prop. 10.73], and the attachment δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. ἐπεὶ οὖν ἡ ΑΗ DG is commensurable (in length) with the (previously) τῆς ΗΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, ἐὰν ἄρα laid down rational (straight-line) AC, and the square on τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΗΔ ἴσον παρὰ τὴν ΑΗ πα- the whole, AG, is greater than (the square on) the at- ραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν tachment, GD, by the (square) on (some straight-line) διαιρεῖ. τετμήσθω οὖν ἡ ΔΗ δίχα κατὰ τὸ Ε· καὶ τῷ ἀπὸ commensurable in length with (AG) [Def. 10.12]. There- τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει fore, since the square on AG is greater than (the square τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ· σύμμετρος ἄρα on) GD by the (square) on (some straight-line) commen- ἐστὶν ἡ ΑΖ τῇ ΖΗ μήκει. καὶ ἡ ΑΗ ἄρα ἑκατέρᾳ τῶν ΑΖ, surable (in length) with (AG), thus if (an area) equal ΖΗ σύμμετρός ἐστι μήκει. ῥητὴ δὲ ἡ ΑΗ καὶ ἀσύμμετρος to the fourth part of the (square) on GD is applied τῇ ΑΓ μήκει· καὶ ἑκατέρα ἄρα τῶν ΑΖ, ΖΗ ῥητή ἐστι καὶ to AG, falling short by a square figure, then it divides ἀσύμμετρος τῇ ΑΓ μήκει· ἑκάτερον ἄρα τῶν ΑΙ, ΖΚ μέσον (AG) into (parts which are) commensurable (in length) ἐστίν. πάλιν, ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΕΗ, καὶ ἡ ΔΗ [Prop. 10.17]. Therefore, let DG have been cut in half at ἄρα ἑκατέρᾳ τῶν ΔΕ, ΕΗ σύμμετρός ἐστιν. ἀλλ᾿ ἡ ΔΗ E. And let (an area) equal to the (square) on EG have σύμμετρός ἐστι τῇ ΑΓ μήκει [ῥητὴ ἄρα καὶ ἑκατέρα τῶν been applied to AG, falling short by a square figure. And ΔΕ, ΕΗ καὶ σύμμετρος τῇ ΑΓ μήκει]. ἑκάτερον ἄρα τῶν let it be the (rectangle contained) by AF and FG. Thus, ΔΘ, ΕΚ ῥητόν ἐστιν. AF is commensurable in length with FG. AG is thus Συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ also commensurable in length with each of AF and FG δὲ ΖΚ ἴσον ἀφῃρήσθω τὸ ΝΞ περὶ τὴν αὐτὴν γωνίαν ὂν τῷ [Prop. 10.15]. And AG (is) a rational (straight-line), and ΛΜ τὴν ὑπὸ τῶν ΛΟΜ· περὶ τὴν αὐτὴν ἄρα ἐστὶ διάμετρον incommensurable in length with AC. AF and FG are τὰ ΛΜ, ΝΞ τετράγωνα. ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ thus also each rational (straight-lines), and incommen- καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν τὰ ΑΙ, ΖΚ μέσα ἐστὶ surable in length with AC [Prop. 10.13]. Thus, AI and καί ἐστιν ἴσα τοῖς ἀπὸ τῶν ΛΟ, ΟΝ, καὶ τὰ ἀπὸ τῶν ΛΟ, FK are each medial (areas) [Prop. 10.21]. Again, since ΟΝ [ἄρα] μέσα ἐστίν· καὶ αἱ ΛΟ, ΟΝ ἄρα μέσαι εἰσὶ δυνάμει DE is commensurable (in length) with EG, thus DG is μόνον σύμμετροι. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΖ, ΖΗ ἴσον ἐστὶ τῷ also commensurable (in length) with each of DE and EG ἀπὸ τῆς ΕΗ, ἔστιν ἄρα ὡς ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἡ ΕΗ [Prop. 10.15]. But, DG is commensurable in length with πρὸς τὴν ΖΗ· ἀλλ᾿ ὡς μὲν ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως τὸ ΑΙ AC [thus, DE and EG are also each rational, and com- πρὸς τὸ ΕΚ· ὡς δὲ ἡ ΕΗ πρὸς τὴν ΖΗ, οὕτως [ἐστὶ] τὸ ΕΚ mensurable in length with AC]. Thus, DH and EK are πρὸς τὸ ΖΚ· τῶν ἄρα ΑΙ, ΖΚ μέσον ἀνάλογόν ἐστι τὸ ΕΚ. each rational (areas) [Prop. 10.19]. ἔστι δὲ καὶ τῶν ΛΜ, ΝΞ τετραγώνων μέσον ἀνάλογον τὸ Therefore, let the square LM , equal to AI, have ΜΝ· καί ἐστιν ἴσον τὸ μὲν ΑΙ τῷ ΛΜ, τὸ δὲ ΖΚ τῷ ΝΞ· καὶ been constructed. And let NO, equal to FK, which is τὸ ΜΝ ἄρα ἴσον ἐστὶ τῷ ΕΚ. ἀλλὰ τῷ μὲν ΕΚ ἴσον [ἐστὶ] about the same angle LPM as LM , have been subtracted τὸ ΔΘ, τῷ δὲ ΜΝ ἴσον τὸ ΛΞ· ὅλον ἄρα τὸ ΔΚ ἴσον ἐστὶ (from LM). Thus, the squares LM and NO are about τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ. ἐπεὶ οὖν ὅλον τὸ ΑΚ ἴσον the same diagonal [Prop. 6.26]. Let PR be their (com- ἐστὶ τοῖς ΛΜ, ΝΞ, ὧν τὸ ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι mon) diagonal, and let the (rest of the) figure have been καὶ τῷ ΝΞ, λοιπὸν ἄρα τὸ ΑΒ ἴσον ἐστὶ τῷ ΤΣ. τὸ δὲ ΤΣ drawn. Therefore, since AI and FK are medial (areas), ἐστι τὸ ἀπὸ τῆς ΛΝ· τὸ ἀπὸ τῆς ΛΝ ἄρα ἴσον ἐστὶ τῷ ΑΒ and are equal to the (squares) on LP and PN (respec- χωρίῳ· ἡ ΛΝ ἄρα δύναται τὸ ΑΒ χωρίον. λέγω [δή], ὅτι ἡ tively), [thus] the (squares) on LP and PN are also me- ΛΝ μέσης ἀποτομή ἐστι πρώτη. dial. Thus, LP and PN are also medial (straight-lines ᾿Επεὶ γὰρ ῥητόν ἐστι τὸ ΕΚ καί ἐστιν ἴσον τῷ ΛΞ, ῥητὸν which are) commensurable in square only.† And since ἄρα ἐστὶ τὸ ΛΞ, τουτέστι τὸ ὑπὸ τῶν ΛΟ, ΟΝ. μέσον δὲ the (rectangle contained) by AF and FG is equal to ἐδείχθη τὸ ΝΞ· ἀσύμμετρον ἄρα ἐστὶ τὸ ΛΞ τῷ ΝΞ· ὡς the (square) on EG, thus as AF is to EG, so EG (is) δὲ τὸ ΛΞ πρὸς τὸ ΝΞ, οὕτως ἐστὶν ἡ ΛΟ πρὸς ΟΝ· αἱ to FG [Prop. 10.17]. But, as AF (is) to EG, so AI ΛΟ, ΟΝ ἄρα ἀσύμμετροί εἰσι μήκει. αἱ ἄρα ΛΟ, ΟΝ μέσαι (is) to EK. And as EG (is) to FG, so EK [is] to FK εἰσὶ δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι· ἡ ΛΝ ἄρα [Prop. 6.1]. Thus, EK is the mean proportional to AI 389 STOIQEIWN iþ. ELEMENTS BOOK 10 μέσης ἀποτομή ἐστι πρώτη· καὶ δύναται τὸ ΑΒ χωρίον. and FK [Prop. 5.11]. And MN is also the mean pro- ῾Η ἄρα τὸ ΑΒ χωρίον δυναμένη μέσης ἀποτομή ἐστι portional to the squares LM and NO [Prop. 10.53 lem.]. πρώτη· ὅπερ ἔδει δεῖξαι. And AI is equal to LM , and FK to NO. Thus, MN is also equal to EK. But, DH [is] equal to EK, and LO equal to MN [Prop. 1.43]. Thus, the whole (of) DK is equal to the gnomon UV W and NO. Therefore, since the whole (of) AK is equal to LM and NO, of which DK is equal to the gnomon UV W and NO, the remainder AB is thus equal to TS. And TS is the (square) on LN . Thus, the (square) on LN is equal to the area AB. LN is thus the square-root of area AB. [So], I say that LN is the first apotome of a medial (straight-line). For since EK is a rational (area), and is equal to LO, LO—that is to say, the (rectangle contained) by LP and PN—is thus a rational (area). And NO was shown (to be) a medial (area). Thus, LO is incommensurable with NO. And as LO (is) to NO, so LP is to PN [Prop. 6.1]. Thus, LP and PN are incommensurable in length [Prop. 10.11]. LP and PN are thus medial (straight-lines which are) commensurable in square only, and which contain a rational (area). Thus, LN is the first apotome of a medial (straight-line) [Prop. 10.74]. And it is the square-root of area AB. Thus, the square root of area AB is the first apotome of a medial (straight-line). (Which is) the very thing it was required to show. † There is an error in the argument here. It should just say that LP and PN are commensurable in square, rather than in square only, since LP and PN are only shown to be incommensurable in length later on.�gþ. Proposition 93 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς τρίτης, If an area is contained by a rational (straight-line) and ἡ τὸ χωρίον δυναμένη μέσης ἀποτομή ἐστι δευτέρα. a third apotome then the square-root of the area is a sec- ond apotome of a medial (straight-line). Φ Α ∆ Ζ Η Γ Β Κ Ο Ν Ξ Ε Θ Ι Σ Λ Ρ Μ Τ ΠΥ Χ C A D B K O NE I S L R M T F G H V P QU W Χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ For let the area AB have been contained by the ratio- ἀποτομῆς τρίτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυ- nal (straight-line) AC and the third apotome AD. I say ναμένη μέσης ἀποτομή ἐστι δευτέρα. that the square-root of area AB is the second apotome of ῎Εστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ· αἱ ΑΗ, ΗΔ a medial (straight-line). ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα τῶν For let DG be an attachment to AD. Thus, AG and ΑΗ, ΗΔ σύμμετρός ἐστι μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ, ἡ GD are rational (straight-lines which are) commensu- δὲ ὅλη ἡ ΑΗ τὴς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται rable in square only [Prop. 10.73], and neither of AG τῷ ἀπὸ συμμέτρου ἑαυτῇ. ἐπεὶ οὖν ἡ ΑΗ τῆς ΗΔ μεῖζον and GD is commensurable in length with the (previ- 390 STOIQEIWN iþ. ELEMENTS BOOK 10 δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, ἐὰν ἄρα τῷ τετάρτῳ μέρει ously) laid down rational (straight-line) AC, and the τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον square on the whole, AG, is greater than (the square on) εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν διελεῖ. τετμήσθω οὖν the attachment, DG, by the (square) on (some straight- ἡ ΔΗ δίχα κατὰ τὸ Ε, καὶ τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν line) commensurable (in length) with (AG) [Def. 10.13]. ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ Therefore, since the square on AG is greater than (the ὑπὸ τῶν ΑΖ, ΖΗ. καὶ ἤχθωσαν διὰ τῶν Ε, Ζ, Η σημείων square on) GD by the (square) on (some straight-line) τῇ ΑΓ παράλληλοι αἱ ΕΘ, ΖΙ, ΗΚ· σύμμετροι ἄρα εἰσὶν αἱ commensurable (in length) with (AG), thus if (an area) ΑΖ, ΖΗ· σύμμετρον ἄρα καὶ τὸ ΑΙ τῷ ΖΚ. καὶ ἐπεὶ αἱ ΑΖ, equal to the fourth part of the square on DG is applied ΖΗ σύμμετροί εἰσι μήκει, καὶ ἡ ΑΗ ἄρα ἑκατέρᾳ τῶν ΑΖ, to AG, falling short by a square figure, then it divides ΖΗ σύμμετρός ἐστι μήκει. ῥητὴ δὲ ἡ ΑΗ καὶ ἀσύμμετρος (AG) into (parts which are) commensurable (in length) τῇ ΑΓ μήκει· ὥστε καὶ αἱ ΑΖ, ΖΗ. ἑκάτερον ἄρα τῶν ΑΙ, [Prop. 10.17]. Therefore, let DG have been cut in half ΖΚ μέσον ἐστίν. πάλιν, ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΕΗ at E. And let (an area) equal to the (square) on EG μήκει, καὶ ἡ ΔΗ ἄρα ἑκατέρᾳ τῶν ΔΕ, ΕΗ σύμμετρός ἐστι have been applied to AG, falling short by a square fig- μήκει. ῥητὴ δὲ ἡ ΗΔ καὶ ἀσύμμετρος τῇ ΑΓ μήκει· ῥητὴ ure. And let it be the (rectangle contained) by AF and ἄρα καὶ ἑκατέρα τῶν ΔΕ, ΕΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει· FG. And let EH , FI, and GK have been drawn through ἑκάτερον ἄρα τῶν ΔΘ, ΕΚ μέσον ἐστίν. καὶ ἐπεὶ αἱ ΑΗ, points E, F , and G (respectively), parallel to AC. Thus, ΗΔ δυνάμει μόνον σύμμετροί εἰσιν, ἀσύμμετρος ἄρα ἐστὶ AF and FG are commensurable (in length). AI (is) thus μήκει ἡ ΑΗ τῇ ΗΔ. ἀλλ᾿ ἡ μὲν ΑΗ τῇ ΑΖ σύμμετρός ἐστι also commensurable with FK [Props. 6.1, 10.11]. And μήκει ἡ δὲ ΔΗ τῇ ΕΗ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΕΗ since AF and FG are commensurable in length, AG is μήκει. ὡς δὲ ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ thus also commensurable in length with each of AF and ΕΚ· ἀσύμμετρον ἄρα ἐστὶ τὸ ΑΙ τῷ ΕΚ. FG [Prop. 10.15]. And AG (is) rational, and incommen- Συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ surable in length with AC. Hence, AF and FG (are) δὲ ΖΚ ἴσον ἀφῇρήσθω τὸ ΝΞ περὶ τὴν αὐτὴν γωνίαν ὂν τῷ also (rational, and incommensurable in length with AC) ΛΜ· περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ. ἔστω [Prop. 10.13]. Thus, AI and FK are each medial (ar- αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ eas) [Prop. 10.21]. Again, since DE is commensurable οὖν τὸ ὑπὸ τῶν ΑΖ, ΖΗ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΗ, ἔστιν ἄρα in length with EG, DG is also commensurable in length ὡς ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΗ. ἀλλ᾿ ὡς with each of DE and EG [Prop. 10.15]. And GD (is) μὲν ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ ΕΚ· ὡς rational, and incommensurable in length with AC. Thus, δὲ ἡ ΕΗ πρὸς τὴν ΖΗ, οὕτως ἐστὶ τὸ ΕΚ πρὸς τὸ ΖΚ· καὶ DE and EG (are) each also rational, and incommensu- ὡς ἄρα τὸ ΑΙ πρὸς τὸ ΕΚ, οὕτως τὸ ΕΚ πρὸς τὸ ΖΚ· τῶν rable in length with AC [Prop. 10.13]. DH and EK are ἄρα ΑΙ, ΖΚ μέσον ἀνάλογόν ἐστι τὸ ΕΚ. ἔστι δὲ καὶ τῶν thus each medial (areas) [Prop. 10.21]. And since AG ΛΜ, ΝΞ τετραγώνων μέσον ἀνάλογον τὸ ΜΝ· καί ἐστιν and GD are commensurable in square only, AG is thus ἴσον τὸ μὲν ΑΙ τῷ ΛΜ, τὸ δὲ ΖΚ τῷ ΝΞ· καὶ τὸ ΕΚ ἄρα incommensurable in length with GD. But, AG is com- ἴσον ἐστὶ τῷ ΜΝ. ἀλλὰ τὸ μὲν ΜΝ ἴσον ἐστὶ τῷ ΛΞ, τὸ mensurable in length with AF , and DG with EG. Thus, δὲ ΕΚ ἴσον [ἐστὶ] τῷ ΔΘ· καὶ ὅλον ἄρα τὸ ΔΚ ἴσον ἐστὶ AF is incommensurable in length with EG [Prop. 10.13]. τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ. ἔστι δὲ καὶ τὸ ΑΚ ἴσον τοῖς And as AF (is) to EG, so AI is to EK [Prop. 6.1]. Thus, ΛΜ, ΝΞ· λοιπὸν ἄρα τὸ ΑΒ ἴσον ἐστὶ τῷ ΣΤ, τουτέστι τῷ AI is incommensurable with EK [Prop. 10.11]. ἀπὸ τῆς ΛΝ τετραγώνῳ· ἡ ΛΝ ἄρα δύναται τὸ ΑΒ χωρίον. Therefore, let the square LM , equal to AI, have λέγω, ὅτι ἡ ΛΝ μέσης ἀποτομή ἐστι δευτέρα. been constructed. And let NO, equal to FK, which is ᾿Επεὶ γὰρ μέσα ἐδείχθη τὰ ΑΙ, ΖΚ καί ἐστιν ἴσα τοῖς ἀπὸ about the same angle as LM , have been subtracted (from τῶν ΛΟ, ΟΝ, μέσον ἄρα καὶ ἑκάτερον τῶν ἀπὸ τῶν ΛΟ, LM). Thus, LM and NO are about the same diagonal ΟΝ· μέση ἄρα ἑκατέρα τῶν ΛΟ, ΟΝ. καὶ ἐπεὶ σύμμετρόν [Prop. 6.26]. Let PR be their (common) diagonal, and ἐστι τὸ ΑΙ τῷ ΖΚ, σύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΛΟ τῷ let the (rest of the) figure have been drawn. Therefore, ἀπὸ τῆς ΟΝ. πάλιν, ἐπεὶ ἀσύμμετρον ἐδείχθη τὸ ΑΙ τῷ ΕΚ, since the (rectangle contained) by AF and FG is equal ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΛΜ τῷ ΜΝ, τουτέστι τὸ ἀπὸ to the (square) on EG, thus as AF is to EG, so EG (is) τῆς ΛΟ τῷ ὑπὸ τῶν ΛΟ, ΟΝ· ὥστε καὶ ἡ ΛΟ ἀσύμμετρός to FG [Prop. 6.17]. But, as AF (is) to EG, so AI is to ἐστι μήκει τῇ ΟΝ· αἱ ΛΟ, ΟΝ ἄρα μέσαι εἰσὶ δυνάμει μόνον EK [Prop. 6.1]. And as EG (is) to FG, so EK is to FK σύμμετροι. λέγω δή, ὅτι καὶ μέσον περιέχουσιν. [Prop. 6.1]. And thus as AI (is) to EK, so EK (is) to ᾿Επεὶ γὰρ μέσον ἐδείχθη τὸ ΕΚ καί ἐστιν ἴσον τῷ ὑπὸ FK [Prop. 5.11]. Thus, EK is the mean proportional to τῶν ΛΟ, ΟΝ, μέσον ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν ΛΟ, ΟΝ· AI and FK. And MN is also the mean proportional to ὥστε αἱ ΛΟ, ΟΝ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον the squares LM and NO [Prop. 10.53 lem.]. And AI is 391 STOIQEIWN iþ. ELEMENTS BOOK 10 περιέχουσαι. ἡ ΛΝ ἄρα μέσης ἀποτομή ἐστι δευτέρα· καὶ equal to LM , and FK to NO. Thus, EK is also equal to δύναται τὸ ΑΒ χωρίον. MN . But, MN is equal to LO, and EK [is] equal to DH ῾Η ἄρα τὸ ΑΒ χωρίον δυναμένη μέσης ἀποτομή ἐστι [Prop. 1.43]. And thus the whole of DK is equal to the δευτέρα· ὅπερ ἔδει δεῖξαι. gnomon UV W and NO. And AK (is) also equal to LM and NO. Thus, the remainder AB is equal to ST—that is to say, to the square on LN . Thus, LN is the square-root of area AB. I say that LN is the second apotome of a medial (straight-line). For since AI and FK were shown (to be) medial (ar- eas), and are equal to the (squares) on LP and PN (re- spectively), the (squares) on each of LP and PN (are) thus also medial. Thus, LP and PN (are) each medial (straight-lines). And since AI is commensurable with FK [Props. 6.1, 10.11], the (square) on LP (is) thus also commensurable with the (square) on PN . Again, since AI was shown (to be) incommensurable with EK, LM is thus also incommensurable with MN—that is to say, the (square) on LP with the (rectangle contained) by LP and PN . Hence, LP is also incommensurable in length with PN [Props. 6.1, 10.11]. Thus, LP and PN are medial (straight-lines which are) commensurable in square only. So, I say that they also contain a medial (area). For since EK was shown (to be) a medial (area), and is equal to the (rectangle contained) by LP and PN , the (rectangle contained) by LP and PN is thus also medial. Hence, LP and PN are medial (straight-lines which are) commensurable in square only, and which contain a me- dial (area). Thus, LN is the second apotome of a medial (straight-line) [Prop. 10.75]. And it is the square-root of area AB. Thus, the square-root of area AB is the second apo- tome of a medial (straight-line). (Which is) the very thing it was required to show.�dþ. Proposition 94 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς τετάρτης, If an area is contained by a rational (straight-line) and ἡ τὸ χωρίον δυναμένη ἐλάσσων ἐστίν. a fourth apotome then the square-root of the area is a minor (straight-line). Φ Α ∆ Ζ Η Γ Β Κ Ο Ν Ξ Ε Θ Ι Σ Λ Ρ Μ Τ ΠΥ Χ C A D B K O NE I S L R M T F G H V P QU W Χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ For let the area AB have been contained by the ra- ἀποτομῆς τετάρτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΒ χωρίον tional (straight-line) AC and the fourth apotome AD. I δυναμένη ἐλάσσων ἐστίν. say that the square-root of area AB is a minor (straight- 392 STOIQEIWN iþ. ELEMENTS BOOK 10 ῎Εστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ· αἱ ἄρα line). For let DG be an attachment to AD. Thus, AG ΑΗ, ΗΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΗ and DG are rational (straight-lines which are) commen- σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ μήκει, ἡ δὲ ὅλη surable in square only [Prop. 10.73], and AG is com- ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται τῷ mensurable in length with the (previously) laid down ra- ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει. ἐπεὶ οὖν ἡ ΑΗ τῆς ΗΔ tional (straight-line) AC, and the square on the whole, μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει, ἐὰν ἄρα AG, is greater than (the square on) the attachment, DG, τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ πα- by the square on (some straight-line) incommensurable ραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν in length with (AG) [Def. 10.14]. Therefore, since the διελεῖ. τετμήσθω οὖν ἡ ΔΗ δίχα κατὰ τὸ Ε, καὶ τῷ ἀπὸ square on AG is greater than (the square on) GD by τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει the (square) on (some straight-line) incommensurable in τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ· ἀσύμμετρος length with (AG), thus if (some area), equal to the fourth ἄρα ἐστὶ μήκει ἡ ΑΖ τῇ ΖΗ. ἤχθωσαν οὖν διὰ τῶν Ε, Ζ, Η part of the (square) on DG, is applied to AG, falling short παράλληλοι ταῖς ΑΓ, ΒΔ αἱ ΕΘ, ΖΙ, ΗΚ. ἐπεὶ οὖν ῥητή ἐστιν by a square figure, then it divides (AG) into (parts which ἡ ΑΗ καὶ σύμμετρος τῇ ΑΓ μήκει, ῥητὸν ἄρα ἐστὶν ὅλον τὸ are) incommensurable (in length) [Prop. 10.18]. There- ΑΚ. πάλιν, ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΔΗ τῇ ΑΓ μήκει, καί fore, let DG have been cut in half at E, and let (some εἰσιν ἀμφότεραι ῥηταί, μέσον ἄρα ἐστὶ τὸ ΔΚ. πάλιν, ἐπεὶ area), equal to the (square) on EG, have been applied ἀσύμμετρός ἐστὶν ἡ ΑΖ τῇ ΖΗ μήκει, ἀσύμμετρον ἄρα καὶ to AG, falling short by a square figure, and let it be the τὸ ΑΙ τῷ ΖΚ. (rectangle contained) by AF and FG. Thus, AF is in- Συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ commensurable in length with FG. Therefore, let EH , δὲ ΖΚ ἴσον ἀφῃρήσθω περὶ τὴν αὐτὴν γωνίαν τὴν ὑπὸ τῶν FI, and GK have been drawn through E, F , and G (re- ΛΟΜ τὸ ΝΞ. περὶ τὴν αὐτὴν ἄρα διάμετόν ἐστι τὰ ΛΜ, ΝΞ spectively), parallel to AC and BD. Therefore, since AG τετράγωνα. ἔστω αὐτῶν διάμετος ἡ ΟΡ, καὶ καταγεγράφθω is rational, and commensurable in length with AC, the τὸ σχῆμα. ἐπεὶ οὖν τὸ ὑπὸ τῶν ΑΖ, ΖΗ ἴσον ἐστὶ τῷ ἀπὸ whole (area) AK is thus rational [Prop. 10.19]. Again, τῆς ΕΗ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως since DG is incommensurable in length with AC, and ἡ ΕΗ πρὸς τὴν ΖΗ. ἀλλ᾿ ὡς μὲν ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως both are rational (straight-lines), DK is thus a medial ἐστὶ τὸ ΑΙ πρὸς τὸ ΕΚ, ὡς δὲ ἡ ΕΗ πρὸς τὴν ΖΗ, οὕτως (area) [Prop. 10.21]. Again, since AF is incommensu- ἐστὶ τὸ ΕΚ πρὸς τὸ ΖΚ· τῶν ἄρα ΑΙ, ΖΚ μέσον ἀνάλογόν rable in length with FG, AI (is) thus also incommensu- ἐστι τὸ ΕΚ. ἔστι δὲ καὶ τῶν ΛΜ, ΝΞ τετραγώνων μέσον rable with FK [Props. 6.1, 10.11]. ἀνάλογον τὸ ΜΝ, καί ἐστιν ἴσον τὸ μὲν ΑΙ τῷ ΛΜ, τὸ δὲ Therefore, let the square LM , equal to AI, have been ΖΚ τῷ ΝΞ· καὶ τὸ ΕΚ ἄρα ἴσον ἐστὶ τῷ ΜΝ. ἀλλὰ τῷ μὲν constructed. And let NO, equal to FK, (and) about the ΕΚ ἴσον ἐστὶ τὸ ΔΘ, τῷ δὲ ΜΝ ἴσον ἐστὶ τὸ ΛΞ· ὅλον same angle, LPM , have been subtracted (from LM). ἄρα τὸ ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ. ἐπεὶ Thus, the squares LM and NO are about the same diag- οὐν ὅλον τὸ ΑΚ ἴσον ἐστὶ τοῖς ΛΜ, ΝΞ τετραγώνοις, ὧν onal [Prop. 6.26]. Let PR be their (common) diagonal, τὸ ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ τετραγώνῳ, and let the (rest of the) figure have been drawn. There- λοιπὸν ἄρα τὸ ΑΒ ἴσον ἐστὶ τῷ ΣΤ, τουτέστι τῷ ἀπὸ τῆς fore, since the (rectangle contained) by AF and FG is ΛΝ τετραγώνῳ· ἡ ΛΝ ἄρα δύναται τὸ ΑΒ χωρίον. λέγω, equal to the (square) on EG, thus, proportionally, as AF ὅτι ἡ ΛΝ ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων. is to EG, so EG (is) to FG [Prop. 6.17]. But, as AF (is) ᾿Επεὶ γὰρ ῥητόν ἐστι τὸ ΑΚ καί ἐστιν ἴσον τοῖς ἀπὸ τῶν to EG, so AI is to EK, and as EG (is) to FG, so EK is ΛΟ, ΟΝ τετράγωνοις, τὸ ἄρα συγκείμενον ἐκ τῶν ἀπὸ τῶν to FK [Prop. 6.1]. Thus, EK is the mean proportional to ΛΟ, ΟΝ ῥητόν ἐστιν. πάλιν, ἐπεὶ τὸ ΔΚ μέσον ἐστίν, καί AI and FK [Prop. 5.11]. And MN is also the mean pro- ἐστιν ἴσον τὸ ΔΚ τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ, τὸ ἄρα δὶς ὑπὸ portional to the squares LM and NO [Prop. 10.13 lem.], τῶν ΛΟ, ΟΝ μέσον ἐστίν. καὶ ἐπεὶ ἀσύμμετρον ἐδείχθη τὸ and AI is equal to LM , and FK to NO. EK is thus ΑΙ τῷ ΖΚ, ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΛΟ τετράγωνον also equal to MN . But, DH is equal to EK, and LO is τῷ ἀπὸ τῆς ΟΝ τετραγώνῳ. αἱ ΛΟ, ΟΝ ἄρα δυνάμει εἰσὶν equal to MN [Prop. 1.43]. Thus, the whole of DK is ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν equal to the gnomon UV W and NO. Therefore, since τετραγώνων ῥητόν, τὸ δὲ δὶς ὑπ᾿ αὐτῶν μέσον. ἡ ΛΝ ἄρα the whole of AK is equal to the (sum of the) squares LM ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων· καὶ δύναται τὸ ΑΒ and NO, of which DK is equal to the gnomon UV W χωρίον. and the square NO, the remainder AB is thus equal to ῾Η ἄρα τὸ ΑΒ χωρίον δυναμένη ἐλάσσων ἐστίν· ὅπερ ST—that is to say, to the square on LN . Thus, LN is the ἔδει δεῖξαι. square-root of area AB. I say that LN is the irrational (straight-line which is) called minor. 393 STOIQEIWN iþ. ELEMENTS BOOK 10 For since AK is rational, and is equal to the (sum of the) squares LP and PN , the sum of the (squares) on LP and PN is thus rational. Again, since DK is me- dial, and DK is equal to twice the (rectangle contained) by LP and PN , thus twice the (rectangle contained) by LP and PN is medial. And since AI was shown (to be) incommensurable with FK, the square on LP (is) thus also incommensurable with the square on PN . Thus, LP and PN are (straight-lines which are) incommensurable in square, making the sum of the squares on them ra- tional, and twice the (rectangle contained) by them me- dial. LN is thus the irrational (straight-line) called minor [Prop. 10.76]. And it is the square-root of area AB. Thus, the square-root of area AB is a minor (straight- line). (Which is) the very thing it was required to show.�eþ. Proposition 95 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς πέμπτης, If an area is contained by a rational (straight-line) and ἡ τὸ χωρίον δυναμένη [ἡ] μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά a fifth apotome then the square-root of the area is that ἐστιν. (straight-line) which with a rational (area) makes a me- dial whole. Φ Α ∆ Ζ Η Γ Β Κ Ο Ν Ξ Ε Θ Ι Σ Λ Ρ Μ Τ ΠΥ Χ C A D B K O NE I S L R M T F G H V P QU W Χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ For let the area AB have been contained by the ra- ἀποτομῆς πέμπτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυ- tional (straight-line) AC and the fifth apotome AD. I ναμένη [ἡ] μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν. say that the square-root of area AB is that (straight-line) ῎Εστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ· αἱ ἄρα which with a rational (area) makes a medial whole. ΑΗ, ΗΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ προ- For let DG be an attachment to AD. Thus, AG and σαρμόζουσα ἡ ΗΔ σύμμετρός ἐστι μήκει τῇ ἐκκειμένῃ DG are rational (straight-lines which are) commensu- ῥητῇ τῇ ΑΓ, ἡ δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς rable in square only [Prop. 10.73], and the attachment ΔΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. ἐὰν ἄρα GD is commensurable in length the the (previously) laid τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ πα- down rational (straight-line) AC, and the square on the ραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν whole, AG, is greater than (the square on) the attach- διελεῖ. τετμήσθω οὖν ἡ ΔΗ δίχα κατὰ τὸ Ε σημεῖον, καὶ ment, DG, by the (square) on (some straight-line) incom- τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον mensurable (in length) with (AG) [Def. 10.15]. Thus, if εἴδει τετραγώνῳ καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ· ἀσύμμετρος (some area), equal to the fourth part of the (square) on ἄρα ἐστὶν ἡ ΑΖ τῇ ΖΗ μήκει. καὶ ἐπεὶ ἀσύμμετρός ἐστὶν ἡ DG, is applied to AG, falling short by a square figure, ΑΗ τῇ ΓΑ μήκει, καί εἰσιν ἀμφότεραι ῥηταί, μέσον ἄρα ἐστὶ then it divides (AG) into (parts which are) incommensu- τὸ ΑΚ. πάλιν, ἐπεὶ ῥητή ἐστιν ἡ ΔΗ καὶ σύμμετρος τῇ ΑΓ rable (in length) [Prop. 10.18]. Therefore, let DG have μήκει, ῥητόν ἐστι τὸ ΔΚ. been divided in half at point E, and let (some area), equal Συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ to the (square) on EG, have been applied to AG, falling δὲ ΖΚ ἴσον τετράγωνον ἀφῃρήσθω τὸ ΝΞ περὶ τὴν αὐτὴν short by a square figure, and let it be the (rectangle con- γωνίαν τὴν ὑπὸ ΛΟΜ· περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι tained) by AF and FG. Thus, AF is incommensurable τὰ ΛΜ, ΝΞ τετράγωνα. ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ in length with FG. And since AG is incommensurable 394 STOIQEIWN iþ. ELEMENTS BOOK 10 καταγεγράφθω τὸ σχῆμα. ὁμοίως δὴ δείξομεν, ὅτι ἡ ΛΝ in length with CA, and both are rational (straight-lines), δύναται τὸ ΑΒ χωρίον. λέγω, ὅτι ἡ ΛΝ ἡ μετὰ ῥητοῦ μέσον AK is thus a medial (area) [Prop. 10.21]. Again, since τὸ ὅλον ποιοῦσά ἐστιν. DG is rational, and commensurable in length with AC, ᾿Επεὶ γὰρ μέσον ἐδείχθη τὸ ΑΚ καί ἐστιν ἴσον τοῖς ἀπὸ DK is a rational (area) [Prop. 10.19]. τῶν ΛΟ, ΟΝ, τὸ ἄρα συγκείμενον ἐκ τῶν ἀπὸ τῶν ΛΟ, ΟΝ Therefore, let the square LM , equal to AI, have been μέσον ἐστίν. πάλιν, ἐπεὶ ῥητόν ἐστι τὸ ΔΚ καί ἐστιν ἴσον constructed. And let the square NO, equal to FK, (and) τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ, καὶ αὑτὸ ῥητόν ἐστιν. καὶ ἐπεὶ about the same angle, LPM , have been subtracted (from ἀσύμμετρόν ἐστι τὸ ΑΙ τῷ ΖΚ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ NO). Thus, the squares LM and NO are about the same ἀπὸ τῆς ΛΟ τῷ ἀπὸ τῆς ΟΝ· αἱ ΛΟ, ΟΝ ἄρα δυνάμει εἰσὶν diagonal [Prop. 6.26]. Let PR be their (common) diag- ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν onal, and let (the rest of) the figure have been drawn. τετραγώνων μέσον, τὸ δὲ δὶς ὑπ᾿ αὐτῶν ῥητόν. ἡ λοιπὴ ἄρα So, similarly (to the previous propositions), we can show ἡ ΛΝ ἄλογός ἐστιν ἡ καλουμένη μετὰ ῥητοῦ μέσον τὸ ὅλον that LN is the square-root of area AB. I say that LN is ποιοῦσα· καὶ δύναται τὸ ΑΒ χωρίον. that (straight-line) which with a rational (area) makes a ῾Η τὸ ΑΒ ἄρα χωρίον δυναμένη μετὰ ῥητοῦ μέσον τὸ medial whole. ὅλον ποιοῦσά ἐστιν· ὅπερ ἔδει δεῖξαι. For since AK was shown (to be) a medial (area), and is equal to (the sum of) the squares on LP and PN , the sum of the (squares) on LP and PN is thus medial. Again, since DK is rational, and is equal to twice the (rectangle contained) by LP and PN , (the latter) is also rational. And since AI is incommensurable with FK, the (square) on LP is thus also incommensurable with the (square) on PN . Thus, LP and PN are (straight-lines which are) incommensurable in square, making the sum of the squares on them medial, and twice the (rectangle contained) by them rational. Thus, the remainder LN is the irrational (straight-line) called that which with a ra- tional (area) makes a medial whole [Prop. 10.77]. And it is the square-root of area AB. Thus, the square-root of area AB is that (straight- line) which with a rational (area) makes a medial whole. (Which is) the very thing it was required to show.��þ. Proposition 96 ᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς ἕκτης, If an area is contained by a rational (straight-line) and ἡ τὸ χωρίον δυναμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά a sixth apotome then the square-root of the area is that ἐστιν. (straight-line) which with a medial (area) makes a medial whole. Φ Α ∆ Ζ Η Γ Β Κ Ο Ν Ξ Ε Θ Ι Σ Λ Ρ Μ Τ ΠΥ Χ C A D B K O NE I S L R M T F G H V P QU W Χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ For let the area AB have been contained by the ra- ἀποτομῆς ἕκτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυ- tional (straight-line) AC and the sixth apotome AD. I ναμένη [ἡ] μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν. say that the square-root of area AB is that (straight-line) ῎Εστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ· αἱ ἄρα ΑΗ, which with a medial (area) makes a medial whole. ΗΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα For let DG be an attachment to AD. Thus, AG and 395 STOIQEIWN iþ. ELEMENTS BOOK 10 αὐτῶν σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ μήκει, ἡ GD are rational (straight-lines which are) commensu- δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται rable in square only [Prop. 10.73], and neither of them is τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει. ἐπεὶ οὖν ἡ ΑΗ τῆς ΗΔ commensurable in length with the (previously) laid down μεῖζον δύναται τῷ ἀπὸ ἁσυμμέτρου ἐαυτῇ μήκει, ἑὰν ἄρα rational (straight-line) AC, and the square on the whole, τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ πα- AG, is greater than (the square on) the attachment, DG, ραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν by the (square) on (some straight-line) incommensurable διελεῖ. τετμήσθω οὖν ἡ ΔΗ δίχα κατὰ τὸ Ε [σημεῖον], καὶ in length with (AG) [Def. 10.16]. Therefore, since the τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον square on AG is greater than (the square on) GD by εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ· ἀσύμμετρος the (square) on (some straight-line) incommensurable in ἄρα ἐστὶν ἡ ΑΖ τῇ ΖΗ μήκει. ὡς δὲ ἡ ΑΖ πρὸς τὴν ΖΗ, length with (AG), thus if (some area), equal to the fourth οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ ΖΚ· ἀσύμμετρον ἄρα ἐστὶ τὸ part of square on DG, is applied to AG, falling short by ΑΙ τῷ ΖΚ. καὶ ἐπεὶ αἱ ΑΗ, ΑΓ ῥηταί εἰσι δυνάμει μόνον a square figure, then it divides (AG) into (parts which σύμμετροι, μέσον ἐστὶ τὸ ΑΚ. πάλιν, ἐπεὶ αἱ ΑΓ, ΔΗ ῥηταί are) incommensurable (in length) [Prop. 10.18]. There- εἰσι καὶ ἀσύμμετροι μήκει, μέσον ἐστὶ καὶ τὸ ΔΚ. ἐπεὶ οὖν fore, let DG have been cut in half at [point] E. And let αἱ ΑΗ, ΗΔ δυνάμει μόνον σύμμετροί εἰσιν, ἀσύμμετρος ἄρα (some area), equal to the (square) on EG, have been ap- ἐστὶν ἡ ΑΗ τῇ ΗΔ μήκει. ὡς δὲ ἡ ΑΗ πρὸς τὴν ΗΔ, οὕτως plied to AG, falling short by a square figure. And let it ἐστὶ τὸ ΑΚ πρὸς τὸ ΚΔ· ἀσύμμετρον ἄρα ἐστὶ τὸ ΑΚ τῷ be the (rectangle contained) by AF and FG. AF is thus ΚΔ. incommensurable in length with FG. And as AF (is) to Συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, FG, so AI is to FK [Prop. 6.1]. Thus, AI is incommen- τῷ δὲ ΖΚ ἴσον ἀφῃρήσθω περὶ τὴν αὐτὴν γωνίαν τὸ ΝΞ· surable with FK [Prop. 10.11]. And since AG and AC περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ τετράγωνα. are rational (straight-lines which are) commensurable in ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα. square only, AK is a medial (area) [Prop. 10.21]. Again, ὁμοίως δὴ τοῖς ἐπάνω δείξομεν, ὅτι ἡ ΛΝ δύναται τὸ ΑΒ since AC and DG are rational (straight-lines which are) χωρίον. λέγω, ὅτι ἡ ΛΝ [ἡ] μετὰ μέσου μέσον τὸ ὅλον incommensurable in length, DK is also a medial (area) ποιοῦσά ἐστιν. [Prop. 10.21]. Therefore, since AG and GD are com- ᾿Επεὶ γὰρ μέσον ἐδείχθη τὸ ΑΚ καί ἐστιν ἴσον τοῖς ἀπὸ mensurable in square only, AG is thus incommensurable τῶν ΛΟ, ΟΝ, τὸ ἄρα συγκείμενον ἐκ τῶν ἀπὸ τῶν ΛΟ, ΟΝ in length with GD. And as AG (is) to GD, so AK is to μέσον ἐστίν. πάλιν, ἐπεὶ μέσον ἐδείχθη τὸ ΔΚ καί ἐστιν KD [Prop. 6.1]. Thus, AK is incommensurable with KD ἴσον τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ, καὶ τὸ δὶς ὑπὸ τῶν ΛΟ, ΟΝ [Prop. 10.11]. μέσον ἐστίν. καὶ ἐπεὶ ἀσύμμετρον ἐδείχθη τὸ ΑΚ τῷ ΔΚ, Therefore, let the square LM , equal to AI, have been ἀσύμμετρα [ἄρα] ἐστὶ καὶ τὰ ἀπὸ τῶν ΛΟ, ΟΝ τετράγωνα constructed. And let NO, equal to FK, (and) about the τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ ΑΙ τῷ same angle, have been subtracted (from LM). Thus, ΖΚ, ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΛΟ τῷ ἀπὸ τῆς ΟΝ· the squares LM and NO are about the same diagonal αἱ ΛΟ, ΟΝ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε [Prop. 6.26]. Let PR be their (common) diagonal, and συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον καὶ τὸ δὶς let (the rest of) the figure have been drawn. So, similarly ὑπ᾿ αὐτῶν μέσον ἔτι τε τὰ ἀπ᾿ αὐτῶν τετράγωνα ἀσύμμετρα to the above, we can show that LN is the square-root of τῷ δὶς ὑπ᾿ αὐτῶν. ἡ ἄρα ΛΝ ἄλογός ἐστιν ἡ καλουμέμη μετὰ area AB. I say that LN is that (straight-line) which with μέσου μέσον τὸ ὅλον ποιοῦσα· καὶ δύναται τὸ ΑΒ χωρίον. a medial (area) makes a medial whole. ῾Η ἄρα τὸ χωρίον δυναμένη μετὰ μέσου μέσον τὸ ὅλον For since AK was shown (to be) a medial (area), and ποιοῦσά ἐστιν· ὅπερ ἔδει δεῖξαι. is equal to the (sum of the) squares on LP and PN , the sum of the (squares) on LP and PN is medial. Again, since DK was shown (to be) a medial (area), and is equal to twice the (rectangle contained) by LP and PN , twice the (rectangle contained) by LP and PN is also medial. And since AK was shown (to be) incommensu- rable with DK, [thus] the (sum of the) squares on LP and PN is also incommensurable with twice the (rect- angle contained) by LP and PN . And since AI is in- commensurable with FK, the (square) on LP (is) thus also incommensurable with the (square) on PN . Thus, LP and PN are (straight-lines which are) incommensu- 396 STOIQEIWN iþ. ELEMENTS BOOK 10 rable in square, making the sum of the squares on them medial, and twice the (rectangle contained) by medial, and, furthermore, the (sum of the) squares on them in- commensurable with twice the (rectangle contained) by them. Thus, LN is the irrational (straight-line) called that which with a medial (area) makes a medial whole [Prop. 10.78]. And it is the square-root of area AB. Thus, the square-root of area (AB) is that (straight- line) which with a medial (area) makes a medial whole. (Which is) the very thing it was required to show.�zþ. Proposition 97 Τὸ ἀπὸ ἀποτομῆς παρὰ ῥητὴν παραβαλλόμενον πλάτος The (square) on an apotome, applied to a rational ποιεῖ ἀποτομὴν πρώτην. (straight-line), produces a first apotome as breadth. N D E L G Z M A B H K JX N D E L M A B C F G K HO ῎Εστω ἀποτομὴ ὴ ΑΒ, ῥητὴ δὲ ἡ ΓΔ, καὶ τῷ ἀπὸ τῆς Let AB be an apotome, and CD a rational (straight- ΑΒ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΕ πλάτος ποιοῦν line). And let CE, equal to the (square) on AB, have τὴν ΓΖ· λέγω, ὅτι ἡ ΓΖ ἀποτομή ἐστι πρώτη. been applied to CD, producing CF as breadth. I say that ῎Εστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ· αἱ ἄρα ΑΗ, CF is a first apotome. ΗΒ ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ τῷ μὲν ἀπὸ For let BG be an attachment to AB. Thus, AG and τῆς ΑΗ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΘ, τῷ δὲ GB are rational (straight-lines which are) commensu- ἀπὸ τῆς ΒΗ τὸ ΚΛ. ὅλον ἄρα τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ rable in square only [Prop. 10.73]. And let CH , equal τῶν ΑΗ, ΗΒ· ὧν τὸ ΓΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ· λοιπὸν to the (square) on AG, and KL, (equal) to the (square) ἄρα τὸ ΖΛ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. τετμήσθω on BG, have been applied to CD. Thus, the whole of ἡ ΖΜ δίχα κατὰ τὸ Ν σημεῖον, καὶ ἤχθω διὰ τοῦ Ν τῇ ΓΔ CL is equal to the (sum of the squares) on AG and GB, παράλληλος ἡ ΝΞ· ἑκάτερον ἄρα τῶν ΖΞ, ΛΝ ἴσον ἐστὶ τῷ of which CE is equal to the (square) on AB. The re- ὑπὸ τῶν ΑΗ, ΗΒ. καὶ ἐπεὶ τὰ ἀπὸ τῶν ΑΗ, ΗΒ ῥητά ἐστιν, mainder FL is thus equal to twice the (rectangle con- καί ἐστι τοῖς ἀπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΔΜ, ῥητὸν ἄρα tained) by AG and GB [Prop. 2.7]. Let FM have been ἐστὶ τὸ ΔΜ. καὶ παρὰ ῥητὴν τὴν ΓΔ παραβέβληται πλάτος cut in half at point N . And let NO have been drawn ποιοῦν τὴν ΓΜ· ῥητὴ ἄρα ἐστὶν ἡ ΓΜ καὶ σύμμετρος τῇ ΓΔ through N , parallel to CD. Thus, FO and LN are each μήκει. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ, καὶ equal to the (rectangle contained) by AG and GB. And τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΖΛ, μέσον ἄρα τὸ ΖΛ. καὶ since the (sum of the squares) on AG and GB is rational, παρὰ ῥητὴν τὴν ΓΔ παράκειται πλάτος ποιοῦν τὴν ΖΜ· ῥητὴ and DM is equal to the (sum of the squares) on AG and ἄρα ἐστὶν ἡ ΖΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ τὰ GB, DM is thus rational. And it has been applied to the μὲν ἀπὸ τῶν ΑΗ, ΗΒ ῥητά ἐστιν, τὸ δὲ δὶς ὑπὸ τῶν ΑΗ, rational (straight-line) CD, producing CM as breadth. ΗΒ μέσον, ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΗ, ΗΒ τῷ δὶς Thus, CM is rational, and commensurable in length with ὑπὸ τῶν ΑΗ, ΗΒ. καὶ τοῖς μὲν ἀπὸ τῶν ΑΗ, ΗΒ ἴσον ἐστὶ CD [Prop. 10.20]. Again, since twice the (rectangle con- τὸ ΓΛ, τῷ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ τὸ ΖΛ· ἀσύμμετρον ἄρα tained) by AG and GB is medial, and FL (is) equal to ἐστὶ τὸ ΔΜ τῷ ΖΛ. ὡς δὲ τὸ ΔΜ πρὸς τὸ ΖΛ, οὕτως ἐστὶν twice the (rectangle contained) by AG and GB, FL (is) ἡ ΓΜ πρὸς τὴν ΖΜ. ἀσύμμετρος ἄρα ἐστὶν ἡ ΓΜ τῇ ΖΜ thus a medial (area). And it is applied to the rational μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ἄρα ΓΜ, ΜΖ ῥηταί εἰσι (straight-line) CD, producing FM as breadth. FM is 397 STOIQEIWN iþ. ELEMENTS BOOK 10 δυνάμει μόνον σύμμετροι· ἡ ΓΖ ἄρα ἀποτομή ἐστιν. λέγω thus rational, and incommensurable in length with CD δή, ὅτι καὶ πρώτη. [Prop. 10.22]. And since the (sum of the squares) on AG ᾿Επεὶ γὰρ τῶν ἀπὸ τῶν ΑΗ, ΗΒ μέσον ἀνάλογόν ἐστι and GB is rational, and twice the (rectangle contained) τὸ ὑπὸ τῶν ΑΗ, ΗΒ, καί ἐστι τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον τὸ by AG and GB medial, the (sum of the squares) on AG ΓΘ, τῷ δὲ ἀπὸ τῆς ΒΗ ἴσον τὸ ΚΛ, τῷ δὲ ὐπὸ τῶν ΑΗ, and GB is thus incommensurable with twice the (rectan- ΗΒ τὸ ΝΛ, καὶ τῶν ΓΘ, ΚΛ ἄρα μέσον ἀνάλογόν ἐστι τὸ gle contained) by AG and GB. And CL is equal to the ΝΛ· ἔστιν ἄρα ὡς τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως τὸ ΝΛ πρὸς (sum of the squares) on AG and GB, and FL to twice the τὸ ΚΛ. ἀλλ᾿ ὡς μὲν τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως ἐστὶν ἡ (rectangle contained) by AG and GB. DM is thus incom- ΓΚ πρὸς τὴν ΝΜ· ὡς δὲ τὸ ΝΛ πρὸς τὸ ΚΛ, οὕτως ἐστὶν mensurable with FL. And as DM (is) to FL, so CM is to ἡ ΝΜ πρὸς τὴν ΚΜ· τὸ ἄρα ὑπὸ τῶν ΓΚ, ΚΜ ἴσον ἐστὶ FM [Prop. 6.1]. CM is thus incommensurable in length τῷ ἀπὸ τῆς ΝΜ, τουτέστι τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς with FM [Prop. 10.11]. And both are rational (straight- ΖΜ. καὶ επεὶ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς lines). Thus, CM and MF are rational (straight-lines ΗΒ, σύμμετρόν [ἐστι] καὶ τὸ ΓΘ τῷ ΚΛ. ὡς δὲ τὸ ΓΘ πρὸς which are) commensurable in square only. CF is thus an τὸ ΚΛ, οὕτως ἡ ΓΚ πρὸς τὴν ΚΜ· σύμμετρος ἄρα ἐστὶν ἡ apotome [Prop. 10.73]. So, I say that (it is) also a first ΓΚ τῇ ΚΜ. ἐπεὶ οὖν δύο εὐθεῖαι ἄνισοί εἰσιν αἱ ΓΜ, ΜΖ, (apotome). καὶ τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ ἴσον παρὰ τὴν ΓΜ For since the (rectangle contained) by AG and GB is παραβέβληται ἐλλεῖπον εἴδει τετραγώνῳ τὸ ὑπὸ τῶν ΓΚ, the mean proportional to the (squares) on AG and GB ΚΜ, καί ἐστι σύμμετρος ἡ ΓΚ τῇ ΚΜ, ἡ ἄρα ΓΜ τῆς ΜΖ [Prop. 10.21 lem.], and CH is equal to the (square) on μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. καί ἐστιν ἡ AG, and KL equal to the (square) on BG, and NL to ΓΜ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ μήκει· ἡ ἄρα ΓΖ the (rectangle contained) by AG and GB, NL is thus ἀποτομή ἐστι πρώτη. also the mean proportional to CH and KL. Thus, as Τὸ ἄρα ἀπὸ ἀποτομῆς παρὰ ῥητὴν παραβαλλόμενον CH is to NL, so NL (is) to KL. But, as CH (is) to πλάτος ποιεῖ ἀποτομὴν πρώτην· ὅπερ ἔδει δεῖξαι. NL, so CK is to NM , and as NL (is) to KL, so NM is to KM [Prop. 6.1]. Thus, the (rectangle contained) by CK and KM is equal to the (square) on NM— that is to say, to the fourth part of the (square) on FM [Prop. 6.17]. And since the (square) on AG is commen- surable with the (square) on GB, CH [is] also commen- surable with KL. And as CH (is) to KL, so CK (is) to KM [Prop. 6.1]. CK is thus commensurable (in length) with KM [Prop. 10.11]. Therefore, since CM and MF are two unequal straight-lines, and the (rectangle con- tained) by CK and KM , equal to the fourth part of the (square) on FM , has been applied to CM , falling short by a square figure, and CK is commensurable (in length) with KM , the square on CM is thus greater than (the square on) MF by the (square) on (some straight-line) commensurable in length with (CM) [Prop. 10.17]. And CM is commensurable in length with the (previously) laid down rational (straight-line) CD. Thus, CF is a first apotome [Def. 10.15]. Thus, the (square) on an apotome, applied to a ratio- nal (straight-line), produces a first apotome as breadth. (Which is) the very thing it was required to show.�hþ. Proposition 98 Τὸ ἀπὸ μέσης ἀποτομῆς πρώτης παρὰ ῥητὴν παρα- The (square) on a first apotome of a medial (straight- βαλλόμενον πλάτος ποιεῖ ἀποτομὴν δευτέραν. line), applied to a rational (straight-line), produces a sec- ῎Εστω μέσης ἀποτομὴ πρώτη ἡ ΑΒ, ῥητὴ δὲ ἡ ΓΔ, καὶ τῷ ond apotome as breadth. ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΕ πλάτος Let AB be a first apotome of a medial (straight-line), 398 STOIQEIWN iþ. ELEMENTS BOOK 10 ποιοῦν τὴν ΓΖ· λέγω, ὅτι ἡ ΓΖ ἀποτομή ἐστι δευτέρα. and CD a rational (straight-line). And let CE, equal to ῎Εστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ· αἱ ἄρα ΑΗ, the (square) on AB, have been applied to CD, producing ΗΒ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι. CF as breadth. I say that CF is a second apotome. καὶ τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω For let BG be an attachment to AB. Thus, AG and τὸ ΓΘ πλάτος ποιοῦν τὴν ΓΚ, τῷ δὲ ἀπὸ τῆς ΗΒ ἴσον τὸ GB are medial (straight-lines which are) commensurable ΚΛ πλάτος ποιοῦν τὴν ΚΜ· ὅλον ἄρα τὸ ΓΛ ἴσον ἐστὶ τοῖς in square only, containing a rational (area) [Prop. 10.74]. ἀπὸ τῶν ΑΗ, ΗΒ· μέσον ἄρα καὶ τὸ ΓΛ. καὶ παρὰ ῥητὴν And let CH , equal to the (square) on AG, have been ap- τὴν ΓΔ παράκειται πλάτος ποιοῦν τὴν ΓΜ· ῥητὴ ἄρα ἐστὶν plied to CD, producing CK as breadth, and KL, equal ἡ ΓΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ τὸ ΓΛ ἴσον to the (square) on GB, producing KM as breadth. Thus, ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, ὧν τὸ ἀπὸ τῆς ΑΒ ἴσον ἐστὶ the whole of CL is equal to the (sum of the squares) τῷ ΓΕ, λοιπὸν ἄρα τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ἴσον ἐστὶ τῷ on AG and GB. Thus, CL (is) also a medial (area) ΖΛ. ῥητὸν δέ [ἐστι] τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ· ῥητὸν ἄρα τὸ [Props. 10.15, 10.23 corr.]. And it is applied to the ratio- ΖΛ. καὶ παρὰ ῥητὴν τὴν ΖΕ παράκειται πλάτος ποιοῦν τὴν nal (straight-line) CD, producing CM as breadth. CM ΖΜ· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΜ καὶ σύμμετρος τῇ ΓΔ μήκει. is thus rational, and incommensurable in length with CD ἐπεὶ οὖν τὰ μὲν ἀπὸ τῶν ΑΗ, ΗΒ, τουτέστι τὸ ΓΛ, μέσον [Prop. 10.22]. And since CL is equal to the (sum of the ἐστίν, τὸ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ, τουτέστι τὸ ΖΛ, ῥητόν squares) on AG and GB, of which the (square) on AB ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΛ τῷ ΖΛ. ὡς δὲ τὸ ΓΛ πρὸς τὸ is equal to CE, the remainder, twice the (rectangle con- ΖΛ, οὕτως ἐστὶν ἡ ΓΜ πρὸς τὴν ΖΜ· ἀσύμμετρος ἄρα ἡ tained) by AG and GB, is thus equal to FL [Prop. 2.7]. ΓΜ τῇ ΖΜ μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ἄρα ΓΜ, And twice the (rectangle contained) by AG and GB [is] ΜΖ ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἡ ΓΖ ἄρα ἀποτομή rational. Thus, FL (is) rational. And it is applied to the ἐστιν. λέγω δή, ὅτι καὶ δευτέρα. rational (straight-line) FE, producing FM as breadth. FM is thus also rational, and commensurable in length with CD [Prop. 10.20]. Therefore, since the (sum of the squares) on AG and GB—that is to say, CL—is medial, and twice the (rectangle contained) by AG and GB— that is to say, FL—(is) rational, CL is thus incommen- surable with FL. And as CL (is) to FL, so CM is to FM [Prop. 6.1]. Thus, CM (is) incommensurable in length with FM [Prop. 10.11]. And they are both ra- tional (straight-lines). Thus, CM and MF are rational (straight-lines which are) commensurable in square only. CF is thus an apotome [Prop. 10.73]. So, I say that (it is) also a second (apotome). N D E L G Z M A B H K JX N D E L M A B C F G K HO Τετμήσθω γὰρ ἡ ΖΜ δίχα κατὰ τὸ Ν, καὶ ἤχθω διὰ τοῦ For let FM have been cut in half at N . And let Ν τῇ ΓΔ παράλληλος ἡ ΝΞ· ἑκάτερον ἄρα τῶν ΖΞ, ΝΛ ἴσον NO have been drawn through (point) N , parallel to ἐστὶ τῷ ὑπὸ τῶν ΑΗ, ΗΒ. καὶ ἐπεὶ τῶν ἀπὸ τῶν ΑΗ, ΗΒ CD. Thus, FO and NL are each equal to the (rectan- τετραγώνων μέσον ἀνάλογόν ἐστι τὸ ὑπὸ τῶν ΑΗ, ΗΒ, καί gle contained) by AG and GB. And since the (rectan- ἐστιν ἴσον τὸ μὲν ἀπὸ τῆς ΑΗ τῷ ΓΘ, τὸ δὲ ὑπὸ τῶν ΑΗ, gle contained) by AG and GB is the mean proportional ΗΒ τῷ ΝΛ, τὸ δὲ ἀπὸ τῆς ΒΗ τῷ ΚΛ, καὶ τῶν ΓΘ, ΚΛ to the squares on AG and GB [Prop. 10.21 lem.], and ἄρα μέσον ἀνάλογόν ἐστι τὸ ΝΛ· ἔστιν ἄρα ὡς τὸ ΓΘ πρὸς the (square) on AG is equal to CH , and the (rectangle τὸ ΝΛ, οὕτως τὸ ΝΛ πρὸς τὸ ΚΛ. ἀλλ᾿ ὡς μὲν τὸ ΓΘ πρὸς contained) by AG and GB to NL, and the (square) on 399 STOIQEIWN iþ. ELEMENTS BOOK 10 τὸ ΝΛ, οὕτως ἐστὶν ἡ ΓΚ πρὸς τὴν ΝΜ, ὡς δὲ τὸ ΝΛ πρὸς BG to KL, NL is thus also the mean proportional to τὸ ΚΛ, οὕτως ἐστὶν ἡ ΝΜ πρὸς τὴν ΜΚ· ὡς ἄρα ἡ ΓΚ CH and KL. Thus, as CH is to NL, so NL (is) to KL πρὸς τὴν ΝΜ, οὕτως ἐστὶν ἡ ΝΜ πρὸς τὴν ΚΜ· τὸ ἄρα [Prop. 5.11]. But, as CH (is) to NL, so CK is to NM , ὑπὸ τῶν ΓΚ, ΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΝΜ, τουτέστι τῷ and as NL (is) to KL, so NM is to MK [Prop. 6.1]. τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ [καὶ ἐπεὶ σύμμετρόν ἐστι τὸ Thus, as CK (is) to NM , so NM is to KM [Prop. 5.11]. ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς ΒΗ, σύμμετρόν ἐστι καὶ τὸ ΓΘ τῷ The (rectangle contained) by CK and KM is thus equal ΚΛ, τουτέστιν ἡ ΓΚ τῇ ΚΜ]. ἐπεὶ οὖν δύο εὐθεῖαι ἄνισοί to the (square) on NM [Prop. 6.17]—that is to say, to εἰσιν αἱ ΓΜ, ΜΖ, καὶ τῷ τετάτρῳ μέρει τοῦ ἀπὸ τῆς ΜΖ the fourth part of the (square) on FM [and since the ἴσον παρὰ τὴν μείζονα τὴν ΓΜ παραβέβληται ἐλλεῖπον εἴδει (square) on AG is commensurable with the (square) on τετραγώνῳ τὸ ὑπὸ τῶν ΓΚ, ΚΜ καὶ εἰς σύμμετρα αὐτὴν BG, CH is also commensurable with KL—that is to say, διαιρεῖ, ἡ ἄρα ΓΜ τῆς ΜΖ μεῖζον δύναται τῷ ἀπὸ συμμέτρου CK with KM]. Therefore, since CM and MF are two ἑαυτῇ μήκει. καί ἐστιν ἡ προσαρμόζουσα ἡ ΖΜ σύμμετρος unequal straight-lines, and the (rectangle contained) by μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ· ἡ ἄρα ΓΖ ἀποτομή ἐστι CK and KM , equal to the fourth part of the (square) δευτέρα. on MF , has been applied to the greater CM , falling Τὸ ἄρα ἀπὸ μέσης ἀποτομῆς πρώτης παρὰ ῥητὴν πα- short by a square figure, and divides it into commensu- ραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν δευτέραν· ὅπερ ἔδει rable (parts), the square on CM is thus greater than (the δεῖξαι. square on) MF by the (square) on (some straight-line) commensurable in length with (CM) [Prop. 10.17]. The attachment FM is also commensurable in length with the (previously) laid down rational (straight-line) CD. CF is thus a second apotome [Def. 10.16]. Thus, the (square) on a first apotome of a medial (straight-line), applied to a rational (straight-line), pro- duces a second apotome as breadth. (Which is) the very thing it was required to show.�jþ. Proposition 99 Τὸ ἀπὸ μέσης ἀποτομῆς δευτέρας παρὰ ῥητὴν παρα- The (square) on a second apotome of a medial βαλλόμενον πλάτος ποιεῖ ἀποτομὴν τρίτην. (straight-line), applied to a rational (straight-line), pro- duces a third apotome as breadth. N D E L G Z M A B H K JX N D E L M A B C F G K HO ῎Εστω μέσης ἀποτομὴ δευτέρα ἡ ΑΒ, ῥητὴ δὲ ἡ ΓΔ, καὶ Let AB be the second apotome of a medial (straight- τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΕ line), and CD a rational (straight-line). And let CE, πλάτος ποιοῦν τὴν ΓΖ· λέγω, ὅτι ἡ ΓΖ ἀποτομή ἐστι τρίτη. equal to the (square) on AB, have been applied to CD, ῎Εστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ· αἱ ἄρα ΑΗ, ΗΒ producing CF as breadth. I say that CF is a third apo- μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον περιέχουσαι. καὶ tome. τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω τὸ ΓΘ For let BG be an attachment to AB. Thus, AG and πλάτος ποιοῦν τὴν ΓΚ, τῷ δὲ ἀπὸ τῆς ΒΗ ἴσον παρὰ τὴν ΚΘ GB are medial (straight-lines which are) commensurable παραβεβλήσθω τὸ ΚΛ πλάτος ποιοῦν τὴν ΚΜ· ὅλον ἄρα τὸ in square only, containing a medial (area) [Prop. 10.75]. ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ [καί ἐστι μέσα τὰ ἀπὸ And let CH , equal to the (square) on AG, have been τῶν ΑΗ, ΗΒ]· μέσον ἄρα καὶ τὸ ΓΛ. καὶ παρὰ ῥητὴν τὴν applied to CD, producing CK as breadth. And let KL, 400 STOIQEIWN iþ. ELEMENTS BOOK 10 ΓΔ παραβέβληται πλάτος ποιοῦν τὴν ΓΜ· ῥητὴ ἄρα ἐστὶν equal to the (square) on BG, have been applied to KH , ἡ ΓΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ ὅλον τὸ ΓΛ producing KM as breadth. Thus, the whole of CL is ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, ὧν τὸ ΓΕ ἴσον ἐστὶ τῷ equal to the (sum of the squares) on AG and GB [and ἀπὸ τῆς ΑΒ, λοιπὸν ἄρα τὸ ΛΖ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν the (sum of the squares) on AG and GB is medial]. CL ΑΗ, ΗΒ. τετμήσθω οὖν ἡ ΖΜ δίχα κατὰ τὸ Ν σημεῖον, (is) thus also medial [Props. 10.15, 10.23 corr.]. And it καὶ τῇ ΓΔ παράλληλος ἤχθω ἡ ΝΞ· ἑκάτερον ἄρα τῶν ΖΞ, has been applied to the rational (straight-line) CD, pro- ΝΛ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΗ, ΗΒ. μέσον δὲ τὸ ὑπὸ τῶν ducing CM as breadth. Thus, CM is rational, and incom- ΑΗ, ΗΒ· μέσον ἄρα ἐστὶ καὶ τὸ ΖΛ. καὶ παρὰ ῥητὴν τὴν mensurable in length with CD [Prop. 10.22]. And since ΕΖ παράκειται πλάτος ποιοῦν τὴν ΖΜ· ῥητὴ ἄρα καὶ ἡ ΖΜ the whole of CL is equal to the (sum of the squares) on καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ αἱ ΑΗ, ΗΒ δυνάμει AG and GB, of which CE is equal to the (square) on μόνον εἰσὶ σύμμετροι, ἀσύμμετρος ἄρα [ἐστὶ] μήκει ἡ ΑΗ AB, the remainder LF is thus equal to twice the (rect- τῇ ΗΒ· ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΗ τῷ ὑπὸ angle contained) by AG and GB [Prop. 2.7]. Therefore, τῶν ΑΗ, ΗΒ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΗ σύμμετρά ἐστι τὰ let FM have been cut in half at point N . And let NO ἀπὸ τῶν ΑΗ, ΗΒ, τῷ δὲ ὑπὸ τῶν ΑΗ, ΗΒ τὸ δὶς ὑπὸ τῶν have been drawn parallel to CD. Thus, FO and NL are ΑΗ, ΗΒ· ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΗ, ΗΒ τῷ δὶς each equal to the (rectangle contained) by AG and GB. ὑπὸ τῶν ΑΗ, ΗΒ. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΑΗ, ΗΒ ἴσον And the (rectangle contained) by AG and GB (is) me- ἐστὶ τὸ ΓΛ, τῷ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ἴσον ἐστὶ τὸ ΖΛ· dial. Thus, FL is also medial. And it is applied to the ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΛ τῷ ΖΛ. ὡς δὲ τὸ ΓΛ πρὸς τὸ rational (straight-line) EF , producing FM as breadth. ΖΛ, οὕτως ἐστὶν ἡ ΓΜ πρὸς τὴν ΖΜ· ἀσύμμετρος ἄρα ἐστὶν FM is thus rational, and incommensurable in length with ἡ ΓΜ τῇ ΖΜ μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ἄρα ΓΜ, CD [Prop. 10.22]. And since AG and GB are commen- ΜΖ ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἀποτομὴ ἄρα ἐστὶν surable in square only, AG [is] thus incommensurable in ἡ ΓΖ. λέγω δή, ὅτι καὶ τρίτη. length with GB. Thus, the (square) on AG is also incom- ᾿Επεὶ γὰρ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς mensurable with the (rectangle contained) by AG and ΗΒ, σύμμετρον ἄρα καὶ τὸ ΓΘ τῷ ΚΛ· ὥστε καὶ ἡ ΓΚ τῇ GB [Props. 6.1, 10.11]. But, the (sum of the squares) ΚΜ. καὶ ἐπεὶ τῶν ἀπὸ τῶν ΑΗ, ΗΒ μέσον ἀνάλογόν ἐστι on AG and GB is commensurable with the (square) on τὸ ὑπὸ τῶν ΑΗ, ΗΒ, καί ἐστι τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον AG, and twice the (rectangle contained) by AG and GB τὸ ΓΘ, τῷ δὲ ἀπὸ τῆς ΗΒ ἴσον τὸ ΚΛ, τῷ δὲ ὑπὸ τῶν with the (rectangle contained) by AG and GB. The ΑΗ, ΗΒ ἴσον τὸ ΝΛ, καὶ τῶν ΓΘ, ΚΛ ἄρα μέσον ἀνάλογόν (sum of the squares) on AG and GB is thus incommen- ἐστι τὸ ΝΛ· ἔστιν ἄρα ὡς τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως τὸ surable with twice the (rectangle contained) by AG and ΝΛ πρὸς τὸ ΚΛ. ἀλλ᾿ ὡς μὲν τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως GB [Prop. 10.13]. But, CL is equal to the (sum of the ἐστὶν ἡ ΓΚ πρὸς τὴν ΝΜ, ὡς δὲ τὸ ΝΛ πρὸς τὸ ΚΛ, οὕτως squares) on AG and GB, and FL is equal to twice the ἐστὶν ἡ ΝΜ πρὸς τὴν ΚΜ· ὡς ἄρα ἡ ΓΚ πρὸς τὴν ΜΝ, (rectangle contained) by AG and GB. Thus, CL is in- οὕτως ἐστὶν ἡ ΜΝ πρὸς τὴν ΚΜ· τὸ ἄρα ὑπὸ τῶν ΓΚ, ΚΜ commensurable with FL. And as CL (is) to FL, so CM ἴσον ἐστὶ τῷ [ἀπὸ τῆς ΜΝ, τουτέστι τῷ] τετάρτῳ μέρει τοῦ is to FM [Prop. 6.1]. CM is thus incommensurable in ἀπὸ τῆς ΖΜ. ἐπεὶ οὖν δύο εὐθεῖαι ἄνισοί εἰσιν αἱ ΓΜ, ΜΖ, length with FM [Prop. 10.11]. And they are both ra- καὶ τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ ἴσον παρὰ τὴν ΓΜ tional (straight-lines). Thus, CM and MF are rational παραβέβληται ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς σύμμετρα (straight-lines which are) commensurable in square only. αὐτὴν διαιρεῖ, ἡ ΓΜ ἄρα τῆς ΜΖ μεῖζον δύναται τῷ ἀπὸ CF is thus an apotome [Prop. 10.73]. So, I say that (it συμμέτρου ἑαυτῇ. καὶ οὐδετέρα τῶν ΓΜ, ΜΖ σύμμετρός is) also a third (apotome). ἐστι μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ· ἡ ἄρα ΓΖ ἀποτομή ἐστι For since the (square) on AG is commensurable with τρίτη. the (square) on GB, CH (is) thus also commensu- Τὸ ἄρα ἀπὸ μέσης ἀποτομῆς δευτέρας παρὰ ῥητὴν παρα- rable with KL. Hence, CK (is) also (commensurable βαλλόμενον πλάτος ποιεῖ ἀποτομὴν τρίτην· ὅπερ ἔδει δεῖξαι. in length) with KM [Props. 6.1, 10.11]. And since the (rectangle contained) by AG and GB is the mean propor- tional to the (squares) on AG and GB [Prop. 10.21 lem.], and CH is equal to the (square) on AG, and KL equal to the (square) on GB, and NL equal to the (rectangle contained) by AG and GB, NL is thus also the mean proportional to CH and KL. Thus, as CH is to NL, so NL (is) to KL. But, as CH (is) to NL, so CK is to NM , and as NL (is) to KL, so NM (is) to KM [Prop. 6.1]. 401 STOIQEIWN iþ. ELEMENTS BOOK 10 Thus, as CK (is) to MN , so MN is to KM [Prop. 5.11]. Thus, the (rectangle contained) by CK and KM is equal to the [(square) on MN—that is to say, to the] fourth part of the (square) on FM [Prop. 6.17]. Therefore, since CM and MF are two unequal straight-lines, and (some area), equal to the fourth part of the (square) on FM , has been applied to CM , falling short by a square figure, and divides it into commensurable (parts), the square on CM is thus greater than (the square on) MF by the (square) on (some straight-line) commensurable (in length) with (CM) [Prop. 10.17]. And neither of CM and MF is commensurable in length with the (pre- viously) laid down rational (straight-line) CD. CF is thus a third apotome [Def. 10.13]. Thus, the (square) on a second apotome of a medial (straight-line), applied to a rational (straight-line), pro- duces a third apotome as breadth. (Which is) the very thing it was required to show.rþ. Proposition 100 Τὸ ἀπὸ ἐλάσσονος παρὰ ῥητὴν παραβαλλόμενον πλάτος The (square) on a minor (straight-line), applied to ποιεῖ ἀποτομὴν τετάρτην. a rational (straight-line), produces a fourth apotome as breadth. N D E L G Z M A B H K JX N D E L M A B C F G K HO ῎Εστω ἐλάσσων ἡ ΑΒ, ῥητὴ δὲ ἡ ΓΔ, καὶ τῷ ἀπὸ τῆς Let AB be a minor (straight-line), and CD a rational ΑΒ ἴσον παρὰ ῥητὴν τὴν ΓΔ παραβεβλήσθω τὸ ΓΕ πλάτος (straight-line). And let CE, equal to the (square) on AB, ποιοῦν τὴν ΓΖ· λέγω, ὅτι ἡ ΓΖ ἀποτομή ἐστι τετάρτη. have been applied to the rational (straight-line) CD, pro- ῎Εστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ· αἱ ἄρα ΑΗ, ducing CF as breadth. I say that CF is a fourth apotome. ΗΒ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον For let BG be an attachment to AB. Thus, AG and ἐκ τῶν ἀπὸ τῶν ΑΗ, ΗΒ τετραγώνων ῥητόν, τὸ δὲ δὶς ὑπὸ GB are incommensurable in square, making the sum of τῶν ΑΗ, ΗΒ μέσον. καὶ τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον παρὰ the squares on AG and GB rational, and twice the (rect- τὴν ΓΔ παραβεβλήσθω τὸ ΓΘ πλάτος ποιοῦν τὴν ΓΚ, τῷ angle contained) by AG and GB medial [Prop. 10.76]. δὲ ἀπὸ τῆς ΒΗ ἴσον τὸ ΚΛ πλάτος ποιοῦν τὴν ΚΜ· ὅλον And let CH , equal to the (square) on AG, have been ap- ἄρα τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ. καί ἐστι τὸ plied to CD, producing CK as breadth, and KL, equal συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΗ, ΗΒ ῥητόν· ῥητὸν ἄρα to the (square) on BG, producing KM as breadth. Thus, ἐστὶ καὶ τὸ ΓΛ. καὶ παρὰ ῥητὴν τὴν ΓΔ παράκειται πλάτος the whole of CL is equal to the (sum of the squares) on ποιοῦν τὴν ΓΜ· ῥητὴ ἄρα καὶ ἡ ΓΜ καὶ σύμμετρος τῇ ΓΔ AG and GB. And the sum of the (squares) on AG and μήκει. καὶ ἐπεὶ ὅλον τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, GB is rational. CL is thus also rational. And it is ap- ὧν τὸ ΓΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ, λοιπὸν ἄρα τὸ ΖΛ ἴσον plied to the rational (straight-line) CD, producing CM ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. τετμήσθω οὖν ἡ ΖΜ δίχα as breadth. Thus, CM (is) also rational, and commen- κατὰ τὸ Ν σημεῖον, καὶ ἤχθω δὶα τοῦ Ν ὁποτέρᾳ τῶν ΓΔ, surable in length with CD [Prop. 10.20]. And since the 402 STOIQEIWN iþ. ELEMENTS BOOK 10 ΜΛ παράλληλος ἡ ΝΞ· ἑκάτερον ἄρα τῶν ΖΞ, ΝΛ ἴσον ἐστὶ whole of CL is equal to the (sum of the squares) on AG τῷ ὑπὸ τῶν ΑΗ, ΗΒ. καὶ ἐπεὶ τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ μέσον and GB, of which CE is equal to the (square) on AB, ἐστὶ καί ἐστιν ἴσον τῷ ΖΛ, καὶ τὸ ΖΛ ἄρα μέσον ἐστίν. καὶ the remainder FL is thus equal to twice the (rectangle παρὰ ῥητὴν τὴν ΖΕ παράκειται πλάτος ποιοῦν τὴν ΖΜ· ῥητὴ contained) by AG and GB [Prop. 2.7]. Therefore, let ἄρα ἐστὶν ἡ ΖΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ ἐπεὶ τὸ FM have been cut in half at point N . And let NO have μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΗ, ΗΒ ῥητόν ἐστιν, τὸ been drawn through N , parallel to either of CD or ML. δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ μέσον, ἀσύμμετρα [ἄρα] ἐστὶ τὰ ἀπὸ Thus, FO and NL are each equal to the (rectangle con- τῶν ΑΗ, ΗΒ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. ἴσον δέ [ἐστι] τὸ ΓΛ tained) by AG and GB. And since twice the (rectangle τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, τῷ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ contained) by AG and GB is medial, and is equal to FL, ΖΛ· ἀσύμμετρον ἄρα [ἐστὶ] τὸ ΓΛ τῷ ΖΛ. ὡς δὲ τὸ ΓΛ πρὸς FL is thus also medial. And it is applied to the ratio- τὸ ΖΛ, οὕτως ἐστὶν ἡ ΓΜ πρὸς τὴν ΜΖ· ἀσύμμετρος ἄρα nal (straight-line) FE, producing FM as breadth. Thus, ἐστὶν ἡ ΓΜ τῇ ΜΖ μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ἄρα FM is rational, and incommensurable in length with CD ΓΜ, ΜΖ ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἀποτομὴ ἄρα [Prop. 10.22]. And since the sum of the (squares) on AG ἐστὶν ἡ ΓΖ. λέγω [δή], ὅτι καὶ τετάρτη. and GB is rational, and twice the (rectangle contained) ᾿Επεὶ γὰρ αἱ ΑΗ, ΗΒ δυνάμει εἰσὶν ἀσύμμετροι, ἀσύμμετ- by AG and GB medial, the (sum of the squares) on AG ρον ἄρα καὶ τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς ΗΒ. καί ἐστι τῷ and GB is [thus] incommensurable with twice the (rect- μὲν ἀπὸ τῆς ΑΗ ἴσον τὸ ΓΘ, τῷ δὲ ἀπὸ τῆς ΗΒ ἴσον τὸ angle contained) by AG and GB. And CL (is) equal to ΚΛ· ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΘ τῷ ΚΛ. ὡς δὲ τὸ ΓΘ πρὸς the (sum of the squares) on AG and GB, and FL equal τὸ ΚΛ, οὕτως ἐστὶν ἡ ΓΚ πρὸς τὴν ΚΜ· ἀσύμμετρος ἄρα to twice the (rectangle contained) by AG and GB. CL ἐστὶν ἡ ΓΚ τῇ ΚΜ μήκει. καὶ ἐπεὶ τῶν ἀπὸ τῶν ΑΗ, ΗΒ [is] thus incommensurable with FL. And as CL (is) to μέσον ἀνάλογόν ἐστι τὸ ὑπὸ τῶν ΑΗ, ΗΒ, καί ἐστιν ἴσον FL, so CM is to MF [Prop. 6.1]. CM is thus incommen- τὸ μὲν ἀπὸ τῆς ΑΗ τῷ ΓΘ, τὸ δὲ ἀπὸ τῆς ΗΒ τῷ ΚΛ, surable in length with MF [Prop. 10.11]. And both are τὸ δὲ ὑπὸ τῶν ΑΗ, ΗΒ τῷ ΝΛ, τῶν ἄρα ΓΘ, ΚΛ μέσον rational (straight-lines). Thus, CM and MF are rational ἀνάλογόν ἐστι τὸ ΝΛ· ἔστιν ἄρα ὡς τὸ ΓΘ πρὸς τὸ ΝΛ, (straight-lines which are) commensurable in square only. οὕτως τὸ ΝΛ πρὸς τὸ ΚΛ. ἀλλ᾿ ὡς μὲν τὸ ΓΘ πρὸς τὸ CF is thus an apotome [Prop. 10.73]. [So], I say that (it ΝΛ, οὕτως ἐστίν ἡ ΓΚ πρὸς τὴν ΝΜ, ὡς δὲ τὸ ΝΛ πρὸς is) also a fourth (apotome). τὸ ΚΛ, οὕτως ἐστὶν ἡ ΝΜ πρὸς τὴν ΚΜ· ὡς ἄρα ἡ ΓΚ For since AG and GB are incommensurable in square, πρὸς τὴν ΜΝ, οὕτως ἐστὶν ἡ ΜΝ πρὸς τὴν ΚΜ· τὸ ἄρα the (square) on AG (is) thus also incommensurable with ὑπὸ τῶν ΓΚ, ΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ, τουτέστι τῷ the (square) on GB. And CH is equal to the (square) on τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ. ἐπεὶ οὖν δύο εὐθεῖαι ἄνισοί AG, and KL equal to the (square) on GB. Thus, CH is εἰσιν αἱ ΓΜ, ΜΖ, καὶ τῷ τετράρτῳ μέρει τοῦ ἀπὸ τῆς ΜΖ incommensurable with KL. And as CH (is) to KL, so ἴσον παρὰ τὴν ΓΜ παραβέβληται ἐλλεῖπον εἴδει τετραγώνῳ CK is to KM [Prop. 6.1]. CK is thus incommensurable τὸ ὑπὸ τῶν ΓΚ, ΚΜ καὶ εἰς ἀσύμμετρα αὐτὴν διαιρεῖ, ἡ ἄρα in length with KM [Prop. 10.11]. And since the (rectan- ΓΜ τῆς ΜΖ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί gle contained) by AG and GB is the mean proportional ἐστιν ὅλη ἡ ΓΜ σύμμετρος μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ· to the (squares) on AG and GB [Prop. 10.21 lem.], and ἡ ἄρα ΓΖ ἀποτομή ἐστι τετάρτη. the (square) on AG is equal to CH , and the (square) Τὸ ἄρα ἀπὸ ἐλάσσονος καὶ τὰ ἑξῆς. on GB to KL, and the (rectangle contained) by AG and GB to NL, NL is thus the mean proportional to CH and KL. Thus, as CH is to NL, so NL (is) to KL. But, as CH (is) to NL, so CK is to NM , and as NL (is) to KL, so NM is to KM [Prop. 6.1]. Thus, as CK (is) to MN , so MN is to KM [Prop. 5.11]. The (rectangle con- tained) by CK and KM is thus equal to the (square) on MN—that is to say, to the fourth part of the (square) on FM [Prop. 6.17]. Therefore, since CM and MF are two unequal straight-lines, and the (rectangle contained) by CK and KM , equal to the fourth part of the (square) on MF , has been applied to CM , falling short by a square figure, and divides it into incommensurable (parts), the square on CM is thus greater than (the square on) MF by the (square) on (some straight-line) incommensurable 403 STOIQEIWN iþ. ELEMENTS BOOK 10 (in length) with (CM) [Prop. 10.18]. And the whole of CM is commensurable in length with the (previously) laid down rational (straight-line) CD. Thus, CF is a fourth apotome [Def. 10.14]. Thus, the (square) on a minor, and so on . . .raþ. Proposition 101 Τὸ ἀπὸ τῆς μετὰ ῥητοῦ μέσον τὸ ὅλον ποιούσης παρὰ The (square) on that (straight-line) which with a ra- ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν πέμπτην. tional (area) makes a medial whole, applied to a rational (straight-line), produces a fifth apotome as breadth. N D E L G Z M A B H K JX N D E L M A B C F G K HO ῎Εστω ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ, ῥητὴ Let AB be that (straight-line) which with a ratio- δὲ ἡ ΓΔ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΓΔ παρα- nal (area) makes a medial whole, and CD a rational βεβλήσθω τὸ ΓΕ πλάτος ποιοῦν τὴν ΓΖ· λέγω, ὅτι ἡ ΓΖ (straight-line). And let CE, equal to the (square) on AB, ἀποτομή ἐστι πέμπτη. have been applied to CD, producing CF as breadth. I ῎Εστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ· αἱ ἄρα ΑΗ, say that CF is a fifth apotome. ΗΒ εὐθεῖαι δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν Let BG be an attachment to AB. Thus, the straight- συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον, τὸ δὲ lines AG and GB are incommensurable in square, mak- δὶς ὑπ᾿ αὐτῶν ῥητόν, καὶ τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον παρὰ ing the sum of the squares on them medial, and twice τὴν ΓΔ παραβεβλήσθω τὸ ΓΘ, τῷ δὲ ἀπὸ τῆς ΗΒ ἵσον the (rectangle contained) by them rational [Prop. 10.77]. τὸ ΚΛ· ὅλον ἄρα τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ. And let CH , equal to the (square) on AG, have been ap- τὸ δὲ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΗ, ΗΒ ἅμα μέσον plied to CD, and KL, equal to the (square) on GB. The ἐστίν· μέσον ἄρα ἐστὶ τὸ ΓΛ. καὶ παρὰ ῥητὴν τὴν ΓΔ whole of CL is thus equal to the (sum of the squares) on παράκειται πλάτος ποιοῦν τὴν ΓΜ· ῥητὴ ἄρα ἐστὶν ἡ ΓΜ καὶ AG and GB. And the sum of the (squares) on AG and ἀσύμμετρος τῇ ΓΔ. καὶ ἐπεὶ ὅλον τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ GB together is medial. Thus, CL is medial. And it has τῶν ΑΗ, ΗΒ, ὧν τὸ ΓΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ, λοιπὸν been applied to the rational (straight-line) CD, produc- ἄρα τὸ ΖΛ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. τετμήσθω ing CM as breadth. CM is thus rational, and incommen- οὖν ἡ ΖΜ δίχα κατὰ τὸ Ν, καὶ ἤχθω διὰ τοῦ Ν ὁποτέρᾳ surable (in length) with CD [Prop. 10.22]. And since τῶν ΓΔ, ΜΛ παράλληλος ἡ ΝΞ· ἑκάτερον ἄρα τῶν ΖΞ, ΝΛ the whole of CL is equal to the (sum of the squares) on ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΗ, ΗΒ, καὶ ἐπεὶ τὸ δὶς ὑπὸ τῶν ΑΗ, AG and GB, of which CE is equal to the (square) on ΗΒ ῥητόν ἐστι καί [ἐστιν] ἴσον τῷ ΖΛ, ῥητὸν ἄρα ἐστὶ τὸ AB, the remainder FL is thus equal to twice the (rect- ΖΛ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται πλάτος ποιοῦν τὴν angle contained) by AG and GB [Prop. 2.7]. Therefore, ΖΜ· ῥητὴ ἄρα ἐστὶν ἡ ΖΜ καὶ σύμμετρος τῇ ΓΔ μήκει. καὶ let FM have been cut in half at N . And let NO have ἐπεὶ τὸ μὲν ΓΛ μέσον ἐστίν, τὸ δὲ ΖΛ ῥητόν, ἀσύμμετρον been drawn through N , parallel to either of CD or ML. ἄρα ἐστὶ τὸ ΓΛ τῷ ΖΛ. ὡς δὲ τὸ ΓΛ πρὸς τὸ ΖΛ, οὕτως Thus, FO and NL are each equal to the (rectangle con- ἡ ΓΜ πρὸς τὴν ΜΖ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΓΜ τῇ ΜΖ tained) by AG and GB. And since twice the (rectangle μήκει. καί εἰσιν ἀμφότεραι ῥηταί· αἱ ἄρα ΓΜ, ΜΖ ῥηταί εἰσι contained) by AG and GB is rational, and [is] equal to δυνάμει μόνον σύμμετροι· ἀποτομὴ ἄρα ἐστὶν ἡ ΓΖ. λὲγω FL, FL is thus rational. And it is applied to the ratio- δή, ὅτι καὶ πέμπτη. nal (straight-line) EF , producing FM as breadth. Thus, ῾Ομοίως γὰρ δείξομεν, ὅτι τὸ ὑπὸ τῶν ΓΚΜ ἴσον ἐστὶ FM is rational, and commensurable in length with CD τῷ ἀπὸ τῆς ΝΜ, τουτέστι τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς [Prop. 10.20]. And since CL is medial, and FL rational, 404 STOIQEIWN iþ. ELEMENTS BOOK 10 ΖΜ. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ τῆς CL is thus incommensurable with FL. And as CL (is) to ΗΒ, ἴσον δὲ τὸ μὲν ἀπὸ τῆς ΑΗ τῷ ΓΘ, τὸ δὲ ἀπὸ τῆς FL, so CM (is) to MF [Prop. 6.1]. CM is thus incom- ΗΒ τῷ ΚΛ, ἀσύμμετρον ἄρα τὸ ΓΘ τῷ ΚΛ. ὡς δὲ τὸ ΓΘ mensurable in length with MF [Prop. 10.11]. And both πρὸς τὸ ΚΛ, οὕτως ἡ ΓΚ πρὸς τὴν ΚΜ· ἀσύμμετρος ἄρα are rational. Thus, CM and MF are rational (straight- ἡ ΓΚ τῇ ΚΜ μήκει. ἐπεὶ οὖν δύο εὐθεῖαι ἄνισοί εἰσιν αἱ lines which are) commensurable in square only. CF is ΓΜ, ΜΖ, καὶ τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΖΜ ἴσον παρὰ thus an apotome [Prop. 10.73]. So, I say that (it is) also τὴν ΓΜ παραβέβληται ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς a fifth (apotome). ἀσύμμετρα αὐτὴν διαιρεῖ, ἡ ἄρα ΓΜ τῆς ΜΖ μεῖζον δύναται For, similarly (to the previous propositions), we can τῷ ἀπὸ ἀσύμμέτρου ἑαυτῇ. καί ἐστιν ἡ προσαρμόζουσα ἡ show that the (rectangle contained) by CKM is equal to ΖΜ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ τῇ ΓΔ· ἡ ἄρα ΓΖ ἀποτομή the (square) on NM—that is to say, to the fourth part ἐστι πέμπτη· ὅπερ ἔδει δεῖξαι. of the (square) on FM . And since the (square) on AG is incommensurable with the (square) on GB, and the (square) on AG (is) equal to CH , and the (square) on GB to KL, CH (is) thus incommensurable with KL. And as CH (is) to KL, so CK (is) to KM [Prop. 6.1]. Thus, CK (is) incommensurable in length with KM [Prop. 10.11]. Therefore, since CM and MF are two un- equal straight-lines, and (some area), equal to the fourth part of the (square) on FM , has been applied to CM , falling short by a square figure, and divides it into incom- mensurable (parts), the square on CM is thus greater than (the square on) MF by the (square) on (some straight-line) incommensurable (in length) with (CM) [Prop. 10.18]. And the attachment FM is commensu- rable with the (previously) laid down rational (straight- line) CD. Thus, CF is a fifth apotome [Def. 10.15]. (Which is) the very thing it was required to show.rbþ. Proposition 102 Τὸ ἀπὸ τῆς μετὰ μέσου μέσον τὸ ὅλον ποιούσης παρὰ The (square) on that (straight-line) which with a me- ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν ἕκτην. dial (area) makes a medial whole, applied to a rational (straight-line), produces a sixth apotome as breadth. N D E L G Z M A B H K JX N D E L M A B C F G K HO ῎Εστω ἡ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ, ῥητὴ Let AB be that (straight-line) which with a me- δὲ ἡ ΓΔ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΓΔ παρα- dial (area) makes a medial whole, and CD a rational βεβλήσθω τὸ ΓΕ πλάτος ποιοῦν τὴν ΓΖ· λέγω, ὅτι ἡ ΓΖ (straight-line). And let CE, equal to the (square) on AB, ἀποτομή ἐστιν ἕκτη. have been applied to CD, producing CF as breadth. I ῎Εστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΗ· αἱ ἄρα ΑΗ, say that CF is a sixth apotome. ΗΒ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον For let BG be an attachment to AB. Thus, AG and ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον καὶ τὸ δὶς ὑπὸ τῶν GB are incommensurable in square, making the sum of ΑΗ, ΗΒ μέσον καὶ ἀσύμμετρον τὰ ἀπὸ τῶν ΑΗ, ΗΒ τῷ the squares on them medial, and twice the (rectangle 405 STOIQEIWN iþ. ELEMENTS BOOK 10 δὶς ὑπὸ τῶν ΑΗ, ΗΒ. παραβεβλήσθω οὖν παρὰ τὴν ΓΔ τῷ contained) by AG and GB medial, and the (sum of the μὲν ἀπὸ τῆς ΑΗ ἴσον τὸ ΓΘ πλάτος ποιοῦν τὴν ΓΚ, τῷ squares) on AG and GB incommensurable with twice δὲ ἀπὸ τῆς ΒΗ τὸ ΚΛ· ὅλον ἄρα τὸ ΓΛ ἴσον ἐστὶ τοῖς ἀπὸ the (rectangle contained) by AG and GB [Prop. 10.78]. τῶν ΑΗ, ΗΒ· μέσον ἄρα [ἐστὶ] καὶ τὸ ΓΛ. καὶ παρὰ ῥητὴν Therefore, let CH , equal to the (square) on AG, have τὴν ΓΔ παράκειται πλάτος ποιοῦν τὴν ΓΜ· ῥητὴ ἄρα ἐστὶν been applied to CD, producing CK as breadth, and KL, ἡ ΓΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. ἐπεὶ οὖν τὸ ΓΛ ἴσον equal to the (square) on BG. Thus, the whole of CL ἐστὶ τοῖς ἀπὸ τῶν ΑΗ, ΗΒ, ὧν τὸ ΓΕ ἴσον τῷ ἀπὸ τῆς ΑΒ, is equal to the (sum of the squares) on AG and GB. λοιπὸν ἄρα τὸ ΖΛ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ. καί CL [is] thus also medial. And it is applied to the ratio- ἐστι τὸ δὶς ὑπὸ τῶν ΑΗ, ΗΒ μέσον· καὶ τὸ ΖΛ ἄρα μέσον nal (straight-line) CD, producing CM as breadth. Thus, ἐστίν. καὶ παρὰ ῥητὴν τὴν ΖΕ παράκειται πλάτος ποιοῦν τὴν CM is rational, and incommensurable in length with ΖΜ· ῥητὴ ἄρα ἐστὶν ἡ ΖΜ καὶ ἀσύμμετρος τῇ ΓΔ μήκει. καὶ CD [Prop. 10.22]. Therefore, since CL is equal to the ἐπεὶ τὰ ἀπὸ τῶν ΑΗ, ΗΒ ἀσύμμετρά ἐστι τῷ δὶς ὑπὸ τῶν (sum of the squares) on AG and GB, of which CE (is) ΑΗ, ΗΒ, καί ἐστι τοῖς μὲν ἀπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΓΛ, equal to the (square) on AB, the remainder FL is thus τῷ δὲ δὶς ὑπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΖΛ, ἀσύμμετρος ἄρα equal to twice the (rectangle contained) by AG and GB [ἐστὶ] τὸ ΓΛ τῷ ΖΛ. ὡς δὲ τὸ ΓΛ πρὸς τὸ ΖΛ, οὕτως ἐστὶν [Prop. 2.7]. And twice the (rectangle contained) by AG ἡ ΓΜ πρὸς τὴν ΜΖ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΓΜ τῇ ΜΖ and GB (is) medial. Thus, FL is also medial. And it is μήκει. καί εἰσιν ἀμφότεραι ῥηταί. αἱ ΓΜ, ΜΖ ἄρα ῥηταί εἰσι applied to the rational (straight-line) FE, producing FM δυνάμει μόνον σύμμετροι· ἀποτομὴ ἄρα ἐστὶν ἡ ΓΖ. λέγω as breadth. FM is thus rational, and incommensurable δή, ὅτι καὶ ἕκτη. in length with CD [Prop. 10.22]. And since the (sum ᾿Επεὶ γὰρ τὸ ΖΛ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΗ, ΗΒ, of the squares) on AG and GB is incommensurable with τετμήσθω δίχα ἡ ΖΜ κατὰ τὸ Ν, καὶ ἤχθω διὰ τοῦ Ν τῇ twice the (rectangle contained) by AG and GB, and CL ΓΔ παράλληλος ἡ ΝΞ· ἑκάτερον ἄρα τῶν ΖΞ, ΝΛ ἴσον ἐστὶ equal to the (sum of the squares) on AG and GB, and τῷ ὑπὸ τῶν ΑΗ, ΗΒ. καὶ ἐπεὶ αἱ ΑΗ, ΗΒ δυνάμει εἰσὶν FL equal to twice the (rectangle contained) by AG and ἀσύμμετροι, ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΗ τῷ ἀπὸ GB, CL [is] thus incommensurable with FL. And as CL τῆς ΗΒ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον ἐστὶ τὸ ΓΘ, τῷ δὲ (is) to FL, so CM is to MF [Prop. 6.1]. Thus, CM is ἀπὸ τῆς ΗΒ ἴσον ἐστὶ τὸ ΚΛ· ἀσύμμετρον ἄρα ἐστὶ τὸ ΓΘ incommensurable in length with MF [Prop. 10.11]. And τῷ ΚΛ. ὡς δὲ τὸ ΓΘ πρὸς τὸ ΚΛ, οὕτως ἐστὶν ἡ ΓΚ πρὸς they are both rational. Thus, CM and MF are rational τὴν ΚΜ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΓΚ τῇ ΚΜ. καὶ ἐπεὶ τῶν (straight-lines which are) commensurable in square only. ἀπὸ τῶν ΑΗ, ΗΒ μέσον ἀνάλογόν ἐστι τὸ ὑπὸ τῶν ΑΗ, CF is thus an apotome [Prop. 10.73]. So, I say that (it ΗΒ, καὶ ἐστι τῷ μὲν ἀπὸ τῆς ΑΗ ἴσον τὸ ΓΘ, τῷ δὲ ἀπὸ is) also a sixth (apotome). τῆς ΗΒ ἴσον τὸ ΚΛ, τῷ δὲ ὑπὸ τῶν ΑΗ, ΗΒ ἴσον τὸ ΝΛ, For since FL is equal to twice the (rectangle con- καὶ τῶν ἄρα ΓΘ, ΚΛ μέσον ἀνάλογόν ἐστι τὸ ΝΛ· ἔστιν ἄρα tained) by AG and GB, let FM have been cut in half ὡς τὸ ΓΘ πρὸς τὸ ΝΛ, οὕτως τὸ ΝΛ πρὸς τὸ ΚΛ. καὶ διὰ at N , and let NO have been drawn through N , parallel τὰ αὐτὰ ἡ ΓΜ τῆς ΜΖ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου to CD. Thus, FO and NL are each equal to the (rect- ἑαυτῇ. καὶ οὐδετέρα αὐτῶν σύμμετρός ἐστι τῇ ἐκκειμένῃ angle contained) by AG and GB. And since AG and GB ῥητῇ τῇ ΓΔ· ἡ ΓΖ ἄρα ἀποτομή ἐστιν ἕκτη· ὅπερ ἔδει δεῖξαι. are incommensurable in square, the (square) on AG is thus incommensurable with the (square) on GB. But, CH is equal to the (square) on AG, and KL is equal to the (square) on GB. Thus, CH is incommensurable with KL. And as CH (is) to KL, so CK is to KM [Prop. 6.1]. Thus, CK is incommensurable (in length) with KM [Prop. 10.11]. And since the (rectangle con- tained) by AG and GB is the mean proportional to the (squares) on AG and GB [Prop. 10.21 lem.], and CH is equal to the (square) on AG, and KL equal to the (square) on GB, and NL equal to the (rectangle con- tained) by AG and GB, NL is thus also the mean pro- portional to CH and KL. Thus, as CH is to NL, so NL (is) to KL. And for the same (reasons as the preced- ing propositions), the square on CM is greater than (the square on) MF by the (square) on (some straight-line) 406 STOIQEIWN iþ. ELEMENTS BOOK 10 incommensurable (in length) with (CM) [Prop. 10.18]. And neither of them is commensurable with the (previ- ously) laid down rational (straight-line) CD. Thus, CF is a sixth apotome [Def. 10.16]. (Which is) the very thing it was required to show.rgþ. Proposition 103 ῾Η τῇ ἀποτομῇ μήκει σύμμετρος ἀποτομή ἐστι καὶ τῇ A (straight-line) commensurable in length with an τάξει ἡ αὐτή. apotome is an apotome, and (is) the same in order. Γ Α ∆ Β Ε Ζ F A D B E C ῎Εστω ἀποτομὴ ἡ ΑΒ, καὶ τῇ ΑΒ μήκει σύμμετρος ἔστω Let AB be an apotome, and let CD be commensu- ἡ ΓΔ· λέγω, ὅτι καὶ ἡ ΓΔ ἀποτομή ἐστι καὶ τῇ τάξει ἡ αὐτὴ rable in length with AB. I say that CD is also an apo- τῇ ΑΒ. tome, and (is) the same in order as AB. ᾿Επεὶ γὰρ ἀποτομή ἐστιν ἡ ΑΒ, ἔστω αὐτῇ προ- For since AB is an apotome, let BE be an attachment σαρμόζουσα ἡ ΒΕ· αἱ ΑΕ, ΕΒ ἄρα ῥηταί εἰσι δυνάμει μόνον to it. Thus, AE and EB are rational (straight-lines which σύμμετροι. καὶ τῷ τῆς ΑΒ πρὸς τὴν ΓΔ λόγῳ ὁ αὐτὸς are) commensurable in square only [Prop. 10.73]. And γεγονέτω ὁ τῆς ΒΕ πρὸς τὴν ΔΖ· καὶ ὡς ἓν ἄρα πρὸς ἕν, let it have been contrived that the (ratio) of BE to DF πάντα [ἐστὶ] πρὸς πάντα· ἔστιν ἄρα καὶ ὡς ὅλη ἡ ΑΕ πρὸς is the same as the ratio of AB to CD [Prop. 6.12]. Thus, ὅλην τὴν ΓΖ, οὕτως ἡ ΑΒ πρὸς τὴν ΓΔ. σύμμετρος δὲ ἡ ΑΒ also, as one is to one, (so) all [are] to all [Prop. 5.12]. τῇ ΓΔ μήκει· σύμμετρος ἄρα καὶ ἡ ΑΕ μὲν τῇ ΓΖ, ἡ δὲ ΒΕ And thus as the whole AE is to the whole CF , so AB τῇ ΔΖ. καὶ αἱ ΑΕ, ΕΒ ῥηταί εἰσι δυνάμει μόνον σύμμετροι· (is) to CD. And AB (is) commensurable in length with καὶ αἱ ΓΖ, ΖΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι [ἀπο- CD. AE (is) thus also commensurable (in length) with τομὴ ἄρα ἐστὶν ἡ ΓΔ. λέγω δή, ὅτι καὶ τῇ τάξει ἡ αὐτὴ τῇ CF , and BE with DF [Prop. 10.11]. And AE and ΑΒ]. BE are rational (straight-lines which are) commensu- ᾿Επεὶ οὖν ἐστιν ὡς ἡ ΑΕ πρὸς τὴν ΓΖ, οὕτως ἡ ΒΕ πρὸς rable in square only. Thus, CF and FD are also rational τὴν ΔΖ, ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΕ πρὸς τὴν ΕΒ, οὕτως (straight-lines which are) commensurable in square only ἡ ΓΖ πρὸς τὴν ΖΔ. ἤτοι δὴ ἡ ΑΕ τῆς ΕΒ μεῖζον δύναται [Prop. 10.13]. [CD is thus an apotome. So, I say that (it τῷ ἀπὸ συμμέτρου ἑαυτῇ ἢ τῷ ἀπὸ ἀσυμμέτρου. εἰ μὲν οὖν is) also the same in order as AB.] ἡ ΑΕ τῆς ΕΒ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ Therefore, since as AE is to CF , so BE (is) to ἡ ΓΖ τῆς ΖΔ μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ. DF , thus, alternately, as AE is to EB, so CF (is) to καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΑΕ τῇ ἐκκειμένῃ ῥητῇ μήκει, FD [Prop. 5.16]. So, the square on AE is greater καὶ ἡ ΓΖ, εἰ δὲ ἡ ΒΕ, καὶ ἡ ΔΖ, εἰ δὲ οὐδετέρα τῶν ΑΕ, than (the square on) EB either by the (square) on ΕΒ, καὶ οὐδετέρα τῶν ΓΖ, ΖΔ. εἰ δὲ ἡ ΑΕ [τῆς ΕΒ] μεῖζον (some straight-line) commensurable, or by the (square) δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ἡ ΓΖ τῆς ΖΔ μεῖζον on (some straight-line) incommensurable, (in length) δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καὶ εἰ μὲν σύμμετρός with (AE). Therefore, if the (square) on AE is greater ἐστιν ἡ ΑΕ τῇ ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΓΖ, εἰ δὲ ἡ ΒΕ, than (the square on) EB by the (square) on (some καὶ ἡ ΔΖ, εἰ δὲ οὐδετέρα τῶν ΑΕ, ΕΒ, οὐδετέρα τῶν ΓΖ, straight-line) commensurable (in length) with (AE) then ΖΔ. the square on CF will also be greater than (the square Ἀποτομὴ ἄρα ἐστὶν ἡ ΓΔ καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ· on) FD by the (square) on (some straight-line) commen- ὅπερ ἔδει δεῖξαι. surable (in length) with (CF ) [Prop. 10.14]. And if AE is commensurable in length with a (previously) laid down rational (straight-line) then so (is) CF [Prop. 10.12], and if BE (is commensurable), so (is) DF , and if nei- ther of AE or EB (are commensurable), neither (are) either of CF or FD [Prop. 10.13]. And if the (square) on AE is greater [than (the square on) EB] by the (square) on (some straight-line) incommensurable (in 407 STOIQEIWN iþ. ELEMENTS BOOK 10 length) with (AE) then the (square) on CF will also be greater than (the square on) FD by the (square) on (some straight-line) incommensurable (in length) with (CF ) [Prop. 10.14]. And if AE is commensurable in length with a (previously) laid down rational (straight- line), so (is) CF [Prop. 10.12], and if BE (is com- mensurable), so (is) DF , and if neither of AE or EB (are commensurable), neither (are) either of CF or FD [Prop. 10.13]. Thus, CD is an apotome, and (is) the same in order as AB [Defs. 10.11—10.16]. (Which is) the very thing it was required to show.rdþ. Proposition 104 ῾Η τῇ μέσης ἀποτομῇ σύμμετρος μέσης ἀποτομή ἐστι A (straight-line) commensurable (in length) with an καὶ τῇ τάξει ἡ αὐτή. apotome of a medial (straight-line) is an apotome of a medial (straight-line), and (is) the same in order. Γ Α ∆ Β Ε Ζ F A D B E C ῎Εστω μέσης ἀποτομὴ ἡ ΑΒ, καὶ τῇ ΑΒ μήκει σύμμετρος Let AB be an apotome of a medial (straight-line), and ἔστω ἡ ΓΔ· λέγω, ὅτι καὶ ἡ ΓΔ μέσης ἀποτομή ἐστι καὶ τῇ let CD be commensurable in length with AB. I say that τάξει ἡ αὐτὴ τῇ ΑΒ. CD is also an apotome of a medial (straight-line), and ᾿Επεὶ γὰρ μέσης ἀποτομή ἐστιν ἡ ΑΒ, ἔστω αὐτῇ προ- (is) the same in order as AB. σαρμόζουσα ἡ ΕΒ. αἱ ΑΕ, ΕΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον For since AB is an apotome of a medial (straight- σύμμετροι. καὶ γεγονέτω ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ line), let EB be an attachment to it. Thus, AE and ΒΕ πρὸς τὴν ΔΖ· σύμμετρος ἄρα [ἐστὶ] καὶ ἡ ΑΕ τῇ ΓΖ, EB are medial (straight-lines which are) commensurable ἡ δὲ ΒΕ τῇ ΔΖ. αἱ δὲ ΑΕ, ΕΒ μέσαι εἰσὶ δυνάμει μόνον in square only [Props. 10.74, 10.75]. And let it have σύμμετροι· καὶ αἱ ΓΖ, ΖΔ ἄρα μέσαι εἰσὶ δυνάμει μόνον been contrived that as AB is to CD, so BE (is) to DF σύμμετροι· μέσης ἄρα ἀποτομή ἐστιν ἡ ΓΔ. λέγω δή, ὅτι [Prop. 6.12]. Thus, AE [is] also commensurable (in καὶ τῇ τάξει ἐστὶν ἡ αὐτὴ τῇ ΑΒ. length) with CF , and BE with DF [Props. 5.12, 10.11]. ᾿Επεὶ [γάρ] ἐστιν ὡς ἡ ΑΕ πρὸς τὴν ΕΒ, οὕτως ἡ ΓΖ And AE and EB are medial (straight-lines which are) πρὸς τὴν ΖΔ [ἀλλ᾿ ὡς μὲν ἡ ΑΕ πρὸς τὴν ΕΒ, οὕτως τὸ commensurable in square only. CF and FD are thus ἀπὸ τῆς ΑΕ πρὸς τὸ ὑπὸ τῶν ΑΕ, ΕΒ, ὡς δὲ ἡ ΓΖ πρὸς τὴν also medial (straight-lines which are) commensurable in ΖΔ, οὕτως τὸ ἀπὸ τῆς ΓΖ πρὸς τὸ ὑπὸ τῶν ΓΖ, ΖΔ], ἔστιν square only [Props. 10.23, 10.13]. Thus, CD is an apo- ἄρα καὶ ὡς τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ὑπὸ τῶν ΑΕ, ΕΒ, οὕτως tome of a medial (straight-line) [Props. 10.74, 10.75]. τὸ ἀπὸ τῆς ΓΖ πρὸς τὸ ὑπὸ τῶν ΓΖ, ΖΔ [καὶ ἐναλλὰξ ὡς So, I say that it is also the same in order as AB. τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ἀπὸ τῆς ΓΖ, οὕτως τὸ ὑπὸ τῶν ΑΕ, [For] since as AE is to EB, so CF (is) to FD ΕΒ πρὸς τὸ ὑπὸ τῶν ΓΖ, ΖΔ]. σύμμετρον δὲ τὸ ἀπὸ τῆς ΑΕ [Props. 5.12, 5.16] [but as AE (is) to EB, so the (square) τῷ ἀπὸ τῆς ΓΖ· σύμμετρον ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν ΑΕ, on AE (is) to the (rectangle contained) by AE and EB, ΕΒ τῷ ὑπὸ τῶν ΓΖ, ΖΔ. εἴτε οὖν ῥητόν ἐστι τὸ ὑπὸ τῶν and as CF (is) to FD, so the (square) on CF (is) to ΑΕ, ΕΒ, ῥητὸν ἔσται καὶ τὸ ὑπὸ τῶν ΓΖ, ΖΔ, εἴτε μέσον the (rectangle contained) by CF and FD], thus as the [ἐστὶ] τὸ ὑπὸ τῶν ΑΕ, ΕΒ, μέσον [ἐστὶ] καὶ τὸ ὑπὸ τῶν ΓΖ, (square) on AE is to the (rectangle contained) by AE ΖΔ. and EB, so the (square) on CF also (is) to the (rectan- Μέσης ἄρα ἀποτομή ἐστιν ἡ ΓΔ καὶ τῇ τάξει ἡ αὐτὴ τῇ gle contained) by CF and FD [Prop. 10.21 lem.] [and, ΑΒ· ὅπερ ἔδει δεῖξαι. alternately, as the (square) on AE (is) to the (square) on CF , so the (rectangle contained) by AE and EB (is) to the (rectangle contained) by CF and FD]. And the (square) on AE (is) commensurable with the (square) 408 STOIQEIWN iþ. ELEMENTS BOOK 10 on CF . Thus, the (rectangle contained) by AE and EB is also commensurable with the (rectangle contained) by CF and FD [Props. 5.16, 10.11]. Therefore, either the (rectangle contained) by AE and EB is rational, and the (rectangle contained) by CF and FD will also be ratio- nal [Def. 10.4], or the (rectangle contained) by AE and EB [is] medial, and the (rectangle contained) by CF and FD [is] also medial [Prop. 10.23 corr.]. Therefore, CD is the apotome of a medial (straight- line), and is the same in order as AB [Props. 10.74, 10.75]. (Which is) the very thing it was required to show.reþ. Proposition 105 ῾Η τῇ ἐλάσσονι σύμμετρος ἐλάσσων ἐστίν. A (straight-line) commensurable (in length) with a minor (straight-line) is a minor (straight-line). Γ Α ∆ Β Ε Ζ F A D B E C ῎Εστω γὰρ ἐλάσσων ἡ ΑΒ καὶ τῇ ΑΒ σύμμετρος ἡ ΓΔ· For let AB be a minor (straight-line), and (let) CD λέγω, ὅτι καὶ ἡ ΓΔ ἐλάσσων ἐστίν. (be) commensurable (in length) with AB. I say that CD Γεγονέτω γὰρ τὰ αὐτά· καὶ ἐπεὶ αἱ ΑΕ, ΕΒ δυνάμει εἰσὶν is also a minor (straight-line). ἀσύμμετροι, καὶ αἱ ΓΖ, ΖΔ ἄρα δυνάμει εἰσὶν ἀσύμμετροι. For let the same things have been contrived (as in ἐπεὶ οὖν ἐστιν ὡς ἡ ΑΕ πρὸς τὴν ΕΒ, οὕτως ἡ ΓΖ πρὸς the former proposition). And since AE and EB are τὴν ΖΔ, ἔστιν ἄρα καὶ ὡς τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ἀπὸ τῆς (straight-lines which are) incommensurable in square ΕΒ, οὕτως τὸ ἀπὸ τῆς ΓΖ πρὸς τὸ ἀπὸ τῆς ΖΔ. συνθέντι [Prop. 10.76], CF and FD are thus also (straight-lines ἄρα ἐστὶν ὡς τὰ ἀπὸ τῶν ΑΕ, ΕΒ πρὸς τὸ ἀπὸ τῆς ΕΒ, which are) incommensurable in square [Prop. 10.13]. οὕτως τὰ ἀπὸ τῶν ΓΖ, ΖΔ πρὸς τὸ ἀπὸ τῆς ΖΔ [καὶ Therefore, since as AE is to EB, so CF (is) to FD ἐναλλάξ]· σύμμετρον δέ ἐστι τὸ ἀπὸ τῆς ΒΕ τῷ ἀπὸ τῆς ΔΖ· [Props. 5.12, 5.16], thus also as the (square) on AE is σύμμετρον ἄρα καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ to the (square) on EB, so the (square) on CF (is) to the τετραγώνων τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τε- (square) on FD [Prop. 6.22]. Thus, via composition, as τραγώνων. ῥητὸν δέ ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν the (sum of the squares) on AE and EB is to the (square) ΑΕ, ΕΒ τετραγώνων· ῥητὸν ἄρα ἐστὶ καὶ τὸ συγκείμενον on EB, so the (sum of the squares) on CF and FD (is) to ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων. πάλιν, ἐπεί ἐστιν ὡς the (square) on FD [Prop. 5.18], [also alternately]. And τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ὑπὸ τῶν ΑΕ, ΕΒ, οὕτως τὸ ἀπὸ the (square) on BE is commensurable with the (square) τῆς ΓΖ πρὸς τὸ ὑπὸ τῶν ΓΖ, ΖΔ, σύμμετρον δὲ τὸ ἀπὸ τῆς on DF [Prop. 10.104]. The sum of the squares on AE ΑΕ τετράγωνον τῷ ἀπὸ τῆς ΓΖ τετραγώνῳ, σύμμετρον ἄρα and EB (is) thus also commensurable with the sum of the ἐστὶ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΒ τῷ ὑπὸ τῶν ΓΖ, ΖΔ. μέσον squares on CF and FD [Prop. 5.16, 10.11]. And the sum δὲ τὸ ὑπὸ τῶν ΑΕ, ΕΒ· μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΓΖ, ΖΔ· of the (squares) on AE and EB is rational [Prop. 10.76]. αἱ ΓΖ, ΖΔ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν Thus, the sum of the (squares) on CF and FD is also συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων ῥητόν, τὸ δ᾿ rational [Def. 10.4]. Again, since as the (square) on ὑπ᾿ αὐτῶν μέσον. AE is to the (rectangle contained) by AE and EB, so ᾿Ελάσσων ἄρα ἐστὶν ἡ ΓΔ· ὅπερ ἔδει δεῖξαι. the (square) on CF (is) to the (rectangle contained) by CF and FD [Prop. 10.21 lem.], and the square on AE (is) commensurable with the square on CF , the (rect- angle contained) by AE and EB is thus also commen- surable with the (rectangle contained) by CF and FD. And the (rectangle contained) by AE and EB (is) me- dial [Prop. 10.76]. Thus, the (rectangle contained) by CF and FD (is) also medial [Prop. 10.23 corr.]. CF and 409 STOIQEIWN iþ. ELEMENTS BOOK 10 FD are thus (straight-lines which are) incommensurable in square, making the sum of the squares on them ratio- nal, and the (rectangle contained) by them medial. Thus, CD is a minor (straight-line) [Prop. 10.76]. (Which is) the very thing it was required to show.r�þ. Proposition 106 ῾Η τῇ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιούσῃ σύμμετρος A (straight-line) commensurable (in length) with a μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν. (straight-line) which with a rational (area) makes a me- dial whole is a (straight-line) which with a rational (area) makes a medial whole. Γ Α ∆ Β Ε Ζ F A D B E C ῎Εστω μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ καὶ τῇ Let AB be a (straight-line) which with a rational ΑΒ σύμμετρος ἡ ΓΔ· λέγω, ὅτι καὶ ἡ ΓΔ μετὰ ῥητοῦ μέσον (area) makes a medial whole, and (let) CD (be) com- τὸ ὅλον ποιοῦσά ἐστιν. mensurable (in length) with AB. I say that CD is also a ῎Εστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΕ· αἱ ΑΕ, ΕΒ ἄρα (straight-line) which with a rational (area) makes a me- δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ dial (whole). τῶν ἀπὸ τῶν ΑΕ, ΕΒ τετραγώνων μέσον, τὸ δ᾿ ὑπ᾿ αὐτῶν For let BE be an attachment to AB. Thus, AE and ῥητόν. καὶ τὰ αὐτὰ κατεσκευάσθω. ὁμοίως δὴ δείξομεν τοῖς EB are (straight-lines which are) incommensurable in πρότερον, ὅτι αἱ ΓΖ, ΖΔ ἐν τῷ αὐτῷ λόγῳ εἰσὶ ταῖς ΑΕ, square, making the sum of the squares on AE and EB ΕΒ, καὶ σύμμετρόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν medial, and the (rectangle contained) by them rational ΑΕ, ΕΒ τετραγώνων τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, [Prop. 10.77]. And let the same construction have been ΖΔ τετραγώνων, τὸ δὲ ὑπὸ τῶν ΑΕ, ΕΒ τῷ ὑπὸ τῶν ΓΖ, made (as in the previous propositions). So, similarly to ΖΔ· ὥστε καὶ αἱ ΓΖ, ΖΔ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι the previous (propositions), we can show that CF and τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων FD are in the same ratio as AE and EB, and the sum of μέσον, τὸ δ᾿ ὑπ᾿ αὐτῶν ῥητόν. the squares on AE and EB is commensurable with the ῾Η ΓΔ ἄρα μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν· sum of the squares on CF and FD, and the (rectangle ὅπερ ἔδει δεῖξαι. contained) by AE and EB with the (rectangle contained) by CF and FD. Hence, CF and FD are also (straight- lines which are) incommensurable in square, making the sum of the squares on CF and FD medial, and the (rect- angle contained) by them rational. CD is thus a (straight-line) which with a rational (area) makes a medial whole [Prop. 10.77]. (Which is) the very thing it was required to show.rzþ. Proposition 107 ῾Η τῇ μετὰ μέσου μέσον τὸ ὅλον ποιούσῃ σύμμετρος A (straight-line) commensurable (in length) with a καὶ αὐτὴ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν. (straight-line) which with a medial (area) makes a me- dial whole is itself also a (straight-line) which with a me- dial (area) makes a medial whole. Γ Α ∆ Β Ε Ζ F A D B E C ῎Εστω μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα ἡ ΑΒ, καὶ τῇ Let AB be a (straight-line) which with a medial (area) 410 STOIQEIWN iþ. ELEMENTS BOOK 10 ΑΒ ἔστω σύμμετρος ἡ ΓΔ· λέγω, ὅτι καὶ ἡ ΓΔ μετὰ μέσου makes a medial whole, and let CD be commensurable (in μέσον τὸ ὅλον ποιοῦσά ἐστιν. length) with AB. I say that CD is also a (straight-line) ῎Εστω γὰρ τῇ ΑΒ προσαρμόζουσα ἡ ΒΕ, καὶ τὰ αὐτὰ which with a medial (area) makes a medial whole. κατεσκευάσθω· αἱ ΑΕ, ΕΒ ἄρα δυνάμει εἱσὶν ἀσύμμετροι For let BE be an attachment to AB. And let the same ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων construction have been made (as in the previous propo- μέσον καὶ τὸ ὑπ᾿ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τὸ sitions). Thus, AE and EB are (straight-lines which συγκέιμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων τῷ ὑπ᾿ αὐτῶν. are) incommensurable in square, making the sum of the καί εἰσιν, ὡς ἐδείχθη, αἱ ΑΕ, ΕΒ σύμμετροι ταῖς ΓΖ, ΖΔ, squares on them medial, and the (rectangle contained) καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τετραγώνων by them medial, and, further, the sum of the squares on τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ, τὸ δὲ ὑπὸ τῶν them incommensurable with the (rectangle contained) by ΑΕ, ΕΒ τῷ ὑπὸ τῶν ΓΖ, ΖΔ· καὶ αἱ ΓΖ, ΖΔ ἄρα δυνάμει them [Prop. 10.78]. And, as was shown (previously), AE εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾿ and EB are commensurable (in length) with CF and FD αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπ᾿ ἀὐτῶν μέσον καὶ (respectively), and the sum of the squares on AE and ἔτι ἀσύμμετρον τὸ συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν [τε- EB with the sum of the squares on CF and FD, and τραγώνων] τῷ ὑπ᾿ αὐτῶν. the (rectangle contained) by AE and EB with the (rect- ῾Η ΓΔ ἄρα μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν· angle contained) by CF and FD. Thus, CF and FD ὅπερ ἔδει δεῖξαι. are also (straight-lines which are) incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them medial, and, fur- ther, the sum of the [squares] on them incommensurable with the (rectangle contained) by them. Thus, CD is a (straight-line) which with a medial (area) makes a medial whole [Prop. 10.78]. (Which is) the very thing it was required to show.rhþ. Proposition 108 Ἀπὸ ῥητοῦ μέσου ἀφαιρουμένου ἡ τὸ λοιπὸν χωρίον A medial (area) being subtracted from a rational δυναμένη μία δύο ἀλόγων γίνεται ἤτοι ἀποτομὴ ἢ ἐλάσσων. (area), one of two irrational (straight-lines) arise (as) the square-root of the remaining area—either an apotome, or a minor (straight-line). ∆ Λ Θ Κ Η Ζ Α Ε Β Γ C L H K G F A E B D Ἀπὸ γὰρ ῥητοῦ τοῦ ΒΓ μέσον ἀφῃρήσθω τὸ ΒΔ· λέγω, For let the medial (area) BD have been subtracted ὅτι ἡ τὸ λοιπὸν δυναμένη τὸ ΕΓ μία δύο ἀλόγων γίνεται from the rational (area) BC. I say that one of two ir- ἤτοι ἀποτομὴ ἢ ἐλάσσων. rational (straight-lines) arise (as) the square-root of the ᾿Εκκείσθω γὰρ ῥητὴ ἡ ΖΗ, καὶ τῷ μὲν ΒΓ ἴσον παρὰ τὴν remaining (area), EC—either an apotome, or a minor ΖΗ παραβεβλήσθω ὀρθογώνιον παραλληλόγραμμον τὸ ΗΘ, (straight-line). τῷ δὲ ΔΒ ἴσον ἀφῃρήσθω τὸ ΗΚ· λοιπὸν ἄρα τὸ ΕΓ ἴσον For let the rational (straight-line) FG have been laid ἐστὶ τῷ ΛΘ. ἐπεὶ οὖν ῥητὸν μέν ἐστι τὸ ΒΓ, μέσον δὲ τὸ out, and let the right-angled parallelogram GH , equal to ΒΔ, ἴσον δὲ τὸ μὲν ΒΓ τῷ ΗΘ, τὸ δὲ ΒΔ τῷ ΗΚ, ῥητὸν μὲν BC, have been applied to FG, and let GK, equal to DB, ἄρα ἐστὶ τὸ ΗΘ, μέσον δὲ τὸ ΗΚ. καὶ παρὰ ῥητὴν τὴν ΖΗ have been subtracted (from GH). Thus, the remainder παράκειται· ῥητὴ μὲν ἄρα ἡ ΖΘ καὶ σύμμετρος τῇ ΖΗ μήκει, EC is equal to LH . Therefore, since BC is a rational ῥητὴ δὲ ἡ ΖΚ καὶ ἀσύμμετρος τῇ ΖΗ μήκει· ἀσύμμετρος ἄρα (area), and BD a medial (area), and BC (is) equal to 411 STOIQEIWN iþ. ELEMENTS BOOK 10 ἐστὶν ἡ ΖΘ τῇ ΖΚ μήκει. αἱ ΖΘ, ΖΚ ἄρα ῥηταί εἰσι δυνάμει GH , and BD to GK, GH is thus a rational (area), and μόνον σύμμετροι· ἀποτομὴ ἄρα ἐστὶν ἡ ΚΘ, προσαρμόζουσα GK a medial (area). And they are applied to the rational δὲ αὐτῇ ἡ ΚΖ. ἤτοι δὴ ἡ ΘΖ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ (straight-line) FG. Thus, FH (is) rational, and commen- συμμέτρου ἢ οὔ. surable in length with FG [Prop. 10.20], and FK (is) Δυνάσθω πρότερον τῷ ἀπὸ συμμέτρου. καί ἐστιν ὅλη ἡ also rational, and incommensurable in length with FG ΘΖ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ ΖΗ· ἀποτομὴ ἄρα [Prop. 10.22]. Thus, FH is incommensurable in length πρώτη ἐστὶν ἡ ΚΘ. τὸ δ᾿ ὑπὸ ῥητῆς καὶ ἀποτομῆς πρώτης with FK [Prop. 10.13]. FH and FK are thus rational περιεχόμενον ἡ δυναμένη ἀποτομή ἐστιν. ἡ ἄρα τὸ ΛΘ, (straight-lines which are) commensurable in square only. τουτέστι τὸ ΕΓ, δυναμένη ἀποτομή ἐστιν. Thus, KH is an apotome [Prop. 10.73], and KF an at- Εἰ δὲ ἡ ΘΖ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου tachment to it. So, the square on HF is greater than ἑαυτῇ, καί ἐστιν ὅλη ἡ ΖΘ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ (the square on) FK by the (square) on (some straight- μήκει τῇ ΖΗ, ἀποτομὴ τετάρτη ἐστὶν ἡ ΚΘ. τὸ δ᾿ ὑπὸ ῥητῆς line which is) either commensurable, or not (commensu- καὶ ἀποτομῆς τετάρτης περιεχόμενον ἡ δυναμένη ἐλάσσων rable), (in length with HF ). ἐστίν· ὅπερ ἔδει δεῖξαι. First, let the square (on it) be (greater) by the (square) on (some straight-line which is) commensurable (in length with HF ). And the whole of HF is com- mensurable in length with the (previously) laid down rational (straight-line) FG. Thus, KH is a first apotome [Def. 10.1]. And the square-root of an (area) contained by a rational (straight-line) and a first apotome is an apo- tome [Prop. 10.91]. Thus, the square-root of LH—that is to say, (of) EC—is an apotome. And if the square on HF is greater than (the square on) FK by the (square) on (some straight-line which is) incommensurable (in length) with (HF ), and (since) the whole of FH is commensurable in length with the (pre- viously) laid down rational (straight-line) FG, KH is a fourth apotome [Prop. 10.14]. And the square-root of an (area) contained by a rational (straight-line) and a fourth apotome is a minor (straight-line) [Prop. 10.94]. (Which is) the very thing it was required to show.rjþ. Proposition 109 Ἀπὸ μέσου ῥητοῦ ἀφαιρουμένου ἄλλαι δύο ἄλογοι A rational (area) being subtracted from a medial γίνονται ἤτοι μέσης ἀποτομὴ πρώτη ἢ μετὰ ῥητοῦ μέσον (area), two other irrational (straight-lines) arise (as the τὸ ὅλον ποιοῦσα. square-root of the remaining area)—either a first apo- Ἀπὸ γὰρ μέσου τοῦ ΒΓ ῥητὸν ἀφῃρήσθω τὸ ΒΔ. λέγω, tome of a medial (straight-line), or that (straight-line) ὅτι ἡ τὸ λοιπὸν τὸ ΕΓ δυναμένη μία δύο ἀλόγων γίνεται which with a rational (area) makes a medial whole. ἤτοι μέσης ἀποτομὴ πρώτη ἢ μετὰ ῥητοῦ μέσον τὸ ὅλον For let the rational (area) BD have been subtracted ποιοῦσα. from the medial (area) BC. I say that one of two ir- ᾿Εκκείσθω γὰρ ῥητὴ ἡ ΖΗ, καὶ παραβεβλήσθω ὁμοίως τὰ rational (straight-lines) arise (as) the square-root of the χωρία. ἔστι δὴ ἀκολούθως ῥητὴ μὲν ἡ ΖΘ καὶ ἀσύμμετρος remaining (area), EC—either a first apotome of a medial τῇ ΖΗ μήκει, ῥητὴ δὲ ἡ ΚΖ καὶ σύμμετρος τῇ ΖΗ μήκει· αἱ (straight-line), or that (straight-line) which with a ratio- ΖΘ, ΖΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἀποτομὴ nal (area) makes a medial whole. ἄρα ἐστὶν ἡ ΚΘ, προσαρμόζουσα δὲ ταύτῃ ἡ ΖΚ. ἤτοι δὴ ἡ For let the rational (straight-line) FG be laid down, ΘΖ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ ἢ τῷ and let similar areas (to the preceding proposition) have ἀπὸ ἀσυμμέτρου. been applied (to it). So, accordingly, FH is rational, and incommensurable in length with FG, and KF (is) also rational, and commensurable in length with FG. Thus, FH and FK are rational (straight-lines which are) com- 412 STOIQEIWN iþ. ELEMENTS BOOK 10 mensurable in square only [Prop. 10.13]. KH is thus an apotome [Prop. 10.73], and FK an attachment to it. So, the square on HF is greater than (the square on) FK either by the (square) on (some straight-line) commensu- rable (in length) with (HF ), or by the (square) on (some straight-line) incommensurable (in length with HF ). ΘΒ Ε ∆Α Γ Η Λ Ζ Κ G B E DA L K C F H Εἰ μὲν οὖν ἡ ΘΖ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ Therefore, if the square on HF is greater than (the συμμέτρου ἑαυτῇ, καί ἐστιν ἡ προσαρμόζουσα ἡ ΖΚ square on) FK by the (square) on (some straight-line) σύμμετρος τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ ΖΗ, ἀποτομὴ δευτέρα commensurable (in length) with (HF ), and (since) the ἐστὶν ἡ ΚΘ. ῥητὴ δὲ ἡ ΖΗ· ὥστε ἡ τὸ ΛΘ, τουτέστι τὸ ΕΓ, attachment FK is commensurable in length with the δυναμένη μέσης ἀποτομὴ πρώτη ἐστίν. (previously) laid down rational (straight-line) FG, KH Εἰ δὲ ἡ ΘΖ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου, is a second apotome [Def. 10.12]. And FG (is) rational. καί ἐστιν ἡ προσαρμόζουσα ἡ ΖΚ σύμμετρος τῇ ἐκκειμένῃ Hence, the square-root of LH—that is to say, (of) EC—is ῥητῇ μήκει τῇ ΖΗ, ἀποτομὴ πέμπτη ἐστὶν ἡ ΚΘ· ὥστε ἡ a first apotome of a medial (straight-line) [Prop. 10.92]. τὸ ΕΓ δυναμένη μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν· And if the square on HF is greater than (the square ὅπερ ἔδει δεῖξαι. on) FK by the (square) on (some straight-line) incom- mensurable (in length with HF ), and (since) the attach- ment FK is commensurable in length with the (previ- ously) laid down rational (straight-line) FG, KH is a fifth apotome [Def. 10.15]. Hence, the square-root of EC is that (straight-line) which with a rational (area) makes a medial whole [Prop. 10.95]. (Which is) the very thing it was required to show.riþ. Proposition 110 Ἀπὸ μέσου μέσου ἀφαιρουμένου ἀσυμμέτρου τῷ ὅλῳ αἱ A medial (area), incommensurable with the whole, λοιπαὶ δύο ἄλογοι γίνονται ἤτοι μέσης ἀποτομὴ δευτέρα ἢ being subtracted from a medial (area), the two remaining μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα. irrational (straight-lines) arise (as) the (square-root of Ἀφῃρήσθω γὰρ ὡς ἐπὶ τῶν προκειμένων καταγραφῶν the area)—either a second apotome of a medial (straight- ἀπὸ μέσου τοῦ ΒΓ μέσον τὸ ΒΔ ἀσύμμετρον τῷ ὅλῳ· λέγω, line), or that (straight-line) which with a medial (area) ὅτι ἡ τὸ ΕΓ δυναμένη μία ἐστὶ δύο ἀλόγων ἤτοι μέσης ἀπο- makes a medial whole. τομὴ δευτέρα ἢ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα. For, as in the previous figures, let the medial (area) BD, incommensurable with the whole, have been sub- tracted from the medial (area) BC. I say that the square- root of EC is one of two irrational (straight-lines)— either a second apotome of a medial (straight-line), or that (straight-line) which with a medial (area) makes a medial whole. 413 STOIQEIWN iþ. ELEMENTS BOOK 10 ΘΒ Ε ∆Α Γ Η Λ Ζ Κ G B E DA L K C F H ᾿Επεὶ γὰρ μέσον ἐστὶν ἑκάτερον τῶν ΒΓ, ΒΔ, καὶ For since BC and BD are each medial (areas), and ἀσύμμετρον τὸ ΒΓ τῷ ΒΔ, ἔσται ἀκολούθως ῥητὴ ἑκατέρα BC (is) incommensurable with BD, accordingly, FH and τῶν ΖΘ, ΖΚ καὶ ἀσύμμετρος τῇ ΖΗ μήκει. καὶ ἐπεὶ FK will each be rational (straight-lines), and incommen- ἀσύμμετρόν ἐστι τὸ ΒΓ τῷ ΒΔ, τουτέστι τὸ ΗΘ τῷ ΗΚ, surable in length with FG [Prop. 10.22]. And since BC ἀσύμμετρος καὶ ἡ ΘΖ τῇ ΖΚ· αἱ ΖΘ, ΖΚ ἄρα ῥηταί εἰσι is incommensurable with BD—that is to say, GH with δυνάμει μόνον σύμμετροι· ἀποτομὴ ἄρα ἐστὶν ἡ ΚΘ [προ- GK—HF (is) also incommensurable (in length) with σαρμόζουσα δὲ ἡ ΖΚ. ἤτοι δὴ ἡ ΖΘ τῆς ΖΚ μεῖζον δύναται FK [Props. 6.1, 10.11]. Thus, FH and FK are ratio- τῷ ἀπὸ συμμέτρου ἢ τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ]. nal (straight-lines which are) commensurable in square Εἰ μὲν δὴ ἡ ΖΘ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ only. KH is thus as apotome [Prop. 10.73], [and FK συμμέτρου ἑαυτῇ, καὶ οὐθετέρα τῶν ΖΘ, ΖΚ σύμμετρός an attachment (to it). So, the square on FH is greater ἐστι τῇ ἐκκειμέμνῃ ῥητῇ μήκει τῇ ΖΗ, ἀποτομὴ τρίτη ἐστὶν than (the square on) FK either by the (square) on ἡ ΚΘ. ῥητὴ δὲ ἡ ΚΛ, τὸ δ᾿ ὑπὸ ῥητῆς καὶ ἀποτομῆς τρίτης (some straight-line) commensurable, or by the (square) περιεχόμενον ὀρθογώνιον ἄλογόν ἐστιν, καὶ ἡ δυναμένη on (some straight-line) incommensurable, (in length) αὐτὸ ἄλογός ἐστιν, καλεῖται δὲ μέσης ἀποτομὴ δευτέρα· with (FH).] ὥστε ἡ τὸ ΛΘ, τουτέστι τὸ ΕΓ, δυναμένη μέσης ἀποτομή So, if the square on FH is greater than (the square ἐστι δευτερά. on) FK by the (square) on (some straight-line) com- Εἰ δὲ ἡ ΖΘ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου mensurable (in length) with (FH), and (since) neither of ἑαυτῇ [μήκει], καὶ οὐθετέρα τῶν ΘΖ, ΖΚ σύμμετρός ἐστι FH and FK is commensurable in length with the (pre- τῇ ΖΗ μήκει, ἀποτομὴ ἕκτη ἐστὶν ἡ ΚΘ. τὸ δ᾿ ὑπὸ ῥητῆς viously) laid down rational (straight-line) FG, KH is a καὶ ἀποτομῆς ἕκτης ἡ δυναμένη ἐστὶ μετὰ μέσου μέσον τὸ third apotome [Def. 10.3]. And KL (is) rational. And ὅλον ποιοῦσα. ἡ τὸ ΛΘ ἄρα, τουτέστι τὸ ΕΓ, δυναμένη the rectangle contained by a rational (straight-line) and μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν· ὅπερ ἔδει δεῖξαι. a third apotome is irrational, and the square-root of it is that irrational (straight-line) called a second apotome of a medial (straight-line) [Prop. 10.93]. Hence, the square- root of LH—that is to say, (of) EC—is a second apotome of a medial (straight-line). And if the square on FH is greater than (the square on) FK by the (square) on (some straight-line) incom- mensurable [in length] with (FH), and (since) neither of HF and FK is commensurable in length with FG, KH is a sixth apotome [Def. 10.16]. And the square-root of the (rectangle contained) by a rational (straight-line) and a sixth apotome is that (straight-line) which with a me- dial (area) makes a medial whole [Prop. 10.96]. Thus, the square-root of LH—that is to say, (of) EC—is that (straight-line) which with a medial (area) makes a me- dial whole. (Which is) the very thing it was required to 414 STOIQEIWN iþ. ELEMENTS BOOK 10 show.riaþ. Proposition 111 ῾Η ἀποτομὴ οὐκ ἔστιν ἡ αὐτὴ τῇ ἐκ δύο ὀνομάτων. An apotome is not the same as a binomial. Ε Α Η∆ Γ Ζ Β F A D EG C B ῎Εστω ἀποτομὴ ἡ ΑΒ· λέγω, ὅτι ἡ ΑΒ οὐκ ἔστιν ἡ αὐτὴ Let AB be an apotome. I say that AB is not the same τῇ ἐκ δύο ὀνομάτων. as a binomial. Εἰ γὰρ δυνατόν, ἔστω· καὶ ἐκκείσθω ῥητὴ ἡ ΔΓ, καὶ τῷ For, if possible, let it be (the same). And let a rational ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΓΔ παραβεβλήσθω ὀρθογώνιον (straight-line) DC be laid down. And let the rectangle τὸ ΓΕ πλάτος ποιοῦν τὴν ΔΕ. ἐπεὶ οὖν ἀποτομή ἐστιν ἡ ΑΒ, CE, equal to the (square) on AB, have been applied to ἀποτομὴ πρώτη ἐστὶν ἡ ΔΕ. ἔστω αὐτῇ προσαρμόζουσα ἡ CD, producing DE as breadth. Therefore, since AB is an ΕΖ· αἱ ΔΖ, ΖΕ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ apotome, DE is a first apotome [Prop. 10.97]. Let EF ἡ ΔΖ τῆς ΖΕ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ be an attachment to it. Thus, DF and FE are rational ΔΖ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ ΔΓ. πάλιν, (straight-lines which are) commensurable in square only, ἐπεὶ ἐκ δύο ὀνομάτων ἐστὶν ἡ ΑΒ, ἐκ δύο ἄρα ὀνομάτων and the square on DF is greater than (the square on) FE πρώτη ἐστὶν ἡ ΔΕ. διῃρήσθω εἰς τὰ ὀνόματα κατὰ τὸ Η, by the (square) on (some straight-line) commensurable καὶ ἔστω μεῖζον ὄνομα τὸ ΔΗ· αἱ ΔΗ, ΗΕ ἄρα ῥηταί εἰσι (in length) with (DF ), and DF is commensurable in δυνάμει μόνον σύμμετροι, καὶ ἡ ΔΗ τῆς ΗΕ μεῖζον δύναται length with the (previously) laid down rational (straight- τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ τὸ μεῖζον ἡ ΔΗ σύμμετρός line) DC [Def. 10.10]. Again, since AB is a binomial, ἐστι τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ ΔΓ. καὶ ἡ ΔΖ ἄρα τῇ ΔΗ DE is thus a first binomial [Prop. 10.60]. Let (DE) have σύμμετρός ἐστι μήκει· καὶ λοιπὴ ἄρα ἡ ΗΖ σύμμετρός ἐστι been divided into its (component) terms at G, and let τῇ ΔΖ μήκει. [ἐπεὶ οῦν σύμμετρός ἐστιν ἡ ΔΖ τῇ ΗΖ, ῥητὴ DG be the greater term. Thus, DG and GE are rational δέ ἐστιν ἡ ΔΖ, ῥητὴ ἄρα ἐστὶ καὶ ἡ ΗΖ. ἐπεὶ οὖν σύμμετρός (straight-lines which are) commensurable in square only, ἐστιν ἡ ΔΖ τῇ ΗΖ μήκει] ἀσύμμετρος δὲ ἡ ΔΖ τῇ ΕΖ μήκει. and the square on DG is greater than (the square on) ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΖΗ τῇ ΕΖ μήκει. αἱ ΗΖ, ΖΕ ἄρα GE by the (square) on (some straight-line) commensu- ῥηταί [εἰσι] δυνάμει μόνον σύμμετροι· ἀποτομὴ ἄρα ἐστὶν ἡ rable (in length) with (DG), and the greater (term) DG ΕΗ. ἀλλὰ καὶ ῥητή· ὅπερ ἐστὶν ἀδύνατον. is commensurable in length with the (previously) laid ῾Η ἄρα ἀποτομὴ οὐκ ἔστιν ἡ αὐτὴ τῇ ἐκ δύο ὀνομάτων· down rational (straight-line) DC [Def. 10.5]. Thus, DF ὅπερ ἔδει δεῖξαι. is also commensurable in length with DG [Prop. 10.12]. The remainder GF is thus commensurable in length with DF [Prop. 10.15]. [Therefore, since DF is commensu- rable with GF , and DF is rational, GF is thus also ra- tional. Therefore, since DF is commensurable in length with GF ,] DF (is) incommensurable in length with EF . Thus, FG is also incommensurable in length with EF [Prop. 10.13]. GF and FE [are] thus rational (straight- lines which are) commensurable in square only. Thus, 415 STOIQEIWN iþ. ELEMENTS BOOK 10 EG is an apotome [Prop. 10.73]. But, (it is) also ratio- nal. The very thing is impossible. Thus, an apotome is not the same as a binomial. (Which is) the very thing it was required to show.[Pìrisma.℄ [Corollary] ῾Η ἀποτομὴ καὶ αἱ μετ᾿ αὐτὴν ἄλογοι οὔτε τῇ μέσῃ οὔτε The apotome and the irrational (straight-lines) after it ἀλλήλαις εἰσὶν αἱ αὐταί. are neither the same as a medial (straight-line) nor (the Τὸ μὲν γὰρ ἀπὸ μέσης παρὰ ῥητὴν παραβαλλόμενον same) as one another. πλάτος ποιεῖ ῥητὴν καὶ ἀσύμμετρον τῇ, παρ᾿ ἣν παράκειται, For the (square) on a medial (straight-line), applied μήκει, τὸ δὲ ἀπὸ ἀποτομῆς παρὰ ῥητὴν παραβαλλόμενον to a rational (straight-line), produces as breadth a ratio- πλάτος ποιεῖ ἀποτομὴν πρώτην, τὸ δὲ ἀπὸ μέσης ἀποτομῆς nal (straight-line which is) incommensurable in length πρώτης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν with the (straight-line) to which (the area) is applied δευτέραν, τὸ δὲ ἀπὸ μέσης ἀποτομῆς δευτέρας παρὰ ῥητὴν [Prop. 10.22]. And the (square) on an apotome, ap- παραβαλλόμενον πλάτος ποιεῖ ἀποτομὴν τρίτην, τὸ δὲ ἀπὸ plied to a rational (straight-line), produces as breadth a ἐλάσσονος παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀπο- first apotome [Prop. 10.97]. And the (square) on a first τομὴν τετάρτην, τὸ δὲ ἀπὸ τῆς μετὰ ῥητοῦ μέσον τὸ ὅλον apotome of a medial (straight-line), applied to a ratio- ποιούσης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀπο- nal (straight-line), produces as breadth a second apotome τομὴν πέμπτην, τὸ δὲ ἀπὸ τῆς μετὰ μέσου μέσον τὸ ὅλον [Prop. 10.98]. And the (square) on a second apotome of ποιούσης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ ἀπο- a medial (straight-line), applied to a rational (straight- τομὴν ἕκτην. ἐπεὶ οὖν τὰ εἰρημένα πλάτη διαφέρει τοῦ line), produces as breadth a third apotome [Prop. 10.99]. τε πρώτου καὶ ἀλλήλων, τοῦ μὲν πρώτου, ὅτι ῥητή ἐστιν, And (square) on a minor (straight-line), applied to a ra- ἀλλήλων δὲ, ἐπεὶ τῇ τάξει οὐκ εἰσὶν αἱ αὐταί, δῆλον, ὡς καὶ tional (straight-line), produces as breadth a fourth apo- αὐταὶ αἱ ἄλογοι διαφέρουσιν ἀλλήλων. καὶ ἐπεὶ δέδεικται tome [Prop. 10.100]. And (square) on that (straight-line) ἡ ἀποτομὴ οὐκ οὖσα ἡ αὐτὴ τῇ ἐκ δύο ὀνομάτων, ποιοῦσι which with a rational (area) produces a medial whole, δὲ πλάτη παρὰ ῥητὴν παραβαλλόμεναι αἱ μετὰ τὴν ἀποτομὴν applied to a rational (straight-line), produces as breadth ἀποτομὰς ἀκολούθως ἑκάστη τῇ τάξει τῇ καθ᾿ αὑτήν, αἱ δὲ a fifth apotome [Prop. 10.101]. And (square) on that μετὰ τὴν ἐκ δύο ὀνομάτων τὰς ἐκ δύο ὀνομάτων καὶ αὐταὶ (straight-line) which with a medial (area) produces a τῇ τάξει ἀκολούθως, ἕτεραι ἄρα εἰσὶν αἱ μετὰ τὴν ἀποτομὴν medial whole, applied to a rational (straight-line), pro- καὶ ἕτεραι αἱ μετὰ τὴν ἐκ δύο ὀνομάτων, ὡς εἶναι τῇ τάξει duces as breadth a sixth apotome [Prop. 10.102]. There- πάσας ἀλόγους ιγ, fore, since the aforementioned breadths differ from the first (breadth), and from one another—from the first, be- cause it is rational, and from one another since they are not the same in order—clearly, the irrational (straight- lines) themselves also differ from one another. And since it has been shown that an apotome is not the same as a binomial [Prop. 10.111], and (that) the (irrational straight-lines) after the apotome, being applied to a ra- tional (straight-line), produce as breadth, each according to its own (order), apotomes, and (that) the (irrational straight-lines) after the binomial themselves also (pro- duce as breadth), according (to their) order, binomials, the (irrational straight-lines) after the apotome are thus different, and the (irrational straight-lines) after the bi- nomial (are also) different, so that there are, in order, 13 irrational (straight-lines) in all: 416 STOIQEIWN iþ. ELEMENTS BOOK 10 Μέσην, ᾿Εκ δύο ὀνομάτων, ᾿Εκ δύο μέσων πρώτην, ᾿Εκ δύο μέσων δευτέραν, Μείζονα, ῾Ρητὸν καὶ μέσον δυναμένην, Δύο μέσα δυναμένην, Ἀποτομήν, Μἑσης ἀποτομὴν πρώτην, Μἑσης ἀποτομὴν δευτέραν, ᾿Ελάσσονα, Μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσαν, Medial, Binomial, First bimedial, Second bimedial, Major, Square-root of a rational plus a medial (area), Square-root of (the sum of) two medial (areas), Apotome, First apotome of a medial, Second apotome of a medial, Minor, That which with a rational (area) produces a medial Μετὰ μέσου μέσον τὸ ὅλον ποιοῦσαν. whole, That which with a medial (area) produces a medial whole.ribþ. Proposition 112† Τὸ ἀπὸ ῥητῆς παρὰ τὴν ἐκ δύο ὀνομάτων παρα- The (square) on a rational (straight-line), applied to βαλλόμενον πλάτος ποιεῖ ἀποτομήν, ἧς τὰ ὀνόματα σύμμετρά a binomial (straight-line), produces as breadth an apo- ἐστι τοῖς τῆς ἐκ δύο ὀνομάτων ὀνόμασι καὶ ἔτι ἐν τῷ αὐτῷ tome whose terms are commensurable (in length) with λόγῳ, καὶ ἔτι ἡ γινομένη ἀποτομὴ τὴν αὐτὴν ἕξει τάξιν τῇ the terms of the binomial, and, furthermore, in the same ἐκ δύο ὀνομάτων. ratio. Moreover, the created apotome will have the same order as the binomial. Ε Β ∆ Γ Κ Ζ Η Θ Α G B D K E F H C A ῎Εστω ῥητὴ μὲν ἡ Α, ἐκ δύο ὀνομάτων δὲ ἡ ΒΓ, ἧς Let A be a rational (straight-line), and BC a binomial μεῖζον ὄνομα ἔστω ἡ ΔΓ, καὶ τῷ ἀπὸ τῆς Α ἴσον ἔστω (straight-line), of which let DC be the greater term. And τὸ ὑπὸ τῶν ΒΓ, ΕΖ· λέγω, ὅτι ἡ ΕΖ ἀποτομή ἐστιν, ἧς τὰ let the (rectangle contained) by BC and EF be equal to ὀνόματα σύμμετρά ἐστι τοῖς ΓΔ, ΔΒ, καὶ ἐν τῷ αὐτῷ λόγῳ, the (square) on A. I say that EF is an apotome whose καὶ ἔτι ἡ ΕΖ τὴν αὐτὴν ἕξει τάξιν τῇ ΒΓ. terms are commensurable (in length) with CD and DB, ῎Εστω γὰρ πάλιν τῷ ἀπὸ τῆς Α ἴσον τὸ ὑπὸ τῶν ΒΔ, and in the same ratio, and, moreover, that EF will have Η. ἐπεὶ οὖν τὸ ὑπὸ τῶν ΒΓ, ΕΖ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΒΔ, the same order as BC. Η, ἔστιν ἄρα ὡς ἡ ΓΒ πρὸς τὴν ΒΔ, οὕτως ἡ Η πρὸς τὴν For, again, let the (rectangle contained) by BD and G ΕΖ. μείζων δὲ ἡ ΓΒ τῆς ΒΔ· μείζων ἄρα ἐστὶ καὶ ἡ Η τῆς be equal to the (square) on A. Therefore, since the (rect- ΕΖ. ἔστω τῇ Η ἴση ἡ ΕΘ· ἔστιν ἄρα ὡς ἡ ΓΒ πρὸς τὴν angle contained) by BC and EF is equal to the (rectan- ΒΔ, οὕτως ἡ ΘΕ πρὸς τὴν ΕΖ· διελόντι ἄρα ἐστὶν ὡς ἡ ΓΔ gle contained) by BD and G, thus as CB is to BD, so G πρὸς τὴν ΒΔ, οὕτως ἡ ΘΖ πρὸς τὴν ΖΕ. γεγονέτω ὡς ἡ (is) to EF [Prop. 6.16]. And CB (is) greater than BD. ΘΖ πρὸς τὴν ΖΕ, οὕτως ἡ ΖΕ πρὸς τὴν ΚΕ· καὶ ὅλη ἄρα ἡ Thus, G is also greater than EF [Props. 5.16, 5.14]. Let ΘΚ πρὸς ὅλην τὴν ΚΖ ἐστιν, ὡς ἡ ΖΚ πρὸς ΚΕ· ὡς γὰρ ἓν EH be equal to G. Thus, as CB is to BD, so HE (is) to τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ EF . Thus, via separation, as CD is to BD, so HF (is) ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα. ὡς δὲ ἡ ΖΚ πρὸς ΚΕ, to FE [Prop. 5.17]. Let it have been contrived that as οὕτως ἐστὶν ἡ ΓΔ πρὸς τὴν ΔΒ· καὶ ὡς ἄρα ἡ ΘΚ πρὸς ΚΖ, HF (is) to FE, so FK (is) to KE. And, thus, the whole οὕτως ἡ ΓΔ πρὸς τὴν ΔΒ. σύμμετρον δὲ τὸ ἀπὸ τῆς ΓΔ τῷ HK is to the whole KF , as FK (is) to KE. For as one ἀπὸ τῆς ΔΒ· σύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΘΚ τῷ of the leading (proportional magnitudes is) to one of the 417 STOIQEIWN iþ. ELEMENTS BOOK 10 ἀπὸ τῆς ΚΖ. καί ἐστιν ὡς τὸ ἀπὸ τῆς ΘΚ πρὸς τὸ ἀπὸ τῆς following, so all of the leading (magnitudes) are to all of ΚΖ, οὕτως ἡ ΘΚ πρὸς τὴν ΚΕ, ἐπεὶ αἱ τρεῖς αἱ ΘΚ, ΚΖ, the following [Prop. 5.12]. And as FK (is) to KE, so CD ΚΕ ἀνάλογόν εἰσιν. σύμμετρος ἄρα ἡ ΘΚ τῇ ΚΕ μήκει. is to DB [Prop. 5.11]. And, thus, as HK (is) to KF , so ὥστε καὶ ἡ ΘΕ τῇ ΕΚ σύμμετρός ἐστι μήκει. καὶ ἐπεὶ τὸ CD is to DB [Prop. 5.11]. And the (square) on CD (is) ἀπὸ τῆς Α ἴσον ἐστὶ τῷ ὑπὸ τῶν ΕΘ, ΒΔ, ῥητὸν δέ ἐστι commensurable with the (square) on DB [Prop. 10.36]. τὸ ἀπὸ τῆς Α, ῥητὸν ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν ΕΘ, ΒΔ. καὶ The (square) on HK is thus also commensurable with παρὰ ῥητὴν τὴν ΒΔ παράκειται· ῥητὴ ἄρα ἐστὶν ἡ ΕΘ καὶ the (square) on KF [Props. 6.22, 10.11]. And as the σύμμετρος τῇ ΒΔ μήκει· ὥστε καὶ ἡ σύμμετρος αὐτῇ ἡ ΕΚ (square) on HK is to the (square) on KF , so HK (is) to ῥητή ἐστι καὶ σύμμετρος τῇ ΒΔ μήκει. ἐπεὶ οὖν ἐστιν ὡς ἡ KE, since the three (straight-lines) HK, KF , and KE ΓΔ πρὸς ΔΒ, οὕτως ἡ ΖΚ πρὸς ΚΕ, αἱ δὲ ΓΔ, ΔΒ δυνάμει are proportional [Def. 5.9]. HK is thus commensurable μόνον εἰσὶ σύμμετροι, καὶ αἱ ΖΚ, ΚΕ δυνάμει μόνον εἰσὶ in length with KE [Prop. 10.11]. Hence, HE is also com- σύμμετροι. ῥητὴ δέ ἐστιν ἡ ΚΕ· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΚ. αἱ mensurable in length with EK [Prop. 10.15]. And since ΖΚ, ΚΕ ἄρα ῥηταὶ δυνάμει μόνον εἰσὶ σύμμετροι· ἀποτομὴ the (square) on A is equal to the (rectangle contained) by ἄρα ἐστὶν ἡ ΕΖ. EH and BD, and the (square) on A is rational, the (rect- ῎Ητοι δὲ ἡ ΓΔ τῆς ΔΒ μεῖζον δύναται τῷ ἀπὸ συμμέτρου angle contained) by EH and BD is thus also rational. ἑαυτῇ ἢ τῷ ἀπὸ ἀσυμμέτρου. And it is applied to the rational (straight-line) BD. Thus, Εἰ μὲν οὖν ἡ ΓΔ τῆς ΔΒ μεῖζον δύναται τῷ ἀπὸ EH is rational, and commensurable in length with BD συμμέτρου [ἑαυτῇ], καὶ ἡ ΖΚ τῆς ΚΕ μεῖζον δυνήσεται τῷ [Prop. 10.20]. And, hence, the (straight-line) commensu- ἀπὸ συμμέτρου ἑαυτῇ. καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΓΔ τῇ rable (in length) with it, EK, is also rational [Def. 10.3], ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΖΚ· εἰ δὲ ἡ ΒΔ, καὶ ἡ ΚΕ· εἰ and commensurable in length with BD [Prop. 10.12]. δὲ οὐδετέρα τῶν ΓΔ, ΔΒ, καὶ οὐδετέρα τῶν ΖΚ, ΚΕ. Therefore, since as CD is to DB, so FK (is) to KE, and Εἰ δὲ ἡ ΓΔ τῆς ΔΒ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου CD and DB are (straight-lines which are) commensu- ἑαυτῇ, καὶ ἡ ΖΚ τῆς ΚΕ μεῖζον δυνήσεται τῷ ἀπὸ rable in square only, FK and KE are also commensu- ἀσυμμέτρου ἑαυτῇ. καὶ εἰ μὲν ἡ ΓΔ σύμμετρός ἐστι τῇ rable in square only [Prop. 10.11]. And KE is rational. ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΖΚ· εἰ δὲ ἡ ΒΔ, καὶ ἡ ΚΕ· εἰ Thus, FK is also rational. FK and KE are thus rational δὲ οὐδετέρα τῶν ΓΔ, ΔΒ, καὶ οὐδετέρα τῶν ΖΚ, ΚΕ· ὥστε (straight-lines which are) commensurable in square only. ἀποτομή ἐστιν ἡ ΖΕ, ἧς τὰ ὀνόματα τὰ ΖΚ, ΚΕ σύμμετρά Thus, EF is an apotome [Prop. 10.73]. ἐστι τοῖς τῆς ἐκ δύο ὀνομάτων ὀνόμασι τοῖς ΓΔ, ΔΒ καὶ ἐν And the square on CD is greater than (the square on) τῷ αὐτῷ λόγῳ, καὶ τὴν αὐτῆν τάξιν ἔχει τῇ ΒΓ· ὅπερ ἔδει DB either by the (square) on (some straight-line) com- δεῖξαι. mensurable, or by the (square) on (some straight-line) incommensurable, (in length) with (CD). Therefore, if the square on CD is greater than (the square on) DB by the (square) on (some straight-line) commensurable (in length) with [CD] then the square on FK will also be greater than (the square on) KE by the (square) on (some straight-line) commensurable (in length) with (FK) [Prop. 10.14]. And if CD is com- mensurable in length with a (previously) laid down ra- tional (straight-line), (so) also (is) FK [Props. 10.11, 10.12]. And if BD (is commensurable), (so) also (is) KE [Prop. 10.12]. And if neither of CD or DB (is com- mensurable), neither also (are) either of FK or KE. And if the square on CD is greater than (the square on) DB by the (square) on (some straight-line) incom- mensurable (in length) with (CD) then the square on FK will also be greater than (the square on) KE by the (square) on (some straight-line) incommensurable (in length) with (FK) [Prop. 10.14]. And if CD is com- mensurable in length with a (previously) laid down ra- tional (straight-line), (so) also (is) FK [Props. 10.11, 10.12]. And if BD (is commensurable), (so) also (is) KE 418 STOIQEIWN iþ. ELEMENTS BOOK 10 [Prop. 10.12]. And if neither of CD or DB (is commen- surable), neither also (are) either of FK or KE. Hence, FE is an apotome whose terms, FK and KE, are com- mensurable (in length) with the terms, CD and DB, of the binomial, and in the same ratio. And (FE) has the same order as BC [Defs. 10.5—10.10]. (Which is) the very thing it was required to show. † Heiberg considers this proposition, and the succeeding ones, to be relatively early interpolations into the original text.rigþ. Proposition 113 Τὸ ἀπὸ ῥητῆς παρὰ ἀποτομὴν παραβαλλόμενον πλάτος The (square) on a rational (straight-line), applied to ποιεῖ τὴν ἐκ δύο ὀνομάτων, ἧς τὰ ὀνόματα σύμμετρά ἐστι an apotome, produces as breadth a binomial whose terms τοῖς τῆς ἀποτομῆς ὀνόμασι καὶ ἐν τῷ αὐτῷ λόγῳ, ἔτι δὲ ἡ are commensurable with the terms of the apotome, and γινομένη ἐκ δύο ὀνομάτων τὴν αὐτὴν τάξιν ἔχει τῇ ἀποτομῇ. in the same ratio. Moreover, the created binomial has the same order as the apotome. Ε Β ∆ Γ Κ Ζ Η Θ Α A B D K E C F H G ῎Εστω ῥητὴ μὲν ἡ Α, ἀποτομὴ δὲ ἡ ΒΔ, καὶ τῷ ἀπὸ τῆς Let A be a rational (straight-line), and BD an apo- Α ἴσον ἔστω τὸ ὑπὸ τῶν ΒΔ, ΚΘ, ὥστε τὸ ἀπὸ τῆς Α ῥητῆς tome. And let the (rectangle contained) by BD and KH παρὰ τὴν ΒΔ ἀποτομὴν παραβαλλόμενον πλάτος ποιεῖ τὴν be equal to the (square) on A, such that the square on the ΚΘ· λέγω, ὅτι ἐκ δύο ὀνομάτων ἐστὶν ἡ ΚΘ, ἧς τὰ ὀνόματα rational (straight-line) A, applied to the apotome BD, σύμμετρά ἐστι τοῖς τῆς ΒΔ ὀνόμασι καὶ ἐν τῷ αὐτῷ λόγῳ, produces KH as breadth. I say that KH is a binomial καὶ ἔτι ἡ ΚΘ τὴν αὐτὴν ἔχει τάξιν τῇ ΒΔ. whose terms are commensurable with the terms of BD, ῎Εστω γὰρ τῇ ΒΔ προσαρμόζουσα ἡ ΔΓ· αἱ ΒΓ, ΓΔ and in the same ratio, and, moreover, that KH has the ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ τῷ ἀπὸ τῆς same order as BD. Α ἴσον ἔστω καὶ τὸ ὑπὸ τῶν ΒΓ, Η. ῥητὸν δὲ τὸ ἀπὸ τῆς For let DC be an attachment to BD. Thus, BC and Α· ῥητὸν ἄρα καὶ τὸ ὑπὸ τῶν ΒΓ, Η. καὶ παρὰ ῥητὴν τὴν CD are rational (straight-lines which are) commensu- ΒΓ παραβέβληται· ῥητὴ ἄρα ἐστὶν ἡ Η καὶ σύμμετρος τῇ ΒΓ rable in square only [Prop. 10.73]. And let the (rectangle μήκει. ἐπεὶ οὖν τὸ ὑπὸ τῶν ΒΓ, Η ἴσον ἐστὶ τῷ ὑπὸ τῶν contained) by BC and G also be equal to the (square) ΒΔ, ΚΘ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΓΒ πρὸς ΒΔ, οὕτως ἡ on A. And the (square) on A (is) rational. The (rect- ΚΘ πρὸς Η. μείζων δὲ ἡ ΒΓ τῆς ΒΔ· μείζων ἄρα καὶ ἡ ΚΘ angle contained) by BC and G (is) thus also rational. τῆς Η. κείσθω τῇ Η ἴση ἡ ΚΕ· σύμμετρος ἄρα ἐστὶν ἡ ΚΕ And it has been applied to the rational (straight-line) τῇ ΒΓ μήκει. καὶ ἐπεί ἐστιν ὡς ἡ ΓΒ πρὸς ΒΔ, οὕτως ἡ BC. Thus, G is rational, and commensurable in length ΘΚ πρὸς ΚΕ, ἀναστρέψαντι ἄρα ἐστὶν ὡς ἡ ΒΓ πρὸς τὴν with BC [Prop. 10.20]. Therefore, since the (rectangle ΓΔ, οὕτως ἡ ΚΘ πρὸς ΘΕ. γεγονέτω ὡς ἡ ΚΘ πρὸς ΘΕ, contained) by BC and G is equal to the (rectangle con- οὕτως ἡ ΘΖ πρὸς ΖΕ· καὶ λοιπὴ ἄρα ἡ ΚΖ πρὸς ΖΘ ἐστιν, tained) by BD and KH , thus, proportionally, as CB is to ὡς ἡ ΚΘ πρὸς ΘΕ, τουτέστιν [ὡς] ἡ ΒΓ πρὸς ΓΔ. αἱ δὲ BD, so KH (is) to G [Prop. 6.16]. And BC (is) greater ΒΓ, ΓΔ δυνάμει μόνον [εἰσὶ] σύμμετροι· καὶ αἱ ΚΖ, ΖΘ ἄρα than BD. Thus, KH (is) also greater than G [Prop. 5.16, δυνάμει μόνον εἰσὶ σύμμετροι· καὶ ἐπεί ἐστιν ὡς ἡ ΚΘ πρὸς 5.14]. Let KE be made equal to G. KE is thus com- ΘΕ, ἡ ΚΖ πρὸς ΖΘ, ἀλλ᾿ ὡς ἡ ΚΘ πρὸς ΘΕ, ἡ ΘΖ πρὸς mensurable in length with BC. And since as CB is to ΖΕ, καὶ ὡς ἄρα ἡ ΚΖ πρὸς ΖΘ, ἡ ΘΖ πρὸς ΖΕ· ὥστε καὶ BD, so HK (is) to KE, thus, via conversion, as BC (is) ὡς ἡ πρώτη πρὸς τὴν τρίτην, τὸ ἀπὸ τῆς πρώτης πρὸς τὸ to CD, so KH (is) to HE [Prop. 5.19 corr.]. Let it have ἀπὸ τῆς δευτέρας· καὶ ὡς ἄρα ἡ ΚΖ πρὸς ΖΕ, οὕτως τὸ ἀπὸ been contrived that as KH (is) to HE, so HF (is) to τῆς ΚΖ πρὸς τὸ ἀπὸ τῆς ΖΘ. σύμμετρον δέ ἐστι τὸ ἀπὸ τῆς FE. And thus the remainder KF is to FH , as KH (is) ΚΖ τῷ ἀπὸ τῆς ΖΘ· αἱ γὰρ ΚΖ, ΖΘ δυνάμει εἰσὶ σύμμετροι· to HE—that is to say, [as] BC (is) to CD [Prop. 5.19]. σύμμετρος ἄρα ἐστὶ καὶ ἡ ΚΖ τῇ ΖΕ μήκει· ὥστε ἡ ΚΖ καὶ And BC and CD [are] commensurable in square only. 419 STOIQEIWN iþ. ELEMENTS BOOK 10 τῇ ΚΕ σύμμετρός [ἐστι] μήκει. ῥητὴ δέ ἐστιν ἡ ΚΕ καὶ KF and FH are thus also commensurable in square only σύμμετρος τῇ ΒΓ μήκει. ῥητὴ ἄρα καὶ ἡ ΚΖ καὶ σύμμετρος [Prop. 10.11]. And since as KH is to HE, (so) KF (is) τῇ ΒΓ μήκει. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς ΓΔ, οὕτως ἡ ΚΖ to FH , but as KH (is) to HE, (so) HF (is) to FE, thus, πρὸς ΖΘ, ἐναλλὰξ ὡς ἡ ΒΓ πρὸς ΚΖ, οὕτως ἡ ΔΓ πρὸς ΖΘ. also as KF (is) to FH , (so) HF (is) to FE [Prop. 5.11]. σύμμετρος δὲ ἡ ΒΓ τῇ ΚΖ· σύμμετρος ἄρα καὶ ἡ ΖΘ τῇ ΓΔ And hence as the first (is) to the third, so the (square) on μήκει. αἱ ΒΓ, ΓΔ δὲ ῥηταί εἰσι δυνάμει μόνον σύμμετροι· the first (is) to the (square) on the second [Def. 5.9]. And καὶ αἱ ΚΖ, ΖΘ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ thus as KF (is) to FE, so the (square) on KF (is) to the δύο ὀνομάτων ἐστὶν ἄρα ἡ ΚΘ. (square) on FH . And the (square) on KF is commen- Εἰ μὲν οὖν ἡ ΒΓ τῆς ΓΔ μεῖζον δύναται τῷ ἀπὸ surable with the (square) on FH . For KF and FH are συμμέτρου ἑαυτῇ, καὶ ἡ ΚΖ τῆς ΖΘ μεῖζον δυνήσεται τῷ commensurable in square. Thus, KF is also commensu- ἀπὸ συμμέτρου ἑαυτῇ. καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΒΓ τῇ rable in length with FE [Prop. 10.11]. Hence, KF [is] ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΚΖ, εἰ δὲ ἡ ΓΔ σύμμετρός ἐστι also commensurable in length with KE [Prop. 10.15]. τῇ ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ΖΘ, εἰ δὲ οὐδετέρα τῶν ΒΓ, And KE is rational, and commensurable in length with ΓΔ, οὐδετέρα τῶν ΚΖ, ΖΘ. BC. Thus, KF (is) also rational, and commensurable in Εἰ δὲ ἡ ΒΓ τῆς ΓΔ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου length with BC [Prop. 10.12]. And since as BC is to ἑαυτῇ, καὶ ἡ ΚΖ τῆς ΖΘ μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρ- CD, (so) KF (is) to FH , alternately, as BC (is) to KF , ου ἑαυτῇ. καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΒΓ τῇ ἐκκειμένῃ so DC (is) to FH [Prop. 5.16]. And BC (is) commen- ῥητῇ μήκει, καὶ ἡ ΚΖ, εἰ δὲ ἡ ΓΔ, καὶ ἡ ΖΘ, εἰ δὲ οὐδετέρα surable (in length) with KF . Thus, FH (is) also com- τῶν ΒΓ, ΓΔ, οὐδετέρα τῶν ΚΖ, ΖΘ. mensurable in length with CD [Prop. 10.11]. And BC ᾿Εκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΚΘ, ἧς τὰ ὀνόματα τὰ and CD are rational (straight-lines which are) commen- ΚΖ, ΖΘ σύμμετρά [ἐστι] τοῖς τῆς ἀποτομῆς ὀνόμασι τοῖς surable in square only. KF and FH are thus also ratio- ΒΓ, ΓΔ καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ ἔτι ἡ ΚΘ τῇ ΒΓ τὴν nal (straight-lines which are) commensurable in square αὐτὴν ἕξει τάξιν· ὅπερ ἔδει δεῖξαι. only [Def. 10.3, Prop. 10.13]. Thus, KH is a binomial [Prop. 10.36]. Therefore, if the square on BC is greater than (the square on) CD by the (square) on (some straight-line) commensurable (in length) with (BC), then the square on KF will also be greater than (the square on) FH by the (square) on (some straight-line) commensurable (in length) with (KF ) [Prop. 10.14]. And if BC is com- mensurable in length with a (previously) laid down ra- tional (straight-line), (so) also (is) KF [Prop. 10.12]. And if CD is commensurable in length with a (previ- ously) laid down rational (straight-line), (so) also (is) FH [Prop. 10.12]. And if neither of BC or CD (are commensurable), neither also (are) either of KF or FH [Prop. 10.13]. And if the square on BC is greater than (the square on) CD by the (square) on (some straight-line) incom- mensurable (in length) with (BC) then the square on KF will also be greater than (the square on) FH by the (square) on (some straight-line) incommensurable (in length) with (KF ) [Prop. 10.14]. And if BC is com- mensurable in length with a (previously) laid down ratio- nal (straight-line), (so) also (is) KF [Prop. 10.12]. And if CD is commensurable, (so) also (is) FH [Prop. 10.12]. And if neither of BC or CD (are commensurable), nei- ther also (are) either of KF or FH [Prop. 10.13]. KH is thus a binomial whose terms, KF and FH , [are] commensurable (in length) with the terms, BC and CD, of the apotome, and in the same ratio. Moreover, 420 STOIQEIWN iþ. ELEMENTS BOOK 10 KH will have the same order as BC [Defs. 10.5—10.10]. (Which is) the very thing it was required to show.ridþ. Proposition 114 ᾿Εὰν χωρίον περιέχηται ὑπὸ ἀποτομῆς καὶ τῆς ἐκ δύο If an area is contained by an apotome, and a binomial ὀνομάτων, ἧς τὰ ὀνόματα σύμμετρά τέ ἐστι τοῖς τῆς ἀπο- whose terms are commensurable with, and in the same τομῆς ὀνόμασι καὶ ἐν τῷ αὐτῷ λόγῳ, ἡ τὸ χωρίον δυναμένη ratio as, the terms of the apotome then the square-root of ῥητή ἐστιν. the area is a rational (straight-line). ΖΑ Β Γ Ε ∆ Η Κ Λ Μ Θ D A B E K L M C G H F Περιεχέσθω γὰρ χωρίον τὸ ὑπὸ τῶν ΑΒ, ΓΔ ὑπὸ For let an area, the (rectangle contained) by AB and ἀποτομῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων τῆς ΓΔ, ἧς CD, have been contained by the apotome AB, and the μεῖζον ὄνομα ἔστω τὸ ΓΕ, καὶ ἔστω τὰ ὀνόματα τῆς ἐκ binomial CD, of which let the greater term be CE. And δύο ὀνομάτων τὰ ΓΕ, ΕΔ σύμμετρά τε τοῖς τῆς ἀποτομῆς let the terms of the binomial, CE and ED, be commen- ὀνόμασι τοῖς ΑΖ, ΖΒ καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ ἔστω ἡ τὸ surable with the terms of the apotome, AF and FB (re- ὑπὸ τῶν ΑΒ, ΓΔ δυναμένη ἡ Η· λέγω, ὅτι ῥητή ἐστιν ἡ Η. spectively), and in the same ratio. And let the square-root ᾿Εκκείσθω γὰρ ῥητὴ ἡ Θ, καὶ τῷ ἀπὸ τῆς Θ ἴσον παρὰ of the (rectangle contained) by AB and CD be G. I say τὴν ΓΔ παραβεβλήσθω πλάτος ποιοῦν τὴν ΚΛ· ἀποτομὴ ἄρα that G is a rational (straight-line). ἐστὶν ἡ ΚΛ, ἧς τὰ ὀνόματα ἔστω τὰ ΚΜ, ΜΛ σύμμετρα τοῖς For let the rational (straight-line) H be laid down. τῆς ἐκ δύο ὀνομάτων ὀνόμασι τοῖς ΓΕ, ΕΔ καὶ ἐν τῷ αὐτῷ And let (some rectangle), equal to the (square) on H , λόγῳ. ἀλλὰ καὶ αἱ ΓΕ, ΕΔ σύμμετροί τέ εἰσι ταῖς ΑΖ, ΖΒ have been applied to CD, producing KL as breadth. καὶ ἐν τῷ αὐτῷ λόγῳ· ἔστιν ἄρα ὡς ἡ ΑΖ πρὸς τὴν ΖΒ, Thus, KL is an apotome, of which let the terms, KM οὕτως ἡ ΚΜ πρὸς ΜΛ. ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΖ πρὸς and ML, be commensurable with the terms of the bino- τὴν ΚΜ, οὕτως ἡ ΒΖ πρὸς τὴν ΛΜ· καὶ λοιπὴ ἄρα ἡ ΑΒ mial, CE and ED (respectively), and in the same ratio πρὸς λοιπὴν τὴν ΚΛ ἐστιν ὡς ἡ ΑΖ πρὸς ΚΜ. σύμμετρος [Prop. 10.112]. But, CE and ED are also commensu- δὲ ἡ ΑΖ τῇ ΚΜ· σύμμετρος ἄρα ἐστὶ καὶ ἡ ΑΒ τῇ ΚΛ. καί rable with AF and FB (respectively), and in the same ra- ἐστιν ὡς ἡ ΑΒ πρὸς ΚΛ, οὕτως τὸ ὑπὸ τῶν ΓΔ, ΑΒ πρὸς tio. Thus, as AF is to FB, so KM (is) to ML. Thus, alter- τὸ ὑπὸ τῶν ΓΔ, ΚΛ· σύμμετρον ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν nately, as AF is to KM , so BF (is) to LM [Prop. 5.16]. ΓΔ, ΑΒ τῷ ὑπὸ τῶν ΓΔ, ΚΛ. ἴσον δὲ τὸ ὑπὸ τῶν ΓΔ, ΚΛ Thus, the remainder AB is also to the remainder KL as τῷ ἀπὸ τῆς Θ· σύμμετρον ἄρα ἐστὶ τὸ ὑπὸ τῶν ΓΔ, ΑΒ τῷ AF (is) to KM [Prop. 5.19]. And AF (is) commensu- ἀπὸ τῆς Θ. τῷ δὲ ὑπὸ τῶν ΓΔ, ΑΒ ἴσον ἐστὶ τὸ ἀπὸ τῆς Η· rable with KM [Prop. 10.12]. AB is thus also commen- σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς Η τῷ ἀπὸ τῆς Θ. ῥητὸν δὲ surable with KL [Prop. 10.11]. And as AB is to KL, τὸ ἀπὸ τῆς Θ· ῥητὸν ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς Η· ῥητὴ ἄρα so the (rectangle contained) by CD and AB (is) to the ἐστὶν ἡ Η. καὶ δύναται τὸ ὑπὸ τῶν ΓΔ, ΑΒ. (rectangle contained) by CD and KL [Prop. 6.1]. Thus, ᾿Εὰν ἄρα χωρίον περιέχηται ὑπὸ ἀποτομῆς καὶ τῆς ἐκ the (rectangle contained) by CD and AB is also com- δύο ὀνομάτων, ἧς τὰ ὀνόματα σύμμετρά ἐστι τοῖς τῆς ἀπο- mensurable with the (rectangle contained) by CD and τομῆς ὀνόμασι καὶ ἐν τῷ αὐτῷ λόγῳ, ἡ τὸ χωρίον δυναμένη KL [Prop. 10.11]. And the (rectangle contained) by CD ῥητή ἐστιν. and KL (is) equal to the (square) on H . Thus, the (rect- angle contained) by CD and AB is commensurable with the (square) on H . And the (square) on G is equal to the (rectangle contained) by CD and AB. The (square) on G 421 STOIQEIWN iþ. ELEMENTS BOOK 10 is thus commensurable with the (square) on H . And the (square) on H (is) rational. Thus, the (square) on G is also rational. G is thus rational. And it is the square-root of the (rectangle contained) by CD and AB. Thus, if an area is contained by an apotome, and a binomial whose terms are commensurable with, and in the same ratio as, the terms of the apotome, then the square-root of the area is a rational (straight-line).Pìrisma. Corollary Καὶ γέγονεν ἡμῖν καὶ διὰ τούτου φανερόν, ὅτι δυνατόν And it has also been made clear to us, through this, ἐστι ῥητὸν χωρίον ὑπὸ ἀλόγων εὐθειῶν περιέχεσθαι. ὅπερ that it is possible for a rational area to be contained by ἔδει δεῖξαι. irrational straight-lines. (Which is) the very thing it was required to show.rieþ. Proposition 115 Ἀπὸ μέσης ἄπειροι ἄλογοι γίνονται, καὶ οὐδεμία οὐδεμιᾷ An infinite (series) of irrational (straight-lines) can be τῶν πρότερον ἡ αὐτή. created from a medial (straight-line), and none of them is the same as any of the preceding (straight-lines). Α Β Γ ∆ C B D A ῎Εστω μέση ἡ Α· λέγω, ὅτι ἀπὸ τῆς Α ἄπειροι ἄλογοι Let A be a medial (straight-line). I say that an infi- γίνονται, καὶ οὐδεμία οὐδεμιᾷ τῶν πρότερον ἡ αὐτή. nite (series) of irrational (straight-lines) can be created ᾿Εκκείσθω ῥητὴ ἡ Β, καὶ τῷ ὑπὸ τῶν Β, Α ἴσον ἔστω τὸ from A, and that none of them is the same as any of the ἀπὸ τῆς Γ· ἄλογος ἄρα ἐστὶν ἡ Γ· τὸ γὰρ ὑπὸ ἀλόγου καὶ preceding (straight-lines). ῥητῆς ἄλογόν ἐστιν. καὶ οὐδεμιᾷ τῶν πρότερον ἡ αὐτή· τὸ Let the rational (straight-line) B be laid down. And γὰρ ἀπ᾿ οὐδεμιᾶς τῶν πρότερον παρὰ ῥητὴν παραβαλλόμενον let the (square) on C be equal to the (rectangle con- πλάτος ποιεῖ μέσην. πάλιν δὴ τῷ ὑπὸ τῶν Β, Γ ἴσον ἔστω τὸ tained) by B and A. Thus, C is irrational [Def. 10.4]. ἀπὸ τῆς Δ· ἄλογον ἄρα ἐστὶ τὸ ἀπὸ τῆς Δ. ἄλογος ἄρα ἐστὶν For an (area contained) by an irrational and a rational ἡ Δ· καὶ οὐδεμιᾷ τῶν πρότερον ἡ αὐτή· τὸ γὰρ ἀπ᾿ οὐδεμιᾶς (straight-line) is irrational [Prop. 10.20]. And (C is) not τῶν πρότερον παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν the same as any of the preceding (straight-lines). For Γ. ὁμοίως δὴ τῆς τοιαύτης τάξεως ἐπ᾿ ἄπειρον προβαινούσης the (square) on none of the preceding (straight-lines), φανερόν, ὅτι ἀπὸ τῆς μέσης ἄπειροι ἄλογοι γίνονται, καὶ applied to a rational (straight-line), produces a medial οὐδεμία οὐδεμιᾷ τῶν πρότερον ἡ αὐτή· ὅπερ ἔδει δεῖξαι. (straight-line) as breadth. So, again, let the (square) on D be equal to the (rectangle contained) by B and C. Thus, the (square) on D is irrational [Prop. 10.20]. D is thus irrational [Def. 10.4]. And (D is) not the same as any of the preceding (straight-lines). For the (square) on none of the preceding (straight-lines), applied to a ratio- nal (straight-line), produces C as breadth. So, similarly, this arrangement being advanced to infinity, it is clear that an infinite (series) of irrational (straight-lines) can be created from a medial (straight-line), and that none of them is the same as any of the preceding (straight-lines). (Which is) the very thing it was required to show. 422 ELEMENTS BOOK 11 Elementary Stereometry 423 STOIQEIWN iaþ. ELEMENTS BOOK 11VOroi. Definitions αʹ. Στερεόν ἐστι τὸ μῆκος καὶ πλάτος καὶ βάθος ἔχον. 1. A solid is a (figure) having length and breadth and βʹ. Στερεοῦ δὲ πέρας ἐπιφάνεια. depth. γʹ. Εὐθεῖα πρὸς ἐπίπεδον ὀρθή ἐστιν, ὅταν πρὸς πάσας 2. The extremity of a solid (is) a surface. τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ [ὑποκειμένῳ] 3. A straight-line is at right-angles to a plane when it ἐπιπέδῳ ὀρθὰς ποιῇ γωνίας. makes right-angles with all of the straight-lines joined to δʹ. ᾿Επίπεδον πρὸς ἐπίπεδον ὀρθόν ἐστιν, ὅταν αἱ τῇ it which are also in the plane. κοινῇ τομῇ τῶν ἐπιπέδων πρὸς ὀρθὰς ἀγόμεναι εὐθεῖαι ἐν 4. A plane is at right-angles to a(nother) plane when ἑνὶ τῶν ἐπιπέδων τῷ λοιπῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν. (all of) the straight-lines drawn in one of the planes, at εʹ. Εὐθείας πρὸς ἐπίπεδον κλίσις ἐστίν, ὅταν ἀπὸ τοῦ right-angles to the common section of the planes, are at μετεώρου πέρατος τῆς εὐθείας ἐπὶ τὸ ἐπίπεδον κάθετος right-angles to the remaining plane. ἀχθῇ, καὶ ἀπὸ τοῦ γενομένου σημείου ἐπὶ τὸ ἐν τῷ ἐπιπέδῳ 5. The inclination of a straight-line to a plane is the πέρας τῆς εὐθείας εὐθεῖα ἐπιζευχθῇ, ἡ περιεχομένη γωνία angle contained by the drawn and standing (straight- ὑπὸ τῆς ἀχθείσης καὶ τῆς ἐφεστώσης. lines), when a perpendicular is lead to the plane from ϛʹ. ᾿Επιπέδου πρὸς ἐπίπεδον κλίσις ἐστὶν ἡ περιεχομένη the end of the (standing) straight-line raised (out of the ὀξεῖα γωνία ὑπὸ τῶν πρὸς ὀρθὰς τῇ κοινῇ τομῇ ἀγομένων plane), and a straight-line is (then) joined from the point πρὸς τῷ αὐτῷ σημείῳ ἐν ἑκατέρῳ τῶν ἐπιπέδων. (so) generated to the end of the (standing) straight-line ζʹ. ᾿Επίπεδον πρὸς ἐπίπεδον ὁμοίως κεκλίσθαι λέγεται (lying) in the plane. καὶ ἕτερον πρὸς ἕτερον, ὅταν αἱ εἰρημέναι τῶν κλίσεων 6. The inclination of a plane to a(nother) plane is the γωνίαι ἴσαι ἀλλήλαις ὦσιν. acute angle contained by the (straight-lines), (one) in ηʹ. Παράλληλα ἐπίπεδά ἐστι τὰ ἀσύμπτωτα. each of the planes, drawn at right-angles to the common θʹ. ῞Ομοια στερεὰ σχήματά ἐστι τὰ ὑπὸ ὁμοίων ἐπιπέδων segment (of the planes), at the same point. περιεχόμενα ἴσων τὸ πλῆθος. 7. A plane is said to have been similarly inclined to a ιʹ. ῎Ισα δὲ καὶ ὅμοια στερεὰ σχήματά ἐστι τὰ ὑπὸ ὁμοίων plane, as another to another, when the aforementioned ἐπιπέδων περιεχόμενα ἴσων τῷ πλήθει καὶ τῷ μεγέθει. angles of inclination are equal to one another. ιαʹ. Στερεὰ γωνία ἐστὶν ἡ ὑπὸ πλειόνων ἢ δύο γραμμῶν 8. Parallel planes are those which do not meet (one ἁπτομένων ἀλλήλων καὶ μὴ ἐν τῇ αὐτῇ ἐπιφανείᾳ οὐσῶν another). πρὸς πάσαις ταῖς γραμμαῖς κλίσις. ἄλλως· στερεὰ γωνία 9. Similar solid figures are those contained by equal ἐστὶν ἡ ὑπὸ πλειόνων ἢ δύο γωνιῶν ἐπιπέδων περιεχομένη numbers of similar planes (which are similarly arranged). μὴ οὐσῶν ἐν τῷ αὐτῷ ἐπιπέδῳ πρὸς ἑνὶ σημείῳ συνι- 10. But equal and similar solid figures are those con- σταμένων. tained by similar planes equal in number and in magni- ιβʹ. Πυραμίς ἐστι σχῆμα στερεὸν ἐπιπέδοις περιχόμενον tude (which are similarly arranged). ἀπὸ ἑνὸς ἐπιπέδου πρὸς ἑνὶ σημείῳ συνεστώς. 11. A solid angle is the inclination (constituted) by ιγʹ. Πρίσμα ἐστὶ σχῆμα στερεὸν ἐπιπέδοις περιεχόμενον, more than two lines joining one another (at the same ὧν δύο τὰ ἀπεναντίον ἴσα τε καὶ ὅμοιά ἐστι καὶ παράλληλα, point), and not being in the same surface, to all of the τὰ δὲ λοιπὰ παραλληλόγραμμα. lines. Otherwise, a solid angle is that contained by more ιδʹ. Σφαῖρά ἐστιν, ὅταν ἡμικυκλίου μενούσης τῆς διαμέτ- than two plane angles, not being in the same plane, and ρου περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ πάλιν ἀποκατα- constructed at one point. σταθῇ, ὅθεν ἤρξατο φέρεσθαι, τὸ περιληφθὲν σχῆμα. 12. A pyramid is a solid figure, contained by planes, ιεʹ. ῎Αξων δὲ τῆς σφαίρας ἐστὶν ἡ μένουσα εὐθεῖα, περὶ (which is) constructed from one plane to one point. ἣν τὸ ἡμικύκλιον στρέφεται. 13. A prism is a solid figure, contained by planes, of ιϛʹ. Κέντρον δὲ τῆς σφαίρας ἐστὶ τὸ αὐτό, ὃ καὶ τοῦ which the two opposite (planes) are equal, similar, and ἡμικυκλίου. parallel, and the remaining (planes are) parallelograms. ιζʹ. Διάμετρος δὲ τῆς σφαίρας ἐστὶν εὐθεῖά τις διὰ τοῦ 14. A sphere is the figure enclosed when, the diam- κέντρου ἠγμένη καὶ περατουμένη ἐφ᾿ ἑκάτερα τὰ μέρη ὑπὸ eter of a semicircle remaining (fixed), the semicircle is τῆς ἐπιφανείας τῆς σφαίρας. carried around, and again established at the same (posi- ιηʹ. Κῶνός ἐστιν, ὅταν ὀρθογωνίου τριγώνου μενούσης tion) from which it began to be moved. μιᾶς πλευρᾶς τῶν περὶ τὴν ὀρθὴν γωνίαν περιενεχθὲν τὸ 15. And the axis of the sphere is the fixed straight-line τρίγωνον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο about which the semicircle is turned. 424 STOIQEIWN iaþ. ELEMENTS BOOK 11 φέρεσθαι, τὸ περιληφθὲν σχῆμα. κἂν μὲν ἡ μένουσα 16. And the center of the sphere is the same as that of εὐθεῖα ἴση ᾖ τῇ λοιπῇ [τῇ] περὶ τὴν ὀρθὴν περιφερομένῃ, the semicircle. ὀρθογώνιος ἔσται ὁ κῶνος, ἐὰν δὲ ἐλάττων, ἀμβλυγώνιος, 17. And the diameter of the sphere is any straight- ἐὰν δὲ μείζων, ὀξυγώνιος. line which is drawn through the center and terminated in ιθʹ. ῎Αξων δὲ τοῦ κώνου ἐστὶν ἡ μένουσα εὐθεῖα, περὶ both directions by the surface of the sphere. ἣν τὸ τρίγωνον στρέφεται. 18. A cone is the figure enclosed when, one of the κʹ. Βάσις δὲ ὁ κύκλος ὁ ὑπὸ τῆς περιφερομένης εὐθείας sides of a right-angled triangle about the right-angle re- γραφόμενος. maining (fixed), the triangle is carried around, and again καʹ.Κύλινδρός ἐστιν, ὅταν ὀρθογωνίου παραλληλογράμ- established at the same (position) from which it began to μου μενούσης μιᾶς πλευρᾶς τῶν περὶ τὴν ὀρθὴν γωνίαν πε- be moved. And if the fixed straight-line is equal to the re- ριενεχθὲν τὸ παραλληλόγραμμον εἰς τὸ αὐτὸ πάλιν ἀποκα- maining (straight-line) about the right-angle, (which is) τασταθῇ, ὅθεν ἤρξατο φέρεσθαι, τὸ περιληφθὲν σχῆμα. carried around, then the cone will be right-angled, and if κβʹ. ῎Αξων δὲ τοῦ κυλίνδρου ἐστὶν ἡ μένουσα εὐθεῖα, less, obtuse-angled, and if greater, acute-angled. περὶ ἣν τὸ παραλληλόγραμμον στρέφεται. 19. And the axis of the cone is the fixed straight-line κγʹ. Βάσεις δὲ οἱ κύκλοι οἱ ὑπὸ τῶν ἀπεναντίον περια- about which the triangle is turned. γομένων δύο πλευρῶν γραφόμενοι. 20. And the base (of the cone is) the circle described κδʹ. ῞Ομοιοι κῶνοι καὶ κύλινδροί εἰσιν, ὧν οἵ τε ἄξονες by the (remaining) straight-line (about the right-angle καὶ αἱ διάμετροι τῶν βάσεων ἀνάλογόν εἰσιν. which is) carried around (the axis). κεʹ.Κύβος ἐστὶ σχῆμα στερεὸν ὑπὸ ἓξ τετραγώνων ἴσων 21. A cylinder is the figure enclosed when, one of περιεχόμενον. the sides of a right-angled parallelogram about the right- κϛʹ. ᾿Οκτάεδρόν ἐστὶ σχῆμα στερεὸν ὑπὸ ὀκτὼ τριγώνων angle remaining (fixed), the parallelogram is carried ἴσων καὶ ἰσοπλεύρων περιεχόμενον. around, and again established at the same (position) κζʹ. Εἰκοσάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ εἴκοσι from which it began to be moved. τριγώνων ἴσων καὶ ἰσοπλεύρων περιεχόμενον. 22. And the axis of the cylinder is the stationary κηʹ.Δωδεκάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ δώδεκα πεν- straight-line about which the parallelogram is turned. ταγώνων ἴσων καὶ ἰσοπλεύρων καὶ ἰσογωνίων περιεχόμενον. 23. And the bases (of the cylinder are) the circles described by the two opposite sides (which are) carried around. 24. Similar cones and cylinders are those for which the axes and the diameters of the bases are proportional. 25. A cube is a solid figure contained by six equal squares. 26. An octahedron is a solid figure contained by eight equal and equilateral triangles. 27. An icosahedron is a solid figure contained by twenty equal and equilateral triangles. 28. A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons.aþ. Proposition 1† Εὐθείας γραμμῆς μέρος μέν τι οὐκ ἔστιν ἐν τῷ ὑπο- Some part of a straight-line cannot be in a reference κειμένῳ ἐπιπέδῳ, μέρος δέ τι ἐν μετεωροτέρῳ. plane, and some part in a more elevated (plane). Εἰ γὰρ δυνατόν, εὐθείας γραμμῆς τῆς ΑΒΓ μέρος μέν For, if possible, let some part, AB, of the straight-line τι τὸ ΑΒ ἔστω ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, μέρος δέ τι τὸ ABC be in a reference plane, and some part, BC, in a ΒΓ ἐν μετεωροτέρῳ. more elevated (plane). ῎Εσται δή τις τῇ ΑΒ συνvεχὴς εὐθεῖα ἐπ᾿ εὐθείας ἐν In the reference plane, there will be some straight-line τῷ ὑποκειμένῳ ἐπιπέδῳ. ἔστω ἡ ΒΔ· δύο ἄρα εὐθειῶν τῶν continuous with, and straight-on to, AB.‡ Let it be BD. ΑΒΓ, ΑΒΔ κοινὸν τμῆμά ἐστιν ἡ ΑΒ· ὅπερ ἐστὶν ἀδύνατον, Thus, AB is a common segment of the two (different) ἐπειδήπερ ἐὰν κέντρῳ τῷ Β καὶ διαστήματι τῷ ΑΒ κύκλον straight-lines ABC and ABD. The very thing is impos- γράψωμεν, αἱ διάμετροι ἀνίσους ἀπολήψονται τοῦ κύκλου sible, inasmuch as if we draw a circle with center B and 425 STOIQEIWN iaþ. ELEMENTS BOOK 11 περιφερείας. radius AB then the diameters (ABD and ABC) will cut off unequal circumferences of the circle. Γ Α Β ∆ D A B C Εὐθείας ἄρα γραμμῆς μέρος μέν τι οὐκ ἔστιν ἐν τῷ ὑπο- Thus, some part of a straight-line cannot be in a refer- κειμένῳ ἐπιπέδῳ, τὸ δὲ ἐν μετεωροτέρῳ· ὅπερ ἔδει δεῖξαι. ence plane, and (some part) in a more elevated (plane). (Which is) the very thing it was required to show. † The proofs of the first three propositions in this book are not at all rigorous. Hence, these three propositions should properly be regarded as additional axioms. ‡ This assumption essentially presupposes the validity of the proposition under discussion.bþ. Proposition 2 ᾿Εὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας, ἐν ἑνί εἰσιν ἐπιπέδῳ, If two straight-lines cut one another then they are in καὶ πᾶν τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ. one plane, and every triangle (formed using segments of both lines) is in one plane. Ζ Γ Θ Κ Β Η Α ∆ Ε D K B A C H F G E Δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ τεμνέτωσαν ἀλλήλας κατὰ For let the two straight-lines AB and CD have cut τὸ Ε σημεῖον. λέγω, ὅτι αἱ ΑΒ, ΓΔ ἐν ἑνί εἰσιν ἐπιπέδῳ, one another at point E. I say that AB and CD are in one καὶ πᾶν τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ. plane, and that every triangle (formed using segments of Εἰλήφθω γὰρ ἐπὶ τῶν ΕΓ, ΕΒ τυχόντα σημεῖα τὰ Ζ, Η, both lines) is in one plane. καὶ ἐπεζεύχθωσαν αἱ ΓΒ, ΖΗ, καὶ διήχθωσαν αἱ ΖΘ, ΗΚ· For let the random points F and G have been taken λέγω πρῶτον, ὅτι τὸ ΕΓΒ τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ. εἰ on EC and EB (respectively). And let CB and FG γάρ ἐστι τοῦ ΕΓΒ τριγώνου μέρος ἤτοι τὸ ΖΘΓ ἢ τὸ ΗΒΚ have been joined, and let FH and GK have been drawn ἐν τῷ ὑποκειμένῳ [ἐπιπέδῳ], τὸ δὲ λοιπὸν ἐν ἄλλῳ, ἔσται καὶ across. I say, first of all, that triangle ECB is in one (ref- μιᾶς τῶν ΕΓ, ΕΒ εὐθειῶν μέρος μέν τι ἐν τῷ ὑποκειμένῳ erence) plane. For if part of triangle ECB, either FHC 426 STOIQEIWN iaþ. ELEMENTS BOOK 11 ἐπιπέδῳ, τὸ δὲ ἐν αλλῳ. εἰ δὲ τοῦ ΕΓΒ τριγώνου τὸ ΖΓΒΗ or GBK, is in the reference [plane], and the remainder μέρος ᾖ ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ λοιπὸν ἐν ἄλλῳ, in a different (plane) then a part of one the straight-lines ἔσται καὶ ἀμφοτέρων τῶν ΕΓ, ΕΒ εὐθειῶν μέρος μέν τι EC and EB will also be in the reference plane, and (a ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ ἐν ἄλλω· ὅπερ ἄτοπον part) in a different (plane). And if the part FCBG of tri- ἐδείχθη. τὸ ἄρα ΕΓΒ τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ. ἐν ᾧ angle ECB is in the reference plane, and the remainder δέ ἐστι τὸ ΕΓΒ τρίγωνον, ἐν τούτῳ καὶ ἑκατέρα τῶν ΕΓ, in a different (plane) then parts of both of the straight- ΕΒ, ἐν ᾧ δὲ ἑκατέρα τῶν ΕΓ, ΕΒ, ἐν τούτῳ καὶ αἱ ΑΒ, lines EC and EB will also be in the reference plane, ΓΔ. αἱ ΑΒ, ΓΔ ἄρα εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ, καὶ πᾶν and (parts) in a different (plane). The very thing was τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ· ὅπερ ἔδει δεῖξαι. shown to be absurb [Prop. 11.1]. Thus, triangle ECB is in one plane. And in whichever (plane) triangle ECB is (found), in that (plane) EC and EB (will) each also (be found). And in whichever (plane) EC and EB (are) each (found), in that (plane) AB and CD (will) also (be found) [Prop. 11.1]. Thus, the straight-lines AB and CD are in one plane, and every triangle (formed using seg- ments of both lines) is in one plane. (Which is) the very thing it was required to show.gþ. Proposition 3 ᾿Εὰν δύο ἐπίπεδα τεμνῇ ἄλληλα, ἡ κοινὴ αὐτῶν τομὴ If two planes cut one another then their common sec- εὐθεῖά ἐστιν. tion is a straight-line. Α Ε Γ Β Ζ ∆ A E B C F D Δύο γὰρ ἐπίπεδα τὰ ΑΒ, ΒΓ τεμνέτω ἄλληλα, κοινὴ δὲ For let the two planes AB and BC cut one another, αὐτῶν τομὴ ἔστω ἡ ΔΒ γραμμή· λέγω, ὅτι ἡ ΔΒ γραμμὴ and let their common section be the line DB. I say that εὐθεῖά ἐστιν. the line DB is straight. Εἰ γὰρ μή, ἐπεζεύχθω ἀπὸ τοῦ Δ ἐπὶ τὸ Β ἐν μὲν τῷ ΑΒ For, if not, let the straight-line DEB have been joined ἐπιπέδῳ εὐθεῖα ἡ ΔΕΒ, ἐν δὲ τῷ ΒΓ ἐπιπέδῳ εὐθεῖα ἡ ΔΖΒ. from D to B in the plane AB, and the straight-line DFB ἔσται δὴ δύο εὐθειῶν τῶν ΔΕΒ, ΔΖΒ τὰ αὐτὰ πέρατα, καὶ in the plane BC. So two straight-lines, DEB and DFB, περιέξουσι δηλαδὴ χωρίον· ὅπερ ἄτοπον. οὔκ ἄρα αἰ ΔΕΒ, will have the same ends, and they will clearly enclose an ΔΖΒ εὐθεῖαί εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἄλλη τις area. The very thing (is) absurd. Thus, DEB and DFB ἀπὸ τοῦ Δ ἐπὶ τὸ Β ἐπιζευγνυμένη εὐθεῖα ἔσται πλὴν τῆς are not straight-lines. So, similarly, we can show than no ΔΒ κοινῆς τομῆς τῶν ΑΒ, ΒΓ ἐπιπέδων. other straight-line can be joined from D to B except DB, ᾿Εὰν ἄρα δύο ἐπίπεδα τέμνῃ ἄλληλα, ἡ κοινὴ αὐτῶν τομὴ the common section of the planes AB and BC. εὐθεῖά ἐστιν· ὅπερ ἔδει δεῖξαι. Thus, if two planes cut one another then their com- mon section is a straight-line. (Which is) the very thing it was required to show. 427 STOIQEIWN iaþ. ELEMENTS BOOK 11dþ. Proposition 4 ᾿Εὰν εὐθεῖα δύο εὐθείαις τεμνούσαις ἀλλήλας πρὸς If a straight-line is set up at right-angles to two ὀρθὰς ἐπὶ τῆς κοινῆς τομῆς ἐπισταθῇ, καὶ τῷ δι᾿ αὐτῶν straight-lines cutting one another, at the common point ἐπιπέδῳ πρὸς ὀρθὰς ἔσται. of section, then it will also be at right-angles to the plane (passing) through them (both). Η Γ Θ Β Ζ ∆ Ε Α A BD E C H F G Εὐθεῖα γάρ τις ἡ ΕΖ δύο εὐθείαις ταῖς ΑΒ, ΓΔ τε- For let some straight-line EF have (been) set up at μνούσαις ἀλλήλας κατὰ τὸ Ε σημεῖον ἀπὸ τοῦ Ε πρὸς ὀρθὰς right-angles to two straight-lines, AB and CD, cutting ἐφεστάτω· λέγω, ὅτι ἡ ΕΖ καὶ τῷ διὰ τῶν ΑΒ, ΓΔ ἐπιπέδῳ one another at point E, at E. I say that EF is also at πρὸς ὀρθάς ἐστιν. right-angles to the plane (passing) through AB and CD. Ἀπειλήφθωσαν γὰρ αἱ ΑΕ, ΕΒ, ΓΕ, ΕΔ ἴσαι ἀλλήλαις, For let AE, EB, CE and ED have been cut off from καὶ διήχθω τις διὰ τοῦ Ε, ὡς ἔτυχεν, ἡ ΗΕΘ, καὶ (the two straight-lines so as to be) equal to one another. ἐπεζεύχθωσαν αἱ ΑΔ, ΓΒ, καὶ ἔτι ἀπὸ τυχόντος τοῦ Ζ And let GEH have been drawn, at random, through E ἐπεζεύχθωσαν αἱ ΖΑ, ΖΗ, ΖΔ, ΖΓ, ΖΘ, ΖΒ. (in the plane passing through AB and CD). And let AD Καὶ ἐπεὶ δύο αἱ ΑΕ, ΕΔ δυσὶ ταῖς ΓΕ, ΕΒ ἴσαι εἰσὶ καὶ and CB have been joined. And, furthermore, let FA, γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΑΔ βάσει τῇ ΓΒ ἴση FG, FD, FC, FH , and FB have been joined from the ἐστίν, καὶ τὸ ΑΕΔ τρίγωνον τῷ ΓΕΒ τριγώνῳ ἴσον ἔσται· random (point) F (on EF ). ὥστε καὶ γωνία ἡ ὑπὸ ΔΑΕ γωνίᾳ τῇ ὑπὸ ΕΒΓ ἴση [ἐστίν]. For since the two (straight-lines) AE and ED are ἔστι δὲ καὶ ἡ ὑπὸ ΑΕΗ γωνία τῇ ὑπὸ ΒΕΘ ἴση. δύο δὴ equal to the two (straight-lines) CE and EB, and they τρίγωνά ἐστι τὰ ΑΗΕ, ΒΕΘ τὰς δύο γωνίας δυσὶ γωνίαις enclose equal angles [Prop. 1.15], the base AD is thus ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ equal to the base CB, and triangle AED will be equal ἴσην τὴν πρὸς ταῖς ἴσαις γωνίαις τὴν ΑΕ τῇ ΕΒ· καὶ τὰς to triangle CEB [Prop. 1.4]. Hence, the angle DAE λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξουσιν. ἴση [is] equal to the angle EBC. And the angle AEG (is) ἄρα ἡ μὲν ΗΕ τῇ ΕΘ, ἡ δὲ ΑΗ τῇ ΒΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ also equal to the angle BEH [Prop. 1.15]. So AGE ΑΕ τῇ ΕΒ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΖΕ, βάσις ἄρα ἡ ΖΑ and BEH are two triangles having two angles equal to βάσει τῇ ΖΒ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΖΓ τῇ ΖΔ ἐστιν two angles, respectively, and one side equal to one side— ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΓΒ, ἔστι δὲ καὶ ἡ ΖΑ τῇ ΖΒ (namely), those by the equal angles, AE and EB. Thus, ἴση, δύο δὴ αἱ ΖΑ, ΑΔ δυσὶ ταῖς ΖΒ, ΒΓ ἴσαι εἰσὶν ἑκατέρα they will also have the remaining sides equal to the re- ἑκατέρᾳ· καὶ βάσις ἡ ΖΔ βάσει τῇ ΖΓ ἐδείχθη ἴση· καὶ γωνία maining sides [Prop. 1.26]. Thus, GE (is) equal to EH , ἄρα ἡ ὑπὸ ΖΑΔ γωνίᾳ τῇ ὑπὸ ΖΒΓ ἴση ἐστίν. καὶ ἐπεὶ πάλιν and AG to BH . And since AE is equal to EB, and FE ἐδείχθη ἡ ΑΗ τῇ ΒΘ ἴση, ἀλλὰ μὴν καὶ ἡ ΖΑ τῇ ΖΒ ἴση, δύο is common and at right-angles, the base FA is thus equal δὴ αἱ ΖΑ, ΑΗ δυσὶ ταῖς ΖΒ, ΒΘ ἴσαι εἰσίν. καὶ γωνία ἡ ὑπὸ to the base FB [Prop. 1.4]. So, for the same (reasons), ΖΑΗ ἐδείχθη ἴση τῇ ὑπὸ ΖΒΘ· βάσις ἄρα ἡ ΖΗ βάσει τῇ ΖΘ FC is also equal to FD. And since AD is equal to CB, ἐστιν ἴση. καὶ ἐπεὶ πάλιν ἴση ἐδείχθη ἡ ΗΕ τῇ ΕΘ, κοινὴ δὲ and FA is also equal to FB, the two (straight-lines) FA ἡ ΕΖ, δύο δὴ αἱ ΗΕ, ΕΖ δυσὶ ταῖς ΘΕ, ΕΖ ἴσαι εἰσίν· καὶ and AD are equal to the two (straight-lines) FB and BC, βάσις ἡ ΖΗ βάσει τῇ ΖΘ ἴση· γωνία ἄρα ἡ ὑπὸ ΗΕΖ γωνίᾳ respectively. And the base FD was shown (to be) equal τῇ ὑπὸ ΘΕΖ ἴση ἐστίν. ὀρθὴ ἄρα ἑκατέρα τῶν ὑπὸ ΗΕΖ, to the base FC. Thus, the angle FAD is also equal to ΘΕΖ γωνιῶν. ἡ ΖΕ ἄρα πρὸς τὴν ΗΘ τυχόντως διὰ τοῦ the angle FBC [Prop. 1.8]. And, again, since AG was Ε ἀχθεῖσαν ὀρθή ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι ἡ ΖΕ καὶ shown (to be) equal to BH , but FA (is) also equal to 428 STOIQEIWN iaþ. ELEMENTS BOOK 11 πρὸς πάσας τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ FB, the two (straight-lines) FA and AG are equal to the ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. εὐθεῖα δὲ πρὸς two (straight-lines) FB and BH (respectively). And the ἐπίπεδον ὀρθή ἐστιν, ὅταν πρὸς πάσας τὰς ἁπτομένας αὐτῆς angle FAG was shown (to be) equal to the angle FBH . εὐθείας καὶ οὔσας ἐν τῷ αὐτῷ ἐπιπέδῳ ὀρθὰς ποιῇ γωνίας· Thus, the base FG is equal to the base FH [Prop. 1.4]. ἡ ΖΕ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τὸ δὲ And, again, since GE was shown (to be) equal to EH , ὑποκείμενον ἐπίπεδόν ἐστι τὸ διὰ τῶν ΑΒ, ΓΔ εὐθειῶν. ἡ and EF (is) common, the two (straight-lines) GE and ΖΕ ἄρα πρὸς ὀρθάς ἐστι τῷ διὰ τῶν ΑΒ, ΓΔ ἐπιπέδῳ. EF are equal to the two (straight-lines) HE and EF ᾿Εὰν ἄρα εὐθεῖα δύο εὐθείαις τεμνούσαις ἀλλήλας πρὸς (respectively). And the base FG (is) equal to the base ὀρθὰς ἐπὶ τῆς κοινῆς τομῆς ἐπισταθῇ, καὶ τῷ δι᾿ αὐτῶν FH . Thus, the angle GEF is equal to the angle HEF ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· ὅπερ ἔδει δεῖξαι. [Prop. 1.8]. Each of the angles GEF and HEF (are) thus right-angles [Def. 1.10]. Thus, FE is at right-angles to GH , which was drawn at random through E (in the reference plane passing though AB and AC). So, simi- larly, we can show that FE will make right-angles with all straight-lines joined to it which are in the reference plane. And a straight-line is at right-angles to a plane when it makes right-angles with all straight-lines joined to it which are in the plane [Def. 11.3]. Thus, FE is at right-angles to the reference plane. And the reference plane is that (passing) through the straight-lines AB and CD. Thus, FE is at right-angles to the plane (passing) through AB and CD. Thus, if a straight-line is set up at right-angles to two straight-lines cutting one another, at the common point of section, then it will also be at right-angles to the plane (passing) through them (both). (Which is) the very thing it was required to show.eþ. Proposition 5 ᾿Εὰν εὐθεῖα τρισὶν εὐθείαις ἁπτομέναις ἀλλήλων πρὸς If a straight-line is set up at right-angles to three ὀρθὰς ἐπὶ τῆς κοινῆς τομῆς ἐπισταθῇ, αἱ τρεῖς εὐθεῖαι ἐν ἑνί straight-lines cutting one another, at the common point εἰσιν ἐπιπέδῳ. of section, then the three straight-lines are in one plane. Α ∆ Ε Γ Ζ Β F D E B A C Εὐθεῖα γάρ τις ἡ ΑΒ τρισὶν εὐθείαις ταῖς ΒΓ, ΒΔ, ΒΕ For let some straight-line AB have been set up at πρὸς ὀρθὰς ἐπὶ τῆς κατὰ τὸ Β ἁφῆς ἐφεστάτω· λέγω, ὅτι αἱ right-angles to three straight-lines BC, BD, and BE, at ΒΓ, ΒΔ, ΒΕ ἐν ἑνί εἰσιν ἐπιπέδῳ. the (common) point of section B. I say that BC, BD, Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, ἔστωσαν αἱ μὲν ΒΔ, ΒΕ and BE are in one plane. ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, ἡ δὲ ΒΓ ἐν μετεωροτέρῳ, καὶ For (if) not, and if possible, let BD and BE be in ἐκβεβλήσθω τὸ δὶα τῶν ΑΒ, ΒΓ ἐπίπεδον· κοινὴν δὴ τομὴν the reference plane, and BC in a more elevated (plane). 429 STOIQEIWN iaþ. ELEMENTS BOOK 11 ποιήσει ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖαν. ποιείτω τὴν And let the plane through AB and BC have been pro- ΒΖ. ἐν ἑνὶ ἄρα εἰσὶν ἐπιπέδῳ τῷ διηγμένῳ διὰ τῶν ΑΒ, duced. So it will make a straight-line as a common sec- ΒΓ αἱ τρεῖς εὐθεῖαι αἱ ΑΒ, ΒΓ, ΒΖ. καὶ ἐπεὶ ἡ ΑΒ ὀρθή tion with the reference plane [Def. 11.3]. Let it make ἐστι πρὸς ἑκατέραν τῶν ΒΔ, ΒΕ, καὶ τῷ διὰ τῶν ΒΔ, BF . Thus, the three straight-lines AB, BC, and BF ΒΕ ἄρα ἐπιπέδῳ ὀρθή ἐστιν ἡ ΑΒ. τὸ δὲ διὰ τῶν ΒΔ, ΒΕ are in one plane—(namely), that drawn through AB and ἐπίπεδον τὸ ὑποκείμενόν ἐστιν· ἡ ΑΒ ἄρα ὀρθή ἐστι πρὸς τὸ BC. And since AB is at right-angles to each of BD and ὑποκείμενον ἐπίπεδον. ὥστε καὶ πρὸς πάσας τὰς ἁπτομένας BE, AB is thus also at right-angles to the plane (passing) αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς through BD and BE [Prop. 11.4]. And the plane (pass- ποιήσει γωνίας ἡ ΑΒ. ἅπτεται δὲ αὐτῆς ἡ ΒΖ οὖσα ἐν τῷ ing) through BD and BE is the reference plane. Thus, ὑποκειμένῳ ἐπιπέδῳ· ἡ ἄρα ὑπὸ ΑΒΖ γωνία ὀρθή ἐστιν. AB is at right-angles to the reference plane. Hence, AB ὑπόκειται δὲ καὶ ἡ ὑπὸ ΑΒΓ ὀρθή· ἴση ἄρα ἡ ὑπὸ ΑΒΖ γωνία will also make right-angles with all straight-lines joined τῇ ὑπὸ ΑΒΓ. καί εἰσιν ἐν ἑνὶ ἐπιπέδῳ· ὅπερ ἐστὶν ἀδύνατον. to it which are also in the reference plane [Def. 11.3]. οὐκ ἄρα ἡ ΒΓ εὐθεῖα ἐν μετεωροτέρῳ ἐστὶν ἐπιπέδῳ· αἱ τρεῖς And BF , which is in the reference plane, is joined to it. ἄρα εὐθεῖαι αἱ ΒΓ, ΒΔ, ΒΕ ἐν ἑνί εἰσιν ἐπιπέδῳ. Thus, the angle ABF is a right-angle. And ABC was also ᾿Εὰν ἄρα εὐθεῖα τρισὶν εὐθείαις ἁπτομέναις ἀλλήλων ἐπὶ assumed to be a right-angle. Thus, angle ABF (is) equal τῆς ἁφῆς πρὸς ὀρθὰς ἐπισταθῇ, αἱ τρεῖς εὐθεῖαι ἐν ἑνί εἰσιν to ABC. And they are in one plane. The very thing is ἐπιπέδῳ· ὅπερ ἔδει δεῖξαι. impossible. Thus, BC is not in a more elevated plane. Thus, the three straight-lines BC, BD, and BE are in one plane. Thus, if a straight-line is set up at right-angles to three straight-lines cutting one another, at the (common) point of section, then the three straight-lines are in one plane. (Which is) the very thing it was required to show.�þ. Proposition 6 ᾿Εὰν δύο εὐθεῖαι τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν, If two straight-lines are at right-angles to the same παράλληλοι ἔσονται αἱ εὐθεῖαι. plane then the straight-lines will be parallel.† B G D E A C D E A B Δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ τῷ ὑποκειμένῳ ἐπιπέδῳ For let the two straight-lines AB and CD be at right- πρὸς ὀρθὰς ἔστωσαν· λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΒ τῇ angles to a reference plane. I say that AB is parallel to ΓΔ. CD. Συμβαλλέτωσαν γὰρ τῷ ὑποκειμένῳ ἐπιπέδῳ κατὰ τὰ Β, For let them meet the reference plane at points B and Δ σημεῖα, καὶ ἐπεζεύχθω ἡ ΒΔ εὐθεῖα, καὶ ἤχθω τῇ ΒΔ D (respectively). And let the straight-line BD have been πρὸς ὀρθὰς ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ἡ ΔΕ, καὶ κείσθω joined. And let DE have been drawn at right-angles to τῇ ΑΒ ἴση ἡ ΔΕ, καὶ ἐπεζεύχθωσαν αἱ ΒΕ, ΑΕ, ΑΔ. BD in the reference plane. And let DE be made equal to Καὶ ἐπεὶ ἡ ΑΒ ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον, AB. And let BE, AE, and AD have been joined. καὶ πρὸς πάσας [ἄρα] τὰς ἁπτομένας αὐτῆς εὐθείας καὶ And since AB is at right-angles to the reference plane, οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. it will [thus] also make right-angles with all straight-lines ἅπτεται δὲ τῆς ΑΒ ἑκατέρα τῶν ΒΔ, ΒΕ οὖσα ἐν τῷ ὑπο- joined to it which are in the reference plane [Def. 11.3]. 430 STOIQEIWN iaþ. ELEMENTS BOOK 11 κειμένῳ ἐπιπέδῳ· ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΑΒΔ, And BD and BE, which are in the reference plane, are ΑΒΕ γωνιῶν. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ὑπὸ ΓΔΒ, each joined to AB. Thus, each of the angles ABD and ΓΔΕ ὀρθή ἐστιν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΔΕ, κοινὴ ABE are right-angles. So, for the same (reasons), each δὲ ἡ ΒΔ, δύο δὴ αἱ ΑΒ, ΒΔ δυσὶ ταῖς ΕΔ, ΔΒ ἴσαι εἰσίν· of the angles CDB and CDE are also right-angles. And καὶ γωνίας ὀρθὰς περιέχουσιν· βάσις ἄρα ἡ ΑΔ βάσει τῇ since AB is equal to DE, and BD (is) common, the ΒΕ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΔΕ, ἀλλὰ καὶ two (straight-lines) AB and BD are equal to the two ἡ ΑΔ τῇ ΒΕ, δύο δὴ αἱ ΑΒ, ΒΕ δυσὶ ταῖς ΕΔ, ΔΑ ἴσαι (straight-lines) ED and DB (respectively). And they εἰσίν· καὶ βάσις αὐτῶν κοινὴ ἡ ΑΕ· γωνία ἄρα ἡ ὑπὸ ΑΒΕ contain right-angles. Thus, the base AD is equal to the γωνιᾴ τῇ ὑπὸ ΕΔΑ ἐστιν ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΑΒΕ· ὀρθὴ base BE [Prop. 1.4]. And since AB is equal to DE, and ἄρα καὶ ἡ ὑπὸ ΕΔΑ· ἡ ΕΔ ἄρα πρὸς τὴν ΔΑ ὀρθή ἐστιν. AD (is) also (equal) to BE, the two (straight-lines) AB ἔστι δὲ καὶ πρὸς ἑκατέραν τῶν ΒΔ, ΔΓ ὀρθή. ἡ ΕΔ ἄρα and BE are thus equal to the two (straight-lines) ED τρισὶν εὐθείαις ταῖς ΒΔ, ΔΑ, ΔΓ πρὸς ὀρθὰς ἐπὶ τῆς ἁφῆς and DA (respectively). And their base AE (is) common. ἐφέστηκεν· αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΒΔ, ΔΑ, ΔΓ ἐν ἑνί εἰσιν Thus, angle ABE is equal to angle EDA [Prop. 1.8]. And ἐπιπέδῳ. ἐν ᾧ δὲ αἱ ΔΒ, ΔΑ, ἐν τούτῳ καὶ ἡ ΑΒ· πᾶν γὰρ ABE (is) a right-angle. Thus, EDA (is) also a right- τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ· αἱ ἄρα ΑΒ, ΒΔ, ΔΓ εὐθεῖαι angle. ED is thus at right-angles to DA. And it is also at ἐν ἑνί εἰσιν ἐπιπέδῳ. καί ἐστιν ὀρθὴ ἑκατέρα τῶν ὑπὸ ΑΒΔ, right-angles to each of BD and DC. Thus, ED is stand- ΒΔΓ γωνιῶν· παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. ing at right-angles to the three straight-lines BD, DA, ᾿Εὰν ἄρα δύο εὐθεῖαι τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν, and DC at the (common) point of section. Thus, the παράλληλοι ἔσονται αἱ εὐθεῖαι· ὅπερ ἔδει δεῖξαι. three straight-lines BD, DA, and DC are in one plane [Prop. 11.5]. And in which(ever) plane DB and DA (are found), in that (plane) AB (will) also (be found). For every triangle is in one plane [Prop. 11.2]. And each of the angles ABD and BDC is a right-angle. Thus, AB is parallel to CD [Prop. 1.28]. Thus, if two straight-lines are at right-angles to the same plane then the straight-lines will be parallel. (Which is) the very thing it was required to show. † In other words, the two straight-lines lie in the same plane, and never meet when produced in either direction.zþ. Proposition 7 ᾿Εὰν ὦσι δύο εὐθεῖαι παράλληλοι, ληφθῇ δὲ ἐφ᾿ ἑκατέρας If there are two parallel straight-lines, and random αὐτῶν τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη points are taken on each of them, then the straight-line εὐθεῖα ἐν τῷ αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς παραλλήλοις. joining the two points is in the same plane as the parallel (straight-lines). H A E B DZG G A E B DC F ῎Εστωσαν δύο εὐθεῖαι παράλληλοι αἱ ΑΒ, ΓΔ, καὶ Let AB and CD be two parallel straight-lines, and let εἰλήφθω ἐφ᾿ ἑκατέρας αὐτῶν τυνχόντα σημεῖα τὰ Ε, Ζ· the random points E and F have been taken on each of λέγω, ὅτι ἡ ἐπὶ τὰ Ε, Ζ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐν τῷ them (respectively). I say that the straight-line joining αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς παραλλήλοις. points E and F is in the same (reference) plane as the Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, ἔστω ἐν μετεωροτέρῳ ὡς ἡ parallel (straight-lines). ΕΗΖ, καὶ διήχθω διὰ τῆς ΕΗΖ ἐπίπεδον· τομὴν δὴ ποιήσει For (if) not, and if possible, let it be in a more elevated 431 STOIQEIWN iaþ. ELEMENTS BOOK 11 ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖαν. ποιείτω ὡς τὴν ΕΖ· δύο (plane), such as EGF . And let a plane have been drawn ἄρα εὐθεῖαι αἱ ΕΗΖ, ΕΖ χωρίον περιέξουσιν· ὅπερ ἐστὶν through EGF . So it will make a straight cutting in the ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ Ε ἐπὶ τὸ Ζ ἐπιζευγνυμένη reference plane [Prop. 11.3]. Let it make EF . Thus, two εὐθεῖαι ἐν μετεωροτέρῳ ἐστὶν ἐπιπέδῳ· ἐν τῷ διὰ τῶν ΑΒ, straight-lines (with the same end-points), EGF and EF , ΓΒ ἄρα παραλλήλων ἐστὶν ἐπιπέδῳ ἡ ἀπὸ τοῦ Ε ἐπὶ τὸ Ζ will enclose an area. The very thing is impossible. Thus, ἐπιζευγνυμένη εὐθεῖα. the straight-line joining E to F is not in a more elevated ᾿Εὰν ἄρα ὦσι δύο εὐθεῖαι παράλληλοι, ληφθῇ δὲ ἐφ᾿ plane. The straight-line joining E to F is thus in the plane ἑκατέρας αὐτῶν τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευ- through the parallel (straight-lines) AB and CD. γνυμένη εὐθεῖα ἐν τῷ αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς παραλλήλοις· Thus, if there are two parallel straight-lines, and ran- ὅπερ ἔδει δεῖξαι. dom points are taken on each of them, then the straight- line joining the two points is in the same plane as the parallel (straight-lines). (Which is) the very thing it was required to show.hþ. Proposition 8 ᾿Εὰν ὦσι δύο εὐθεῖαι παράλληλοι, ἡ δὲ ἑτέρα αὐτῶν If two straight-lines are parallel, and one of them is at ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ λοιπὴ τῷ αὐτῷ ἐπιπέδῳ right-angles to some plane, then the remaining (one) will πρὸς ὀρθὰς ἔσται. also be at right-angles to the same plane. D A E B G D A E B C ῎Εστωσαν δύο εὐθεῖαι παράλληλοι αἱ ΑΒ, ΓΔ, ἡ δὲ ἑτέρα Let AB and CD be two parallel straight-lines, and let αὐτῶν ἡ ΑΒ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἔστω· one of them, AB, be at right-angles to a reference plane. λέγω, ὅτι καὶ ἡ λοιπὴ ἡ ΓΔ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς I say that the remaining (one), CD, will also be at right- ἔσται. angles to the same plane. Συμβαλλέτωσαν γὰρ αἱ ΑΒ, ΓΔ τῷ ὑποκειμένῳ ἐπιπέδῳ For let AB and CD meet the reference plane at points κατὰ τὰ Β, Δ σημεῖα, καὶ ἐπεζέυχθω ἡ ΒΔ· αἱ ΑΒ, ΓΔ, ΒΔ B and D (respectively). And let BD have been joined. ἄρα ἐν ἑνί εἰσιν ἐπιπέδῳ. ἤχθω τῇ ΒΑ πρὸς ὀρθὰς ἐν τῷ AB, CD, and BD are thus in one plane [Prop. 11.7]. ὑποκειμένῳ ἐπιπέδῳ ἡ ΔΕ, καὶ κείσθω τῇ ΑΒ ἴση ἡ ΔΕ, Let DE have been drawn at right-angles to BD in the καὶ ἐπεζεύχθωσαν αἱ ΒΕ, ΑΕ, ΑΔ. reference plane, and let DE be made equal to AB, and Καὶ ἐπεὶ ἡ ΑΒ ὁρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον, let BE, AE, and AD have been joined. καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας And since AB is at right-angles to the reference ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν ἡ ΑΒ· ὀρθὴ ἄρα plane, AB is thus also at right-angles to all of the [ἐστὶν] ἑκατέρα τῶν ὑπὸ ΑΒΔ, ΑΒΕ γωνιῶν. καὶ ἐπεὶ εἰς straight-lines joined to it which are in the reference plane παραλλήλους τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπέπτωκεν ἡ ΒΔ, αἱ ἄρα [Def. 11.3]. Thus, the angles ABD and ABE [are] each ὑπὸ ΑΒΔ, ΓΔΒ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. ὀρθὴ δὲ ἡ right-angles. And since the straight-line BD has met the ὑπὸ ΑΒΔ· ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΓΔΒ· ἡ ΓΔ ἄρα πρὸς τὴν ΒΔ parallel (straight-lines) AB and CD, the (sum of the) ὀρθή ἐστιν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΔΕ, κοινὴ δὲ ἡ ΒΔ, angles ABD and CDB is thus equal to two right-angles 432 STOIQEIWN iaþ. ELEMENTS BOOK 11 δύο δὴ αἱ ΑΒ, ΒΔ δυσὶ ταῖς ΕΔ, ΔΒ ἴσαι εἰσίν· καὶ γωνία [Prop. 1.29]. And ABD (is) a right-angle. Thus, CDB ἡ ὑπὸ ΑΒΔ γωνίᾳ τῇ ὑπὸ ΕΔΒ ἴση· ὀρθὴ γὰρ ἑκατέρα· (is) also a right-angle. CD is thus at right-angles to BD. βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΕ ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ And since AB is equal to DE, and BD (is) common, μὲν ΑΒ τῇ ΔΕ, ἡ δὲ ΒΕ τῇ ΑΔ, δύο δὴ αἱ ΑΒ, ΒΕ δυσὶ the two (straight-lines) AB and BD are equal to the two ταῖς ΕΔ, ΔΑ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ. καὶ βάσις αὐτῶν (straight-lines) ED and DB (respectively). And angle κοινὴ ἡ ΑΕ· γωνία ἄρα ἡ ὑπὸ ΑΒΕ γωνίᾳ τῇ ὑπὸ ΕΔΑ ABD (is) equal to angle EDB. For each (is) a right- ἐστιν ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΑΒΕ· ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΕΔΑ· angle. Thus, the base AD (is) equal to the base BE ἡ ΕΔ ἄρα πρὸς τὴν ΑΔ ὀρθή ἐστιν. ἔστι δὲ καὶ πρὸς τὴν [Prop. 1.4]. And since AB is equal to DE, and BE to ΔΒ ὀρθή· ἡ ΕΔ ἄρα καὶ τῲ διὰ τῶν ΒΔ, ΔΑ ἐπιπέδῳ ὀρθή AD, the two (sides) AB, BE are equal to the two (sides) ἐστιν. καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ ED, DA, respectively. And their base AE is common. οὔσας ἐν τῷ διὰ τῶν ΒΔΑ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας ἡ Thus, angle ABE is equal to angle EDA [Prop. 1.8]. ΕΔ. ἐν δὲ τῷ διὰ τῶν ΒΔΑ ἐπιπέδῳ ἐστὶν ἡ ΔΓ, ἐπειδήπερ And ABE (is) a right-angle. EDA (is) thus also a right- ἐν τῷ διὰ τῶν ΒΔΑ ἐπιπέδῳ ἐστὶν αἱ ΑΒ, ΒΔ, ἐν ᾧ δὲ angle. Thus, ED is at right-angles to AD. And it is also αἱ ΑΒ, ΒΔ, ἐν τούτῳ ἐστὶ καὶ ἡ ΔΓ. ἡ ΕΔ ἄρα τῇ ΔΓ at right-angles to DB. Thus, ED is also at right-angles πρὸς ὀρθάς ἐστιν· ὥστε καὶ ἡ ΓΔ τῇ ΔΕ πρὸς ὀρθάς ἐστιν. to the plane through BD and DA [Prop. 11.4]. And ἔστι δὲ καὶ ἡ ΓΔ τῇ ΒΔ πρὸς ὀρθάς. ἡ ΓΔ ἄρα δύο εὐθείαις ED will thus make right-angles with all of the straight- τεμνούσαις ἀλλήλας ταῖς ΔΕ, ΔΒ ἀπὸ τῆς κατὰ τὸ Δ τομῆς lines joined to it which are also in the plane through πρὸς ὀρθὰς ἐφέστηκεν· ὥστε ἡ ΓΔ καὶ τῷ διὰ τῶν ΔΕ, ΔΒ BDA. And DC is in the plane through BDA, inas- ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τὸ δὲ διὰ τῶν ΔΕ, ΔΒ ἐπίπεδον much as AB and BD are in the plane through BDA τὸ ὑποκείμενόν ἐστιν· ἡ ΓΔ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ [Prop. 11.2], and in which(ever plane) AB and BD (are πρὸς ὀρθάς ἐστιν. found), DC is also (found). Thus, ED is at right-angles ᾿Εὰν ἄρα ὦσι δύο εὐθεῖαι παράλληλοι, ἡ δὲ μία αὐτῶν to DC. Hence, CD is also at right-angles to DE. And ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ λοιπὴ τῷ αὐτῷ ἐπιπέδῳ CD is also at right-angles to BD. Thus, CD is standing πρὸς ὀρθὰς ἔσται· ὅπερ ἔδει δεῖξαι. at right-angles to two straight-lines, DE and DB, which meet one another, at the (point) of section, D. Hence, CD is also at right-angles to the plane through DE and DB [Prop. 11.4]. And the plane through DE and DB is the reference (plane). CD is thus at right-angles to the reference plane. Thus, if two straight-lines are parallel, and one of them is at right-angles to some plane, then the remain- ing (one) will also be at right-angles to the same plane. (Which is) the very thing it was required to show.jþ. Proposition 9 Αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ μὴ οὖσαι αὐτῇ ἐν τῷ (Straight-lines) parallel to the same straight-line, and αὐτῷ ἐπιπέδῳ καὶ ἀλλήλαις εἰσὶ παράλληλοι. which are not in the same plane as it, are also parallel to one another. E B Z H A D GK J E B A D K F C H G ῎Εστω γὰρ ἑκατέρα τῶν ΑΒ, ΓΔ τῇ ΕΖ παράλληλος For let AB and CD each be parallel to EF , not being μὴ οὖσαι αὐτῇ ἐν τῷ αὐτῷ ἐπιπέδῳ· λέγω, ὅτι παράλληλός in the same plane as it. I say that AB is parallel to CD. 433 STOIQEIWN iaþ. ELEMENTS BOOK 11 ἐστιν ἡ ΑΒ τῇ ΓΔ. For let some point G have been taken at random on Εἰλήφθω γὰρ ἐπὶ τῆς ΕΖ τυχὸν σημεῖον τὸ Η, καὶ ἀπ᾿ EF . And from it let GH have been drawn at right-angles αὐτοῦ τῇ ΕΖ ἐν μὲν τῷ διὰ τῶν ΕΖ, ΑΒ ἐπιπέδῳ πρὸς ὀρθὰς to EF in the plane through EF and AB. And let GK ἤχθω ἡ ΗΘ, ἐν δὲ τῷ διὰ τῶν ΖΕ, ΓΔ τῇ ΕΖ πάλιν πρὸς have been drawn, again at right-angles to EF , in the ὀρθὰς ἤχθω ἡ ΗΚ. plane through FE and CD. Καὶ ἐπεὶ ἡ ΕΖ πρὸς ἑκατέραν τῶν ΗΘ, ΗΚ ὀρθή ἐστιν, And since EF is at right-angles to each of GH and ἡ ΕΖ ἄρα καὶ τῷ διὰ τῶν ΗΘ, ΗΚ ἐπιπέδῳ πρὸς ὀρθάς GK, EF is thus also at right-angles to the plane through ἐστιν. καί ἐστιν ἡ ΕΖ τῇ ΑΒ παράλληλος· καὶ ἡ ΑΒ ἄρα GH and GK [Prop. 11.4]. And EF is parallel to AB. τῷ διὰ τῶν ΘΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. διὰ τὰ αὐτὰ Thus, AB is also at right-angles to the plane through δὴ καὶ ἡ ΓΔ τῷ διὰ τῶν ΘΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν· HGK [Prop. 11.8]. So, for the same (reasons), CD is ἑκατέρα ἄρα τῶν ΑΒ, ΓΔ τῷ διὰ τῶν ΘΗΚ ἐπιπέδῳ πρὸς also at right-angles to the plane through HGK. Thus, ὀρθάς ἐστιν. ἐὰν δὲ δύο εὐθεῖαι τῷ αὐτῷ ἐπιπέδῳ πρὸς AB and CD are each at right-angles to the plane through ὀρθὰς ὦσιν, παράλληλοί εἰσιν αἱ εὐθεῖαι· παράλληλος ἄρα HGK. And if two straight-lines are at right–angles ἐστὶν ἡ ΑΒ τῇ ΓΔ· ὅπερ ἔδει δεῖξαι. to the same plane then the straight-lines are parallel [Prop. 11.6]. Thus, AB is parallel to CD. (Which is) the very thing it was required to show.iþ. Proposition 10 ᾿Εὰν δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας If two straight-lines joined to one another are (respec- ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ, ἵσας tively) parallel to two straight-lines joined to one another, γωνίας περιέξουσιν. (but are) not in the same plane, then they will contain equal angles. A D ZE B G C D E B A F Δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΒΓ ἁπτόμεναι ἀλλήλων παρὰ For let the two straight-lines joined to one another, δύο εὐθείας τὰς ΔΕ, ΕΖ ἁπτομένας ἀλλήλων ἔστωσαν μὴ AB and BC, be (respectively) parallel to the two ἐν τῷ αὐτῷ ἐπιπέδῳ· λέγω, ὅτι ἴση ἐστὶν ἡ ὑπὸ ΑΒΓ γωνία straight-lines joined to one another, DE and EF , (but) τῇ ὑπὸ ΔΕΖ. not in the same plane. I say that angle ABC is equal to Ἀπειλήφθωσαν γὰρ αἱ ΒΑ, ΒΓ, ΕΔ, ΕΖ ἴσαι ἀλλήλαις, (angle) DEF . καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΓΖ, ΒΕ, ΑΓ, ΔΖ. For let BA, BC, ED, and EF have been cut off (so Καὶ ἐπεὶ ἡ ΒΑ τῇ ΕΔ ἴση ἐστὶ καὶ παράλληλος, καὶ ἡ as to be, respectively) equal to one another. And let AD, ΑΔ ἄρα τῇ ΒΕ ἴση ἐστὶ καὶ παράλληλος. διὰ τὰ αὐτὰ δὴ CF , BE, AC, and DF have been joined. καὶ ἡ ΓΖ τῇ ΒΕ ἴση ἐστὶ καὶ παράλληλος· ἑκατέρα ἄρα τῶν And since BA is equal and parallel to ED, AD is thus ΑΔ, ΓΖ τῇ ΒΕ ἴση ἐστὶ καὶ παράλληλος. αἱ δὲ τῇ αὐτῇ also equal and parallel to BE [Prop. 1.33]. So, for the εὐθείᾳ παράλληλοι καὶ μὴ οὖσαι αὐτῇ ἐν τῷ αὐτῷ ἐπιπέδῳ same reasons, CF is also equal and parallel to BE. Thus, καὶ ἀλλήλαις εἰσὶ παράλληλοι· παράλληλος ἄρα ἐστὶν ἡ ΑΔ AD and CF are each equal and parallel to BE. And τῇ ΓΖ καὶ ἴση. καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΑΓ, ΔΖ· καὶ straight-lines parallel to the same straight-line, and which ἡ ΑΓ ἄρα τῇ ΔΖ ἴση ἐστὶ καὶ παράλληλος. καὶ ἐπεὶ δύο αἱ are not in the same plane as it, are also parallel to one an- ΑΒ, ΒΓ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσίν, καὶ βάσις ἡ ΑΓ βάσει other [Prop. 11.9]. Thus, AD is parallel and equal to CF . τῇ ΔΖ ἴση, γωνία ἄρα ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΔΕΖ ἐστιν And AC and DF join them. Thus, AC is also equal and 434 STOIQEIWN iaþ. ELEMENTS BOOK 11 ἴση. parallel to DF [Prop. 1.33]. And since the two (straight- ᾿Εὰν ἄρα δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο lines) AB and BC are equal to the two (straight-lines) εὐθείας ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ, DE and EF (respectvely), and the base AC (is) equal to ἵσας γωνίας περιέξουσιν· ὅπερ ἔδει δεῖξαι. the base DF , the angle ABC is thus equal to the (angle) DEF [Prop. 1.8]. Thus, if two straight-lines joined to one another are (respectively) parallel to two straight-lines joined to one another, (but are) not in the same plane, then they will contain equal angles. (Which is) the very thing it was required to show.iaþ. Proposition 11 Ἀπὸ τοῦ δοθέντος σημείου μετεώρου ἐπὶ τὸ δοθὲν To draw a perpendicular straight-line from a given ἐπίπεδον κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν. raised point to a given plane. Z G J DBH E A H D B E A F C G ῎Εστω τὸ μὲν δοθὲν σημεῖον μετέωρον τὸ Α, τὸ δὲ δοθὲν Let A be the given raised point, and the given plane ἐπίπεδον τὸ ὑποκείμενον· δεῖ δὴ ἀπὸ τοῦ Α σημείου ἐπὶ τὸ the reference (plane). So, it is required to draw a perpen- ὑποκείμενον ἐπίπεδον κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν. dicular straight-line from point A to the reference plane. Διήχθω γάρ τις ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖα, ὡς Let some random straight-line BC have been drawn ἔτυχεν, ἡ ΒΓ, καὶ ἤχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὴν ΒΓ across in the reference plane, and let the (straight-line) κάθετος ἡ ΑΔ. εἰ μὲν οὖν ἡ ΑΔ κάθετός ἐστι καὶ ἐπὶ τὸ AD have been drawn from point A perpendicular to BC ὑποκείμενον ἐπίπεδον, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. εἰ δὲ [Prop. 1.12]. If, therefore, AD is also perpendicular to οὔ, ἤχθω ἀπὸ τοῦ Δ σημείου τῇ ΒΓ ἐν τῷ ὑποκειμένῳ the reference plane then that which was prescribed will ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΔΕ, καὶ ἤχθω ἀπὸ τοῦ Α ἐπὶ τὴν ΔΕ have occurred. And, if not, let DE have been drawn in κάθετος ἡ ΑΖ, καὶ διὰ τοῦ Ζ σημείου τῇ ΒΓ παράλληλος the reference plane from point D at right-angles to BC ἤχθω ἡ ΗΘ. [Prop. 1.11], and let the (straight-line) AF have been Καὶ ἐπεὶ ἡ ΒΓ ἑκατέρᾳ τῶν ΔΑ, ΔΕ πρὸς ὀρθάς ἐστιν, drawn from A perpendicular to DE [Prop. 1.12], and let ἡ ΒΓ ἄρα καὶ τῷ διὰ τῶν ΕΔΑ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. GH have been drawn through point F , parallel to BC καί ἐστιν αὐτῇ παράλληλος ἡ ΗΘ· ἐὰν δὲ ὦσι δύο εὐθεῖαι [Prop. 1.31]. παράλληλοι, ἡ δὲ μία αὐτῶν ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ And since BC is at right-angles to each of DA and λοιπὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· καὶ ἡ ΗΘ ἄρα τῷ DE, BC is thus also at right-angles to the plane through διὰ τῶν ΕΔ, ΔΑ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. καὶ πρὸς πάσας EDA [Prop. 11.4]. And GH is parallel to it. And if two ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ διὰ τῶν straight-lines are parallel, and one of them is at right- ΕΔ, ΔΑ ἐπιπέδῳ ὀρθή ἐστιν ἡ ΗΘ. ἅπτεται δὲ αὐτῆς ἡ ΑΖ angles to some plane, then the remaining (straight-line) οὖσα ἐν τῷ διὰ τῶν ΕΔ, ΔΑ ἐπιπέδῳ· ἡ ΗΘ ἄρα ὀρθή ἐστι will also be at right-angles to the same plane [Prop. 11.8]. πρὸς τὴν ΖΑ· ὥστε καὶ ἡ ΖΑ ὀρθή ἐστι πρὸς τὴν ΘΗ. ἔστι Thus, GH is also at right-angles to the plane through 435 STOIQEIWN iaþ. ELEMENTS BOOK 11 δὲ ἡ ΑΖ καὶ πρὸς τὴν ΔΕ ὀρθή· ἡ ΑΖ ἄρα πρὸς ἑκατέραν ED and DA. And GH is thus at right-angles to all of τῶν ΗΘ, ΔΕ ὀρθή ἐστιν. ἐὰν δὲ εὐθεῖα δυσὶν εὐθείαις the straight-lines joined to it which are also in the plane τεμνούσαις ἀλλήλας ἐπὶ τῆς τομῆς πρὸς ὀρθὰς ἐπισταθῇ, through ED and AD [Def. 11.3]. And AF , which is in the καὶ τῷ δι᾿ αὐτῶν ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· ἡ ΖΑ ἄρα τῷ plane through ED and DA, is joined to it. Thus, GH is at διὰ τῶν ΕΔ, ΗΘ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τὸ δὲ διὰ right-angles to FA. Hence, FA is also at right-angles to τῶν ΕΔ, ΗΘ ἐπίπεδόν ἐστι τὸ ὑποκείμενον· ἡ ΑΖ ἄρα τῷ HG. And AF is also at right-angles to DE. Thus, AF is ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. at right-angles to each of GH and DE. And if a straight- Ἀπὸ τοῦ ἄρα δοθέντος σημείου μετεώρου τοῦ Α ἐπὶ τὸ line is set up at right-angles to two straight-lines cutting ὑποκείμενον ἐπίπεδον κάθετος εὐθεῖα γραμμὴ ἦκται ἡ ΑΖ· one another, at the point of section, then it will also be ὅπερ ἔδει ποιῆσαι. at right-angles to the plane through them [Prop. 11.4]. Thus, FA is at right-angles to the plane through ED and GH . And the plane through ED and GH is the refer- ence (plane). Thus, AF is at right-angles to the reference plane. Thus, the straight-line AF has been drawn from the given raised point A perpendicular to the reference plane. (Which is) the very thing it was required to do.ibþ. Proposition 12 Τῷ δοθέντι ἐπιπέδῳ ἀπὸ τοῦ πρὸς αὐτῷ δοθέντος To set up a straight-line at right-angles to a given σημείου πρὸς ὀρθὰς εὐθεῖαν γραμμὴν ἀναστῆσαι. plane from a given point in it. GA D B B A D C ῎Εστω τὸ μὲν δοθὲν ἐπίπεδον τὸ ὑποκείμενον, τὸ δὲ Let the given plane be the reference (plane), and A a πρὸς αὐτῷ σημεῖον τὸ Α· δεῖ δὴ ἀπὸ τοῦ Α σημείου τῷ ὑπο- point in it. So, it is required to set up a straight-line at κειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς εὐθεῖαν γραμμὴν ἀναστῆσαι. right-angles to the reference plane at point A. Νενοήσθω τι σημεῖον μετέωρον τὸ Β, καὶ ἀπὸ τοῦ Β ἐπὶ Let some raised point B have been assumed, and let τὸ ὑποκείμενον ἐπίπεδον κάθετος ἤχθω ἡ ΒΓ, καὶ διὰ τοῦ the perpendicular (straight-line) BC have been drawn Α σημείου τῇ ΒΓ παράλληλος ἤχθω ἡ ΑΔ. from B to the reference plane [Prop. 11.11]. And ᾿Επεὶ οὖν δύο εὐθεῖαι παράλληλοί εἰσιν αἱ ΑΔ, ΓΒ, ἡ δὲ let AD have been drawn from point A parallel to BC μία αὐτῶν ἡ ΒΓ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, [Prop. 1.31]. καὶ ἡ λοιπὴ ἄρα ἡ ΑΔ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς Therefore, since AD and CB are two parallel straight- ἐστιν. lines, and one of them, BC, is at right-angles to the refer- Τῷ ἄρα δοθέντι ἐπιπέδῳ ἀπὸ τοῦ πρὸς αὐτῷ σημείου ence plane, the remaining (one) AD is thus also at right- τοῦ Α πρὸς ὀρθὰς ἀνέσταται ἡ ΑΔ· ὅπερ ἔδει ποιῆσαι. angles to the reference plane [Prop. 11.8]. 436 STOIQEIWN iaþ. ELEMENTS BOOK 11 Thus, AD has been set up at right-angles to the given plane, from the point in it, A. (Which is) the very thing it was required to do.igþ. Proposition 13 Ἀπὸ τοῦ αὐτοῦ σημείου τῷ αὐτῷ ἐπιπέδῳ δύο εὐθεῖαι Two (different) straight-lines cannot be set up at the πρὸς ὀρθὰς οὐκ ἀναστήσονται ἐπὶ τὰ αὐτὰ μέρη. same point at right-angles to the same plane, on the same side. D A B E G C A B E D Εἰ γὰρ δυνατόν, ἀπὸ τοῦ αὐτοῦ σημείου τοῦ Α τῷ For, if possible, let the two straight-lines AB and AC ὑποκειμένῳ ἐπιπέδῳ δύο εὐθεῖαι αἱ ΑΒ, ΒΓ πρὸς ὀρθὰς have been set up at the same point A at right-angles ἀνεστάτωσαν ἐπὶ τὰ αὐτὰ μέρη, καὶ διήχθω τὸ διὰ τῶν to the reference plane, on the same side. And let the ΒΑ, ΑΓ ἐπὶπεδον· τομὴν δὴ ποιήσει διὰ τοῦ Α ἐν τῷ ὑπο- plane through BA and AC have been drawn. So it will κειμένῳ ἐπιπέδῳ εὐθεῖαν. ποιείτω τὴν ΔΑΕ· αἱ ἄρα ΑΒ, make a straight cutting (passing) through (point) A in ΑΓ, ΔΑΕ εὐθεῖαι ἐν ἑνι εἰσιν ἐπιπέδῳ. καὶ ἐπεὶ ἡ ΓΑ τῷ the reference plane [Prop. 11.3]. Let it have made DAE. ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, καὶ πρὸς πάσας ἄρα Thus, AB, AC, and DAE are straight-lines in one plane. τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ And since CA is at right-angles to the reference plane, it ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. ἅπτεται δὲ αὐτῆς ἡ ΔΑΕ will thus also make right-angles with all of the straight- οὖσα ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ· ἡ ἄρα ὑπὸ ΓΑΕ γωνία lines joined to it which are also in the reference plane ὀρθή ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΒΑΕ ὀρθή ἐστιν· ἴση [Def. 11.3]. And DAE, which is in the reference plane, is ἄρα ἡ ὑπὸ ΓΑΕ τῇ ὑπὸ ΒΑΕ καί εἰσιν ἐν ἑνὶ ἐπιπέδῳ· ὅπερ joined to it. Thus, angle CAE is a right-angle. So, for the ἐστὶν ἀδύνατον. same (reasons), BAE is also a right-angle. Thus, CAE Οὐκ ἄρα ἀπὸ τοῦ αὐτοῦ σημείου τῷ αὐτῷ ἐπιπέδῳ δύο (is) equal to BAE. And they are in one plane. The very εὐθεῖαι πρὸς ὀρθὰς ἀνασταθήσονται ἐπὶ τὰ αὐτὰ μέρη· ὅπερ thing is impossible. ἔδει δεῖξαι. Thus, two (different) straight-lines cannot be set up at the same point at right-angles to the same plane, on the same side. (Which is) the very thing it was required to show.idþ. Proposition 14 Πρὸς ἃ ἐπίπεδα ἡ αὐτὴ εὐθεῖα ὀρθή ἐστιν, παράλληλα Planes to which the same straight-line is at right- ἔσται τὰ ἐπίπεδα. angles will be parallel planes. Εὐθεῖα γάρ τις ἡ ΑΒ πρὸς ἑκάτερον τῶν ΓΔ, ΕΖ For let some straight-line AB be at right-angles to ἐπιπέδων πρὸς ὀρθὰς ἔστω· λέγω, ὅτι παράλληλά ἐστι τὰ each of the planes CD and EF . I say that the planes ἐπίπεδα. are parallel. 437 STOIQEIWN iaþ. ELEMENTS BOOK 11 E A B HG ZD J K C A B D K E F H G Εἰ γὰρ μή, ἐκβαλλόμενα συμπεσοῦνται. συμπιπτέτ- For, if not, being produced, they will meet. Let them ωσαν· ποιήσουσι δὴ κοινὴν τομὴν εὐθεῖαν. ποιείτωσαν τὴν have met. So they will make a straight-line as a common ΗΘ, καὶ εἰλήφθω ἐπὶ τῆς ΗΘ τυχὸν σημεῖον τὸ Κ, καὶ section [Prop. 11.3]. Let them have made GH . And let ἐπεζεύχθωσαν αἱ ΑΚ, ΒΚ. some random point K have been taken on GH . And let Καὶ ἐπεὶ ἡ ΑΒ ὀρθή ἐστι πρὸς τὸ ΕΖ ἐπίπεδον, καὶ πρὸς AK and BK have been joined. τὴν ΒΚ ἄρα εὐθεῖαν οὖσαν ἐν τῷ ΕΖ ἐκβληθέντι ἐπιπέδῳ And since AB is at right-angles to the plane EF , AB ὀρθή ἐστιν ἡ ΑΒ· ἡ ἄρα ὑπὸ ΑΒΚ γωνία ὀρθή ἐστιν. διὰ is thus also at right-angles to BK, which is a straight-line τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΒΑΚ ὀρθή ἐστιν. τριγώνου δὴ τοῦ in the produced plane EF [Def. 11.3]. Thus, angle ABK ΑΒΚ αἱ δύο γωνίαι αἱ ὑπὸ ΑΒΚ, ΒΑΚ δυσὶν ὀρθαῖς εἰσιν is a right-angle. So, for the same (reasons), BAK is also ἴσαι· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὰ ΓΔ, ΕΖ ἐπίπεδα a right-angle. So the (sum of the) two angles ABK and ἐκβαλλόμενα συμπεσοῦνται· παράλληλα ἄρα ἐστὶ τὰ ΓΔ, BAK in the triangle ABK is equal to two right-angles. ΕΖ ἐπίπεδα. The very thing is impossible [Prop. 1.17]. Thus, planes Πρὸς ἃ ἐπίπεδα ἄρα ἡ αὐτὴ εὐθεῖα ὀρθή ἐστιν, παράλληλά CD and EF , being produced, will not meet. Planes CD ἐστι τὰ ἐπίπεδα· ὅπερ ἔδει δεῖξαι. and EF are thus parallel [Def. 11.8]. Thus, planes to which the same straight-line is at right-angles are parallel planes. (Which is) the very thing it was required to show.ieþ. Proposition 15 ᾿Εὰν δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας If two straight-lines joined to one another are parallel ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι, (respectively) to two straight-lines joined to one another, παράλληλά ἐστι τὰ δι᾿ αὐτῶν ἐπίπεδα. which are not in the same plane, then the planes through Δύο γὰρ εὐθεῖαι ἁπτόμεναι ἀλλήλων αἱ ΑΒ, ΒΓ παρὰ them are parallel (to one another). δύο εὐθείας ἁπτομένας ἀλλήλων τὰς ΔΕ, ΕΖ ἔστωσαν μὴ For let the two straight-lines joined to one another, ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι· λέγω, ὅτι ἐκβαλλόμενα τὰ διὰ AB and BC, be parallel to the two straight-lines joined to τῶν ΑΒ, ΒΓ, ΔΕ, ΕΖ ἐπίπεδα οὐ συμπεσεῖται ἀλλήλοις. one another, DE and EF (respectively), not being in the ῎Ηχθω γὰρ ἀπὸ τοῦ Β σημείου ἐπὶ τὸ διὰ τῶν ΔΕ, ΕΖ same plane. I say that the planes through AB, BC and ἐπίπεδον κάθετος ἡ ΒΗ καὶ συμβαλλέτω τῷ ἐπιπέδῳ κατὰ DE, EF will not meet one another (when) produced. τὸ Η σημεῖον, καὶ διὰ τοῦ Η τῇ μὲν ΕΔ παράλληλος ἤχθω For let BG have been drawn from point B perpendic- ἡ ΗΘ, τῇ δὲ ΕΖ ἡ ΗΚ. ular to the plane through DE and EF [Prop. 11.11], and let it meet the plane at point G. And let GH have been drawn through G parallel to ED, and GK (parallel) to EF [Prop. 1.31]. 438 STOIQEIWN iaþ. ELEMENTS BOOK 11 H ZD B A G E KJ GD B A E K C H F Καὶ ἐπεὶ ἡ ΒΗ ὀρθή ἐστι πρὸς τὸ διὰ τῶν ΔΕ, ΕΖ And since BG is at right-angles to the plane through ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας DE and EF , it will thus also make right-angles with all καὶ οὔσας ἐν τῷ διὰ τῶν ΔΕ, ΕΖ ἐπιπέδῳ ὀρθὰς ποιήσει of the straight-lines joined to it, which are also in the γωνίας. ἅπτεται δὲ αὐτῆς ἑκατέρα τῶν ΗΘ, ΗΚ οὖσα ἐν plane through DE and EF [Def. 11.3]. And each of τῷ διὰ τῶν ΔΕ, ΕΖ ἐπιπέδῳ· ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν GH and GK, which are in the plane through DE and ὑπὸ ΒΗΘ, ΒΗΚ γωνιῶν. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΒΑ EF , are joined to it. Thus, each of the angles BGH and τῇ ΗΘ, αἱ ἄρα ὑπὸ ΗΒΑ, ΒΗΘ γωνίαι δυσὶν ὀρθαῖς ἴσαι BGK are right-angles. And since BA is parallel to GH εἰσίν. ὀρθὴ δὲ ἡ ὑπὸ ΒΗΘ· ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΗΒΑ· ἡ ΗΒ [Prop. 11.9], the (sum of the) angles GBA and BGH is ἄρα τῇ ΒΑ πρὸς ὀρθάς ἐστιν. διὰ τὰ αὐτὰ δὴ ἡ ΗΒ καὶ τῇ equal to two right-angles [Prop. 1.29]. And BGH (is) ΒΓ ἐστι πρὸς ὀρθάς. ἐπεὶ οὖν εὐθεῖα ἡ ΗΒ δυσὶν εὐθείαις a right-angle. GBA (is) thus also a right-angle. Thus, ταῖς ΒΑ, ΒΓ τεμνούσαις ἀλλήλας πρὸς ὀρθὰς ἐφέστηκεν, ἡ GB is at right-angles to BA. So, for the same (reasons), ΗΒ ἄρα καὶ τῷ διὰ τῶν ΒΑ, ΒΓ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. GB is also at right-angles to BC. Therefore, since the [διὰ τὰ αὐτὰ δὴ ἡ ΒΗ καὶ τῷ διὰ τῶν ΗΘ, ΗΚ ἐπιπέδῳ straight-line GB has been set up at right-angles to two πρὸς ὀρθάς ἐστιν. τὸ δὲ διὰ τῶν ΗΘ, ΗΚ ἐπίπεδόν ἐστι τὸ straight-lines, BA and BC, cutting one another, GB is διὰ τῶν ΔΕ, ΕΖ· ἡ ΒΗ ἄρα τῷ διὰ τῶν ΔΕ, ΕΖ ἐπιπέδῳ thus at right-angles to the plane through BA and BC ἐστὶ πρὸς ὀρθάς. ἐδείχθη δὲ ἡ ΗΒ καὶ τῷ διὰ τῶν ΑΒ, ΒΓ [Prop. 11.4]. [So, for the same (reasons), BG is also ἐπιπέδῳ πρὸς ὀρθάς]. πρὸς ἃ δὲ ἐπίπεδα ἡ αὐτὴ εὐθεῖα ὀρθή at right-angles to the plane through GH and GK. And ἐστιν, παράλληλά ἐστι τὰ ἐπίπεδα· παράλληλον ἄρα ἐστὶ τὸ the plane through GH and GK is the (plane) through διὰ τῶν ΑΒ, ΒΓ ἐπίπεδον τῷ διὰ τῶν ΔΕ, ΕΖ. DE and EF . And it was also shown that GB is at right- ᾿Εὰν ἄρα δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο angles to the plane through AB and BC.] And planes εὐθείας ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ, to which the same straight-line is at right-angles are par- παράλληλά ἐστι τὰ δι᾿ αὐτῶν ἐπίπεδα· ὅπερ ἔδει δεῖξαι. allel planes [Prop. 11.14]. Thus, the plane through AB and BC is parallel to the (plane) through DE and EF . Thus, if two straight-lines joined to one another are parallel (respectively) to two straight-lines joined to one another, which are not in the same plane, then the planes through them are parallel (to one another). (Which is) the very thing it was required to show.i�þ. Proposition 16 ᾿Εὰν δύο ἐπίπεδα παράλληλα ὑπὸ ἐπιπέδου τινὸς τέμνηται, If two parallel planes are cut by some plane then their αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν. common sections are parallel. Δύο γὰρ ἐπίπεδα παράλληλα τὰ ΑΒ, ΓΔ ὑπὸ ἐπιπέδου For let the two parallel planes AB and CD have been τοῦ ΕΖΗΘ τεμνέσθω, κοιναὶ δὲ αὐτῶν τομαὶ ἔστωσαν αἱ cut by the plane EFGH . And let EF and GH be their ΕΖ, ΗΘ· λέγω, ὅτι παράλληλός ἐστιν ἡ ΕΖ τῇ ΗΘ. common sections. I say that EF is parallel to GH . 439 STOIQEIWN iaþ. ELEMENTS BOOK 11 D J K E H G A ZB F K E A B D G C H Εἰ γὰρ μή, ἐκβαλλόμεναι αἱ ΕΖ, ΗΘ ἤτοι ἐπὶ τὰ Ζ, Θ For, if not, being produced, EF and GH will meet ei- μέρη ἢ ἐπὶ τὰ Ε, Η συμπεσοῦνται. ἐκβεβλήσθωσαν ὡς ἐπὶ ther in the direction of F , H , or of E, G. Let them be τὰ Ζ, Θ μέρη καὶ συμπιπτέτωσαν πρότερον κατὰ τὸ Κ. καὶ produced, as in the direction of F , H , and let them, first ἐπεὶ ἡ ΕΖΚ ἐν τῷ ΑΒ ἐστιν ἐπιπέδῳ, καὶ πάντα ἄρα τὰ ἐπὶ of all, have met at K. And since EFK is in the plane τῆς ΕΖΚ σημεῖα ἐν τῷ ΑΒ ἐστιν ἐπιπέδῳ. ἓν δὲ τῶν ἐπὶ τῆς AB, all of the points on EFK are thus also in the plane ΕΖΚ εὐθείας σημείων ἐστὶ τὸ Κ· τὸ Κ ἄρα ἐν τῷ ΑΒ ἐστιν AB [Prop. 11.1]. And K is one of the points on EFK. ἐπιπέδῳ. διὰ τὰ αὐτὰ δὴ τὸ Κ καὶ ἐν τῷ ΓΔ ἐστιν ἐπιπέδῳ· Thus, K is in the plane AB. So, for the same (reasons), τὰ ΑΒ, ΓΔ ἄρα ἐπίπεδα ἐκβαλλόμενα συμπεσοῦνται. οὐ K is also in the plane CD. Thus, the planes AB and CD, συμπίπτουσι δὲ διὰ τὸ παράλληλα ὑποκεῖσθαι· οὐκ ἄρα being produced, will meet. But they do not meet, on ac- αἱ ΕΖ, ΗΘ εὐθεῖαι ἐκβαλλόμεναι ἐπὶ τὰ Ζ, Θ μέρη συμ- count of being (initially) assumed (to be mutually) paral- πεσοῦνται. ὁμοίως δὴ δείξομεν, ὅτι αἱ ΕΖ, ΗΘ εὐθεῖαι lel. Thus, the straight-lines EF and GH , being produced οὐδέ ἐπὶ τὰ Ε, Η μέρη ἐκβαλλόμεναι συμπεσοῦνται. αἱ in the direction of F , H , will not meet. So, similarly, we δὲ ἐπὶ μηδέτερα τὰ μέρη συμπίπτουσαι παράλληλοί εἰσιν. can show that the straight-lines EF and GH , being pro- παράλληλος ἄρα ἐστὶν ἡ ΕΖ τῇ ΗΘ. duced in the direction of E, G, will not meet either. And ᾿Εὰν ἄρα δύο ἐπίπεδα παράλληλα ὑπὸ ἐπιπέδου τινὸς (straight-lines in one plane which), being produced, do τέμνηται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν· ὅπερ ἔδει not meet in either direction are parallel [Def. 1.23]. EF δεῖξαι. is thus parallel to GH . Thus, if two parallel planes are cut by some plane then their common sections are parallel. (Which is) the very thing it was required to show.izþ. Proposition 17 ᾿Εὰν δύο εὐθεῖαι ὑπὸ παραλλήλων ἐπιπεδων τέμνωνται, If two straight-lines are cut by parallel planes then εἰς τοὺς αὐτοὺς λόγους τμηθήσονται. they will be cut in the same ratios. Δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ ὑπὸ παραλλήλων ἐπιπέδων For let the two straight-lines AB and CD be cut by the τῶν ΗΘ, ΚΛ, ΜΝ τεμνέσθωσαν κατὰ τὰ Α, Ε, Β, Γ, Ζ, parallel planes GH , KL, and MN at the points A, E, B, Δ σημεῖα· λέγω, ὅτι ἐστὶν ὡς ἡ ΑΕ εὐθεῖα πρὸς τὴν ΕΒ, and C, F , D (respectively). I say that as the straight-line οὕτως ἡ ΓΖ πρὸς τὴν ΖΔ. AE is to EB, so CF (is) to FD. ᾿Επεζεύχθωσαν γὰρ αἱ ΑΓ, ΒΔ, ΑΔ, καὶ συμβαλλέτω ἡ For let AC, BD, and AD have been joined, and let ΑΔ τῷ ΚΛ ἐπιπέδῳ κατὰ τὸ Ξ σημεῖον, καὶ ἐπεζεύχθωσαν AD meet the plane KL at point O, and let EO and OF αἱ ΕΞ, ΞΖ. have been joined. Καὶ ἐπεὶ δύο ἐπίπεδα παράλληλα τὰ ΚΛ, ΜΝ ὑπὸ And since two parallel planes KL and MN are cut ἐπιπέδου τοῦ ΕΒΔΞ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ αἱ by the plane EBDO, their common sections EO and BD ΕΞ, ΒΔ παράλληλοί εἰσιν. διὰ τὰ αὐτὰ δὴ ἐπεὶ δύο ἐπίπεδα are parallel [Prop. 11.16]. So, for the same (reasons), παράλληλα τὰ ΗΘ, ΚΛ ὑπὸ ἐπιπέδου τοῦ ΑΞΖΓ τέμνεται, since two parallel planes GH and KL are cut by the αἱ κοιναὶ αὐτῶν τομαὶ αἱ ΑΓ, ΞΖ παράλληλοί εἰσιν. καὶ ἐπεὶ plane AOFC, their common sections AC and OF are τριγώνου τοῦ ΑΒΔ παρὰ μίαν τῶν πλευρῶν τὴν ΒΔ εὐθεῖα parallel [Prop. 11.16]. And since the straight-line EO ἦκται ἡ ΕΞ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΑΕ πρὸς ΕΒ, οὕτως has been drawn parallel to one of the sides BD of trian- 440 STOIQEIWN iaþ. ELEMENTS BOOK 11 ἡ ΑΞ πρὸς ΞΔ. πάλιν ἐπεὶ τριγώνου τοῦ ΑΔΓ παρὰ μίαν gle ABD, thus, proportionally, as AE is to EB, so AO τῶν πλευρῶν τὴν ΑΓ εὐθεῖα ἦκται ἡ ΞΖ, ἀνάλογόν ἐστιν (is) to OD [Prop. 6.2]. Again, since the straight-line OF ὡς ἡ ΑΞ πρὸς ΞΔ, οὕτως ἡ ΓΖ πρὸς ΖΔ. ἐδείχθη δὲ καὶ has been drawn parallel to one of the sides AC of trian- ὡς ἡ ΑΞ πρὸς ΞΔ, οὕτως ἡ ΑΕ πρὸς ΕΒ· καὶ ὡς ἄρα ἡ ΑΕ gle ADC, proportionally, as AO is to OD, so CF (is) to πρὸς ΕΒ, οὕτως ἡ ΓΖ πρὸς ΖΔ. FD [Prop. 6.2]. And it was also shown that as AO (is) to OD, so AE (is) to EB. And thus as AE (is) to EB, so CF (is) to FD [Prop. 5.11]. ZK NB DM H JA LE G X F K N B D M A L E C G H O ᾿Εὰν ἄρα δύο εὐθεῖαι ὑπὸ παραλλήλων ἐπιπέδων τέμνων- Thus, if two straight-lines are cut by parallel planes ται, εἰς τοὺς αὐτοὺς λόγους τμηθήσονται· ὅπερ ἔδει δειξαι. then they will be cut in the same ratios. (Which is) the very thing it was required to show.ihþ. Proposition 18 ᾿Εὰν εὐθεῖα ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ πάντα τὰ δι᾿ If a straight-line is at right-angles to some plane then αὐτῆς ἐπίπεδα τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται. all of the planes (passing) through it will also be at right- Εὐθεῖα γάρ τις ἡ ΑΒ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς angles to the same plane. ἔστω· λέγω, ὅτι καὶ πάντα τὰ διὰ τῆς ΑΒ ἐπίπεδα τῷ ὑπο- For let some straight-line AB be at right-angles to κειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. a reference plane. I say that all of the planes (pass- ᾿Εκβεβλήσθω γὰρ διὰ τῆς ΑΒ ἐπίπεδον τὸ ΔΕ, καὶ ἔστω ing) through AB are also at right-angles to the reference κοινὴ τομὴ τοῦ ΔΕ ἐπιπέδου καὶ τοῦ ὑποκειμένου ἡ ΓΕ, καὶ plane. εἰλήφθω ἐπὶ τῆς ΓΕ τυχὸν σημεῖον τὸ Ζ, καὶ ἀπὸ τοῦ Ζ τῇ For let the plane DE have been produced through ΓΕ πρὸς ὀρθὰς ἤχθω ἐν τῷ ΔΕ ἐπιπέδῳ ἡ ΖΗ. AB. And let CE be the common section of the plane Καὶ ἐπεὶ ἡ ΑΒ πρὸς τὸ ὑποκείμενον ἐπίπεδον ὀρθή DE and the reference (plane). And let some random ἐστιν, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ point F have been taken on CE. And let FG have been οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθή ἐστιν ἡ ΑΒ· ὥστε drawn from F , at right-angles to CE, in the plane DE καὶ πρὸς τὴν ΓΕ ὀρθή ἐστιν· ἡ ἄρα ὑπὸ ΑΒΖ γωνία ὀρθή [Prop. 1.11]. ἐστιν. ἔστι δὲ καὶ ἡ ὑπὸ ΗΖΒ ὀρθὴ· παράλληλος ἄρα ἐστὶν And since AB is at right-angles to the reference plane, ἡ ΑΒ τῇ ΖΗ. ἡ δὲ ΑΒ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς AB is thus also at right-angles to all of the straight- ἐστιν· καὶ ἡ ΖΗ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς lines joined to it which are also in the reference plane ἐστιν. καὶ ἐπίπεδον πρὸς ἐπίπεδον ὀρθόν ἐστιν, ὅταν αἱ τῇ [Def. 11.3]. Hence, it is also at right-angles to CE. Thus, κοινῇ τομῇ τῶν ἐπιπέδων πρὸς ὀρθὰς ἀγόμεναι εὐθεῖαι ἐν angle ABF is a right-angle. And GFB is also a right- ἑνὶ τῶν ἐπιπέδων τῷ λοιπῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν. καὶ angle. Thus, AB is parallel to FG [Prop. 1.28]. And AB τῇ κοινῇ τομῇ τῶν ἐπιπέδων τῇ ΓΕ ἐν ἑνὶ τῶν ἐπιπέδων is at right-angles to the reference plane. Thus, FG is also 441 STOIQEIWN iaþ. ELEMENTS BOOK 11 τῷ ΔΕ πρὸς ὀρθὰς ἀχθεῖσα ἡ ΖΗ ἐδείχθη τῷ ὑποκειμένῳ at right-angles to the reference plane [Prop. 11.8]. And ἐπιπέδῳ πρὸς ὀρθάς· τὸ ἄρα ΔΕ ἐπίπεδον ὀρθόν ἐστι πρὸς a plane is at right-angles to a(nother) plane when the τὸ ὑποκείμενον. ὁμοίως δὴ δειχθήσεται καὶ πάντα τὰ διὰ straight-lines drawn at right-angles to the common sec- τῆς ΑΒ ἐπίπεδα ὀρθὰ τυγχανοντα πρὸς τὸ ὑποκείμενον tion of the planes, (and lying) in one of the planes, are ἐπίπεδον. at right-angles to the remaining plane [Def. 11.4]. And FG, (which was) drawn at right-angles to the common section of the planes, CE, in one of the planes, DE, was shown to be at right-angles to the reference plane. Thus, plane DE is at right-angles to the reference (plane). So, similarly, it can be shown that all of the planes (passing) at random through AB (are) at right-angles to the refer- ence plane. Z D H A EG B GD A E B C F ᾿Εὰν ἄρα εὐθεῖα ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ πάντα τὰ Thus, if a straight-line is at right-angles to some plane δι᾿ αὐτῆς ἐπίπεδα τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· ὅπερ then all of the planes (passing) through it will also be at ἔδει δεῖξαι. right-angles to the same plane. (Which is) the very thing it was required to show.ijþ. Proposition 19 ᾿Εὰν δύο ἐπίπεδα τέμνοντα ἄλληλα ἐπιπέδῳ τινὶ πρὸς If two planes cutting one another are at right-angles ὀρθὰς ᾖ, καὶ ἡ κοινὴ αὐτῶν τομὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς to some plane then their common section will also be at ὀρθὰς ἔσται. right-angles to the same plane. D A G ZE B F A E B D C Δύο γὰρ ἐπίπεδα τὰ ΑΒ, ΒΓ τῷ ὑποκειμένῳ ἐπιπέδῳ For let the two planes AB and BC be at right-angles πρὸς ὀρθὰς ἔστω, κοινὴ δὲ αὐτῶν τομὴ ἔστω ἡ ΒΔ· λέγω, to a reference plane, and let their common section be ὅτι ἡ ΒΔ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. BD. I say that BD is at right-angles to the reference 442 STOIQEIWN iaþ. ELEMENTS BOOK 11 Μὴ γάρ, καὶ ἤχθωσαν ἀπὸ τοῦ Δ σημείου ἐν μὲν τῷ plane. ΑΒ ἐπιπέδῳ τῇ ΑΔ εὐθείᾳ πρὸς ὀρθὰς ἡ ΔΕ, ἐν δὲ τῷ ΒΓ For (if) not, let DE also have been drawn from point ἐπιπέδῳ τῇ ΓΔ πρὸς ὀρθὰς ἡ ΔΖ. D, in the plane AB, at right-angles to the straight-line Καὶ ἐπεὶ τὸ ΑΒ ἐπίπεδον ὀρθόν ἐστι πρὸς τὸ ὑποκείμενον, AD, and DF , in the plane BC, at right-angles to CD. καὶ τῇ κοινῇ αὐτῶν τομῇ τῇ ΑΔ πρὸς ὀρθὰς ἐν τῷ And since the plane AB is at right-angles to the refer- ΑΒ ἐπιπέδῳ ἦκται ἡ ΔΕ, ἡ ΔΕ ἄρα ὀρθή ἐστι πρὸς τὸ ence (plane), and DE has been drawn at right-angles to ὑποκείμενον ἐπίπεδον. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΔΖ their common section AD, in the plane AB, DE is thus at ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον. ἀπὸ τοῦ αὐτοῦ right-angles to the reference plane [Def. 11.4]. So, simi- ἄρα σημείου τοῦ Δ τῷ ὑποκειμένῳ ἐπιπέδῳ δύο εὐθεῖα larly, we can show that DF is also at right-angles to the πρὸς ὀρθὰς ἀνεσταμέναι εἰσὶν ἐπὶ τὰ αὐτὰ μέρη· ὅπερ ἐστὶν reference plane. Thus, two (different) straight-lines are ἀδύνατον. οὐκ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ ἀπὸ τοῦ Δ set up, at the same point D, at right-angles to the refer- σημείου ἀνασταθήσεται πρὸς ὀρθὰς πλὴν τῆς ΔΒ κοινῆς ence plane, on the same side. The very thing is impossible τομῆς τῶν ΑΒ, ΒΓ ἐπιπέδων. [Prop. 11.13]. Thus, no (other straight-line) except the ᾿Εὰν ἄρα δύο ἐπίπεδα τέμνοντα ἄλληλα ἐπιπέδῳ τινὶ πρὸς common section DB of the planes AB and BC can be set ὀρθὰς ᾖ, καὶ ἡ κοινὴ αὐτῶν τομὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς up at point D, at right-angles to the reference plane. ὀρθὰς ἔσται· ὅπερ ἔδει δεῖξαι. Thus, if two planes cutting one another are at right- angles to some plane then their common section will also be at right-angles to the same plane. (Which is) the very thing it was required to show.kþ. Proposition 20 ᾿Εὰν στερεὰ γωνία ὑπὸ τριῶν γωνιῶν ἐπιπέδων περιέχη- If a solid angle is contained by three plane angles then ται, δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμ- (the sum of) any two (angles) is greater than the remain- βανόμεναι. ing (one), (the angles) being taken up in any (possible way). E GB D A EB D A C Στερεὰ γὰρ γωνία ἡ πρὸς τῷ Α ὑπὸ τριῶν γωνιῶν For let the solid angle A have been contained by the ἐπιπέδων τῶν ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ περιεχέσθω· λέγω, three plane angles BAC, CAD, and DAB. I say that (the ὅτι τῶν ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ γωνιῶν δύο ὁποιαιοῦν τῆς sum of) any two of the angles BAC, CAD, and DAB λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι. is greater than the remaining (one), (the angles) being Εἰ μὲν οὖν αἱ ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ γωνίαι ἴσαι ἀλλήλαις taken up in any (possible way). εἰσίν, φανερόν, ὅτι δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσιν. For if the angles BAC, CAD, and DAB are equal to εἰ δὲ οὔ, ἔστω μείζων ἡ ὑπὸ ΒΑΓ, καὶ συνεστάτω πρὸς τῇ one another then (it is) clear that (the sum of) any two ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΔΑΒ is greater than the remaining (one). But, if not, let BAC γωνίᾳ ἐν τῷ διὰ τῶν ΒΑΓ ἐπιπέδῳ ἴση ἡ ὑπὸ ΒΑΕ, καὶ be greater (than CAD or DAB). And let (angle) BAE, κείσθω τῇ ΑΔ ἴση ἡ ΑΕ, καὶ διὰ τοῦ Ε σημείου διαχθεῖσα equal to the angle DAB, have been constructed in the ἡ ΒΕΓ τεμνέτω τὰς ΑΒ, ΑΓ εὐθείας κατὰ τὰ Β, Γ σημεῖα, plane through BAC, on the straight-line AB, at the point καὶ ἐπεζεύχθωσαν αἱ ΔΒ, ΔΓ. A on it. And let AE be made equal to AD. And BEC be- Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΕ, κοινὴ δὲ ἡ ΑΒ, δύο ing drawn across through point E, let it cut the straight- δυσὶν ἴσαι· καὶ γωνία ἡ ὑπὸ ΔΑΒ γωνίᾳ τῇ ὑπὸ ΒΑΕ ἴση· lines AB and AC at points B and C (respectively). And βάσις ἄρα ἡ ΔΒ βάσει τῇ ΒΕ ἐστιν ἴση. καὶ ἐπεὶ δύο αἱ ΒΔ, let DB and DC have been joined. ΔΓ τῆς ΒΓ μείζονές εἰσιν, ὧν ἡ ΔΒ τῇ ΒΕ ἐδείχθη ἴση, And since DA is equal to AE, and AB (is) common, 443 STOIQEIWN iaþ. ELEMENTS BOOK 11 λοιπὴ ἄρα ἡ ΔΓ λοιπῆς τῆς ΕΓ μείζων ἐστίν. καὶ ἐπεὶ ἴση the two (straight-lines AD and AB are) equal to the ἐστὶν ἡ ΔΑ τῇ ΑΕ, κοινὴ δὲ ἡ ΑΓ, καὶ βάσις ἡ ΔΓ βάσεως two (straight-lines EA and AB, respectively). And an- τῆς ΕΓ μείζων ἐστίν, γωνία ἄρα ἡ ὑπὸ ΔΑΓ γωνάις τῆς ὑπὸ gle DAB (is) equal to angle BAE. Thus, the base DB ΕΑΓ μείζων ἐστίν. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΔΑΒ τῇ ὑπὸ ΒΑΕ is equal to the base BE [Prop. 1.4]. And since the (sum ἴση· αἱ ἄρα ὑπὸ ΔΑΒ, ΔΑΓ τῆς ὑπὸ ΒΑΓ μείζονές εἰσιν. of the) two (straight-lines) BD and DC is greater than ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ λοιπαὶ σύνδυο λαμβανόμεναι BC [Prop. 1.20], of which DB was shown (to be) equal τῆς λοιπῆς μείζονές εἰσιν. to BE, the remainder DC is thus greater than the re- ᾿Εὰν ἄρα στερεὰ γωνία ὑπὸ τριῶν γωνιῶν ἐπιπέδων mainder EC. And since DA is equal to AE, but AC περιέχηται, δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσι πάντῃ (is) common, and the base DC is greater than the base μεταλαμβανόμεναι· ὅπερ ἔδει δεῖξαι. EC, the angle DAC is thus greater than the angle EAC [Prop. 1.25]. And DAB was also shown (to be) equal to BAE. Thus, (the sum of) DAB and DAC is greater than BAC. So, similarly, we can also show that the remain- ing (angles), being taken in pairs, are greater than the remaining (one). Thus, if a solid angle is contained by three plane an- gles then (the sum of) any two (angles) is greater than the remaining (one), (the angles) being taken up in any (possible way). (Which is) the very thing it was required to show.kaþ. Proposition 21 ῞Απασα στερεὰ γωνία ὑπὸ ἐλασσόνων [ἢ] τεσσάρων Any solid angle is contained by plane angles (whose ὀρθῶν γωνιῶν ἐπιπέδων περιέχεται. sum is) less [than] four right-angles.†G A DB A D C B ῎Εστω στερεὰ γωνία ἡ πρὸς τῷ Α περιεχομένη ὑπὸ Let the solid angle A be contained by the plane angles ἐπιπέδων γωνιῶν τῶν ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ· λέγω, ὅτι αἱ BAC, CAD, and DAB. I say that (the sum of) BAC, ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ τεσσάρων ὀρθῶν ἐλάσσονές εἰσιν. CAD, and DAB is less than four right-angles. Εἰλήφθω γὰρ ἐφ᾿ ἑκάστης τῶν ΑΒ, ΑΓ, ΑΔ τυχόντα For let the random points B, C, and D have been σημεῖα τὰ Β, Γ, Δ, καὶ ἐπεζεύχθωσαν αἱ ΒΓ, ΓΔ, ΔΒ. καὶ taken on each of (the straight-lines) AB, AC, and AD ἐπεὶ στερεὰ γωνία ἡ πρὸς τῷ Β ὑπὸ τριῶν γωνιῶν ἐπιπέδων (respectively). And let BC, CD, and DB have been περιέχεται τῶν ὑπὸ ΓΒΑ, ΑΒΔ, ΓΒΔ, δύο ὁποιαιοῦν τῆς joined. And since the solid angle at B is contained λοιπῆς μείζονές εἰσιν· αἱ ἄρα ὑπὸ ΓΒΑ, ΑΒΔ τῆς ὑπὸ ΓΒΔ by the three plane angles CBA, ABD, and CBD, (the μείζονές εἰσιν. διὰ τὰ αὐτὰ δὴ καὶ αἱ μὲν ὑπὸ ΒΓΑ, ΑΓΔ sum of) any two is greater than the remaining (one) τῆς ὑπὸ ΒΓΔ μείζονές εἰσιν, αἱ δὲ ὑπὸ ΓΔΑ, ΑΔΒ τῆς ὑπὸ [Prop. 11.20]. Thus, (the sum of) CBA and ABD is ΓΔΒ μείζονές εἰσιν· αἱ ἓξ ἄρα γωνίαι αἱ ὑπὸ ΓΒΑ, ΑΒΔ, greater than CBD. So, for the same (reasons), (the sum ΒΓΑ, ΑΓΔ, ΓΔΑ, ΑΔΒ τριῶν τῶν ὑπὸ ΓΒΔ, ΒΓΑ, ΓΔΒ of) BCA and ACD is also greater than BCD, and (the μείζονές εἰσιν. ἀλλὰ αἱ τρεῖς αἱ ὑπὸ ΓΒΔ, ΒΔΓ, ΒΓΔ δυσὶν sum of) CDA and ADB is greater than CDB. Thus, ὀρθαῖς ἴσαι εἰσίν· αἱ ἓξ ἄρα αἱ ὑπὸ ΓΒΑ, ΑΒΔ, ΒΓΑ, ΑΓΔ, the (sum of the) six angles CBA, ABD, BCA, ACD, ΓΔΑ, ΑΔΒ δύο ὀρθῶν μείζονές εἰσιν. καὶ ἐπεὶ ἑκάστου τῶν CDA, and ADB is greater than the (sum of the) three ΑΒΓ, ΑΓΔ, ΑΔΒ τριγώνων αἱ τρεῖς γωνίαι δυσὶν ὀρθαῖς (angles) CBD, BCD, and CDB. But, the (sum of the) ἴσαι εἰσίν, αἱ ἄρα τῶν τριῶν τριγώνων ἐννέα γωνίαι αἱ ὑπὸ three (angles) CBD, BDC, and BCD is equal to two 444 STOIQEIWN iaþ. ELEMENTS BOOK 11 ΓΒΑ, ΑΓΒ, ΒΑΓ, ΑΓΔ, ΓΔΑ, ΓΑΔ, ΑΔΒ, ΔΒΑ, ΒΑΔ ἓξ right-angles [Prop. 1.32]. Thus, the (sum of the) six an- ὀρθαῖς ἴσαι εἰσίν, ὧν αἱ ὑπὸ ΑΒΓ, ΒΓΑ, ΑΓΔ, ΓΔΑ, ΑΔΒ, gles CBA, ABD, BCA, ACD, CDA, and ADB is greater ΔΒΑ ἓξ γωνίαι δύο ὀρθῶν εἰσι μείζονες· λοιπαὶ ἄρα αἱ ὑπὸ than two right-angles. And since the (sum of the) three ΒΑΓ, ΓΑΔ, ΔΑΒ τρεῖς [γωνίαι] περιέχουσαι τὴν στερεὰν angles of each of the triangles ABC, ACD, and ADB γωνίαν τεσσάρων ὀρθῶν ἐλάσσονές εἰσιν. is equal to two right-angles, the (sum of the) nine angles ῞Απασα ἄρα στερεὰ γωνία ὑπὸ ἐλασσόνων [ἤ] τεσσάρων CBA, ACB, BAC, ACD, CDA, CAD, ADB, DBA, and ὀρθῶν γωνιῶν ἐπιπέδων περιέχεται· ὅπερ ἔδει δεῖξαι. BAD of the three triangles is equal to six right-angles, of which the (sum of the) six angles ABC, BCA, ACD, CDA, ADB, and DBA is greater than two right-angles. Thus, the (sum of the) remaining three [angles] BAC, CAD, and DAB, containing the solid angle, is less than four right-angles. Thus, any solid angle is contained by plane angles (whose sum is) less [than] four right-angles. (Which is) the very thing it was required to show. † This proposition is only proved for the case of a solid angle contained by three plane angles. However, the generalization to a solid angle contained by more than three plane angles is straightforward.kbþ. Proposition 22 ᾿Εὰν ὧσι τρεῖς γωνίαι ἐπίπεδοι, ὧν αἱ δύο τῆς λοιπῆς If there are three plane angles, of which (the sum of μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, περιέχωσι δὲ αὐτὰς any) two is greater than the remaining (one), (the an- ἴσαι εὐθεῖαι, δυνατόν ἐστιν ἐκ τῶν ἐπιζευγνυουσῶν τὰς ἴσας gles) being taken up in any (possible way), and if equal εὐθείας τρίγωνον συστήσασθαι. straight-lines contain them, then it is possible to construct a triangle from (the straight-lines created by) joining the (ends of the) equal straight-lines. L KA Z E H J G D B G KA E D B L C F H ῎Εστωσαν τρεῖς γωνίαι ἐπίπεδοι αἱ ὑπὸ ΑΒΓ, ΔΕΖ, Let ABC, DEF , and GHK be three plane angles, of ΗΘΚ, ὧν αἱ δύο τῆς λοιπῆς μείζονές εἰσι πάντῃ μετα- which the sum of any) two is greater than the remain- λαμβανόμεναι, αἱ μὲν ὑπὸ ΑΒΓ, ΔΕΖ τῆς ὑπὸ ΗΘΚ, αἱ ing (one), (the angles) being taken up in any (possible δὲ ὑπὸ ΔΕΖ, ΗΘΚ τῆς ὑπὸ ΑΒΓ, καὶ ἔτι αἱ ὑπὸ ΗΘΚ, way)—(that is), ABC and DEF (greater) than GHK, ΑΒΓ τῆς ὑπὸ ΔΕΖ, καὶ ἔστωσαν ἴσαι αἱ ΑΒ, ΒΓ, ΔΕ, DEF and GHK (greater) than ABC, and, further, GHK ΕΖ, ΗΘ, ΘΚ εὐθεῖαι, καὶ ἐπεζεύχθωσαν αἱ ΑΓ, ΔΖ, ΗΚ· and ABC (greater) than DEF . And let AB, BC, DE, λέγω, ὅτι δυνατόν ἐστιν ἐκ τῶν ἴσων ταῖς ΑΓ, ΔΖ, ΗΚ EF , GH , and HK be equal straight-lines. And let AC, τρίγωνον συστήσασθαι, τουτέστιν ὅτι τῶν ΑΓ, ΔΖ, ΗΚ DF , and GK have been joined. I say that that it is possi- δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσιν. ble to construct a triangle out of (straight-lines) equal to Εἰ μὲν οὖν αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ γωνίαι ἴσαι AC, DF , and GK—that is to say, that (the sum of) any ἀλλήλαις εἰσίν, φανερόν, ὅτι καὶ τῶν ΑΓ, ΔΖ, ΗΚ ἴσων two of AC, DF , and GK is greater than the remaining γινομένων δυνατόν ἐστιν ἐκ τῶν ἴσων ταῖς ΑΓ, ΔΖ, ΗΚ (one). τρίγωνον συστήσασθαι. εἰ δὲ οὔ, ἔστωσαν ἄνισοι, καὶ συ- Now, if the angles ABC, DEF , and GHK are equal νεστάτω πρὸς τῇ ΘΚ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ to one another then (it is) clear that, (with) AC, DF , Θ τῇ ὑπὸ ΑΒΓ γωνίᾳ ἴση ἡ ὑπὸ ΚΘΛ· καὶ κείσθω μιᾷ τῶν and GK also becoming equal, it is possible to construct a ΑΒ, ΒΓ, ΔΕ, ΕΖ, ΗΘ, ΘΚ ἴση ἡ ΘΛ, καὶ ἐπεζεύχθωσαν triangle from (straight-lines) equal to AC, DF , and GK. αἱ ΚΛ, ΗΛ. καὶ ἐπεὶ δύο αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΚΘ, ΘΛ ἴσαι And if not, let them be unequal, and let KHL, equal to εἰσίν, καὶ γωνία ἡ πρὸς τῷ Β γωνίᾳ τῇ ὑπὸ ΚΘΛ ἴση, βάσις angle ABC, have been constructed on the straight-line ἄρα ἡ ΑΓ βάσει τῇ ΚΛ ἴση. καὶ ἐπεὶ αἱ ὑπὸ ΑΒΓ, ΗΘΚ τῆς HK, at the point H on it. And let HL be made equal to 445 STOIQEIWN iaþ. ELEMENTS BOOK 11 ὑπὸ ΔΕΖ μείζονές εἰσιν, ἴση δὲ ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΚΘΛ, one of AB, BC, DE, EF , GH , and HK. And let KL ἡ ἄρα ὑπὸ ΗΘΛ τῆς ὑπὸ ΔΕΖ μείζων ἐστίν. καὶ ἐπεὶ δύο and GL have been joined. And since the two (straight- αἱ ΗΘ, ΘΛ δύο ταῖς ΔΕ, ΕΖ ἴσαι εἰσίν, καὶ γωνία ἡ ὑπὸ lines) AB and BC are equal to the two (straight-lines) ΗΘΛ γωνίας τῆς ὑπὸ ΔΕΖ μείζων, βάσις ἄρα ἡ ΗΛ βάσεως KH and HL (respectively), and the angle at B (is) equal τῆς ΔΖ μείζων ἐστίν. ἀλλὰ αἱ ΗΚ, ΚΛ τῆς ΗΛ μείζονές to KHL, the base AC is thus equal to the base KL εἰσιν. πολλῷ ἄρα αἱ ΗΚ, ΚΛ τῆς ΔΖ μείζονές εἰσιν. ἴση δὲ [Prop. 1.4]. And since (the sum of) ABC and GHK ἡ ΚΛ τῇ ΑΓ· αἱ ΑΓ, ΗΚ ἄρα τῆς λοιπῆς τῆς ΔΖ μείζονές is greater than DEF , and ABC equal to KHL, GHL εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ μὲν ΑΓ, ΔΖ τῆς ΗΚ is thus greater than DEF . And since the two (straight- μείζονές εἰσιν, καὶ ἔτι αἱ ΔΖ, ΗΚ τῆς ΑΓ μείζονές εἰσιν. lines) GH and HL are equal to the two (straight-lines) δυνατὸν ἄρα ἐστὶν ἐκ τῶν ἴσων ταῖς ΑΓ, ΔΖ, ΗΚ τρίγωνον DE and EF (respectively), and angle GHL (is) greater συστήσασθαι· ὅπερ ἔδει δεῖξαι. than DEF , the base GL is thus greater than the base DF [Prop. 1.24]. But, (the sum of) GK and KL is greater than GL [Prop. 1.20]. Thus, (the sum of) GK and KL is much greater than DF . And KL (is) equal to AC. Thus, (the sum of) AC and GK is greater than the remaining (straight-line) DF . So, similarly, we can show that (the sum of) AC and DF is greater than GK, and, further, that (the sum of) DF and GK is greater than AC. Thus, it is possible to construct a triangle from (straight-lines) equal to AC, DF , and GK. (Which is) the very thing it was required to show.kgþ. Proposition 23 ᾿Εκ τριῶν γωνιῶν ἐπιπέδων, ὧν αἱ δύο τῆς λοιπῆς To construct a solid angle from three (given) plane μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, στερεὰν γωνίαν angles, (the sum of) two of which is greater than the re- συστήσασθαι· δεῖ δὴ τὰς τρεῖς τεσσάρων ὀρθῶν ἐλάσς- maining (one, the angles) being taken up in any (possible ονας εἶναι. way). So, it is necessary for the (sum of the) three (an- gles) to be less than four right-angles [Prop. 11.21].B KA Z E H J G D G KA E D B C F H ῎Εστωσαν αἱ δοθεῖσαι τρεῖς γωνίαι ἐπίπεδοι αἱ ὑπὸ Let ABC, DEF , and GHK be the three given plane ΑΒΓ, ΔΕΖ, ΗΘΚ, ὧν αἱ δύο τῆς λοιπῆς μείζονες ἔστωσαν angles, of which let (the sum of) two be greater than the πάντῃ μεταλαμβανόμεναι, ἔτι δὲ αἱ τρεῖς τεσσάρων ὀρθῶν remaining (one, the angles) being taken up in any (pos- ἐλάσσονες· δεῖ δὴ ἐκ τῶν ἴσων ταῖς ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ sible way), and, further, (let) the (sum of the) three (be) στερεὰν γωνίαν συστήσασθαι. less than four right-angles. So, it is necessary to construct Ἀπειλήφθωσαν ἴσαι αἱ ΑΒ, ΒΓ, ΔΕ, ΕΖ, ΗΘ, ΘΚ, καὶ a solid angle from (plane angles) equal to ABC, DEF , ἐπεζεύχθωσαν αἱ ΑΓ, ΔΖ, ΗΚ· δυνατὸν ἄρα ἐστὶν ἐκ τῶν and GHK. ἴσων ταῖς ΑΓ, ΔΖ, ΗΚ τρίγωνον συστήσασθαι. συνεστάτω Let AB, BC, DE, EF , GH , and HK be cut off (so τὸ ΛΜΝ, ὥστε ἴσην εἶναι τὴν μὲν ΑΓ τῇ ΛΜ, τὴν δὲ ΔΖ as to be) equal (to one another). And let AC, DF , and τῇ ΜΝ, καὶ ἔτι τὴν ΗΚ τῇ ΝΛ, καὶ περιγεγράφθω περὶ GK have been joined. It is, thus, possible to construct a τὸ ΛΜΝ τρίγωνον κύκλος ὁ ΛΜΝ, καὶ εἰλήφθω αὐτοῦ τὸ triangle from (straight-lines) equal to AC, DF , and GK κέντρον καὶ ἔστω τὸ Ξ, καὶ ἐπεζεύχθωσαν αἱ ΛΞ, ΜΞ, ΝΞ· [Prop. 11.22]. Let (such a triangle), LMN , have be con- structed, such that AC is equal to LM , DF to MN , and, further, GK to NL. And let the circle LMN have been circumscribed about triangle LMN [Prop. 4.5]. And let 446 STOIQEIWN iaþ. ELEMENTS BOOK 11 its center have been found, and let it be (at) O. And let LO, MO, and NO have been joined. P MR X NL O R M N L Q P O Λέγω, ὅτι ἡ ΑΒ μείζων ἐστὶ τῆς ΛΞ. εἰ γὰρ μή, ἤτοι I say that AB is greater than LO. For, if not, AB is ἴση ἐστὶν ἡ ΑΒ τῇ ΛΞ ἢ ἐλάττων. ἔστω πρότερον ἴση. καὶ either equal to, or less than, LO. Let it, first of all, be ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΛΞ, ἀλλὰ ἡ μὲν ΑΒ τῇ ΒΓ ἐστιν equal. And since AB is equal to LO, but AB is equal to ἴση, ἡ δὲ ΞΛ τῇ ΞΜ, δύο δὴ αἱ ΑΒ, ΒΓ δύο ταῖς ΛΞ, ΞΜ BC, and OL to OM , so the two (straight-lines) AB and ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ βάσις ἡ ΑΓ βάσει τῇ ΛΜ BC are equal to the two (straight-lines) LO and OM , re- ὑπόκειται ἴση· γωνία ἄρα ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΛΞΜ spectively. And the base AC was assumed (to be) equal ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ μὲν ὑπὸ ΔΕΖ τῇ ὑπὸ ΜΞΝ to the base LM . Thus, angle ABC is equal to angle ἐστιν ἴση, καὶ ἔτι ἡ ὑπὸ ΗΘΚ τῇ ὑπὸ ΝΞΛ· αἱ ἄρα τρεῖς αἱ LOM [Prop. 1.8]. So, for the same (reasons), DEF is ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ γωνίαι τρισὶ ταῖς ὑπὸ ΛΞΜ, ΜΞΝ, also equal to MON , and, further, GHK to NOL. Thus, ΝΞΛ εἰσιν ἴσαι. ἀλλὰ αἱ τρεῖς αἱ ὑπὸ ΛΞΜ, ΜΞΝ, ΝΞΛ the three angles ABC, DEF , and GHK are equal to the τέτταρσιν ὀρθαῖς εἰσιν ἴσαι· καὶ αἱ τρεῖς ἄρα αἱ ὑπὸ ΑΒΓ, three angles LOM , MON , and NOL, respectively. But, ΔΕΖ, ΗΘΚ τέτταρσιν ὀρθαῖς ἴσαι εἰσίν. ὑπόκεινται δὲ καὶ the (sum of the) three angles LOM , MON , and NOL is τεσσάρων ὀρθῶν ἐλάσσονες· ὅπερ ἄτοπον. οὐκ ἄρα ἡ ΑΒ equal to four right-angles. Thus, the (sum of the) three τῇ ΛΞ ἴση ἐστίν. λέγω δή, ὅτι οὑδὲ ἐλάττων ἐστὶν ἡ ΑΒ τῆς angles ABC, DEF , and GHK is also equal to four right- ΛΞ. εἰ γὰρ δυνατόν, ἔστω· καὶ κείσθω τῇ μὲν ΑΒ ἴση ἡ ΞΟ, angles. And it was also assumed (to be) less than four τῇ δὲ ΒΓ ἴση ἡ ΞΠ, καὶ ἐπεζεύχθω ἡ ΟΠ. καὶ ἐπεὶ ἴση ἐστὶν right-angles. The very thing (is) absurd. Thus, AB is ἡ ΑΒ τῇ ΒΓ, ἴση ἐστὶ καὶ ἡ ΞΟ τῇ ΞΠ· ὥστε καὶ λοιπὴ ἡ not equal to LO. So, I say that AB is not less than LO ΛΟ τῇ ΠΜ ἐστιν ἴση. παράλληλος ἄρα ἐστὶν ἡ ΛΜ τῇ ΟΠ, either. For, if possible, let it be (less). And let OP be καὶ ἰσογώνιον τὸ ΛΜΞ τῷ ΟΠΞ· ἔστιν ἄρα ὡς ἡ ΞΛ πρὸς made equal to AB, and OQ equal to BC, and let PQ ΛΜ, οὕτως ἡ ΞΟ πρὸς ΟΠ· ἐναλλὰξ ὡς ἡ ΛΞ πρὸς ΞΟ, have been joined. And since AB is equal to BC, OP οὕτως ἡ ΛΜ πρὸς ΟΠ. μείζων δὲ ἡ ΛΞ τῆς ΞΟ· μείζων ἄρα is also equal to OQ. Hence, the remainder LP is also καὶ ἡ ΛΜ τῆς ΟΠ. ἀλλὰ ἡ ΛΜ κεῖται τῇ ΑΓ ἴση· καὶ ἡ ΑΓ equal to (the remainder) QM . LM is thus parallel to PQ ἄρα τῆς ΟΠ μείζων ἐστίν. ἐπεὶ οὖν δύο αἱ ΑΒ, ΒΓ δυσὶ ταῖς [Prop. 6.2], and (triangle) LMO (is) equiangular with ΟΞ, ΞΠ ἴσαι εἰσίν, καὶ βάσις ἡ ΑΓ βάσεως τῆς ΟΠ μείζων (triangle) PQO [Prop. 1.29]. Thus, as OL is to LM , so ἐστίν, γωνία ἄρα ἡ ὑπὸ ΑΒΓ γωνίας τῆς ὑπὸ ΟΞΠ μεῖζων OP (is) to PQ [Prop. 6.4]. Alternately, as LO (is) to OP , ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ μὲν ὑπὸ ΔΕΖ τῆς ὑπὸ so LM (is) to PQ [Prop. 5.16]. And LO (is) greater than ΜΞΝ μείζων ἐστίν, ἡ δὲ ὑπὸ ΗΘΚ τῆς ὑπὸ ΝΞΛ. αἱ ἄρα OP . Thus, LM (is) also greater than PQ [Prop. 5.14]. τρεῖς γωνίαι αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ τριῶν τῶν ὑπὸ ΛΞΜ, But LM was made equal to AC. Thus, AC is also greater ΜΞΝ, ΝΞΛ μείζονές εἰσιν. ἀλλὰ αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ than PQ. Therefore, since the two (straight-lines) AB τεσσάρων ὀρθῶν ἐλάσσονες ὑπόκεινται· πολλῷ ἄρα αἱ ὑπὸ and BC are equal to the two (straight-lines) PO and OQ ΛΞΜ, ΜΞΝ, ΝΞΛ τεσσάρων ὀρθῶν ἐλάσσονές εἰσιν. ἀλλὰ (respectively), and the base AC is greater than the base καὶ ἴσαι· ὅπερ ἐστὶν ἄτοπον. οὐκ ἄρα ἡ ΑΒ ἐλάσσων ἐστὶ PQ, the angle ABC is thus greater than the angle POQ τῆς ΛΞ. ἐδείχθη δέ, ὅτι οὐδὲ ἴση· μείζων ἄρα ἡ ΑΒ τῆς ΛΞ. [Prop. 1.25]. So, similarly, we can show that DEF is Ἀνεστάτω δὴ ἀπὸ τοῦ Ξ σημείου τῷ τοῦ ΛΜΝ κύκλου also greater than MON , and GHK than NOL. Thus, ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΞΡ, καὶ ᾧ μεῖζόν ἐστι τὸ ἀπὸ τῆς the (sum of the) three angles ABC, DEF , and GHK is ΑΒ τετράγωνον τοῦ ἀπὸ τῆς ΛΞ, ἐκείνῳ ἴσον ἔστω τὸ ἀπὸ greater than the (sum of the) three angles LOM , MON , 447 STOIQEIWN iaþ. ELEMENTS BOOK 11 τῆς ΞΡ, καὶ ἐπεζεύχθωσαν αἱ ΡΛ, ΡΜ, ΡΝ. and NOL. But, (the sum of) ABC, DEF , and GHK was Καὶ ἐπεὶ ἡ ΡΞ ὀρθὴ ἐστι πρὸς τὸ τοῦ ΛΜΝ κύκλου assumed (to be) less than four right-angles. Thus, (the ἐπίπεδον, καὶ πρὸς ἑκάστην ἄρα τῶν ΛΞ, ΜΞ, ΝΞ ὀρθή sum of) LOM , MON , and NOL is much less than four ἐστιν ἡ ΡΞ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΛΞ τῇ ΞΜ, κοινὴ δὲ καὶ right-angles. But, (it is) also equal (to four right-angles). πρὸς ὀρθὰς ἡ ΞΡ, βάσις ἄρα ἡ ΡΛ βάσει τῂ ΡΜ ἐστιν ἴση. The very thing is absurd. Thus, AB is not less than LO. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΡΝ ἑκατέρᾳ τῶν ΡΛ, ΡΜ ἐστιν ἴση· And it was shown (to be) not equal either. Thus, AB (is) αἱ τρεῖς ἄρα αἱ ΡΛ, ΡΜ, ΡΝ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ greater than LO. ᾧ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΛΞ, ἐκείνῳ ἴσον So let OR have been set up at point O at right- ὑπόκειται τὸ ἀπὸ τῆς ΞΡ, τὸ ἄρα ἀπὸ τῆς ΑΒ ἴσον ἐστὶ τοῖς angles to the plane of circle LMN [Prop. 11.12]. And ἀπὸ τῶν ΛΞ, ΞΡ. τοῖς δὲ ἀπὸ τῶν ΛΞ, ΞΡ ἴσον ἐστὶ τὸ ἀπὸ let the (square) on OR be equal to that (area) by which τῆς ΛΡ· ὀρθὴ γὰρ ἡ ὑπὸ ΛΞΡ· τὸ ἄρα ἀπὸ τῆς ΑΒ ἴσον the square on AB is greater than the (square) on LO ἐστὶ τῷ ἀπὸ τῆς ΡΛ· ἴση ἄρα ἡ ΑΒ τῇ ΡΛ. ἀλλὰ τῇ μὲν ΑΒ [Prop. 11.23 lem.]. And let RL, RM , and RN have been ἴση ἐστὶν ἑκάστη τῶν ΒΓ, ΔΕ, ΕΖ, ΗΘ, ΘΚ, τῇ δὲ ΡΛ ἴση joined. ἑκατέρα τῶν ΡΜ, ΡΝ· ἑκάστη ἄρα τῶν ΑΒ, ΒΓ, ΔΕ, ΕΖ, And since RO is at right-angles to the plane of cir- ΗΘ, ΘΚ ἑκάστῃ τῶν ΡΛ, ΡΜ, ΡΝ ἴση ἐστίν. καὶ ἐπεὶ δύο cle LMN , RO is thus also at right-angles to each of LO, αἱ ΛΡ, ΡΜ δυσὶ ταῖς ΑΒ, ΒΓ ἴσαι εἰσίν, καὶ βάσις ἡ ΛΜ MO, and NO. And since LO is equal to OM , and OR βάσει τῇ ΑΓ ὑπόκειται ἴση, γωνία ἄρα ἡ ὑπὸ ΛΡΜ γωνίᾳ is common and at right-angles, the base RL is thus equal τῇ ὑπὸ ΑΒΓ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ μὲν ὑπὸ ΜΡΝ to the base RM [Prop. 1.4]. So, for the same (reasons), τῇ ὑπὸ ΔΕΖ ἐστιν ἴση, ἡ δὲ ὑπὸ ΛΡΝ τῇ ὑπὸ ΗΘΚ. RN is also equal to each of RL and RM . Thus, the three ᾿Εκ τριῶν ἄρα γωνιῶν ἐπιπέδων τῶν ὑπὸ ΛΡΜ, ΜΡΝ, (straight-lines) RL, RM , and RN are equal to one an- ΛΡΝ, αἵ εἰσιν ἴσαι τρισὶ ταῖς δοθείσαις ταῖς ὑπὸ ΑΒΓ, ΔΕΖ, other. And since the (square) on OR was assumed to ΗΘΚ, στερεὰ γωνία συνέσταται ἡ πρὸς τῷ Ρ περιεχομένη be equal to that (area) by which the (square) on AB is ὑπὸ τῶν ΛΡΜ, ΜΡΝ, ΛΡΝ γωνιῶν· ὅπερ ἔδει ποιῆσαι. greater than the (square) on LO, the (square) on AB is thus equal to the (sum of the squares) on LO and OR. And the (square) on LR is equal to the (sum of the squares) on LO and OR. For LOR (is) a right-angle [Prop. 1.47]. Thus, the (square) on AB is equal to the (square) on RL. Thus, AB (is) equal to RL. But, each of BC, DE, EF , GH , and HK is equal to AB, and each of RM and RN equal to RL. Thus, each of AB, BC, DE, EF , GH , and HK is equal to each of RL, RM , and RN . And since the two (straight-lines) LR and RM are equal to the two (straight-lines) AB and BC (respec- tively), and the base LM was assumed (to be) equal to the base AC, the angle LRM is thus equal to the angle ABC [Prop. 1.8]. So, for the same (reasons), MRN is also equal to DEF , and LRN to GHK. Thus, the solid angle R, contained by the angles LRM , MRN , and LRN , has been constructed out of the three plane angles LRM , MRN , and LRN , which are equal to the three given (plane angles) ABC, DEF , and GHK (respectively). (Which is) the very thing it was required to do. 448 STOIQEIWN iaþ. ELEMENTS BOOK 11 A B G A B C L¨mma. Lemma ῝Ον δὲ τρόπον, ᾧ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ And we can demonstrate, thusly, in which manner to τῆς ΛΞ, ἐκείνῳ ἴσον λαβεῖν ἔστι τὸ ἀπὸ τῆς ΞΡ, δείξομεν take the (square) on OR equal to that (area) by which οὕτως. ἐκκείσθωσαν αἱ ΑΒ, ΛΞ εὐθεῖαι, καὶ ἔστω μείζων ἡ the (square) on AB is greater than the (square) on LO. ΑΒ, καὶ γεγράφθω ἐπ᾿ αὐτῆς ἡμικύκλιον τὸ ΑΒΓ, καὶ εἰς τὸ Let the straight-lines AB and LO be set out, and let AB ΑΒΓ ἡμικύκλιον ἐνηρμόσθω τῇ ΛΞ εὐθείᾳ μὴ μείζονι οὔσῃ be greater, and let the semicircle ABC have been drawn τῆς ΑΒ διαμέτρου ἴση ἡ ΑΓ, καὶ ἐπεζεύχθω ἡ ΓΒ. ἐπεὶ οὖν around it. And let AC, equal to the straight-line LO, ἐν ἡμικυκλίῳ τῷ ΑΓΒ γωνία ἐστὶν ἡ ὑπὸ ΑΓΒ, ὀρθὴ ἄρα which is not greater than the diameter AB, have been ἐστὶν ἡ ὑπὸ ΑΓΒ. τὸ ἄρα ἀπὸ τῆς ΑΒ ἴσον ἐστὶ τοῖς ἀπὸ inserted into the semicircle ABC [Prop. 4.1]. And let τῶν ΑΓ, ΓΒ. ὥστε τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΑΓ μεῖζόν CB have been joined. Therefore, since the angle ACB ἐστι τῷ ἀπὸ τῆς ΓΒ. ἴση δὲ ἡ ΑΓ τῇ ΛΞ. τὸ ἄρα ἀπὸ τῆς is in the semicircle ACB, ACB is thus a right-angle ΑΒ τοῦ ἀπὸ τῆς ΛΞ μεῖζόν ἐστι τῷ ἀπὸ τῆς ΓΒ. ἐὰν οὖν [Prop. 3.31]. Thus, the (square) on AB is equal to the τῇ ΒΓ ἴσην τὴν ΞΡ ἀπολάβωμεν, ἔσται τὸ ἀπὸ τῆς ΑΒ τοῦ (sum of the) squares on AC and CB [Prop. 1.47]. Hence, ἀπὸ τῆς ΛΞ μεῖζον τῷ ἀπὸ τῆς ΞΡ· ὅπερ προέκειτο ποιῆσαι. the (square) on AB is greater than the (square) on AC by the (square) on CB. And AC (is) equal to LO. Thus, the (square) on AB is greater than the (square) on LO by the (square) on CB. Therefore, if we take OR equal to BC then the (square) on AB will be greater than the (square) on LO by the (square) on OR. (Which is) the very thing it was prescribed to do.kdþ. Proposition 24 ᾿Εὰν στερεὸν ὑπὸ παραλλήλων ἐπιπέδων περιέχηται, τὰ If a solid (figure) is contained by (six) parallel planes ἀπεναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά then its opposite planes are both equal and parallelo- ἐστιν. grammic. G E H D A Z JB H E A B FC G D Στερεὸν γὰρ τὸ ΓΔΘΗ ὑπὸ παραλλήλων ἐπιπέδων πε- For let the solid (figure) CDHG have been contained ριεχέσθω τῶν ΑΓ, ΗΖ, ΑΘ, ΔΖ, ΒΖ, ΑΕ· λέγω, ὅτι τὰ ἀπε- by the parallel planes AC, GF , and AH , DF , and BF , ναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά ἐστιν. AE. I say that its opposite planes are both equal and ᾿Επεὶ γὰρ δύο ἐπίπεδα παράλληλα τὰ ΒΗ, ΓΕ ὑπὸ parallelogrammic. ἐπιπέδου τοῦ ΑΓ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί For since the two parallel planes BG and CE are εἰσιν. παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΔΓ. πάλιν, ἑπεὶ cut by the plane AC, their common sections are parallel δύο ἐπίπεδα παράλληλα τὰ ΒΖ, ΑΕ ὑπὸ ἐπιπέδου τοῦ [Prop. 11.16]. Thus, AB is parallel to DC. Again, since ΑΓ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν. the two parallel planes BF and AE are cut by the plane 449 STOIQEIWN iaþ. ELEMENTS BOOK 11 παράλληλος ἄρα ἐστὶν ἡ ΒΓ τῇ ΑΔ. ἐδείχθη δὲ καὶ ἡ ΑΒ AC, their common sections are parallel [Prop. 11.16]. τῇ ΔΓ παράλληλος· παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΓ. Thus, BC is parallel to AD. And AB was also shown (to ὁμοίως δὴ δείξομεν, ὅτι καὶ ἕκαστον τῶν ΔΖ, ΖΗ, ΗΒ, ΒΖ, be) parallel to DC. Thus, AC is a parallelogram. So, sim- ΑΕ παραλληλόγραμμόν ἐστιν. ilarly, we can also show that DF , FG, GB, BF , and AE ᾿Επεζεύχθωσαν αἱ ΑΘ, ΔΖ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ are each parallelograms. μὲν ΑΒ τῇ ΔΓ, ἡ δὲ ΒΘ τῇ ΓΖ, δύο δὴ αἱ ΑΒ, ΒΘ ἁπτόμεναι Let AH and DF have been joined. And since AB is ἀλλήλων παρὰ δύο εὐθείας τὰς ΔΓ, ΓΖ ἁπτομένας ἀλλήλων parallel to DC, and BH to CF , so the two (straight-lines) εἰσὶν οὐκ ἐν τῷ αὐτῷ ἐπιπέδῳ· ἴσας ἄρα γωνίας περιέξουσιν· joining one another, AB and BH , are parallel to the two ἴση ἄρα ἡ ὑπὸ ΑΒΘ γωνία τῇ ὑπὸ ΔΓΖ. καὶ ἐπεὶ δύο αἱ ΑΒ, straight-lines joining one another, DC and CF (respec- ΒΘ δυσὶ ταῖς ΔΓ, ΓΖ ἴσαι εἰσίν, καὶ γωνία ἡ ὑπὸ ΑΒΘ γωνίᾳ tively), not (being) in the same plane. Thus, they will τῇ ὑπὸ ΔΓΖ ἐστιν ἴση, βάσις ἄρα ἡ ΑΘ βάσει τῇ ΔΖ ἐστιν contain equal angles [Prop. 11.10]. Thus, angle ABH ἴση, καὶ τὸ ΑΒΘ τρίγωνον τῷ ΔΓΖ τριγώνῳ ἴσον ἐστίν. καί (is) equal to (angle) DCF . And since the two (straight- ἐστι τοῦ μὲν ΑΒΘ διπλάσιον τὸ ΒΗ παραλληλόγραμμον, lines) AB and BH are equal to the two (straight-lines) τοῦ δὲ ΔΓΖ διπλάσιον τὸ ΓΕ παραλληλόγραμμον· ἴσον DC and CF (respectively) [Prop. 1.34], and angle ABH ἄρα τὸ ΒΗ παραλληλόγραμμον τῷ ΓΕ παραλληλογράμμῳ· is equal to angle DCF , the base AH is thus equal to the ὁμοίως δὴ δείξομεν, ὅτι καὶ τὸ μὲν ΑΓ τῷ ΗΖ ἐστιν ἴσον, base DF , and triangle ABH is equal to triangle DCF τὸ δὲ ΑΕ τῷ ΒΖ. [Prop. 1.4]. And parallelogram BG is double (triangle) ᾿Εὰν ἄρα στερεὸν ὑπὸ παραλλήλων ἐπιπέδων περιέχηται, ABH , and parallelogram CE double (triangle) DCF τὰ ἀπεναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά [Prop. 1.34]. Thus, parallelogram BG (is) equal to paral- ἐστιν· ὅπερ ἔδει δεῖξαι. lelogram CE. So, similarly, we can show that AC is also equal to GF , and AE to BF . Thus, if a solid (figure) is contained by (six) parallel planes then its opposite planes are both equal and paral- lelogrammic. (Which is) the very thing it was required to show.keþ. Proposition 25 ᾿Εὰν στερεὸν παραλληλεπίπεδον ἐπιπέδῳ τμηθῇ πα- If a parallelipiped solid is cut by a plane which is par- ραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔσται ὡς ἡ βάσις allel to the opposite planes (of the parallelipiped) then as πρὸς τὴν βάσιν, οὕτως τὸ στερεὸν πρὸς τὸ στερεόν. the base (is) to the base, so the solid will be to the solid. Y R U D W T H I K A E N P FO Z G J S M Q L X B R U D TI K A E NML X O H YQ G V F C SWP B Στερεὸν γὰρ παραλληλεπίπεδον τὸ ΑΒΓΔ ἐπιπέδῳ τῷ For let the parallelipiped solid ABCD have been cut ΖΗ τετμήσθω παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις by the plane FG which is parallel to the opposite planes τοῖς ΡΑ, ΔΘ· λέγω, ὅτι ἐστὶν ὡς ἡ ΑΕΖΦ βάσις πρὸς τὴν RA and DH . I say that as the base AEFV (is) to the base ΕΘΓΖ βάσιν, οὕτως τὸ ΑΒΖΥ στερεὸν πρὸς τὸ ΕΗΓΔ EHCF , so the solid ABFU (is) to the solid EGCD. στερεόν. For let AH have been produced in each direction. And ᾿Εκβεβλήσθω γὰρ ἡ ΑΘ ἐφ᾿ ἑκάτερα τὰ μέρη, καὶ let any number whatsoever (of lengths), AK and KL, κείσθωσαν τῇ μὲν ΑΕ ἴσαι ὁσαιδηποτοῦν αἱ ΑΚ, ΚΛ, τῇ δὲ be made equal to AE, and any number whatsoever (of ΕΘ ἴσαι ὁσαιδηποτοῦν αἱ ΘΜ, ΜΝ, καὶ συμπεπληρώσθω lengths), HM and MN , equal to EH . And let the paral- τὰ ΛΟ, ΚΦ, ΘΧ, ΜΣ παραλληλόγραμμα καὶ τὰ ΛΠ, ΚΡ, lelograms LP , KV , HW , and MS have been completed, 450 STOIQEIWN iaþ. ELEMENTS BOOK 11 ΔΜ, ΜΤ στερεά. and the solids LQ, KR, DM , and MT . Καὶ ἐπεὶ ἴσαι εἰσὶν αἱ ΛΚ, ΚΑ, ΑΕ εὐθεῖαι ἀλλήλαις, ἴσα And since the straight-lines LK, KA, and AE are ἐστὶ καὶ τὰ μὲν ΛΟ, ΚΦ, ΑΖ παραλληλόγραμμα ἀλλήλοις, τὰ equal to one another, the parallelograms LP , KV , and δὲ ΚΞ, ΚΒ, ΑΗ ἀλλήλοις καὶ ἔτι τὰ ΛΨ, ΚΠ, ΑΡ ἀλλήλοις· AF are also equal to one another, and KO, KB, and AG ἀπεναντίον γάρ. διὰ τὰ αὐτὰ δὴ καὶ τὰ μὲν ΕΓ, ΘΧ, ΜΣ (are equal) to one another, and, further, LX , KQ, and παραλληλόγραμμα ἴσα εἰσὶν ἀλλήλοις, τὰ δὲ ΘΗ, ΘΙ, ΙΝ ἴσα AR (are equal) to one another. For (they are) opposite εἰσὶν ἀλλήλοις, καὶ ἔτι τὰ ΔΘ, ΜΩ, ΝΤ· τρία ἄρα ἐπίπεδα [Prop. 11.24]. So, for the same (reasons), the parallelo- τῶν ΛΠ, ΚΡ, ΑΥ στερεῶν τρισὶν ἐπιπέδοις ἐστὶν ἴσα. ἀλλὰ grams EC, HW , and MS are also equal to one another, τὰ τρία τρισὶ τοῖς ἀπεναντίον ἐστὶν ἴσα· τὰ ἄρα τρία στερεὰ and HG, HI, and IN are equal to one another, and, τὰ ΛΠ, ΚΡ, ΑΥ ἴσα ἀλλήλοις ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ τὰ further, DH , MY , and NT (are equal to one another). τρία στερεὰ τὰ ΕΔ, ΔΜ, ΜΤ ἴσα ἀλλήλοις ἐστίν· ὁσα- Thus, three planes of (one of) the solids LQ, KR, and πλασίων ἄρα ἐστὶν ἡ ΛΖ βάσις τῆς ΑΖ βάσεως, τοσαυ- AU are equal to the (corresponding) three planes (of the ταπλάσιόν ἐστι καὶ τὸ ΛΥ στερεὸν τοῦ ΑΥ στερεοῦ. διὰ others). But, the three planes (in one of the soilds) are τὰ αὐτὰ δὴ ὁσαπλασίων ἐστὶν ἡ ΝΖ βάσις τῆς ΖΘ βάσεως, equal to the three opposite planes [Prop. 11.24]. Thus, τοσαυταπλάσιόν ἐστι καὶ τὸ ΝΥ στερεὸν τοῦ ΘΥ στερεοῦ. the three solids LQ, KR, and AU are equal to one an- καὶ εἰ ἴση ἐστὶν ἡ ΛΖ βάσις τῇ ΝΖ βάσει, ἴσον ἐστὶ καὶ τὸ other [Def. 11.10]. So, for the same (reasons), the three ΛΥ στερεὸν τῷ ΝΥ στερεῷ, καὶ εἰ ὑπερέχει ἡ ΛΖ βάσις τῆς solids ED, DM , and MT are also equal to one another. ΝΖ βάσεως, ὑπερέχει καὶ τὸ ΛΥ στερεὸν τοῦ ΝΥ στερεοῦ, Thus, as many multiples as the base LF is of the base AF , καὶ εἰ ἐλλείπει, ἐλλείπει. τεσσάρων δὴ ὄντων μεγεθῶν, δύο so many multiples is the solid LU also of the the solid AU . μὲν βάσεων τῶν ΑΖ, ΖΘ, δύο δὲ στερεῶν τῶν ΑΥ, ΥΘ, So, for the same (reasons), as many multiples as the base εἴληπται ἰσάκις πολλαπλάσια τῆς μὲν ΑΖ βάσεως καὶ τοῦ NF is of the base FH , so many multiples is the solid NU ΑΥ στερεοῦ ἥ τε ΛΖ βάσις καὶ τὸ ΛΥ στερεόν, τῆς δὲ ΘΖ also of the solid HU . And if the base LF is equal to the βάσεως καὶ τοῦ ΘΥ στερεοῦ ἥ τε ΝΖ βάσις καὶ τὸ ΝΥ base NF then the solid LU is also equal to the solid NU .† στερεόν, καὶ δέδεικται, ὅτι εἱ ὑπερέχει ἡ ΛΖ βάσις τῆς ΖΝ And if the base LF exceeds the base NF then the solid βάσεως, ὑπερέχει καὶ τὸ ΛΥ στερεὸν τοῦ ΝΥ [στερεοῦ], καὶ LU also exceeds the solid NU . And if (LF ) is less than εἰ ἴση, ἴσον, καὶ εἰ ἐλλείπει, ἐλλείπει. ἔστιν ἄρα ὡς ἡ ΑΖ (NF ) then (LU) is (also) less than (NU). So, there are βάσις πρὸς τὴν ΖΘ βάσιν, οὕτως τὸ ΑΥ στερεὸν πρὸς τὸ four magnitudes, the two bases AF and FH , and the two ΥΘ στερεόν· ὅπερ ἔδει δεῖξαι. solids AU and UH , and equal multiples have been taken of the base AF and the solid AU— (namely), the base LF and the solid LU—and of the base HF and the solid HU—(namely), the base NF and the solid NU . And it has been shown that if the base LF exceeds the base FN then the solid LU also exceeds the [solid] NU , and if (LF is) equal (to FN) then (LU is) equal (to NU), and if (LF is) less than (FN) then (LU is) less than (NU). Thus, as the base AF is to the base FH , so the solid AU (is) to the solid UH [Def. 5.5]. (Which is) the very thing it was required to show. † Here, Euclid assumes that LF T NF implies LU T NU . This is easily demonstrated.k�þ. Proposition 26 Πρὸς τῇ δοθείσῃ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῇ To construct a solid angle equal to a given solid angle δοθείσῃ στερεᾷ γωνίᾳ ἴσην στερεὰν γωνίαν συστήσασθαι. on a given straight-line, and at a given point on it. ῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ πρὸς αὐτῇ Let AB be the given straight-line, and A the given δοθὲν σημεῖον τὸ Α, ἡ δὲ δοθεῖσα στερεὰ γωνία ἡ πρὸς point on it, and D the given solid angle, contained by the τῷ Δ περιεχομένη ὑπὸ τῶν ὑπὸ ΕΔΓ, ΕΔΖ, ΖΔΓ γωνιῶν plane angles EDC, EDF , and FDC. So, it is necessary ἐπιπέδων· δεῖ δὴ πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ to construct a solid angle equal to the solid angle D on σημείῳ τῷ Α τῇ πρὸς τῷ Δ στερεᾷ γωνίᾳ ἴσην στερεὰν the straight-line AB, and at the point A on it. γωνίαν συστήσασθαι. 451 STOIQEIWN iaþ. ELEMENTS BOOK 11 HD G BA K Z L E J B A KL E H F C GD Εἰλήφθω γὰρ ἐπὶ τῆς ΔΖ τυχὸν σημεῖον τὸ Ζ, καὶ ἤχθω For let some random point F have been taken on DF , ἀπὸ τοῦ Ζ ἐπὶ τὸ διὰ τῶν ΕΔ, ΔΓ ἐπίπεδον κάθετος ἡ ΖΗ, and let FG have been drawn from F perpendicular to καὶ συμβαλλέτω τῷ ἐπιπέδῳ κατὰ τὸ Η, καὶ ἐπεζεύχθω ἡ the plane through ED and DC [Prop. 11.11], and let it ΔΗ, καὶ συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ meet the plane at G, and let DG have been joined. And σημείῳ τῷ Α τῇ μὲν ὑπὸ ΕΔΓ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΛ, τῇ δὲ let BAL, equal to the angle EDC, and BAK, equal to ὑπὸ ΕΔΗ ἴση ἡ ὑπὸ ΒΑΚ, καὶ κείσθω τῇ ΔΗ ἴση ἡ ΑΚ, καὶ EDG, have been constructed on the straight-line AB at ἀνεστάτω ἀπὸ τοῦ Κ σημείου τῷ διὰ τῶν ΒΑΛ ἐπιπέδῳ πρὸς the point A on it [Prop. 1.23]. And let AK be made equal ὀρθὰς ἡ ΚΘ, καὶ κείσθω ἴση τῇ ΗΖ ἡ ΚΘ, καὶ ἐπεζεύχθω to DG. And let KH have been set up at the point K ἡ ΘΑ· λέγω, ὅτι ἡ πρὸς τῷ Α στερεὰ γωνία περιεχομένη at right-angles to the plane through BAL [Prop. 11.12]. ὑπὸ τῶν ΒΑΛ, ΒΑΘ, ΘΑΛ γωνιῶν ἴση ἐστὶ τῇ πρὸς τῷ Δ And let KH be made equal to GF . And let HA have στερεᾷ γωνίᾳ τῇ περιεχομένῃ ὑπὸ τῶν ΕΔΓ, ΕΔΖ, ΖΔΓ been joined. I say that the solid angle at A, contained by γωνιῶν. the (plane) angles BAL, BAH , and HAL, is equal to the Ἀπειλήφθωσαν γὰρ ἴσαι αἱ ΑΒ, ΔΕ, καὶ ἐπεζεύχθωσαν solid angle at D, contained by the (plane) angles EDC, αἱ ΘΒ, ΚΒ, ΖΕ, ΗΕ. καὶ ἐπεὶ ἡ ΖΗ ὀρθή ἐστι πρὸς τὸ EDF , and FDC. ὑποκείμενον ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας For let AB and DE have been cut off (so as to be) αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς equal, and let HB, KB, FE, and GE have been joined. ποιήσει γωνίας· ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΖΗΔ, And since FG is at right-angles to the reference plane ΖΗΕ γωνιῶν. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ὑπὸ ΘΚΑ, (EDC), it will also make right-angles with all of the ΘΚΒ γωνιῶν ὀρθή ἐστιν. καὶ ἐπεὶ δύο αἱ ΚΑ, ΑΒ δύο straight-lines joined to it which are also in the reference ταῖς ΗΔ, ΔΕ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ γωνίας ἴσας plane [Def. 11.3]. Thus, the angles FGD and FGE περιέχουσιν, βάσις ἄρα ἡ ΚΒ βάσει τῇ ΗΕ ἴση ἐστίν. ἔστι are right-angles. So, for the same (reasons), the an- δὲ καὶ ἡ ΚΘ τῇ ΗΖ ἴση· καὶ γωνίας ὀρθὰς περιέχουσιν· ἴση gles HKA and HKB are also right-angles. And since ἄρα καὶ ἡ ΘΒ τῇ ΖΕ. πάλιν ἐπεὶ δύο αἱ ΑΚ, ΚΘ δυσὶ ταῖς the two (straight-lines) KA and AB are equal to the two ΔΗ, ΗΖ ἴσαι εἰσίν, καὶ γωνίας ὀρθὰς περιέχουσιν, βάσις ἄρα (straight-lines) GD and DE, respectively, and they con- ἡ ΑΘ βάσει τῇ ΖΔ ἴση ἐστίν. ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΕ ἴση· tain equal angles, the base KB is thus equal to the base δύο δὴ αἱ ΘΑ, ΑΒ δύο ταῖς ΔΖ, ΔΕ ἴσαι εἰσίν. καὶ βάσις ἡ GE [Prop. 1.4]. And KH is also equal to GF . And they ΘΒ βάσει τῇ ΖΕ ἴση· γωνία ἄρα ἡ ὑπὸ ΒΑΘ γωνίᾳ τῇ ὑπὸ contain right-angles (with the respective bases). Thus, ΕΔΖ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΘΑΛ τῇ ὑπὸ ΖΔΓ HB (is) also equal to FE [Prop. 1.4]. Again, since the ἐστιν ἴση. ἔστι δὲ καὶ ἡ ὑπὸ ΒΑΛ τῇ ὑπὸ ΕΔΓ ἴση. two (straight-lines) AK and KH are equal to the two Πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ (straight-lines) DG and GF (respectively), and they con- σημείῳ τῷ Α τῇ δοθείσῃ στερεᾷ γωνίᾳ τῇ πρὸς τῷ Δ ἴση tain right-angles, the base AH is thus equal to the base συνέσταται· ὅπερ ἔδει ποιῆσαι. FD [Prop. 1.4]. And AB (is) also equal to DE. So, the two (straight-lines) HA and AB are equal to the two (straight-lines) DF and DE (respectively). And the base HB (is) equal to the base FE. Thus, the angle BAH is equal to the angle EDF [Prop. 1.8]. So, for the same (reasons), HAL is also equal to FDC. And BAL is also equal to EDC. Thus, (a solid angle) has been constructed, equal to the given solid angle at D, on the given straight-line AB, 452 STOIQEIWN iaþ. ELEMENTS BOOK 11 at the given point A on it. (Which is) the very thing it was required to do.kzþ. Proposition 27 Ἀπὸ τῆς δοθείσης εὐθείας τῷ δοθέντι στερεῷ παραλλη- To describe a parallelepiped solid similar, and simi- λεπιπέδῳ ὅμοιόν τε καὶ ὁμοίως κείμενον στερεὸν παραλλη- larly laid out, to a given parallelepiped solid on a given λεπίπεδον ἀναγράψαι. straight-line. ῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν στερεὸν Let the given straight-line be AB, and the given par- παραλληλεπίπεδον τὸ ΓΔ· δεῖ δὴ ἀπὸ τῆς δοθείσης εὐθείας allelepiped solid CD. So, it is necessary to describe a τῆς ΑΒ τῷ δοθέντι στερεῷ παραλληλεπιπέδῳ τῷ ΓΔ parallelepiped solid similar, and similarly laid out, to the ὅμοιόν τε καὶ ὁμοίως κείμενον στερεὸν παραλληλεπίπεδον given parallelepiped solid CD on the given straight-line ἀναγράψαι. AB. Συνεστάτω γὰρ πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ For, let a (solid angle) contained by the (plane angles) σημείῳ τῷ Α τῇ πρὸς τῷ Γ στερεᾷ γωνίᾳ ἴση ἡ περιεχομένη BAH , HAK, and KAB have been constructed, equal to ὑπὸ τῶν ΒΑΘ, ΘΑΚ, ΚΑΒ, ὥστε ἴσην εἶναι τὴν μὲν ὑπὸ solid angle at C, on the straight-line AB at the point A on ΒΑΘ γωνίαν τῇ ὑπὸ ΕΓΖ, τὴν δὲ ὑπὸ ΒΑΚ τῇ ὑπὸ ΕΓΗ, it [Prop. 11.26], such that angle BAH is equal to ECF , τὴν δὲ ὑπὸ ΚΑΘ τῇ ὑπὸ ΗΓΖ· καὶ γεγονέτω ὡς μὲν ἡ ΕΓ and BAK to ECG, and KAH to GCF . And let it have πρὸς τὴν ΓΗ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΚ, ὡς δὲ ἡ ΗΓ πρὸς been contrived that as EC (is) to CG, so BA (is) to AK, τὴν ΓΖ, οὕτως ἡ ΚΑ πρὸς τὴν ΑΘ. καὶ δι᾿ ἴσου ἄρα ἐστὶν and as GC (is) to CF , so KA (is) to AH [Prop. 6.12]. ὡς ἡ ΕΓ πρὸς τὴν ΓΖ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΘ. καὶ συμ- And thus, via equality, as EC is to CF , so BA (is) to AH πεπληρώσθω τὸ ΘΒ παραλληλόγραμμον καὶ τὸ ΑΛ στερεόν. [Prop. 5.22]. And let the parallelogram HB have been completed, and the solid AL. K EG A B D L H M Z J K E A B D L M F C H G Καὶ ἐπεί ἐστιν ὡς ἡ ΕΓ πρὸς τὴν ΓΗ, οὕτως ἡ ΒΑ πρὸς And since as EC is to CG, so BA (is) to AK, and τὴν ΑΚ, καὶ περὶ ἴσας γωνίας τὰς ὑπὸ ΕΓΗ, ΒΑΚ αἱ πλευραὶ the sides about the equal angles ECG and BAK are ἀνάλογόν εἰσιν, ὅμοιον ἄρα ἐστὶ τὸ ΗΕ παραλληλόγραμμον (thus) proportional, the parallelogram GE is thus simi- τῷ ΚΒ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ μὲν ΚΘ lar to the parallelogram KB. So, for the same (reasons), παραλληλόγραμμον τῷ ΗΖ παραλληλογράμμῳ ὅμοιόν ἐστι the parallelogram KH is also similar to the parallelogram καὶ ἔτι τὸ ΖΕ τῷ ΘΒ· τρία ἄρα παραλληλόγραμμα τοῦ ΓΔ GF , and, further, FE (is similar) to HB. Thus, three στερεοῦ τρισὶ παραλληλογράμμοις τοῦ ΑΛ στερεοῦ ὅμοιά of the parallelograms of solid CD are similar to three of ἐστιν. ἀλλὰ τὰ μὲν τρία τρισὶ τοῖς ἀπεναντίον ἴσα τέ ἐστι the parallelograms of solid AL. But, the (former) three καὶ ὅμοια, τὰ δὲ τρία τρισὶ τοῖς ἀπεναντίον ἴσα τέ ἐστι καὶ are equal and similar to the three opposite, and the (lat- ὅμοια· ὅλον ἄρα τὸ ΓΔ στερεὸν ὅλῳ τῷ ΑΛ στερεῷ ὅμοιόν ter) three are equal and similar to the three opposite. ἐστιν. Thus, the whole solid CD is similar to the whole solid Ἀπὸ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τῷ δοθέντι AL [Def. 11.9]. στερεῷ παραλληλεπιπέδῳ τῷ ΓΔ ὅμοιόν τε καὶ ὁμοίως Thus, AL, similar, and similarly laid out, to the given κείμενον ἀναγέγραπται τὸ ΑΛ· ὅπερ ἔδει ποιῆσαι. parallelepiped solid CD, has been described on the given straight-lines AB. (Which is) the very thing it was re- quired to do.khþ. Proposition 28 ᾿Εὰν στερεὸν παραλληλεπίπεδον ἐπιπέδῳ τμηθῇ κατὰ If a parallelepiped solid is cut by a plane (passing) 453 STOIQEIWN iaþ. ELEMENTS BOOK 11 τὰς διαγωνίους τῶν ἀπεναντίον ἐπιπέδων, δίχα τμηθήσεται through the diagonals of (a pair of) opposite planes then τὸ στερεὸν ὑπὸ τοῦ ἐπιπέδου. the solid will be cut in half by the plane.ZB EG D A J H B E D A H F G C Στερεὸν γὰρ παραλληλεπίπεδον τὸ ΑΒ ἐπιπέδῳ τῷ For let the parallelepiped solid AB have been cut by ΓΔΕΖ τετμήσθω κατὰ τὰς διαγωνίους τῶν ἀπεναντίον the plane CDEF (passing) through the diagonals of the ἐπιπέδων τὰς ΓΖ, ΔΕ· λέγω, ὅτι δίχα τμηθήσεται τὸ ΑΒ opposite planes CF and DE.† I say that the solid AB will στερεὸν ὑπὸ τοῦ ΓΔΕΖ ἐπιπέδου. be cut in half by the plane CDEF . ᾿Επεὶ γὰρ ἴσον ἐστὶ τὸ μὲν ΓΗΖ τρίγωνον τῷ ΓΖΒ For since triangle CGF is equal to triangle CFB, and τριγώνῳ, τὸ δὲ ΑΔΕ τῷ ΔΕΘ, ἔστι δὲ καὶ τὸ μὲν ΓΑ πα- ADE (is equal) to DEH [Prop. 1.34], and parallelo- ραλληλόγραμμον τῷ ΕΒ ἴσον· ἀπεναντίον γάρ· τὸ δὲ ΗΕ gram CA is also equal to EB—for (they are) opposite τῷ ΓΘ, καὶ τὸ πρίσμα ἄρα τὸ περιεχόμενον ὑπὸ δύο μὲν [Prop. 11.24]—and GE (equal) to CH , thus the prism τριγώνων τῶν ΓΗΖ, ΑΔΕ, τριῶν δὲ παραλληλογράμμων contained by the two triangles CGF and ADE, and the τῶν ΗΕ, ΑΓ, ΓΕ ἴσον ἐστὶ τῷ πρίσματι τῷ περιεχομένῳ three parallelograms GE, AC, and CE, is also equal to ὑπὸ δύο μὲν τριγώνων τῶν ΓΖΒ, ΔΕΘ, τριῶν δὲ παραλ- the prism contained by the two triangles CFB and DEH , ληλογράμμων τῶν ΓΘ, ΒΕ, ΓΕ· ὑπὸ γὰρ ἴσων ἐπιπέδων and the three parallelograms CH , BE, and CE. For they περιέχονται τῷ τε πλήθει καὶ τῷ μεγέθει. ὥστε ὅλον τὸ are contained by planes (which are) equal in number and ΑΒ στερεὸν δίχα τέτμηται ὑπὸ τοῦ ΓΔΕΖ ἐπιπέδου· ὅπερ in magnitude [Def. 11.10].‡ Thus, the whole of solid AB ἔδει δεῖξαι. is cut in half by the plane CDEF . (Which is) the very thing it was required to show. † Here, it is assumed that the two diagonals lie in the same plane. The proof is easily supplied. ‡ However, strictly speaking, the prisms are not similarly arranged, being mirror images of one another.kjþ. Proposition 29 Τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλληλεπίπεδα Parallelepiped solids which are on the same base, and καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι ἐπὶ τῶν αὐτῶν εἰσιν (have) the same height, and in which the (ends of the εὐθειῶν, ἴσα ἀλλήλοις ἐστίν. straight-lines) standing up are on the same straight-lines, are equal to one another. Z E H G A L M N J ED B KE A L M N D B G F C H ῎Εστω ἐπὶ τῆς αὐτῆς βάσεως τῆς ΑΒ στερεὰ παραλλη- For let the parallelepiped solids CM and CN be on 454 STOIQEIWN iaþ. ELEMENTS BOOK 11 λεπίπεδα τὰ ΓΜ, ΓΝ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι the same base AB, and (have) the same height, and let αἱ ΑΗ, ΑΖ, ΛΜ, ΛΝ, ΓΔ, ΓΕ, ΒΘ, ΒΚ ἐπὶ τῶν αὐτῶν the (ends of the straight-lines) standing up in them, AG, εὐθειῶν ἔστωσαν τῶν ΖΝ, ΔΚ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΓΜ AF , LM , LN , CD, CE, BH , and BK, be on the same στερεὸν τῷ ΓΝ στερεῷ. straight-lines, FN and DK. I say that solid CM is equal ᾿Επεὶ γὰρ παραλληλόγραμμόν ἐστιν ἑκάτερον τῶν ΓΘ, to solid CN . ΓΚ, ἴση ἐστὶν ἡ ΓΒ ἑκατέρᾳ τῶν ΔΘ, ΕΚ· ὥστε καὶ ἡ For since CH and CK are each parallelograms, CB ΔΘ τῇ ΕΚ ἐστιν ἴση. κοινὴ ἀφῃρήσθω ἡ ΕΘ· λοιπὴ ἄρα is equal to each of DH and EK [Prop. 1.34]. Hence, ἡ ΔΕ λοιπῇ τῇ ΘΚ ἐστιν ἴση. ὥστε καὶ τὸ μὲν ΔΓΕ DH is also equal to EK. Let EH have been subtracted τρίγωνον τῷ ΘΒΚ τριγώνῳ ἴσον ἐστίν, τὸ δὲ ΔΗ παραλ- from both. Thus, the remainder DE is equal to the re- ληλόγραμμον τῷ ΘΝ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ καὶ mainder HK. Hence, triangle DCE is also equal to tri- τὸ ΑΖΗ τρίγωνον τῷ ΜΛΝ τριγώνῳ ἴσον ἐστίν. ἔστι δὲ καὶ angle HBK [Props. 1.4, 1.8], and parallelogram DG to τὸ μὲν ΓΖ παραλληλόγραμμον τῷ ΒΜ παραλληλογράμμῳ parallelogram HN [Prop. 1.36]. So, for the same (rea- ἴσον, τὸ δὲ ΓΗ τῷ ΒΝ· ἀπεναντίον γάρ· καὶ τὸ πρίσμα ἄρα sons), traingle AFG is also equal to triangle MLN . And τὸ περιεχόμενον ὑπὸ δύο μὲν τριγώνων τῶν ΑΖΗ, ΔΓΕ, parallelogram CF is also equal to parallelogram BM , τριῶν δὲ παραλληλογράμμων τῶν ΑΔ, ΔΗ, ΓΗ ἴσον ἐστὶ and CG to BN [Prop. 11.24]. For they are opposite. τῷ πρίσματι τῷ περιεχομένῳ ὑπὸ δύο μὲν τριγώνων τῶν Thus, the prism contained by the two triangles AFG and ΜΛΝ, ΘΒΚ, τριῶν δὲ παραλληλογράμμων τῶν ΒΜ, ΘΝ, DCE, and the three parallelograms AD, DG, and CG, is ΒΝ. κοινὸν προσκείσθω τὸ στερεὸν, οὗ βάσις μὲν τὸ ΑΒ equal to the prism contained by the two triangles MLN παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΗΕΘΜ· ὅλον ἄρα τὸ and HBK, and the three parallelograms BM , HN , and ΓΜ στερεὸν παραλληλεπίπεδον ὅλῳ τῷ ΓΝ στερεῷ παραλ- BN . Let the solid whose base (is) parallelogram AB, and ληλεπιπέδῳ ἴσον ἐστίν. (whose) opposite (face is) GEHM , have been added to Τὰ ἄρα ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλλη- both (prisms). Thus, the whole parallelepiped solid CM λεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι ἐπὶ τῶν is equal to the whole parallelepiped solid CN . αὐτῶν εἰσιν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι. Thus, parallelepiped solids which are on the same base, and (have) the same height, and in which the (ends of the straight-lines) standing up (are) on the same straight-lines, are equal to one another. (Which is) the very thing it was required to show.lþ. Proposition 30 Τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλληλεπίπεδα Parallelepiped solids which are on the same base, and καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν (have) the same height, and in which the (ends of the αὐτῶν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν. straight-lines) standing up are not on the same straight- lines, are equal to one another. L A G Z O K R B N MX EH J D P O A K R B N M EG P F C H D L Q ῎Εστω ἐπὶ τῆς αὐτῆς βάσεως τῆς ΑΒ στερεὰ παραλλη- Let the parallelepiped solids CM and CN be on the λεπίπεδα τὰ ΓΜ, ΓΝ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ same base, AB, and (have) the same height, and let the ΑΖ, ΑΗ, ΛΜ, ΛΝ, ΓΔ, ΓΕ, ΒΘ, ΒΚ μὴ ἔστωσαν ἐπὶ τῶν (ends of the straight-lines) standing up in them, AF , AG, 455 STOIQEIWN iaþ. ELEMENTS BOOK 11 αὐτῶν εὐθειῶν· λέγω, ὅτι ἴσον ἐστὶ τὸ ΓΜ στερεὸν τῷ ΓΝ LM , LN , CD, CE, BH , and BK, not be on the same στερεῷ. straight-lines. I say that the solid CM is equal to the ᾿Εκβεβλήσθωσαν γὰρ αἱ ΝΚ, ΔΘ καὶ συμπιπτέτωσαν solid CN . ἀλλήλαις κατὰ τὸ Ρ, καὶ ἔτι ἐκβεβλήσθωσαν αἱ ΖΜ, ΗΕ For let NK and DH have been produced, and let ἐπὶ τὰ Ο, Π, καὶ ἐπεζεύχθωσαν αἱ ΑΞ, ΛΟ, ΓΠ, ΒΡ. ἴσον them have joined one another at R. And, further, let FM δή ἐστι τὸ ΓΜ στερεόν, οὗ βάσις μὲν τὸ ΑΓΒΛ παραλ- and GE have been produced to P and Q (respectively). ληλόγραμμον, ἀπεναντίον δὲ τὸ ΖΔΘΜ, τῷ ΓΟ στερεῷ, οὗ And let AO, LP , CQ, and BR have been joined. So, solid βάσις μὲν τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ CM , whose base (is) parallelogram ACBL, and oppo- ΞΠΡΟ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΑΓΒΛ καὶ site (face) FDHM , is equal to solid CP , whose base (is) ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΖ, ΑΞ, ΛΜ, ΛΟ, parallelogram ACBL, and opposite (face) OQRP . For ΓΔ, ΓΠ, ΒΘ, ΒΡ ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν τῶν ΖΟ, ΔΡ. they are on the same base, ACBL, and (have) the same ἀλλὰ τὸ ΓΟ στερεόν, οὗ βάσις μέν ἐστι τὸ ΑΓΒΛ παραλ- height, and the (ends of the straight-lines) standing up in ληλόγραμμον, ἀπεναντίον δὲ τὸ ΞΠΡΟ, ἴσον ἐστὶ τῷ ΓΝ them, AF , AO, LM , LP , CD, CQ, BH , and BR, are on στερεῷ, οὗ βάσις μὲν τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπε- the same straight-lines, FP and DR [Prop. 11.29]. But, ναντίον δὲ τὸ ΗΕΚΝ· ἐπί τε γὰρ πάλιν τῆς αὐτῆς βάσεώς solid CP , whose base is parallelogram ACBL, and oppo- εἰσι τῆς ΑΓΒΛ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ site (face) OQRP , is equal to solid CN , whose base (is) ΑΗ, ΑΞ, ΓΕ, ΓΠ, ΛΝ, ΛΟ, ΒΚ, ΒΡ ἐπὶ τῶν αὐτῶν εἰσιν parallelogram ACBL, and opposite (face) GEKN . For, εὐθειῶν τῶν ΗΠ, ΝΡ. ὥστε καὶ τὸ ΓΜ στερεὸν ἴσον ἐστὶ again, they are on the same base, ACBL, and (have) τῷ ΓΝ στερεῷ. the same height, and the (ends of the straight-lines) Τὰ ἄρα ἐπὶ τῆς αὐτῆς βάσεως στερεὰ παραλληλεπίπεδα standing up in them, AG, AO, CE, CQ, LN , LP , BK, καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν and BR, are on the same straight-lines, GQ and NR αὐτῶν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι. [Prop. 11.29]. Hence, solid CM is also equal to solid CN . Thus, parallelepiped solids (which are) on the same base, and (have) the same height, and in which the (ends of the straight-lines) standing up are not on the same straight-lines, are equal to one another. (Which is) the very thing it was required to show.laþ. Proposition 31 Τὰ ἐπὶ ἴσων βάσεων ὄντα στερεὰ παραλληλεπίπεδα καὶ Parallelepiped solids which are on equal bases, and ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν. (have) the same height, are equal to one another. ῎Εστω ἐπὶ ἴσων βάσεων τῶν ΑΒ, ΓΔ στερεὰ παραλλη- Let the parallelepiped solids AE and CF be on the λεπίπεδα τὰ ΑΕ, ΓΖ ὑπὸ τὸ αὐτὸ ὕψος. λέγω, ὅτι ἴσον ἐστὶ equal bases AB and CD (respectively), and (have) the τὸ ΑΕ στερεὸν τῷ ΓΖ στερεῷ. same height. I say that solid AE is equal to solid CF . ῎Εστωσαν δὴ πρότερον αἱ ἐφεστηκυῖαι αἱ ΘΚ, ΒΕ, ΑΗ, So, let the (straight-lines) standing up, HK, BE, AG, ΛΜ, ΟΠ, ΔΖ, ΓΞ, ΡΣ πρὸς ὀρθὰς ταῖς ΑΒ, ΓΔ βάσεσιν, LM , PQ, DF , CO, and RS, first of all, be at right-angles καὶ ἐκβεβλήσθω ἐπ᾿ εὐθείας τῇ ΓΡ εὐθεῖα ἡ ΡΤ, καὶ συ- to the bases AB and CD. And let RT have been produced νεστάτω πρὸς τῇ ΡΤ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ in a straight-line with CR. And let (angle) TRU , equal Ρ τῇ ὑπὸ ΑΛΒ γωνίᾳ ἴση ἡ ὑπὸ ΤΡΥ, καὶ κείσθω τῇ μὲν to angle ALB, have been constructed on the straight-line ΑΛ ἴση ἡ ΡΤ, τῇ δὲ ΛΒ ἴση ἡ ΡΥ, καὶ συμπεπληρώσθω ἥ RT , at the point R on it [Prop. 1.23]. And let RT be τε ΡΧ βάσις καὶ τὸ ΨΥ στερεόν. made equal to AL, and RU to LB. And let the base RW , and the solid XU , have been completed. 456 STOIQEIWN iaþ. ELEMENTS BOOK 11 � H Z I G W Q R T �K E H LA BJ Uo �~ a M O D F X S Y M I S R T K E LA B F C O D a Y U b W H G Q cde X V P Καὶ ἐπεὶ δύο αἱ ΤΡ, ΡΥ δυσὶ ταῖς ΑΛ, ΛΒ ἴσαι εἰσίν, And since the two (straight-lines) TR and RU are καὶ γωνίας ἴσας περιέχουσιν, ἴσον ἄρα καὶ ὅμοιον τὸ ΡΧ equal to the two (straight-lines) AL and LB (respec- παραλληλόγραμμον τῷ ΘΛ παραλληλογράμμῳ. καὶ ἐπεὶ tively), and they contain equal angles, parallelogram πάλιν ἴση μὲν ἡ ΑΛ τῇ ΡΤ, ἡ δὲ ΛΜ τῇ ΡΣ, καὶ γωνίας RW is thus equal and similar to parallelogram HL ὀρθὰς περιέχουσιν, ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΡΨ παραλ- [Prop. 6.14]. And, again, since AL is equal to RT , ληλόγραμμον τῷ ΑΜ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ and LM to RS, and they contain right-angles, paral- καὶ τὸ ΛΕ τῷ ΣΥ ἴσον τέ ἐστι καὶ ὅμοιον· τρία ἄρα πα- lelogram RX is thus equal and similar to parallelogram ραλληλόγραμμα τοῦ ΑΕ στερεοῦ τρισὶ παραλληλογράμμοις AM [Prop. 6.14]. So, for the same (reasons), LE is also τοῦ ΨΥ στερεοῦ ἴσα τέ ἐστι καὶ ὅμοια. ἀλλὰ τὰ μὲν τρία equal and similar to SU . Thus, three parallelograms of τρισὶ τοῖς ἀπεναντίον ἴσα τέ ἐστι καὶ ὅμοια, τὰ δὲ τρία solid AE are equal and similar to three parallelograms τρισὶ τοῖς ἀπεναντίον· ὅλον ἄρα τὸ ΑΕ στερεὸν παραλλη- of solid XU . But, the three (faces of the former solid) λεπίπεδον ὅλῳ τῷ ΨΥ στερεῷ παραλληλεπιπέδῳ ἴσον ἐστίν. are equal and similar to the three opposite (faces), and διήχθωσαν αἱ ΔΡ, ΧΥ καὶ συμπιπτέτωσαν ἀλλήλαις κατὰ the three (faces of the latter solid) to the three opposite τὸ Ω, καὶ διὰ τοῦ Τ τῇ ΔΩ παράλληλος ἤχθω ἡ αΤϡ, καὶ (faces) [Prop. 11.24]. Thus, the whole parallelepiped ἐκβεβλήσθω ἡ ΟΔ κατὰ τὸ α, καὶ συμπεπληρώσθω τὰ ΩΨ, solid AE is equal to the whole parallelepiped solid XU ΡΙ στερεά. ἴσον δή ἐστι τὸ ΨΩ στερεόν, οὗ βάσις μέν [Def. 11.10]. Let DR and WU have been drawn across, ἐστι τὸ ΡΨ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ Ωϟ, τῷ and let them have met one another at Y . And let aT b ΨΥ στερεῷ, οὗ βάσις μὲν τὸ ΡΨ παραλληλόγραμμον, ἀπε- have been drawn through T parallel to DY . And let PD ναντίον δὲ τὸ ΥΦ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς have been produced to a. And let the solids Y X and ΡΨ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΡΩ, ΡΥ, RI have been completed. So, solid XY , whose base is Τϡ, ΤΧ, Σϛ, Σο̃, Ψϟ, ΨΦ ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν τῶν parallelogram RX , and opposite (face) Y c, is equal to ΩΧ, ϛΦ. ἀλλὰ τὸ ΨΥ στερεὸν τῷ ΑΕ ἐστιν ἴσον· καὶ τὸ ΨΩ solid XU , whose base (is) parallelogram RX , and oppo- ἄρα στερεὸν τῷ ΑΕ στερεῷ ἐστιν ἴσον. καὶ ἐπεὶ ἴσον ἐστὶ site (face) UV . For they are on the same base RX , and τὸ ΡΥΧΤ παραλληλόγραμμον τῷ ΩΤ παραλληλογράμμῳ· (have) the same height, and the (ends of the straight- ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΡΤ καὶ ἐν ταῖς αὐταῖς lines) standing up in them, RY , RU , Tb, TW , Se, Sd, παραλλήλοις ταῖς ΡΤ, ΩΧ· ἀλλὰ τὸ ΡΥΧΤ τῷ ΓΔ ἐστιν Xc and XV , are on the same straight-lines, Y W and ἴσον, ἐπεὶ καὶ τῷ ΑΒ, καὶ τὸ ΩΤ ἄρα παραλληλόγραμμον eV [Prop. 11.29]. But, solid XU is equal to AE. Thus, 457 STOIQEIWN iaþ. ELEMENTS BOOK 11 τῷ ΓΔ ἐστιν ἴσον. ἄλλο δὲ τὸ ΔΤ· ἔστιν ἄρα ὡς ἡ ΓΔ solid XY is also equal to solid AE. And since parallel- βάσις πρὸς τὴν ΔΤ, οὕτως ἡ ΩΤ πρὸς τὴν ΔΤ. καὶ ἐπεὶ ogram RUWT is equal to parallelogram Y T . For they στερεὸν παραλληλεπίπεδον τὸ ΓΙ ἐπιπέδῳ τῷ ΡΖ τέτμηται are on the same base RT , and between the same par- παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔστιν ὡς ἡ ΓΔ allels RT and Y W [Prop. 1.35]. But, RUWT is equal βάσις πρὸς τὴν ΔΤ βάσιν, οὕτως τὸ ΓΖ στερεὸν πρὸς τὸ ΡΙ to CD, since (it is) also (equal) to AB. Parallelogram στερεόν. διὰ τὰ αὐτὰ δή, ἐπεὶ στερεὸν παραλληλεπίπεδον Y T is thus also equal to CD. And DT is another (par- τὸ ΩΙ ἐπιπέδῳ τῷ ΡΨ τέτμηται παραλλήλῳ ὄντι τοῖς ἀπε- allelogram). Thus, as base CD is to DT , so Y T (is) to ναντίον ἐπιπέδοις, ἔστιν ὡς ἡ ΩΤ βάσις πρὸς τὴν ΤΛ βάσιν, DT [Prop. 5.7]. And since the parallelepiped solid CI οὕτως τὸ ΩΨ στερεὸν πρὸς τὸ ΡΙ. ἀλλ᾿ ὡς ἡ ΓΔ βάσις has been cut by the plane RF , which is parallel to the πρὸς τὴν ΔΤ, οὕτως ἡ ΩΤ πρὸς τὴν ΔΤ· καὶ ὡς ἄρα τὸ opposite planes (of CI), as base CD is to base DT , so ΓΖ στερεὸν πρὸς τὸ ΡΙ στερεόν, οὕτως τὸ ΩΨ στερεὸν solid CF (is) to solid RI [Prop. 11.25]. So, for the same πρὸς τὸ ΡΙ. ἑκάτερον ἄρα τῶν ΓΖ, ΩΨ στερεῶν πρὸς τὸ (reasons), since the parallelepiped solid Y I has been cut ΡΙ τὸν αὐτὸν ἔχει λόγον· ἴσον ἄρα ἐστὶ τὸ ΓΖ στερεὸν τῷ by the plane RX , which is parallel to the opposite planes ΩΨ στερεῷ. ἀλλὰ τὸ ΩΨ τῷ ΑΕ ἐδείχθη ἴσον· καὶ τὸ ΑΕ (of Y I), as base Y T is to base TD, so solid Y X (is) to ἄρα τῷ ΓΖ ἐστιν ἴσον. solid RI [Prop. 11.25]. But, as base CD (is) to DT , so Y T (is) to DT . And, thus, as solid CF (is) to solid RI, so solid Y X (is) to solid RI. Thus, solids CF and Y X each have the same ratio to RI [Prop. 5.11]. Thus, solid CF is equal to solid Y X [Prop. 5.9]. But, Y X was show (to be) equal to AE. Thus, AE is also equal to CF . N KE L B W IR U Q DA H J P X G Z T M O Y S F X K E L B I R U D A O F C H G W Q T Y V N M P S Μὴ ἔστωσαν δὴ αἱ ἐφεστηκυῖαι αἱ ΑΗ, ΘΚ, ΒΕ, ΛΜ, And so let the (straight-lines) standing up, AG, HK, ΓΞ, ΟΠ, ΔΖ, ΡΣ πρὸς ὀρθὰς ταῖς ΑΒ, ΓΔ βάσεσιν· λέγω BE, LM , CO, PQ, DF , and RS, not be at right-angles πάλιν, ὅτι ἵσον τὸ ΑΕ στερεὸν τῷ ΓΖ στερεῷ. ἤχθωσαν γὰρ to the bases AB and CD. Again, I say that solid AE ἀπὸ τῶν Κ, Ε, Η, Μ, Π, Ζ, Ξ, Σ σημείων ἐπὶ τὸ ὑποκείμενον (is) equal to solid CF . For let KN , ET , GU , MV , QW , ἐπίπεδον κάθετοι αἱ ΚΝ, ΕΤ, ΗΥ, ΜΦ, ΠΧ, ΖΨ, ΞΩ, ΣΙ, FX , OY , and SI have been drawn from points K, E, καὶ συμβαλλέτωσαν τῷ ἐπιπέδῳ κατὰ τὰ Ν, Τ, Υ, Φ, Χ, Ψ, G, M , Q, F , O, and S (respectively) perpendicular to Ω, Ι σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΝΤ, ΝΥ, ΥΦ, ΤΦ, ΧΨ, the reference plane (i.e., the plane of the bases AB and ΧΩ, ΩΙ, ΙΨ. ἴσον δή ἐστι τὸ ΚΦ στερεὸν τῷ ΠΙ στερεῷ· CD), and let them have met the plane at points N , T , ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΚΜ, ΠΣ καὶ ὑπὸ τὸ αὐτὸ U , V , W , X , Y , and I (respectively). And let NT , NU , ὕψος, ὧν αἱ ἐφεστῶσαι πρὸς ὀρθάς εἰσι ταῖς βάσεσιν. ἀλλὰ UV , TV , WX , WY , Y I, and IX have been joined. So τὸ μὲν ΚΦ στερεὸν τῷ ΑΕ στερεῷ ἐστιν ἴσον, τὸ δὲ ΠΙ τῷ solid KV is equal to solid QI. For they are on the equal ΓΖ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι καὶ ὑπὸ τὸ αὐτὸ ὕψος, bases KM and QS, and (have) the same height, and the ὧν αἱ ἐφεστῶσαι οὔκ εἰσιν ἐπὶ τῶν αὐτῶν εὐθειῶν. καὶ τὸ (straight-lines) standing up in them are at right-angles ΑΕ ἄρα στερεὸν τῷ ΓΖ στερεῷ ἐστιν ἴσον. to their bases (see first part of proposition). But, solid Τὰ ἄρα ἐπὶ ἴσων βάσεων ὄντα στερεὰ παραλληλεπίπεδα KV is equal to solid AE, and QI to CF . For they are καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι. on the same base, and (have) the same height, and the (straight-lines) standing up in them are not on the same straight-lines [Prop. 11.30]. Thus, solid AE is also equal to solid CF . Thus, parallelepiped solids which are on equal bases, 458 STOIQEIWN iaþ. ELEMENTS BOOK 11 and (have) the same height, are equal to one another. (Which is) the very thing it was required to show.lbþ. Proposition 32 Τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα στερεὰ παραλληλεπίπεδα Parallelepiped solids which (have) the same height πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις. are to one another as their bases. L A E B G H J D Z K L B A E D K F C HG ῎Εστω ὑπὸ τὸ αὐτὸ ὕψος στερεὰ παραλληλεπίπεδα τὰ Let AB and CD be parallelepiped solids (having) the ΑΒ, ΓΔ· λέγω, ὅτι τὰ ΑΒ, ΓΔ στερεὰ παραλληλεπίπεδα same height. I say that the parallelepiped solids AB and πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις, τουτέστιν ὅτι ἐστὶν ὡς ἡ CD are to one another as their bases. That is to say, as ΑΕ βάσις πρὸς τὴν ΓΖ βάσιν, οὕτως τὸ ΑΒ στερεὸν πρὸς base AE is to base CF , so solid AB (is) to solid CD. τὸ ΓΔ στερεόν. For let FH , equal to AE, have been applied to FG (in Παραβεβλήσθω γὰρ παρὰ τὴν ΖΗ τῷ ΑΕ ἴσον τὸ ΖΘ, the angle FGH equal to angle LCG) [Prop. 1.45]. And καὶ ἀπὸ βάσεως μὲν τῆς ΖΘ, ὕψους δὲ τοῦ αὐτοῦ τῷ ΓΔ let the parallelepiped solid GK, (having) the same height στερεὸν παραλληλεπίπεδον συμπεπληρώσθω τὸ ΗΚ. ἴσον as CD, have been completed on the base FH . So solid δή ἐστι τὸ ΑΒ στερεὸν τῷ ΗΚ στερεῷ· ἐπί τε γὰρ ἴσων AB is equal to solid GK. For they are on the equal bases βάσεών εἰσι τῶν ΑΕ, ΖΘ καὶ ὑπὸ τὸ αὐτο ὕψος. καὶ ἐπεὶ AE and FH , and (have) the same height [Prop. 11.31]. στερεὸν παραλληλεπίπεδον τὸ ΓΚ ἐπιπέδῳ τῷ ΔΗ τέτμηται And since the parallelepiped solid CK has been cut by παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔστιν ἄρα ὡς ἡ the plane DG, which is parallel to the opposite planes (of ΓΖ βάσις πρὸς τὴν ΖΘ βάσιν, οὕτως τὸ ΓΔ στερεὸν πρὸς τὸ CK), thus as the base CF is to the base FH , so the solid ΔΘ στερεόν. ἴση δὲ ἡ μὲν ΖΘ βάσις τῇ ΑΕ βάσει, τὸ δὲ ΗΚ CD (is) to the solid DH [Prop. 11.25]. And base FH (is) στερεὸν τῷ ΑΒ στερεῷ· ἔστιν ἄρα καὶ ὡς ἡ ΑΕ βάσις πρὸς equal to base AE, and solid GK to solid AB. And thus τὴν ΓΖ βάσιν, οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΓΔ στερεόν. as base AE is to base CF , so solid AB (is) to solid CD. Τὰ ἄρα ὑπὸ τὸ αὐτὸ ὕψος ὄντα στερεὰ παραλληλεπίπεδα Thus, parallelepiped solids which (have) the same πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις· ὅπερ ἔδει δεῖξαι. height are to one another as their bases. (Which is) the very thing it was required to show.lgþ. Proposition 33 Τὰ ὅμοια στερεὰ παραλληλεπίπεδα πρὸς ἄλληλα ἐν τρι- Similar parallelepiped solids are to one another as the πλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. cubed ratio of their corresponding sides. ῎Εστω ὅμοια στερεὰ παραλληλεπίπεδα τὰ ΑΒ, ΓΔ, Let AB and CD be similar parallelepiped solids, and ὁμόλογος δὲ ἔστω ἡ ΑΕ τῇ ΓΖ· λέγω, ὅτι τὸ ΑΒ στερεὸν let AE correspond to CF . I say that solid AB has to solid πρὸς τὸ ΓΔ στερεὸν τριπλασίονα λόγον ἔχει, ἤπερ ἡ ΑΕ CD the cubed ratio that AE (has) to CF . πρὸς τὴν ΓΖ. For let EK, EL, and EM have been produced in a ᾿Εκβεβλήσθωσαν γὰρ ἐπ᾿ εὐθείας ταῖς ΑΕ, ΗΕ, ΘΕ αἱ straight-line with AE, GE, and HE (respectively). And ΕΚ, ΕΛ, ΕΜ, καὶ κείσθω τῇ μὲν ΓΖ ἴση ἡ ΕΚ, τῇ δὲ ΖΝ let EK be made equal to CF , and EL equal to FN , and, ἴση ἡ ΕΛ, καὶ ἔτι τῇ ΖΡ ἴση ἡ ΕΜ, καὶ συμπεπληρώσθω τὸ further, EM equal to FR. And let the parallelogram KL ΚΛ παραλληλόγραμμον καὶ τὸ ΚΟ στερεόν. have been completed, and the solid KP . 459 STOIQEIWN iaþ. ELEMENTS BOOK 11 D G Z N R D N R C F Καὶ ἐπεὶ δύο αἱ ΚΕ, ΕΛ δυσὶ ταῖς ΓΖ, ΖΝ ἴσαι εἰσίν, And since the two (straight-lines) KE and EL are ἀλλὰ καὶ γωνία ἡ ὑπὸ ΚΕΛ γωνίᾳ τῇ ὑπὸ ΓΖΝ ἐστιν ἴση, equal to the two (straight-lines) CF and FN , but angle ἐπειδήπερ καὶ ἡ ὑπὸ ΑΕΗ τῇ ὑπὸ ΓΖΝ ἐστιν ἴση διὰ τὴν KEL is also equal to angle CFN , inasmuch as AEG is ὁμοιότητα τῶν ΑΒ, ΓΔ στερεῶν, ἴσον ἄρα ἐστὶ [καὶ ὅμοιον] also equal to CFN , on account of the similarity of the τὸ ΚΛ παραλληλόγραμμον τῷ ΓΝ παραλληλογράμμῳ. διὰ solids AB and CD, parallelogram KL is thus equal [and τὰ αὐτὰ δὴ καὶ τὸ μὲν ΚΜ παραλληλόγραμμον ἴσον ἐστὶ καὶ similar] to parallelogram CN . So, for the same (reasons), ὅμοιον τῷ ΓΡ [παραλληλογράμμῳ] καὶ ἔτι τὸ ΕΟ τῷ ΔΖ· parallelogram KM is also equal and similar to [parallel- τρία ἄρα παραλληλόγραμμα τοῦ ΚΟ στερεοῦ τρισὶ παραλ- ogram] CR, and, further, EP to DF . Thus, three par- ληλογράμμοις τοῦ ΓΔ στερεοῦ ἴσα ἐστὶ καὶ ὅμοια. ἀλλὰ τὰ allelograms of solid KP are equal and similar to three μὲν τρία τρισὶ τοῖς ἀπεναντίον ἴσα ἐστὶ καὶ ὅμοια, τὰ δὲ τρία parallelograms of solid CD. But the three (former par- τρισὶ τοῖς ἀπεναντίον ἴσα ἐστὶ καὶ ὅμοια· ὅλον ἄρα τὸ ΚΟ allelograms) are equal and similar to the three opposite στερεὸν ὅλῳ τῷ ΓΔ στερεῷ ἴσον ἐστὶ καὶ ὅμοιον. συμπε- (parallelograms), and the three (latter parallelograms) πληρώσθω τὸ ΗΚ παραλληλόγραμμον, καὶ ἀπὸ βάσεων μὲν are equal and similar to the three opposite (parallelo- τῶν ΗΚ, ΚΛ παραλληλόγραμμων, ὕψους δὲ τοῦ αὐτοῦ τῷ grams) [Prop. 11.24]. Thus, the whole of solid KP is ΑΒ στερεὰ συμπεπληρώσθω τὰ ΕΞ, ΛΠ. καὶ ἐπεὶ διὰ τὴν equal and similar to the whole of solid CD [Def. 11.10]. ὁμοιότητα τῶν ΑΒ, ΓΔ στερεῶν ἐστιν ὡς ἡ ΑΕ πρὸς τὴν Let parallelogram GK have been completed. And let the ΓΖ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΝ, καὶ ἡ ΕΘ πρὸς τὴν ΖΡ, ἴση the solids EO and LQ, with bases the parallelograms GK δὲ ἡ μὲν ΓΖ τῇ ΕΚ, ἡ δὲ ΖΝ τῇ ΕΛ, ἡ δὲ ΖΡ τῇ ΕΜ, ἔστιν and KL (respectively), and with the same height as AB, ἄρα ὡς ἡ ΑΕ πρὸς τὴν ΕΚ, οὕτως ἡ ΗΕ πρὸς τὴν ΕΛ καὶ ἡ have been completed. And since, on account of the sim- ΘΕ πρὸς τὴν ΕΜ. ἀλλ᾿ ὡς μὲν ἡ ΑΕ πρὸς τὴν ΕΚ, οὕτως τὸ ilarity of solids AB and CD, as AE is to CF , so EG (is) ΑΗ [παραλληλόγραμμον] πρὸς τὸ ΗΚ παραλληλόγραμμον, to FN , and EH to FR [Defs. 6.1, 11.9], and CF (is) ὡς δὲ ἡ ΗΕ πρὸς τὴν ΕΛ, οὕτως τὸ ΗΚ πρὸς τὸ ΚΛ, ὡς equal to EK, and FN to EL, and FR to EM , thus as δὲ ἡ ΘΕ πρὸς ΕΜ, οὕτως τὸ ΠΕ πρὸς τὸ ΚΜ· καὶ ὡς ἄρα AE is to EK, so GE (is) to EL, and HE to EM . But, τὸ ΑΗ παραλληλόγραμμον πρὸς τὸ ΗΚ, οὕτως τὸ ΗΚ πρὸς as AE (is) to EK, so [parallelogram] AG (is) to paral- τὸ ΚΛ καὶ τὸ ΠΕ πρὸς τὸ ΚΜ. ἀλλ᾿ ὡς μὲν τὸ ΑΗ πρὸς lelogram GK, and as GE (is) to EL, so GK (is) to KL, τὸ ΗΚ, οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΕΞ στερεόν, ὡς δὲ and as HE (is) to EM , so QE (is) to KM [Prop. 6.1]. τὸ ΗΚ πρὸς τὸ ΚΛ, οὕτως τὸ ΞΕ στερεὸν πρὸς τὸ ΠΛ And thus as parallelogram AG (is) to GK, so GK (is) στερεόν, ὡς δὲ τὸ ΠΕ πρὸς τὸ ΚΜ, οὕτως τὸ ΠΛ στερεὸν to KL, and QE (is) to KM . But, as AG (is) to GK, so πρὸς τὸ ΚΟ στερεόν· καὶ ὡς ἄρα τὸ ΑΒ στερεὸν πρὸς τὸ solid AB (is) to solid EO, and as GK (is) to KL, so solid ΕΞ, οὕτως τὸ ΕΞ πρὸς τὸ ΠΛ καὶ τὸ ΠΛ πρὸς τὸ ΚΟ. OE (is) to solid QL, and as QE (is) to KM , so solid QL ἐὰν δὲ τέσσαρα μεγέθη κατὰ τὸ συνεχὲς ἀνάλογον ᾖ, τὸ (is) to solid KP [Prop. 11.32]. And, thus, as solid AB πρῶτον πρὸς τὸ τέταρτον τριπλασίονα λόγον ἔχει ἤπερ πρὸς is to EO, so EO (is) to QL, and QL to KP . And if four τὸ δεύτερον· τὸ ΑΒ ἄρα στερεὸν πρὸς τὸ ΚΟ τριπλασίονα magnitudes are continuously proportional then the first λόγον ἔχει ἤπερ τὸ ΑΒ πρὸς τὸ ΕΞ. ἀλλ᾿ ὡς τὸ ΑΒ πρὸς has to the fourth the cubed ratio that (it has) to the sec- τὸ ΕΞ, οὕτως τὸ ΑΗ παραλληλόγραμμον πρὸς τὸ ΗΚ καὶ ἡ ond [Def. 5.10]. Thus, solid AB has to KP the cubed ΑΕ εὐθεῖα πρὸς τὴν ΕΚ· ὥστε καὶ τὸ ΑΒ στερεὸν πρὸς τὸ ratio which AB (has) to EO. But, as AB (is) to EO, so ΚΟ τριπλασίονα λόγον ἔχει ἤπερ ἡ ΑΕ πρὸς τὴν ΕΚ. ἴσον parallelogram AG (is) to GK, and the straight-line AE δὲ τὸ [μὲν] ΚΟ στερεὸν τῷ ΓΔ στερεῷ, ἡ δὲ ΕΚ εὐθεῖα to EK [Prop. 6.1]. Hence, solid AB also has to KP the τῇ ΓΖ· καὶ τὸ ΑΒ ἄρα στερεὸν πρὸς τὸ ΓΔ στερεὸν τρι- cubed ratio that AE (has) to EK. And solid KP (is) 460 STOIQEIWN iaþ. ELEMENTS BOOK 11 πλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος αὐτοῦ πλευρὰ ἡ ΑΕ equal to solid CD, and straight-line EK to CF . Thus, πρὸς τὴν ὁμόλογον πλευρὰν τὴν ΓΖ. solid AB also has to solid CD the cubed ratio which its corresponding side AE (has) to the corresponding side CF . P A H X M L JB O KE Q A M L B E O P G K H Τὰ ἄρα ὅμοια στερεὰ παραλληλεπίπεδα ἐν τριπλασίονι Thus, similar parallelepiped solids are to one another λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν· ὅπερ ἔδει δεῖξαι. as the cubed ratio of their corresponding sides. (Which is) the very thing it was required to show.Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν τέσσαρες εὐθεῖαι So, (it is) clear, from this, that if four straight-lines are ἀνάλογον ὦσιν, ἔσται ὡς ἡ πρώτη πρὸς τὴν τετάρτην, οὕτω (continuously) proportional then as the first is to the τὸ ἀπὸ τῆς πρώτης στερεὸν παραλληλεπίπεδον πρὸς τὸ ἀπὸ fourth, so the parallelepiped solid on the first will be to τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον, ἐπείπερ the similar, and similarly described, parallelepiped solid καὶ ἡ πρώτη πρὸς τὴν τετάρτην τριπλασίονα λόγον ἔχει ἤπερ on the second, since the first also has to the fourth the πρὸς τὴν δευτέραν. cubed ratio that (it has) to the second.ldþ. Proposition 34† Τῶν ἴσων στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν The bases of equal parallelepiped solids are recip- αἱ βάσεις τοῖς ὕψεσιν· καὶ ὧν στερεῶν παραλληλεπιπέδων rocally proportional to their heights. And those paral- ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, ἵσα ἐστὶν ἐκεῖνα. lelepiped solids whose bases are reciprocally proportional ῎Εστω ἴσα στερεὰ παραλληλεπίπεδα τὰ ΑΒ, ΓΔ· λέγω, to their heights are equal. ὅτι τῶν ΑΒ, ΓΔ στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν Let AB and CD be equal parallelepiped solids. I say αἱ βάσεις τοῖς ὕψεσιν, καί ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν that the bases of the parallelepiped solids AB and CD ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ are reciprocally proportional to their heights, and (so) as στερεοῦ ὕψος. base EH is to base NQ, so the height of solid CD (is) to ῎Εστωσαν γὰρ πρότερον αἱ ἐφεστηκυῖαι αἱ ΑΗ, ΕΖ, ΛΒ, the height of solid AB. ΘΚ, ΓΜ, ΝΞ, ΟΔ, ΠΡ πρὸς ὀρθὰς ταῖς βάσεσιν αὐτῶν· For, first of all, let the (straight-lines) standing up, λέγω, ὅτι ἐστὶν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως AG, EF , LB, HK, CM , NO, PD, and QR, be at right- ἡ ΓΜ πρὸς τὴν ΑΗ. angles to their bases. I say that as base EH is to base Εἰ μὲν οὖν ἴση ἐστὶν ἡ ΕΘ βάσιν τῇ ΝΠ βάσει, ἔστι δὲ NQ, so CM (is) to AG. καὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ ἴσον, ἔσται καὶ ἡ ΓΜ τῇ Therefore, if base EH is equal to base NQ, and solid ΑΗ ἴση. τὰ γὰρ ὑπὸ τὸ αὐτὸ ὕψος στερεὰ παραλληλεπίπεδα AB is also equal to solid CD, CM will also be equal to πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις. καὶ ἔσται ὡς ἡ ΕΘ βάσις AG. For parallelepiped solids of the same height are to πρὸς τὴν ΝΠ, οὕτως ἡ ΓΜ πρὸς τὴν ΑΗ, καὶ φανερόν, ὅτι one another as their bases [Prop. 11.32]. And as base 461 STOIQEIWN iaþ. ELEMENTS BOOK 11 τῶν ΑΒ, ΓΔ στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ EH (is) to NQ, so CM will be to AG. And (so it is) clear βάσεις τοῖς ὕψεσιν. that the bases of the parallelepiped solids AB and CD are reciprocally proportional to their heights. P K Z B DMR A EJ G NO FTH X L O K B D M R A E N T F P V C H G L Q Μὴ ἔστω δὴ ἴση ἡ ΕΘ βάσις τῇ ΝΠ βάσει, ἀλλ᾿ ἔστω So let base EH not be equal to base NQ, but let EH μείζων ἡ ΕΘ. ἔστι δὲ καὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ be greater. And solid AB is also equal to solid CD. Thus, ἴσον· μείζων ἄρα ἐστὶ καὶ ἡ ΓΜ τῆς ΑΗ. κείσθω οὖν τῇ CM is also greater than AG. Therefore, let CT be made ΑΗ ἴση ἡ ΓΤ, καὶ συμπεπληρώσθω ἀπὸ βάσεως μὲν τῆς equal to AG. And let the parallelepiped solid V C have ΝΠ, ὕψους δὲ τοῦ ΓΤ, στερεὸν παραλληλεπίπεδον τὸ ΦΓ. been completed on the base NQ, with height CT . And καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ, ἔξωθεν δὲ since solid AB is equal to solid CD, and CV (is) extrinsic τὸ ΓΦ, τὰ δὲ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον, ἔστιν (to them), and equal (magnitudes) have the same ratio to ἄρα ὡς τὸ ΑΒ στερεὸν πρὸς τὸ ΓΦ στερεόν, οὕτως τὸ ΓΔ the same (magnitude) [Prop. 5.7], thus as solid AB is to στερεὸν πρὸς τὸ ΓΦ στερεόν. ἀλλ᾿ ὡς μὲν τὸ ΑΒ στερεὸν solid CV , so solid CD (is) to solid CV . But, as solid AB πρὸς τὸ ΓΦ στερεόν, οὕτως ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν· (is) to solid CV , so base EH (is) to base NQ. For the ἰσοϋψῆ γὰρ τὰ ΑΒ, ΓΦ στερεά· ὡς δὲ τὸ ΓΔ στερεὸν πρὸς solids AB and CV (are) of equal height [Prop. 11.32]. τὸ ΓΦ στερεόν, οὕτως ἡ ΜΠ βάσις πρὸς τὴν ΤΠ βάσιν And as solid CD (is) to solid CV , so base MQ (is) to base καὶ ἡ ΓΜ πρὸς τὴν ΓΤ· καὶ ὡς ἄρα ἡ ΕΘ βάσις πρὸς τὴν TQ [Prop. 11.25], and CM to CT [Prop. 6.1]. And, thus, ΝΠ βάσιν, οὕτως ἡ ΜΓ πρὸς τὴν ΓΤ. ἴση δὲ ἡ ΓΤ τῇ ΑΗ· as base EH is to base NQ, so MC (is) to AG. And CT καὶ ὡς ἄρα ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΜΓ (is) equal to AG. And thus as base EH (is) to base NQ, πρὸς τὴν ΑΗ. τῶν ΑΒ, ΓΔ ἄρα στερεῶν παραλληλεπιπέδων so MC (is) to AG. Thus, the bases of the parallelepiped ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν. solids AB and CD are reciprocally proportional to their Πάλιν δὴ τῶν ΑΒ, ΓΔ στερεῶν παραλληλεπιπέδων ἀντι- heights. πεπονθέτωσαν αἱ βάσεις τοῖς ὕψεσιν, καὶ ἔστω ὡς ἡ ΕΘ So, again, let the bases of the parallelepipid solids AB βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος and CD be reciprocally proportional to their heights, and πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒ let base EH be to base NQ, as the height of solid CD (is) στερεὸν τῷ ΓΔ στερεῷ. to the height of solid AB. I say that solid AB is equal to ῎Εστωσαν [γὰρ] πάλιν αἱ ἐφεστηκυῖαι πρὸς ὀρθὰς ταῖς solid CD. [For] let the (straight-lines) standing up again βάσεσιν. καὶ εἰ μὲν ἴση ἐστὶν ἡ ΕΘ βάσις τῇ ΝΠ βάσει, καί be at right-angles to the bases. And if base EH is equal ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ to base NQ, and as base EH is to base NQ, so the height ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος, ἴσον ἄρα of solid CD (is) to the height of solid AB, the height of ἐστὶ καὶ τὸ τοῦ ΓΔ στερεοῦ ὕψος τῷ τοῦ ΑΒ στερεοῦ ὕψει. solid CD is thus also equal to the height of solid AB. τὰ δὲ ἐπὶ ἴσων βάσεων στερεά παραλληλεπίπεδα καὶ ὑπὸ τὸ And parallelepiped solids on equal bases, and also with αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν· ἴσον ἄρα ἐστὶ τὸ ΑΒ στερεὸν the same height, are equal to one another [Prop. 11.31]. τῷ ΓΔ στερεῷ. Thus, solid AB is equal to solid CD. Μὴ ἔστω δὴ ἡ ΕΘ βάσις τῇ ΝΠ [βάσει] ἴση, ἀλλ᾿ ἔστω So, let base EH not be equal to [base] NQ, but let μείζων ἡ ΕΘ· μεῖζον ἄρα ἐστὶ καὶ τὸ τοῦ ΓΔ στερεοῦ ὕψος EH be greater. Thus, the height of solid CD is also τοῦ τοῦ ΑΒ στερεοῦ ὕψους, τουτέστιν ἡ ΓΜ τῆς ΑΗ. greater than the height of solid AB, that is to say CM κείσθω τῇ ΑΗ ἴση πάλιν ἡ ΓΤ, καὶ συμπεπληρώσθω ὁμοίως (greater) than AG. Let CT again be made equal to AG, τὸ ΓΦ στερεόν. ἐπεί ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ and let the solid CV have been similarly completed. βάσιν, οὕτως ἡ ΜΓ πρὸς τὴν ΑΗ, ἴση δὲ ἡ ΑΗ τῇ ΓΤ, Since as base EH is to base NQ, so MC (is) to AG, 462 STOIQEIWN iaþ. ELEMENTS BOOK 11 ἔστιν ἄρα ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΓΜ and AG (is) equal to CT , thus as base EH (is) to base πρὸς τὴν ΓΤ. ἀλλ᾿ ὡς μὲν ἡ ΕΘ [βάσις] πρὸς τὴν ΝΠ βάσιν, NQ, so CM (is) to CT . But, as [base] EH (is) to base οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΓΦ στερεόν· ἰσοϋψῆ γάρ NQ, so solid AB (is) to solid CV . For solids AB and CV ἐστι τὰ ΑΒ, ΓΦ στερεά· ὡς δὲ ἡ ΓΜ πρὸς τὴν ΓΤ, οὕτως are of equal heights [Prop. 11.32]. And as CM (is) to ἥ τε ΜΠ βάσις πρὸς τὴν ΠΤ βάσιν καὶ τὸ ΓΔ στερεὸν CT , so (is) base MQ to base QT [Prop. 6.1], and solid πρὸς τὸ ΓΦ στερεόν. καὶ ὡς ἄρα τὸ ΑΒ στερεὸν πρὸς τὸ CD to solid CV [Prop. 11.25]. And thus as solid AB (is) ΓΦ στερεόν, οὕτως τὸ ΓΔ στερεὸν πρὸς τὸ ΓΦ στερεόν· to solid CV , so solid CD (is) to solid CV . Thus, AB and ἑκάτερον ἄρα τῶν ΑΒ, ΓΔ πρὸς τὸ ΓΦ τὸν αὐτὸν ἔχει CD each have the same ratio to CV . Thus, solid AB is λόγον. ἴσον ἄρα ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ. equal to solid CD [Prop. 5.9]. Z A HK B E R P NX L S GM Y X J U O WD FT F A K B E R N L S M H G Q C a U D Y P O V W T X Μὴ ἔστωσαν δὴ αἱ ἐφεστηκυῖαι αἱ ΖΕ, ΒΛ, ΗΑ, ΚΘ, So, let the (straight-lines) standing up, FE, BL, GA, ΞΝ, ΔΟ, ΜΓ, ΡΠ πρὸς ὀρθὰς ταῖς βάσεσιν αὐτῶν, καὶ KH , ON , DP , MC, and RQ, not be at right-angles to ἤχθωσαν ἀπὸ τῶν Ζ, Η, Β, Κ, Ξ, Μ, Ρ, Δ σημείων ἐπὶ their bases. And let perpendiculars have been drawn to τὰ διὰ τῶν ΕΘ, ΝΠ ἐπίπεδα κάθετοι καὶ συμβαλλέτωσαν the planes through EH and NQ from points F , G, B, K, τοῖς ἐπιπέδοις κατὰ τὰ Σ, Τ, Υ, Φ, Χ, Ψ, Ω, ς, καὶ συμ- O, M , R, and D, and let them have joined the planes at πεπληρώσθω τὰ ΖΦ, ΞΩ στερεά· λέγω, ὅτι καὶ οὕτως ἴσων (points) S, T , U , V , W , X , Y , and a (respectively). And ὄντων τῶν ΑΒ, ΓΔ στερεῶν ἀντιπεπόνθασιν αἱ βάσεις τοῖς let the solids FV and OY have been completed. In this ὕψεσιν, καί ἐστιν ὡς ἡ ΕΘ βἁσιν πρὸς τὴν ΝΠ βάσιν, οὕτως case, also, I say that the solids AB and CD being equal, τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος. their bases are reciprocally proportional to their heights, ᾿Επεὶ ἴσον ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ, ἀλλὰ τὸ and (so) as base EH is to base NQ, so the height of solid μὲν ΑΒ τῷ ΒΤ ἐστιν ἴσον· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς CD (is) to the height of solid AB. εἰσι τῆς ΖΚ καὶ ὑπὸ τὸ αὐτὸ ὕψος· τὸ δὲ ΓΔ στερεὸν τῷ Since solid AB is equal to solid CD, but AB is equal ΔΨ ἐστιν ἴσον· ἐπί τε γὰρ πάλιν τῆς αὐτῆς βάσεώς εἰσι τῆς to BT . For they are on the same base FK, and (have) the ΡΞ καὶ ὑπὸ τὸ αὐτὸ ὕψος· καὶ τὸ ΒΤ ἄρα στερεὸν τῷ ΔΨ same height [Props. 11.29, 11.30]. And solid CD is equal στερεῷ ἴσον ἐστίν. ἔστιν ἄρα ὡς ἡ ΖΚ βάσις πρὸς τὴν ΞΡ is equal to DX . For, again, they are on the same base RO, βάσιν, οὕτως τὸ τοῦ ΔΨ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΤ and (have) the same height [Props. 11.29, 11.30]. Solid στερεοῦ ὕψος. ἴση δὲ ἡ μὲν ΖΚ βάσις τῇ ΕΘ βάσει, ἡ δὲ BT is thus also equal to solid DX . Thus, as base FK (is) ΞΡ βάσις τῇ ΝΠ βάσει· ἔστιν ἄρα ὡς ἡ ΕΘ βάσις πρὸς τὴν to base OR, so the height of solid DX (is) to the height ΝΠ βάσιν, οὕτως τὸ τοῦ ΔΨ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΤ of solid BT (see first part of proposition). And base FK στερεοῦ ὕψος. τὰ δ᾿ αὐτὰ ὕψη ἐστὶ τῶν ΔΨ, ΒΤ στερεῶν (is) equal to base EH , and base OR to NQ. Thus, as καὶ τῶν ΔΓ, ΒΑ· ἔστιν ἄρα ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ base EH is to base NQ, so the height of solid DX (is) to 463 STOIQEIWN iaþ. ELEMENTS BOOK 11 βάσιν, οὕτως τὸ τοῦ ΔΓ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στε- the height of solid BT . And solids DX , BT are the same ρεοῦ ὕψος. τῶν ΑΒ, ΓΔ ἄρα στερεῶν παραλληλεπιπέδων height as (solids) DC, BA (respectively). Thus, as base ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν. EH is to base NQ, so the height of solid DC (is) to the Πάλιν δὴ τῶν ΑΒ, ΓΔ στερεῶν παραλληλεπιπέδων ἀντι- height of solid AB. Thus, the bases of the parallelepiped πεπονθέτωσαν αἱ βάσεις τοῖς ὕψεσιν, καὶ ἔστω ὡς ἡ ΕΘ solids AB and CD are reciprocally proportional to their βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος heights. πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒ So, again, let the bases of the parallelepiped solids στερεὸν τῷ ΓΔ στερεῷ. AB and CD be reciprocally proportional to their heights, Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ ΕΘ and (so) let base EH be to base NQ, as the height of βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος solid CD (is) to the height of solid AB. I say that solid πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος, ἴση δὲ ἡ μὲν ΕΘ βάσις τῇ AB is equal to solid CD. ΖΚ βάσει, ἡ δὲ ΝΠ τῇ ΞΡ, ἔστιν ἄρα ὡς ἡ ΖΚ βάσις πρὸς For, with the same construction (as before), since as τὴν ΞΡ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ base EH is to base NQ, so the height of solid CD (is) to τοῦ ΑΒ στερεοῦ ὕψος. τὰ δ᾿ αὐτὰ ὕψη ἐστὶ τῶν ΑΒ, ΓΔ the height of solid AB, and base EH (is) equal to base στερεῶν καὶ τῶν ΒΤ, ΔΨ· ἔστιν ἄρα ὡς ἡ ΖΚ βάσις πρὸς FK, and NQ to OR, thus as base FK is to base OR, τὴν ΞΡ βάσιν, οὕτως τὸ τοῦ ΔΨ στερεοῦ ὕψος πρὸς τὸ τοῦ so the height of solid CD (is) to the height of solid AB. ΒΤ στερεοῦ ὕψος. τῶν ΒΤ, ΔΨ ἄρα στερεῶν παραλληλε- And solids AB, CD are the same height as (solids) BT , πιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν· ἴσον ἄρα ἐστὶ DX (respectively). Thus, as base FK is to base OR, so τὸ ΒΤ στερεὸν τῷ ΔΨ στερεῷ. ἀλλὰ τὸ μὲν ΒΤ τῷ ΒΑ the height of solid DX (is) to the height of solid BT . ἴσον ἐστίν· ἐπί τε γὰρ τῆς αὐτῆς βάσεως [εἰσι] τῆς ΖΚ καὶ Thus, the bases of the parallelepiped solids BT and DX ὑπὸ τὸ αὐτὸ ὕψος. τὸ δὲ ΔΨ στερεὸν τῷ ΔΓ στερεῷ ἴσον are reciprocally proportional to their heights. Thus, solid ἐστίν. καὶ τὸ ΑΒ ἄρα στερεὸν τῷ ΓΔ στερεῷ ἐστιν ἴσον· BT is equal to solid DX (see first part of proposition). ὅπερ ἔδει δεῖξαι. But, BT is equal to BA. For [they are] on the same base FK, and (have) the same height [Props. 11.29, 11.30]. And solid DX is equal to solid DC [Props. 11.29, 11.30]. Thus, solid AB is also equal to solid CD. (Which is) the very thing it was required to show. † This proposition assumes that (a) if two parallelepipeds are equal, and have equal bases, then their heights are equal, and (b) if the bases of two equal parallelepipeds are unequal, then that solid which has the lesser base has the greater height.leþ. Proposition 35 ᾿Εὰν ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι, ἐπὶ δὲ τῶν κο- If there are two equal plane angles, and raised ρυφῶν αὐτῶν μετέωροι εὐθεῖαι ἐπισταθῶσιν ἴσας γωνίας straight-lines are stood on the apexes of them, containing περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, equal angles respectively with the original straight-lines ἐπὶ δὲ τῶν μετεώρων ληφθῇ τυχόντα σημεῖα, καὶ ἀπ᾿ αὐτῶν (forming the angles), and random points are taken on ἐπὶ τὰ ἐπίπεδα, ἐν οἷς εἰσιν αἱ ἐξ ἀρχῆς γωνίαι, κάθετοι the raised (straight-lines), and perpendiculars are drawn ἀχθῶσιν, ἀπὸ δὲ τῶν γενομένων σημείων ἐν τοῖς ἐπιπέδοις from them to the planes in which the original angles are, ἐπὶ τὰς ἐξ ἀρχῆς γωνίας ἐπιζευχθῶσιν εὐθεῖαι, ἴσας γωνίας and straight-lines are joined from the points created in περιέξουσι μετὰ τῶν μετεώρων. the planes to the (vertices of the) original angles, then ῎Εστωσαν δύο γωνίαι εὐθύγραμμοι ἴσαι αἱ ὑπὸ ΒΑΓ, they will enclose equal angles with the raised (straight- ΕΔΖ, ἀπὸ δὲ τῶν Α, Δ σημείων μετέωροι εὐθεῖαι ἐφεστάτ- lines). ωσαν αἱ ΑΗ, ΔΜ ἴσας γωνίας περιέχουσιν μετὰ τῶν ἐξ Let BAC and EDF be two equal rectilinear angles. ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, τὴν μὲν ὑπὸ ΜΔΕ τῇ And let the raised straight-lines AG and DM have been ὑπὸ ΗΑΒ, τὴν δὲ ὑπὸ ΜΔΖ τῇ ὑπὸ ΗΑΓ, καὶ εἰλήφθω stood on points A and D, containing equal angles respec- ἐπὶ τῶν ΑΗ, ΔΜ τυχόντα σημεῖα τὰ Η, Μ, καὶ ἤχθωσαν tively with the original straight-lines. (That is) MDE ἀπὸ τῶν Η, Μ σημείων ἐπὶ τὰ διὰ τῶν ΒΑΓ, ΕΔΖ ἐπίπεδα (equal) to GAB, and MDF (to) GAC. And let the ran- κάθετοι αἱ ΗΛ, ΜΝ, καὶ συμβαλλέτωσαν τοῖς ἐπιπέδοις dom points G and M have been taken on AG and DM κατὰ τὰ Λ, Ν, καὶ ἐπεζεύχθωσαν αἱ ΛΑ, ΝΔ· λέγω, ὅτι ἴση (respectively). And let the GL and MN have been drawn ἐστὶν ἡ ὑπὸ ΗΑΛ γωνία τῇ ὑπὸ ΜΔΝ γωνίᾳ. from points G and M perpendicular to the planes through 464 STOIQEIWN iaþ. ELEMENTS BOOK 11 BAC and EDF (respectively). And let them have joined the planes at points L and N (respectively). And let LA and ND have been joined. I say that angle GAL is equal to angle MDN . ND M GA Z L E B J H K ND M A L E C F B H G K Κείσθω τῇ ΔΜ ἴση ἡ ΑΘ, καὶ ἤχθω διὰ τοῦ Θ σημείου Let AH be made equal to DM . And let HK have been τῇ ΗΛ παράλληλος ἡ ΘΚ. ἡ δὲ ΗΛ κάθετός ἐστιν ἐπὶ τὸ drawn through point H parallel to GL. And GL is per- διὰ τῶν ΒΑΓ ἐπίπεδον· καὶ ἡ ΘΚ ἄρα κάθετός ἐστιν ἐπὶ τὸ pendicular to the plane through BAC. Thus, HK is also διὰ τῶν ΒΑΓ ἐπίπεδον. ἤχθωσαν ἀπὸ τῶν Κ, Ν σημείων perpendicular to the plane through BAC [Prop. 11.8]. ἐπὶ τὰς ΑΓ, ΔΖ, ΑΒ, ΔΕ εὐθείας κάθετοι αἱ ΚΓ, ΝΖ, ΚΒ, And let KC, NF , KB, and NE have been drawn from ΝΕ, καὶ ἐπεζεύχθωσαν αἱ ΘΓ, ΓΒ, ΜΖ, ΖΕ. ἐπεὶ τὸ ἀπὸ points K and N perpendicular to the straight-lines AC, τῆς ΘΑ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΘΚ, ΚΑ, τῷ δὲ ἀπὸ τῆς ΚΑ DF , AB, and DE. And let HC, CB, MF , and FE have ἴσα ἐστὶ τὰ ἀπὸ τῶν ΚΓ, ΓΑ, καὶ τὸ ἀπὸ τῆς ΘΑ ἄρα ἴσον been joined. Since the (square) on HA is equal to the ἐστὶ τοῖς ἀπὸ τῶν ΘΚ, ΚΓ, ΓΑ. τοῖς δὲ ἀπὸ τῶν ΘΚ, ΚΓ (sum of the squares) on HK and KA [Prop. 1.47], and ἴσον ἐστὶ τὸ ἀπὸ τῆς ΘΓ· τὸ ἄρα ἀπὸ τῆς ΘΑ ἴσον ἐστὶ τοῖς the (sum of the squares) on KC and CA is equal to the ἀπὸ τῶν ΘΓ, ΓΑ. ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΘΓΑ γωνία. διὰ τὰ (square) on KA [Prop. 1.47], thus the (square) on HA αὐτὰ δὴ καὶ ἡ ὑπὸ ΔΖΜ γωνία ὀρθή ἐστιν. ἴση ἄρα ἐστὶν is equal to the (sum of the squares) on HK, KC, and ἡ ὑπὸ ΑΓΘ γωνία τῇ ὑπὸ ΔΖΜ. ἔστι δὲ καὶ ἡ ὑπὸ ΘΑΓ CA. And the (square) on HC is equal to the (sum of τῇ ὑπὸ ΜΔΖ ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΜΔΖ, ΘΑΓ δύο the squares) on HK and KC [Prop. 1.47]. Thus, the γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν (square) on HA is equal to the (sum of the squares) πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν on HC and CA. Thus, angle HCA is a right-angle ἴσων γωνιῶν τὴν ΘΑ τῇ ΜΔ· καὶ τὰς λοιπὰς ἄρα πλευρὰς [Prop. 1.48]. So, for the same (reasons), angle DFM ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει ἑκατέραν ἑκαρέρᾳ. ἴση ἄρα is also a right-angle. Thus, angle ACH is equal to (an- ἐστὶν ἡ ΑΓ τῇ ΔΖ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΑΒ τῇ gle) DFM . And HAC is also equal to MDF . So, MDF ΔΕ ἐστιν ἴση. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΓ τῇ ΔΖ, ἡ δὲ and HAC are two triangles having two angles equal to ΑΒ τῇ ΔΕ, δύο δὴ αἱ ΓΑ, ΑΒ δυσὶ ταῖς ΖΔ, ΔΕ ἴσαι εἰσίν. two angles, respectively, and one side equal to one side— ἀλλὰ καὶ γωνία ἡ ὑπὸ ΓΑΒ γωνίᾳ τῇ ὑπὸ ΖΔΕ ἐστιν ἴση· (namely), that subtending one of the equal angles —(that βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστὶ καὶ τὸ τρίγωνον τῷ is), HA (equal) to MD. Thus, they will also have the re- τριγώνῳ καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις· ἴση ἄρα ἡ maining sides equal to the remaining sides, respectively ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΔΖΕ. ἔστι δὲ καὶ ὀρθὴ ἡ ὑπὸ ΑΓΚ [Prop. 1.26]. Thus, AC is equal to DF . So, similarly, we ὀρθῇ τῇ ὑπὸ ΔΖΝ ἴση· καὶ λοιπὴ ἄρα ἡ ὑπὸ ΒΓΚ λοιπῇ τῇ can show that AB is also equal to DE. Therefore, since ὑπὸ ΕΖΝ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΒΚ τῇ ὑπὸ AC is equal to DF , and AB to DE, so the two (straight- ΖΕΝ ἐστιν ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΒΓΚ, ΕΖΝ [τὰς] lines) CA and AB are equal to the two (straight-lines) δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ FD and DE (respectively). But, angle CAB is also equal μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν πρὸς ταῖς ἴσαις γωνίαις to angle FDE. Thus, base BC is equal to base EF , and τὴν ΒΓ τῇ ΕΖ· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς triangle (ACB) to triangle (DFE), and the remaining πλευραῖς ἴσας ἕξουσιν. ἴση ἄρα ἐστὶν ἡ ΓΚ τῇ ΖΝ. ἔστι δὲ angles to the remaining angles (respectively) [Prop. 1.4]. 465 STOIQEIWN iaþ. ELEMENTS BOOK 11 καὶ ἡ ΑΓ τῇ ΔΖ ἴση· δύο δὴ αἱ ΑΓ, ΓΚ δυσὶ ταῖς ΔΖ, ΖΝ Thus, angle ACB (is) equal to DFE. And the right-angle ἴσαι εἰσίν· καὶ ὀρθὰς γωνίας περιέχουσιν. βάσις ἄρα ἡ ΑΚ ACK is also equal to the right-angle DFN . Thus, the βάσει τῇ ΔΝ ἴση ἐστίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΘ τῇ ΔΜ, remainder BCK is equal to the remainder EFN . So, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΘ τῷ ἀπὸ τῆς ΔΜ. ἀλλὰ τῷ μὲν for the same (reasons), CBK is also equal to FEN . ἀπὸ τῆς ΑΘ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΚ, ΚΘ· ὀρθὴ γὰρ ἡ ὑπὸ So, BCK and EFN are two triangles having two an- ΑΚΘ· τῷ δὲ ἀπὸ τῆς ΔΜ ἴσα τὰ ἀπὸ τῶν ΔΝ, ΝΜ· ὀρθὴ gles equal to two angles, respectively, and one side equal γὰρ ἡ ὑπὸ ΔΝΜ· τὰ ἄρα ἀπὸ τῶν ΑΚ, ΚΘ ἴσα ἐστὶ τοῖς to one side—(namely), that by the equal angles—(that ἀπὸ τῶν ΔΝ, ΝΜ, ὧν τὸ ἀπὸ τῆς ΑΚ ἴσον ἐστὶ τῷ ἀπὸ τῆς is), BC (equal) to EF . Thus, they will also have the re- ΔΝ· λοιπὸν ἄρα τὸ ἀπὸ τῆς ΚΘ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΝΜ· maining sides equal to the remaining sides (respectively) ἴση ἄρα ἡ ΘΚ τῇ ΜΝ. καὶ ἐπεὶ δύο αἱ ΘΑ, ΑΚ δυσὶ ταῖς [Prop. 1.26]. Thus, CK is equal to FN . And AC (is) also ΜΔ, ΔΝ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ βάσις ἡ ΘΚ βάσει equal to DF . So, the two (straight-lines) AC and CK are τῇ ΜΝ ἐδείχθη ἴση, γωνία ἄρα ἡ ὑπὸ ΘΑΚ γωνίᾳ τῇ ὑπὸ equal to the two (straight-lines) DF and FN (respec- ΜΔΝ ἐστιν ἴση. tively). And they enclose right-angles. Thus, base AK is ᾿Εὰν ἄρα ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι καὶ τὰ ἑξῆς τῆς equal to base DN [Prop. 1.4]. And since AH is equal to προτάσεως [ὅπερ ἔδει δεῖξαι]. DM , the (square) on AH is also equal to the (square) on DM . But, the the (sum of the squares) on AK and KH is equal to the (square) on AH . For angle AKH (is) a right-angle [Prop. 1.47]. And the (sum of the squares) on DN and NM (is) equal to the square on DM . For an- gle DNM (is) a right-angle [Prop. 1.47]. Thus, the (sum of the squares) on AK and KH is equal to the (sum of the squares) on DN and NM , of which the (square) on AK is equal to the (square) on DN . Thus, the remaining (square) on KH is equal to the (square) on NM . Thus, HK (is) equal to MN . And since the two (straight-lines) HA and AK are equal to the two (straight-lines) MD and DN , respectively, and base HK was shown (to be) equal to base MN , angle HAK is thus equal to angle MDN [Prop. 1.8]. Thus, if there are two equal plane angles, and so on of the proposition. [(Which is) the very thing it was re- quired to show].Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι, ἑὰν ὦσι δύο γωνίαι ἐπίπεδοι So, it is clear, from this, that if there are two equal ἴσαι, ἐπισταθῶσι δὲ ἐπ᾿ αὐτῶν μετέωροι εὐθεῖαι ἴσαι ἴσας plane angles, and equal raised straight-lines are stood γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν on them (at their apexes), containing equal angles re- ἑκατέρᾳ, αἱ ἀπ᾿ αὐτῶν κάθετοι ἀγόμεναι ἐπὶ τὰ ἐπίπεδα, ἐν spectively with the original straight-lines (forming the οἷς εἰσιν αἱ ἐξ ἀρχῆς γωνίαι, ἴσαι ἀλλήλαις εἰσίν. ὅπερ ἔδει angles), then the perpendiculars drawn from (the raised δεῖξαι. ends of) them to the planes in which the original angles lie are equal to one another. (Which is) the very thing it was required to show.l�þ. Proposition 36 ᾿Εὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ἐκ τῶν τριῶν If three straight-lines are (continuously) proportional στερεὸν παραλληλεπίπεδον ἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης then the parallelepiped solid (formed) from the three στερεῷ παραλληλεπιπέδῳ ἰσοπλεύρῳ μέν, ἰσογωνίῳ δὲ τῷ (straight-lines) is equal to the equilateral parallelepiped προειρημένῳ. solid on the middle (straight-line which is) equiangular to the aforementioned (parallelepiped solid). 466 STOIQEIWN iaþ. ELEMENTS BOOK 11 LM N BA E DX G H J Z K LM N B A H O F E C D G K ῎Εστωσαν τρεῖς εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, ὡς ἡ Α Let A, B, and C be three (continuously) proportional πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ· λέγω, ὅτι τὸ ἐκ τῶν straight-lines, (such that) as A (is) to B, so B (is) to C. Α, Β, Γ στερεὸν ἴσον ἐστὶ τῷ ἀπὸ τῆς Β στερεῷ ἰσοπλεύρῳ I say that the (parallelepiped) solid (formed) from A, B, μέν, ἰσογωνίῳ δὲ τῷ προειρημένῳ. and C is equal to the equilateral solid on B (which is) ᾿Εκκείσθω στερεὰ γωνία ἡ πρὸς τῷ Ε περιεχομένη ὑπὸ equiangular with the aforementioned (solid). τῶν ὑπὸ ΔΕΗ, ΗΕΖ, ΖΕΔ, καὶ κείσθω τῇ μὲν Β ἴση ἑκάστη Let the solid angle at E, contained by DEG, GEF , τῶν ΔΕ, ΗΕ, ΕΖ, καὶ συμπεπληρώσθω τὸ ΕΚ στερεὸν πα- and FED, be set out. And let DE, GE, and EF each ραλληλεπίπεδον, τῇ δὲ Α ἴση ἡ ΛΜ, καὶ συνεστάτω πρὸς be made equal to B. And let the parallelepiped solid τῇ ΛΜ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Λ τῇ πρὸς τῷ EK have been completed. And (let) LM (be made) Ε στερεᾷ γωνίᾳ ἴση στερεὰ γωνία ἡ περειχομένη ὑπὸ τῶν equal to A. And let the solid angle contained by NLO, ΝΛΞ, ΞΛΜ, ΜΛΝ, καὶ κείσθω τῇ μὲν Β ἴση ἡ ΛΞ, τῇ δὲ OLM , and MLN have been constructed on the straight- Γ ἴση ἡ ΛΝ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β line LM , and at the point L on it, (so as to be) equal πρὸς τὴν Γ, ἴση δὲ ἡ μὲν Α τῇ ΛΜ, ἡ δὲ Β ἑκατέρᾳ τῶν ΛΞ, to the solid angle E [Prop. 11.23]. And let LO be made ΕΔ, ἡ δὲ Γ τῇ ΛΝ, ἔστιν ἄρα ὡς ἡ ΛΜ πρὸς τὴν ΕΖ, οὕτως equal to B, and LN equal to C. And since as A (is) ἡ ΔΕ πρὸς τὴν ΛΝ. καὶ περὶ ἴσας γωνίας τὰς ὑπὸ ΝΛΜ, to B, so B (is) to C, and A (is) equal to LM , and B ΔΕΖ αἱ πλευραὶ ἀντιπεπόνθασιν· ἴσον ἄρα ἐστὶ τὸ ΜΝ πα- to each of LO and ED, and C to LN , thus as LM (is) ραλληλόγραμμον τῷ ΔΖ παραλληλογραμάμμῳ. καὶ ἐπεὶ δύο to EF , so DE (is) to LN . And (so) the sides around γωνίαι ἐπίπεδοι εὐθύγραμμοι ἴσαι εἰσὶν αἱ ὑπὸ ΔΕΖ, ΝΛΜ, the equal angles NLM and DEF are reciprocally pro- καὶ ἐπ᾿ αὐτῶν μετέωροι εὐθεῖαι ἐφεστᾶσιν αἱ ΛΞ, ΕΗ ἴσαι portional. Thus, parallelogram MN is equal to parallel- τε ἀλλήλαις καὶ ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς ogram DF [Prop. 6.14]. And since the two plane recti- εὐθειῶν ἑκατέραν ἑκατέρᾳ, αἱ ἄρα ἀπὸ τῶν Η, Ξ σημείων linear angles DEF and NLM are equal, and the raised κάθετοι ἀγόμεναι ἐπὶ τὰ διὰ τῶν ΝΛΜ, ΔΕΖ ἐπίπεδα ἴσαι straight-lines stood on them (at their apexes), LO and ἀλλήλαις εἰσίν· ὥστε τὰ ΛΘ, ΕΚ στερεὰ ὑπὸ τὸ αὐτὸ ὕψος EG, are equal to one another, and contain equal angles ἐστίν. τὰ δὲ ἐπὶ ἴσων βάσεων στερεὰ παραλληλεπίπεδα καὶ respectively with the original straight-lines (forming the ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν· ἴσον ἄρα ἐστὶ τὸ ΘΛ angles), the perpendiculars drawn from points G and O στερεὸν τῷ ΕΚ στερεῷ. καί ἐστι τὸ μὲν ΛΘ τὸ ἐκ τῶν Α, to the planes through NLM and DEF (respectively) are Β, Γ στερεόν, τὸ δὲ ΕΚ τὸ ἀπὸ τῆς Β στερεόν· τὸ ἄρα ἐκ thus equal to one another [Prop. 11.35 corr.]. Thus, the τῶν Α, Β, Γ στερεὸν παραλληλεπίπεδον ἴσον ἐστὶ τῷ ἀπὸ solids LH and EK (have) the same height. And paral- τῆς Β στερεῷ ἰσοπλεύρῳ μέν, ἰσογωνίῳ δὲ τῷ προειρημένῳ· lelepiped solids on equal bases, and with the same height, ὅπερ ἔδει δεῖξαι. are equal to one another [Prop. 11.31]. Thus, solid HL is equal to solid EK. And LH is the solid (formed) from A, B, and C, and EK the solid on B. Thus, the par- allelepiped solid (formed) from A, B, and C is equal to the equilateral solid on B (which is) equiangular with the aforementioned (solid). (Which is) the very thing it was required to show.lzþ. Proposition 37† ᾿Εὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, καὶ τὰ ἀπ᾿ αὐτῶν If four straight-lines are proportional then the similar, 467 STOIQEIWN iaþ. ELEMENTS BOOK 11 στερεὰ παραλληλεπίπεδα ὅμοιά τε καὶ ὁμοίως ἀναγραφόμενα and similarly described, parallelepiped solids on them ἀνάλογον ἔσται· καὶ ἐὰν τὰ ἀπ᾿ αὐτῶν στερεὰ παραλλη- will also be proportional. And if the similar, and similarly λεπίπεδα ὅμοιά τε καὶ ὁμοίως ἀναγραφόμενα ἀνάλογον ᾖ, described, parallelepiped solids on them are proportional καὶ αὐταὶ αἱ εὐθεῖαι ἀνάλογον ἔσονται. then the straight-lines themselves will be proportional. JE Z M A K L G DB H N C E M A K L DB N F G H ῎Εστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ ΑΒ, ΓΔ, ΕΖ, Let AB, CD, EF , and GH , be four proportional ΗΘ, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, καὶ straight-lines, (such that) as AB (is) to CD, so EF (is) ἀναγεγράφθωσαν ἀπὸ τῶν ΑΒ, ΓΔ, ΕΖ, ΗΘ ὅμοιά τε καὶ to GH . And let the similar, and similarly laid out, par- ὁμοίως κείμενα στερεὰ παραλληλεπίπεδα τὰ ΚΑ, ΛΓ, ΜΕ, allelepiped solids KA, LC, ME and NG have been de- ΝΗ· λέγω, ὅτι ἐστὶν ὡς τὸ ΚΑ πρὸς τὸ ΛΓ, οὕτως τὸ ΜΕ scribed on AB, CD, EF , and GH (respectively). I say πρὸς τὸ ΝΗ. that as KA is to LC, so ME (is) to NG. ᾿Επεὶ γὰρ ὅμοιόν ἐστι τὸ ΚΑ στερεὸν παραλληλεπίπεδον For since the parallelepiped solid KA is similar to LC, τῷ ΛΓ, τὸ ΚΑ ἄρα πρὸς τὸ ΛΓ τριπλασίονα λόγον ἔχει ἤπερ KA thus has to LC the cubed ratio that AB (has) to CD ἡ ΑΒ πρὸς τὴν ΓΔ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΜΕ πρὸς τὸ ΝΗ [Prop. 11.33]. So, for the same (reasons), ME also has to τριπλασίονα λόγον ἔχει ἤπερ ἡ ΕΖ πρὸς τὴν ΗΘ. καί ἐστιν NG the cubed ratio that EF (has) to GH [Prop. 11.33]. ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. καὶ ὡς And since as AB is to CD, so EF (is) to GH , thus, also, ἄρα τὸ ΑΚ πρὸς τὸ ΛΓ, οὕτως τὸ ΜΕ πρὸς τὸ ΝΗ. as AK (is) to LC, so ME (is) to NG. Ἀλλὰ δὴ ἔστω ὡς τὸ ΑΚ στερεὸν πρὸς τὸ ΛΓ στερεόν, And so let solid AK be to solid LC, as solid ME (is) οὕτως τὸ ΜΕ στερεὸν πρὸς τὸ ΝΗ· λέγω, ὅτι ἐστὶν ὡς ἡ to NG. I say that as straight-line AB is to CD, so EF (is) ΑΒ εὐθεῖα πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. to GH . ᾿Επεὶ γὰρ πάλιν τὸ ΚΑ πρὸς τὸ ΛΓ τριπλασίονα λόγον For, again, since KA has to LC the cubed ratio that ἔχει ἤπερ ἡ ΑΒ πρὸς τὴν ΓΔ, ἔχει δὲ καὶ τὸ ΜΕ πρὸς τὸ AB (has) to CD [Prop. 11.33], and ME also has to NG ΝΗ τριπλασίονα λόγον ἤπερ ἡ ΕΖ πρὸς τὴν ΗΘ, καί ἐστιν the cubed ratio that EF (has) to GH [Prop. 11.33], and ὡς τὸ ΚΑ πρὸς τὸ ΛΓ, οὕτως τὸ ΜΕ πρὸς τὸ ΝΗ, καὶ ὡς as KA is to LC, so ME (is) to NG, thus, also, as AB (is) ἄρα ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. to CD, so EF (is) to GH . ᾿Εὰν ἄρα τέσσαρες εὐθεῖαι ἀνάλογον ὦσι καὶ τὰ ἑξῆς Thus, if four straight-lines are proportional, and so τῆς προτάσεως· ὅπερ ἔδει δεῖξαι. on of the proposition. (Which is) the very thing it was required to show. † This proposition assumes that if two ratios are equal then the cube of the former is also equal to the cube of the latter, and vice versa.lhþ. Proposition 38 ᾿Εὰν κύβου τῶν ἀπεναντίον ἐπιπέδων αἱ πλευραὶ δίχα If the sides of the opposite planes of a cube are cut τμηθῶσιν, διὰ δὲ τῶν τομῶν ἐπίπεδα ἐκβληθῇ, ἡ κοινὴ τομὴ in half, and planes are produced through the pieces, then τῶν ἐπιπέδων καὶ ἡ τοῦ κύβου διάμετρος δίχα τέμνουσιν the common section of the (latter) planes and the diam- ἀλλήλας. eter of the cube cut one another in half. 468 STOIQEIWN iaþ. ELEMENTS BOOK 11 J D Z OX G HN M T U L E S RP A K B H D N T U L E S R A C F G K O P Q MB Κύβου γὰρ τοῦ ΑΖ τῶν ἀπεναντίον ἐπιπέδων τῶν ΓΖ, For let the opposite planes CF and AH of the cube ΑΘ αἱ πλευραὶ δίχα τετμήσθωσαν κατὰ τὰ Κ, Λ, Μ, Ν, Ξ, AF have been cut in half at the points K, L, M , N , O, Π, Ο, Ρ σημεῖα, διὰ δὲ τῶν τομῶν ἐπίπεδα ἐκβεβλήσθω τὰ Q, P , and R. And let the planes KN and OR have been ΚΝ, ΞΡ, κοινὴ δὲ τομὴ τῶν ἐπιπέδων ἔστω ἡ ΥΣ, τοῦ δὲ produced through the pieces. And let US be the common ΑΖ κύβου διαγώνιος ἡ ΔΗ. λέγω, ὅτι ἴση ἐστὶν ἡ μὲν ΥΤ section of the planes, and DG the diameter of cube AF . τῇ ΤΣ, ἡ δὲ ΔΤ τῇ ΤΗ. I say that UT is equal to TS, and DT to TG. ᾿Επεζεύχθωσαν γὰρ αἱ ΔΥ, ΥΕ, ΒΣ, ΣΗ. καὶ ἐπεὶ For let DU , UE, BS, and SG have been joined. And παράλληλός ἐστιν ἡ ΔΞ τῇ ΟΕ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ since DO is parallel to PE, the alternate angles DOU and ΔΞΥ, ΥΟΕ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν UPE are equal to one another [Prop. 1.29]. And since ΔΞ τῇ ΟΕ, ἡ δὲ ΞΥ τῇ ΥΟ, καὶ γωνίας ἴσας περιέχουσιν, DO is equal to PE, and OU to UP , and they contain βάσις ἄρα ἡ ΔΥ τῇ ΥΕ ἐστιν ἴση, καὶ τὸ ΔΞΥ τρίγωνον τῷ equal angles, base DU is thus equal to base UE, and tri- ΟΥΕ τριγώνῳ ἐστὶν ἴσον καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς angle DOU is equal to triangle PUE, and the remaining γωνίαις ἴσαι· ἴση ἄρα ἡ ὑπὸ ΞΥΔ γωνία τῇ ὑπὸ ΟΥΕ γωνίᾳ. angles (are) equal to the remaining angles [Prop. 1.4]. διὰ δὴ τοῦτο εὐθεῖά ἐστιν ἡ ΔΥΕ. διὰ τὰ αὐτὰ δὴ καὶ ΒΣΗ Thus, angle OUD (is) equal to angle PUE. So, for this εὐθεῖά ἐστιν, καὶ ἴση ἡ ΒΣ τῇ ΣΗ. καὶ ἐπεὶ ἡ ΓΑ τῇ ΔΒ (reason), DUE is a straight-line [Prop. 1.14]. So, for ἴση ἐστὶ καὶ παράλληλος, ἀλλὰ ἡ ΓΑ καὶ τῇ ΕΗ ἴση τέ the same (reason), BSG is also a straight-line, and BS ἐστι καὶ παράλληλος, καὶ ἡ ΔΒ ἄρα τῇ ΕΗ ἴση τέ ἐστι equal to SG. And since CA is equal and parallel to DB, καὶ παράλληλος. καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ΔΕ, but CA is also equal and parallel to EG, DB is thus also ΒΗ· παράλληλος ἄρα ἐστὶν ἡ ΔΕ τῇ ΒΗ. ἴση ἄρα ἡ μὲν ὑπὸ equal and parallel to EG [Prop. 11.9]. And the straight- ΕΔΤ γωνία τῇ ὑπὸ ΒΗΤ· ἐναλλὰξ γάρ· ἡ δὲ ὑπὸ ΔΤΥ τῇ lines DE and BG join them. DE is thus parallel to BG ὑπὸ ΗΤΣ. δύο δὴ τρίγωνά ἐστι τὰ ΔΤΥ, ΗΤΣ τὰς δύο [Prop. 1.33]. Thus, angle EDT (is) equal to BGT . For γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ (they are) alternate [Prop. 1.29]. And (angle) DTU (is πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν equal) to GTS [Prop. 1.15]. So, DTU and GTS are two τὴν ΔΥ τῇ ΗΣ· ἡμίσειαι γάρ εἰσι τῶν ΔΕ, ΒΗ· καὶ τὰς triangles having two angles equal to two angles, and one λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει. ἴση ἄρα ἡ side equal to one side—(namely), that subtended by one μὲν ΔΤ τῇ ΤΗ, ἡ δὲ ΥΤ τῇ ΤΣ. of the equal angles—(that is), DU (equal) to GS. For ᾿Εὰν ἄρα κύβου τῶν ἀπεναντίον ἐπιπέδων αἱ πλευραὶ δίχα they are halves of DE and BG (respectively). (Thus), τμηθῶσιν, διὰ δὲ τῶν τομῶν ἐπίπεδα ἐκβληθῇ, ἡ κοινὴ τομὴ they will also have the remaining sides equal to the re- τῶν ἐπιπέδων καὶ ἡ τοῦ κύβου διάμετρος δίχα τέμνουσιν maining sides [Prop. 1.26]. Thus, DT (is) equal to TG, ἀλλήλας· ὅπερ ἔδει δεῖξαι. and UT to TS. Thus, if the sides of the opposite planes of a cube are cut in half, and planes are produced through the pieces, then the common section of the (latter) planes and the diameter of the cube cut one another in half. (Which is) the very thing it was required to show. 469 STOIQEIWN iaþ. ELEMENTS BOOK 11 ljþ. Proposition 39 ᾿Εὰν ᾖ δύο πρίσματα ἰσοϋψῆ, καὶ τὸ μὲν ἔχῇ βάσιν πα- If there are two equal height prisms, and one has a ραλληλόγραμμον, τὸ δὲ τρίγωνον, διπλάσιον δὲ ᾖ τὸ παραλ- parallelogram, and the other a triangle, (as a) base, and ληλόγραμμον τοῦ τριγώνου, ἴσα ἔσται τὰ πρίσματα. the parallelogram is double the triangle, then the prisms will be equal. N KE B D G H OM LJXZ A A KE B D M L C G P H NO F ῎Εστω δύο πρίσματα ἰσοϋψῆ τὰ ΑΒΓΔΕΖ, ΗΘΚΛΜΝ, Let ABCDEF and GHKLMN be two equal height καὶ τὸ μὲν ἐχέτω βάσιν τὸ ΑΖ παραλληλόγραμμον, τὸ prisms, and let the former have the parallelogram AF , δὲ τὸ ΗΘΚ τρίγωνον, διπλάσιον δὲ ἔστω τὸ ΑΖ παραλ- and the latter the triangle GHK, as a base. And let par- ληλόγραμμον τοῦ ΗΘΚ τριγώνου· λέγω, ὅτι ἴσον ἐστὶ τὸ allelogram AF be twice triangle GHK. I say that prism ΑΒΓΔΕΖ πρίσμα τῷ ΗΘΚΛΜΝ πρίσματι. ABCDEF is equal to prism GHKLMN . Συμπεπληρώσθω γὰρ τὰ ΑΞ, ΗΟ στερεά. ἐπεὶ διπλάσιόν For let the solids AO and GP have been com- ἐστι τὸ ΑΖ παραλληλόγραμμον τοῦ ΗΘΚ τριγώνου, ἔστι pleted. Since parallelogram AF is double triangle GHK, δὲ καὶ τὸ ΘΚ παραλληλόγραμμον διπλάσιον τοῦ ΗΘΚ and parallelogram HK is also double triangle GHK τριγώνου, ἴσον ἄρα ἐστὶ τὸ ΑΖ παραλληλόγραμμον τῷ ΘΚ [Prop. 1.34], parallelogram AF is thus equal to paral- παραλληλογράμμῳ. τὰ δὲ ἐπὶ ἴσων βάσεων ὄντα στερεὰ lelogram HK. And parallelepiped solids which are on παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν· equal bases, and (have) the same height, are equal to ἴσον ἄρα ἐστὶ τὸ ΑΞ στερεὸν τῷ ΗΟ στερεῷ. καί ἐστι one another [Prop. 11.31]. Thus, solid AO is equal to τοῦ μὲν ΑΞ στερεοῦ ἥμισυ τὸ ΑΒΓΔΕΖ πρίσμα, τοῦ δὲ solid GP . And prism ABCDEF is half of solid AO, and ΗΟ στερεοῦ ἥμισυ τὸ ΗΘΚΛΜΝ πρίσμα· ἴσον ἄρα ἐστὶ τὸ prism GHKLMN half of solid GP [Prop. 11.28]. Prism ΑΒΓΔΕΖ πρίσμα τῷ ΗΘΚΛΜΝ πρίσματι. ABCDEF is thus equal to prism GHKLMN . ᾿Εὰν ἄρα ᾖ δύο πρίσματα ἰσοϋψῆ, καὶ τὸ μὲν ἔχῇ βάσιν Thus, if there are two equal height prisms, and one παραλληλόγραμμον, τὸ δὲ τρίγωνον, διπλάσιον δὲ ᾖ τὸ πα- has a parallelogram, and the other a triangle, (as a) base, ραλληλόγραμμον τοῦ τριγώνου, ἴσα ἔστὶ τὰ πρίσματα· ὅπερ and the parallelogram is double the triangle, then the ἔδει δεῖξαι. prisms are equal. (Which is) the very thing it was re- quired to show. 470 ELEMENTS BOOK 12 Proportional Stereometry† †The novel feature of this book is the use of the so-called method of exhaustion (see Prop. 10.1), a precursor to integration which is generally attributed to Eudoxus of Cnidus. 471 STOIQEIWN ibþ. ELEMENTS BOOK 12aþ. Proposition 1 Τὰ ἐν τοῖς κύκλοις ὅμοια πολύγωνα πρὸς ἄλληλά ἐστιν Similar polygons (inscribed) in circles are to one an- ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα. other as the squares on the diameters (of the circles). NM EB A Z H K L G JD D M E B A K L N C H G F ῎Εστωσαν κύκλοι οἱ ΑΒΓ, ΖΗΘ, καὶ ἐν αὐτοῖς ὅμοια Let ABC and FGH be circles, and let ABCDE and πολύγωνα ἔστω τὰ ΑΒΓΔΕ, ΖΗΘΚΛ, διάμετροι δὲ τῶν FGHKL be similar polygons (inscribed) in them (re- κύκλων ἔστωσαν ΒΜ, ΗΝ· λέγω, ὅτι ἐστὶν ὡς τὸ ἀπὸ τῆς spectively), and let BM and GN be the diameters of the ΒΜ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΝ τετράγωνον, οὕτως circles (respectively). I say that as the square on BM is to τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον. the square on GN , so polygon ABCDE (is) to polygon ᾿Επεζεύχθωσαν γὰρ αἱ ΒΕ, ΑΜ, ΗΛ, ΖΝ. καὶ ἐπεὶ FGHKL. ὅμοιον τὸ ΑΒΓΔΕ πολύγωνον τῷ ΖΗΘΚΛ πολυγώνῳ, ἴση For let BE, AM , GL, and FN have been joined. And ἐστὶ καὶ ἡ ὑπὸ ΒΑΕ γωνία τῇ ὑπὸ ΗΖΛ, καί ἐστιν ὡς ἡ ΒΑ since polygon ABCDE (is) similar to polygon FGHKL, πρὸς τὴν ΑΕ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΛ. δύο δὴ τρίγωνά angle BAE is also equal to (angle) GFL, and as BA ἐστι τὰ ΒΑΕ, ΗΖΛ μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχοντα τὴν is to AE, so GF (is) to FL [Def. 6.1]. So, BAE and ὑπὸ ΒΑΕ τῇ ὑπὸ ΗΖΛ, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς GFL are two triangles having one angle equal to one ἀνάλογον· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΕ τρίγωνον τῷ ΖΗΛ angle, (namely), BAE (equal) to GFL, and the sides τριγώνῳ. ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΕΒ γωνία τῇ ὑπὸ ΖΛΗ. ἀλλ᾿ around the equal angles proportional. Triangle ABE is ἡ μὲν ὑπὸ ΑΕΒ τῇ ὑπὸ ΑΜΒ ἐστιν ἴση· ἐπὶ γὰρ τῆς αὐτῆς thus equiangular with triangle FGL [Prop. 6.6]. Thus, περιφερείας βεβήκασιν· ἡ δὲ ὑπὸ ΖΛΗ τῇ ὑπὸ ΖΝΗ· καὶ ἡ angle AEB is equal to (angle) FLG. But, AEB is equal ὑπὸ ΑΜΒ ἄρα τῇ ὑπὸ ΖΝΗ ἐστιν ἴση. ἔστι δὲ καὶ ὀρθὴ to AMB, and FLG to FNG, for they stand on the same ἡ ὑπὸ ΒΑΜ ὀρθῇ τῇ ὑπὸ ΗΖΝ ἴση· καὶ ἡ λοιπὴ ἄρα τῇ circumference [Prop. 3.27]. Thus, AMB is also equal λοιπῇ ἐστιν ἴση. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΜ τρίγωνον τῷ to FNG. And the right-angle BAM is also equal to the ΖΗΝ τρίγωνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΜ πρὸς τὴν right-angle GFN [Prop. 3.31]. Thus, the remaining (an- ΗΝ, οὕτως ἡ ΒΑ πρὸς τὴν ΗΖ. ἀλλὰ τοῦ μὲν τῆς ΒΜ gle) is also equal to the remaining (angle) [Prop. 1.32]. πρὸς τὴν ΗΝ λόγον διπλασίων ἐστὶν ὁ τοῦ ἀπὸ τῆς ΒΜ Thus, triangle ABM is equiangular with triangle FGN . τετραγώνου πρὸς τὸ ἀπὸ τῆς ΗΝ τετράγωνον, τοῦ δὲ τῆς Thus, proportionally, as BM is to GN , so BA (is) to GF ΒΑ πρὸς τὴν ΗΖ διπλασίων ἐστὶν ὁ τοῦ ΑΒΓΔΕ πολυγώνου [Prop. 6.4]. But, the (ratio) of the square on BM to the πρὸς τὸ ΖΗΘΚΛ πολύγωνον· καὶ ὡς ἄρα τὸ ἁπὸ τῆς ΒΜ square on GN is the square of the ratio of BM to GN , τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΝ τετράγωνον, οὕτως τὸ and the (ratio) of polygon ABCDE to polygon FGHKL ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον. is the square of the (ratio) of BA to GF [Prop. 6.20]. Τὰ ἄρα ἐν τοῖς κύκλοις ὅμοια πολύγωνα πρὸς ἄλληλά And, thus, as the square on BM (is) to the square on ἐστιν ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα· ὅπερ ἔδει δεῖξαι. GN , so polygon ABCDE (is) to polygon FGHKL. Thus, similar polygons (inscribed) in circles are to one another as the squares on the diameters (of the circles). (Which is) the very thing it was required to show.bþ. Proposition 2 Οἱ κύκλοι πρὸς ἀλλήλους εἰσὶν ὡς τὰ ἀπὸ τῶν διαμέτρων Circles are to one another as the squares on (their) τετράγωνα. diameters. ῎Εστωσαν κύκλοι οἱ ΑΒΓΔ, ΕΖΗΘ, διάμετροι δὲ αὐτῶν Let ABCD and EFGH be circles, and [let] BD and 472 STOIQEIWN ibþ. ELEMENTS BOOK 12 [ἔστωσαν] αἱ ΒΔ, ΖΘ· λέγω, ὅτι ἐστὶν ὡς ὁ ΑΒΓΔ κύκλος FH [be] their diameters. I say that as circle ABCD is to πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως τὸ ἀπὸ τῆς ΒΔ τετράγωνον circle EFGH , so the square on BD (is) to the square on πρὸς τὸ ἀπὸ τῆς ΖΘ τετράγωνον. FH . M J K EX R D P S T B Z N GO A L H M K E R D S T B N A L C F G H O P Q Εἰ γὰρ μή ἐστιν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ, For if the circle ABCD is not to the (circle) EFGH , οὕτως τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, as the square on BD (is) to the (square) on FH , then as ἔσται ὡς τὸ ἀπὸ τῆς ΒΔ πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ the (square) on BD (is) to the (square) on FH , so circle ΑΒΓΔ κύκλος ἤτοι πρὸς ἔλασσόν τι τοῦ ΕΖΗΘ κύκλου ABCD will be to some area either less than, or greater χωρίον ἢ πρὸς μεῖζον. ἔστω πρότερον πρὸς ἔλασσον τὸ than, circle EFGH . Let it, first of all, be (in that ratio) to Σ. και ἐγγεγράφθω εἰς τὸν ΕΖΗΘ κύκλον τετράγωνον τὸ (some) lesser (area), S. And let the square EFGH have ΕΖΗΘ. τὸ δὴ ἐγγεγραμμένον τετράγωνον μεῖζόν ἐστιν ἢ τὸ been inscribed in circle EFGH [Prop. 4.6]. So the in- ἥμισυ τοῦ ΕΖΗΘ κύκλου, ἐπειδήπερ ἐὰν διὰ τῶν Ε, Ζ, Η, Θ scribed square is greater than half of circle EFGH , inas- σημείων ἐφαπτομένας [εὐθείας] τοῦ κύκλου ἀγάγωμεν, τοῦ much as if we draw tangents to the circle through the περιγραφομένου περὶ τὸν κύκλον τετραγώνου ἥμισύ ἐστι points E, F , G, and H , then square EFGH is half of the τὸ ΕΖΗΘ τετράγωνον, τοῦ δὲ περιγραφέντος τετραγώνου square circumscribed about the circle [Prop. 1.47], and ἐλάττων ἐστὶν ὁ κύκλος· ὥστε τὸ ΕΖΗΘ ἐγγεγραμμένον the circle is less than the circumscribed square. Hence, τετράγωνον μεῖζόν ἐστι τοῦ ἡμίσεως τοῦ ΕΖΗΘ κύκλου. the inscribed square EFGH is greater than half of cir- τετμήσθωσαν δίχα αἱ ΕΖ, ΖΗ, ΗΘ, ΘΕ περιφέρειαι κατὰ cle EFGH . Let the circumferences EF , FG, GH , and τὰ Κ, Λ, Μ, Ν σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΕΚ, ΚΖ, HE have been cut in half at points K, L, M , and N ΖΛ, ΛΗ, ΗΜ, ΜΘ, ΘΝ, ΝΕ· καὶ ἕκαστον ἄρα τῶν ΕΚΖ, (respectively), and let EK, KF , FL, LG, GM , MH , ΖΛΗ, ΗΜΘ, ΘΝΕ τριγώνων μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ HN , and NE have been joined. And, thus, each of καθ᾿ ἑαυτὸ τμήματος τοῦ κύκλου, ἐπειδήπερ ἐὰν διὰ τῶν the triangles EKF , FLG, GMH , and HNE is greater Κ, Λ, Μ, Ν σημείων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν than half of the segment of the circle about it, inasmuch καὶ ἀναπληρώσωμεν τὰ ἐπὶ τῶν ΕΖ, ΖΗ, ΗΘ, ΘΕ εὐθειῶν as if we draw tangents to the circle through points K, παραλληλόγραμμα, ἕκαστον τῶν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ L, M , and N , and complete the parallelograms on the τριγώνων ἥμισυ ἔσται τοῦ καθ᾿ ἑαυτὸ παραλληλογράμμου, straight-lines EF , FG, GH , and HE, then each of the ἀλλὰ τὸ καθ᾿ ἑαυτὸ τμῆμα ἔλαττόν ἐστι τοῦ παραλλη- triangles EKF , FLG, GMH , and HNE will be half λογράμμου· ὥστε ἕκαστον τῶν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ of the parallelogram about it, but the segment about it τριγώνων μεῖζόν ἐστι τοῦ ἡμίσεως τοῦ καθ᾿ ἑαυτὸ τμήματος is less than the parallelogram. Hence, each of the tri- τοῦ κύκλου. τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας angles EKF , FLG, GMH , and HNE is greater than δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ τοῦτο ἀεὶ ποιοῦντες κα- half of the segment of the circle about it. So, by cutting ταλείψομέν τινα ἀποτμήματα τοῦ κύκλου, ἃ ἔσται ἐλάσσονα the circumferences remaining behind in half, and joining τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ ΕΖΗΘ κύκλος τοῦ Σ χωρίου. straight-lines, and doing this continually, we will (even- 473 STOIQEIWN ibþ. ELEMENTS BOOK 12 ἐδείχθη γὰρ ἐν τῷ πρώτῳ θεωρήματι τοῦ δεκάτου βιβλίου, tually) leave behind some segments of the circle whose ὅτι δύο μεγεθῶν ἀνίσων ἐκκειμένων, ἐὰν ἀπὸ τοῦ μείζονος (sum) will be less than the excess by which circle EFGH ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου μεῖζον exceeds the area S. For we showed in the first theo- ἢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ γίγνηται, λειφθήσεταί τι μέγεθος, rem of the tenth book that if two unequal magnitudes ὃ ἔσται ἔλασσον τοῦ ἐκκειμένου ἐλάσσονος μεγέθους. are laid out, and if (a part) greater than a half is sub- λελείφθω οὖν, καὶ ἔστω τὰ ἐπὶ τῶν ΕΚ, ΚΖ, ΖΛ, ΛΗ, tracted from the greater, and (if from) the remainder (a ΗΜ, ΜΘ, ΘΝ, ΝΕ τμήματα τοῦ ΕΖΗΘ κύκλου ἐλάττονα part) greater than a half (is subtracted), and this hap- τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ ΕΖΗΘ κύκλος τοῦ Σ χωρίου. pens continually, then some magnitude will (eventually) λοιπὸν ἄρα τὸ ΕΚΖΛΗΜΘΝ πολύγωνον μεῖζόν ἐστι τοῦ Σ be left which will be less than the lesser laid out mag- χωρίου. ἐγγεγράφθω καὶ εἰς τὸν ΑΒΓΔ κύκλον τῷ ΕΚΖ- nitude [Prop. 10.1]. Therefore, let the (segments) have ΛΗΜΘΝ πολυγώνῳ ὅμοιον πολύγωνον τὸ ΑΞΒΟΓΠΔΡ· been left, and let the (sum of the) segments of the circle ἔστιν ἄρα ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς EFGH on EK, KF , FL, LG, GM , MH , HN , and NE ΖΘ τετράγωνον, οὕτως τὸ ΑΞΒΟΓΠΔΡ πολύγωνον πρὸς be less than the excess by which circle EFGH exceeds τὸ ΕΚΖΛΗΜΘΝ πολύγωνον. ἀλλὰ καὶ ὡς τὸ ἀπὸ τῆς ΒΔ area S. Thus, the remaining polygon EKFLGMHN is τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος greater than area S. And let the polygon AOBPCQDR, πρὸς τὸ Σ χωρίον· καὶ ὡς ἄρα ὁ ΑΒΓΔ κύκλος πρὸς τὸ Σ similar to the polygon EKFLGMHN , have been in- χωρίον, οὕτως τὸ ΑΞΒΟΓΠΔΡ πολύγωνον πρὸς τὸ ΕΚΖ- scribed in circle ABCD. Thus, as the square on BD is ΛΗΜΘΝ πολύγωνον· ἐναλλὰξ ἄρα ὡς ὁ ΑΒΓΔ κύκλος to the square on FH , so polygon AOBPCQDR (is) to πρὸς τὸ ἐν αὐτῷ πολύγωνον, οὕτως τὸ Σ χωρίον πρὸς polygon EKFLGMHN [Prop. 12.1]. But, also, as the τὸ ΕΚΖΛΗΜΘΝ πολύγωνον. μείζων δὲ ὁ ΑΒΓΔ κύκλος square on BD (is) to the square on FH , so circle ABCD τοῦ ἐν αὐτῷ πολυγώνου· μεῖζον ἄρα καὶ τὸ Σ χωρίον τοῦ (is) to area S. And, thus, as circle ABCD (is) to area S, ΕΚΖΛΗΜΘΝ πολυγώνου. ἀλλὰ καὶ ἔλαττον· ὅπερ ἐστὶν so polygon AOBPGQDR (is) to polygon EKFLGMHN ἀδύνατον. οὐκ ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον [Prop. 5.11]. Thus, alternately, as circle ABCD (is) to πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς ἔλασσόν the polygon (inscribed) within it, so area S (is) to poly- τι τοῦ ΕΖΗΘ κύκλου χωρίον. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ gon EKFLGMHN [Prop. 5.16]. And circle ABCD (is) ὡς τὸ ἀπὸ ΖΘ πρὸς τὸ ἀπὸ ΒΔ, οὕτως ὁ ΕΖΗΘ κύκλος greater than the polygon (inscribed) within it. Thus, area πρὸς ἔλασσόν τι τοῦ ΑΒΓΔ κύκλου χωρίον. S is also greater than polygon EKFLGMHN . But, (it is) Λέγω δή, ὅτι οὐδὲ ὡς τὸ ἀπὸ τῆς ΒΔ πρὸς τὸ ἀπὸ also less. The very thing is impossible. Thus, the square τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς μεῖζόν τι τοῦ ΕΖΗΘ on BD is not to the (square) on FH , as circle ABCD (is) κύκλου χωρίον. to some area less than circle EFGH . So, similarly, we can Εἰ γὰρ δυνατόν, ἔστω πρὸς μεῖζον τὸ Σ. ἀνάπαλιν ἄρα show that the (square) on FH (is) not to the (square) on [ἐστὶν] ὡς τὸ ἀπὸ τῆς ΖΘ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΔΒ, BD as circle EFGH (is) to some area less than circle οὕτως τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον. ἀλλ᾿ ὡς τὸ ABCD either. Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος So, I say that neither (is) the (square) on BD to πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου χωρίον· καὶ ὡς ἄρα τὸ the (square) on FH , as circle ABCD (is) to some area ἀπὸ τῆς ΖΘ πρὸς τὸ ἀπὸ τῆς ΒΔ, οὕτως ὁ ΕΖΗΘ κύκλος greater than circle EFGH . πρὸς ἔλασσόν τι τοῦ ΑΒΓΔ κύκλου χωρίον· ὅπερ ἀδύνατον For, if possible, let it be (in that ratio) to (some) ἐδείχθη. οὐκ ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς greater (area), S. Thus, inversely, as the square on FH τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς μεῖζόν τι τοῦ [is] to the (square) on DB, so area S (is) to circle ABCD ΕΖΗΘ κύκλου χωρίον. ἐδείχθη δέ, ὅτι οὐδὲ πρὸς ἔλασσον· [Prop. 5.7 corr.]. But, as area S (is) to circle ABCD, so ἔστιν ἄρα ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς circle EFGH (is) to some area less than circle ABCD ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον. (see lemma). And, thus, as the (square) on FH (is) to Οἱ ἄρα κύκλοι πρὸς ἀλλήλους εἰσὶν ὡς τὰ ἀπὸ τῶν the (square) on BD, so circle EFGH (is) to some area διαμέτρων τετράγωνα· ὅπερ ἔδει δεῖξαι. less than circle ABCD [Prop. 5.11]. The very thing was shown (to be) impossible. Thus, as the square on BD is to the (square) on FH , so circle ABCD (is) not to some area greater than circle EFGH . And it was shown that neither (is it in that ratio) to (some) lesser (area). Thus, as the square on BD is to the (square) on FH , so circle ABCD (is) to circle EFGH . Thus, circles are to one another as the squares on 474 STOIQEIWN ibþ. ELEMENTS BOOK 12 (their) diameters. (Which is) the very thing it was re- quired to show.L¨mma. Lemma Λέγω δή, ὅτι τοῦ Σ χωρίου μείζονος ὄντος τοῦ ΕΖΗΘ So, I say that, area S being greater than circle EFGH , κύκλου ἐστὶν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, as area S is to circle ABCD, so circle EFGH (is) to some οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου area less than circle ABCD. χωρίον. For let it have been contrived that as area S (is) to Γεγονέτω γὰρ ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, circle ABCD, so circle EFGH (is) to area T . I say that οὕτως ὁ ΕΖΗΘ κύκλος πρὸς τὸ Τ χωρίον. λέγω, ὅτι area T is less than circle ABCD. For since as area S is ἔλαττόν ἐστι τὸ Τ χωρίον τοῦ ΑΒΓΔ κύκλου. ἐπεὶ γάρ to circle ABCD, so circle EFGH (is) to area T , alter- ἐστιν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ nately, as area S is to circle EFGH , so circle ABCD (is) ΕΖΗΘ κύκλος πρὸς τὸ Τ χωρίον, ἐναλλάξ ἐστιν ὡς τὸ Σ to area T [Prop. 5.16]. And area S (is) greater than circle χωρίον πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΒΓΔ κύκλος EFGH . Thus, circle ABCD (is) also greater than area πρὸς τὸ Τ χωρίον. μεῖζον δὲ τὸ Σ χωρίον τοῦ ΕΖΗΘ T [Prop. 5.14]. Hence, as area S is to circle ABCD, so κύκλου· μείζων ἄρα καὶ ὁ ΑΒΓΔ κύκλος τοῦ Τ χωρίου. circle EFGH (is) to some area less than circle ABCD. ὥστε ἐστὶν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, (Which is) the very thing it was required to show. οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου χωρίον· ὅπερ ἔδει δεῖξαι. gþ. Proposition 3 Πᾶσα πυραμὶς τρίγωνον ἔχουσα βάσιν διαιρεῖται εἰς δύο Any pyramid having a triangular base is divided into πυραμίδας ἴσας τε καὶ ὁμοίας ἀλλήλαις καὶ [ὁμοίας] τῇ ὅλῇ two pyramids having triangular bases (which are) equal, τριγώνους ἐχουσας βάσεις καὶ εἰς δύο πρίσματα ἴσα· καὶ τὰ similar to one another, and [similar] to the whole, and δύο πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος. into two equal prisms. And the (sum of the) two prisms is greater than half of the whole pyramid.D BZ GH L E J A K B L FG H C EA K D ῎Εστω πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΒΓ τρίγωνον, κο- Let there be a pyramid whose base is triangle ABC, ρυφὴ δὲ τὸ Δ σημεῖον· λέγω, ὅτι ἡ ΑΒΓΔ πυραμὶς διαιρεῖται and (whose) apex (is) point D. I say that pyramid εἰς δύο πυραμίδας ἴσας ἀλλήλαις τριγώνους βάσεις ἐχούσας ABCD is divided into two pyramids having triangular καὶ ὁμοίας τῇ ὅλῇ καὶ εἰς δύο πρίσματα ἴσα· καὶ τὰ δύο bases (which are) equal to one another, and similar to πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος. the whole, and into two equal prisms. And the (sum of Τετμήσθωσαν γὰρ αἱ ΑΒ, ΒΓ, ΓΑ, ΑΔ, ΔΒ, ΔΓ δίχα the) two prisms is greater than half of the whole pyramid. κατὰ τὰ Ε, Ζ, Η, Θ, Κ, Λ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ For let AB, BC, CA, AD, DB, and DC have been ΘΕ, ΕΗ, ΗΘ, ΘΚ, ΚΛ, ΛΘ, ΚΖ, ΖΗ. ἐπεὶ ἴση ἐστὶν ἡ μὲν cut in half at points E, F , G, H , K, and L (respectively). ΑΕ τῇ ΕΒ, ἡ δὲ ΑΘ τῇ ΔΘ, παράλληλος ἄρα ἐστὶν ἡ ΕΘ And let HE, EG, GH , HK, KL, LH , KF , and FG have τῇ ΔΒ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΘΚ τῇ ΑΒ παράλληλός ἐστιν. been joined. Since AE is equal to EB, and AH to DH , 475 STOIQEIWN ibþ. ELEMENTS BOOK 12 παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΕΒΚ· ἴση ἄρα ἐστὶν ἡ ΘΚ EH is thus parallel to DB [Prop. 6.2]. So, for the same τῇ ΕΒ. ἀλλὰ ἡ ΕΒ τῇ ΕΑ ἐστιν ἴση· καὶ ἡ ΑΕ ἄρα τῇ ΘΚ (reasons), HK is also parallel to AB. Thus, HEBK is ἐστιν ἴση. ἔστι δὲ καὶ ἡ ΑΘ τῇ ΘΔ ἴση· δύο δὴ αἱ ΕΑ, ΑΘ a parallelogram. Thus, HK is equal to EB [Prop. 1.34]. δυσὶ ταῖς ΚΘ, ΘΔ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνία ἡ But, EB is equal to EA. Thus, AE is also equal to HK. ὑπὸ ΕΑΘ γωνίᾳ τῇ ὑπὸ ΚΘΔ ἴση· βάσις ἄρα ἡ ΕΘ βάσει τῇ And AH is also equal to HD. So the two (straight-lines) ΚΔ ἐστιν ἴση. ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΑΕΘ τρίγωνον EA and AH are equal to the two (straight-lines) KH τῷ ΘΚΔ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΘΗ τρίγωνον and HD, respectively. And angle EAH (is) equal to an- τῷ ΘΛΔ τριγώνῳ ἴσον τέ ἐστι καὶ ὅμοιον. καὶ ἐπεὶ δύο gle KHD [Prop. 1.29]. Thus, base EH is equal to base εὐθεῖαι ἁπτόμεναι ἀλλήλων αἱ ΕΘ, ΘΗ παρὰ δύο εὐθείας KD [Prop. 1.4]. Thus, triangle AEH is equal and simi- ἁπτομένας ἀλλήλων τὰς ΚΔ, ΔΛ εἰσιν οὐκ ἐν τῷ αὐτῷ lar to triangle HKD [Prop. 1.4]. So, for the same (rea- ἐπιπέδῳ οὖσαι, ἴσας γωνίας περιέξουσιν. ἴση ἄρα ἐστὶν ἡ ὑπὸ sons), triangle AHG is also equal and similar to trian- ΕΘΗ γωνία τῇ ὑπὸ ΚΔΛ γωνίᾳ. καὶ ἐπεὶ δύο εὐθεῖαι αἱ ΕΘ, gle HLD. And since EH and HG are two straight-lines ΘΗ δυσὶ ταῖς ΚΔ, ΔΛ ἴσαι εἰσὶν ἑκατέρα εκατέρᾳ, καὶ γωνία joining one another (which are respectively) parallel to ἡ ὑπὸ ΕΘΗ γωνίᾳ τῇ ὑπὸ ΚΔΛ ἐστιν ἴση, βάσις ἄρα ἡ ΕΗ two straight-lines joining one another, KD and DL, not βάσει τῇ ΚΛ [ἐστιν] ἴση· ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΕΘΗ being in the same plane, they will contain equal angles τρίγωνον τῷ ΚΔΛ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΕΗ [Prop. 11.10]. Thus, angle EHG is equal to angle KDL. τρίγωνον τῷ ΘΚΛ τριγώνῳ ἴσον τε καὶ ὅμοιόν ἐστιν. ἡ ἄρα And since the two straight-lines EH and HG are equal πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ to the two straight-lines KD and DL, respectively, and Θ σημεῖον, ἴση καὶ ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ angle EHG is equal to angle KDL, base EG [is] thus ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον. καὶ ἐπεὶ τριγώνου equal to base KL [Prop. 1.4]. Thus, triangle EHG is τοῦ ΑΔΒ παρὰ μίαν τῶν πλευρῶν τὴν ΑΒ ἦκται ἡ ΘΚ, equal and similar to triangle KDL. So, for the same (rea- ἰσογώνιόν ἐστι τὸ ΑΔΒ τρίγωνον τῷ ΔΘΚ τριγώνῳ, καὶ sons), triangle AEG is also equal and similar to triangle τὰς πλευρὰς ἀνάλογον ἔχουσιν· ὅμοιον ἄρα ἐστὶ τὸ ΑΔΒ HKL. Thus, the pyramid whose base is triangle AEG, τρίγωνον τῷ ΔΘΚ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ μὲν and apex the point H , is equal and similar to the pyra- ΔΒΓ τρίγωνον τῷ ΔΚΛ τριγώνῳ ὅμοιόν ἐστιν, τὸ δὲ ΑΔΓ mid whose base is triangle HKL, and apex the point D τῷ ΔΛΘ. καὶ ἐπεὶ δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων αἱ ΒΑ, [Def. 11.10]. And since HK has been drawn parallel to ΑΓ παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων τὰς ΚΘ, ΘΛ εἰσιν one of the sides, AB, of triangle ADB, triangle ADB οὐκ ἐν τῷ αὐτῷ ἐπιπέδῳ, ἴσας γωνίας περιέξουσιν. ἴση ἄρα is equiangular to triangle DHK [Prop. 1.29], and they ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΚΘΛ. καί ἐστιν ὡς ἡ ΒΑ have proportional sides. Thus, triangle ADB is similar to πρὸς τὴν ΑΓ, οὕτως ἡ ΚΘ πρὸς τὴν ΘΛ· ὅμοιον ἄρα ἐστὶ triangle DHK [Def. 6.1]. So, for the same (reasons), tri- τὸ ΑΒΓ τρίγωνον τῷ ΘΚΛ τριγώνῳ. καὶ πυραμὶς ἄρα, ἧς angle DBC is also similar to triangle DKL, and ADC to βάσις μέν ἐστι τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, DLH . And since two straight-lines joining one another, ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΘΚΛ τρίγωνον, BA and AC, are parallel to two straight-lines joining one κορυφὴ δὲ τὸ Δ σημεῖον. ἀλλὰ πυραμίς, ἧς βάσις μέν [ἐστι] another, KH and HL, not in the same plane, they will τὸ ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ὁμοία ἐδείχθη contain equal angles [Prop. 11.10]. Thus, angle BAC is πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ equal to (angle) KHL. And as BA is to AC, so KH (is) τὸ Θ σημεῖον. ἑκατέρα ἄρα τῶν ΑΕΗΘ, ΘΚΛΔ πυραμίδων to HL. Thus, triangle ABC is similar to triangle HKL ὁμοία ἐστὶ τῇ ὅλῃ τῇ ΑΒΓΔ πυραμίδι. [Prop. 6.6]. And, thus, the pyramid whose base is trian- Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΖ τῇ ΖΓ, διπλάσιόν ἐστι τὸ gle ABC, and apex the point D, is similar to the pyra- ΕΒΖΗ παραλληλόγραμμον τοῦ ΗΖΓ τριγώνου. καὶ ἐπεὶ, mid whose base is triangle HKL, and apex the point D ἐὰν ᾖ δύο πρίσματα ἰσοϋψῆ, καὶ τὸ μὲν ἔχῃ βάσιν παραλ- [Def. 11.9]. But, the pyramid whose base [is] triangle ληλόγραμμον, τὸ δὲ τρίγωνον, διπλάσιον δὲ ᾖ τὸ παραλ- HKL, and apex the point D, was shown (to be) similar ληλόγραμμον τοῦ τριγώνου, ἴσα ἐστὶ τὰ πρίσματα, ἴσον to the pyramid whose base is triangle AEG, and apex the ἄρα ἐστὶ τὸ πρίσμα τὸ περιεχόμενον ὑπὸ δύο μὲν τριγώνων point H . Thus, each of the pyramids AEGH and HKLD τῶν ΒΚΖ, ΕΘΗ, τριῶν δὲ παραλληλογράμμων τῶν ΕΒΖΗ, is similar to the whole pyramid ABCD. ΕΒΚΘ, ΘΚΖΗ τῷ πρισματι τῷ περιεχομένῳ ὑπὸ δύο μὲν And since BF is equal to FC, parallelogram EBFG τριγώνων τῶν ΗΖΓ, ΘΚΛ, τριῶν δὲ παραλληλογράμμων is double triangle GFC [Prop. 1.41]. And since, if two τῶν ΚΖΓΛ, ΛΓΗΘ, ΘΚΖΗ. καὶ φανερόν, ὅτι ἑκάτρον τῶν prisms (have) equal heights, and the former has a par- πρισμάτων, οὗ τε βάσις τὸ ΕΒΖΗ παραλληλόγραμμον, ἀπε- allelogram as a base, and the latter a triangle, and the ναντίον δὲ ἡ ΘΚ εὐθεῖα, καὶ οὗ βάσις τὸ ΗΖΓ τρίγωνον, parallelogram (is) double the triangle, then the prisms ἀπεναντίον δὲ τὸ ΘΚΛ τρίγωνον, μεῖζόν ἐστιν ἑκατέρας are equal [Prop. 11.39], the prism contained by the two 476 STOIQEIWN ibþ. ELEMENTS BOOK 12 τῶν πυραμίδων, ὧν βάσεις μὲν τὰ ΑΕΗ, ΘΚΛ τρίγωνα, κο- triangles BKF and EHG, and the three parallelograms ρυφαὶ, δὲ τὰ Θ, Δ σημεῖα, ἐπειδήπερ [καί] ἐὰν ἐπιζεύξωμεν EBFG, EBKH , and HKFG, is thus equal to the prism τὰς ΕΖ, ΕΚ εὐθείας, τὸ μὲν πρίσμα, οὗ βάσις τὸ ΕΒΖΗ πα- contained by the two triangles GFC and HKL, and ραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΘΚ εὐθεῖα, μεῖζόν ἐστι the three parallelograms KFCL, LCGH , and HKFG. τῆς πυραμίδος, ἧς βάσις τὸ ΕΒΖ τρίγωνον, κορυφὴ δὲ τὸ And (it is) clear that each of the prisms whose base (is) Κ σημεῖον. ἀλλ᾿ ἡ πυραμίς, ἧς βάσις τὸ ΕΒΖ τρίγωνον, parallelogram EBFG, and opposite (side) straight-line κορυφὴ δὲ τὸ Κ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις τὸ HK, and whose base (is) triangle GFC, and opposite ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον· ὑπὸ γὰρ ἴσων (plane) triangle HKL, is greater than each of the pyra- καὶ ὁμοίων ἐπιπέδων περιέχονται. ὥστε καὶ τὸ πρίσμα, οὗ mids whose bases are triangles AEG and HKL, and βάσις μὲν τὸ ΕΒΖΗ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ apexes the points H and D (respectively), inasmuch as, ΘΚ εὐθεῖα, μεῖζόν ἐστι πυραμίδος, ἧς βάσις μὲν τὸ ΑΕΗ if we [also] join the straight-lines EF and EK then the τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον. ἴσον δὲ τὸ μὲν πρίσμα, prism whose base (is) parallelogram EBFG, and oppo- οὗ βάσις τὸ ΕΒΖΗ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΘΚ site (side) straight-line HK, is greater than the pyramid εὐθεῖα, τῷ πρίσματι, οὗ βάσις μὲν τὸ ΗΖΓ τρίγωνον, ἀπε- whose base (is) triangle EBF , and apex the point K. But ναντίον δὲ τὸ ΘΚΛ τρίγωνον· ἡ δὲ πυραμίς, ἧς βάσις τὸ the pyramid whose base (is) triangle EBF , and apex the ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον, ἴση ἐστὶ πυραμίδι, point K, is equal to the pyramid whose base is triangle ἧς βάσις τὸ ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον. τὰ AEG, and apex point H . For they are contained by equal ἄρα εἰρημένα δύο πρίσματα μείζονά ἐστι τῶν εἰρημένων δύο and similar planes. And, hence, the prism whose base πυραμίδων, ὧν βάσεις μὲν τὰ ΑΕΗ, ΘΚΛ τρίγωνα, κορυφαὶ (is) parallelogram EBFG, and opposite (side) straight- δὲ τὰ Θ, Δ σημεῖα. line HK, is greater than the pyramid whose base (is) ῾Η ἄρα ὅλη πυραμίς, ἧς βάσις τὸ ΑΒΓ τρίγωνον, κο- triangle AEG, and apex the point H . And the prism ρυφὴ δὲ τὸ Δ σημεῖον, διῄρηται εἴς τε δύο πυραμίδας ἴσας whose base is parallelogram EBFG, and opposite (side) ἀλλήλαις [καὶ ὁμοίας τῇ ὅλῃ] καὶ εἰς δύο πρίσματα ἴσα, καὶ τὰ straight-line HK, (is) equal to the prism whose base (is) δύο πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος· triangle GFC, and opposite (plane) triangle HKL. And ὅπερ ἔδει δεῖξαι. the pyramid whose base (is) triangle AEG, and apex the point H , is equal to the pyramid whose base (is) trian- gle HKL, and apex the point D. Thus, the (sum of the) aforementioned two prisms is greater than the (sum of the) aforementioned two pyramids, whose bases (are) tri- angles AEG and HKL, and apexes the points H and D (respectively). Thus, the whole pyramid, whose base (is) triangle ABC, and apex the point D, has been divided into two pyramids (which are) equal to one another [and similar to the whole], and into two equal prisms. And the (sum of the) two prisms is greater than half of the whole pyra- mid. (Which is) the very thing it was required to show.dþ. Proposition 4 ᾿Εὰν ὦσι δύο πυραμίδες ὑπὸ τὸ αὐτὸ ὕψος τριγώνους If there are two pyramids with the same height, hav- ἔχουσαι βάσεις, διαιρεθῇ δὲ ἑκατέρα αὐτῶν εἴς τε δύο ing trianglular bases, and each of them is divided into two πυραμίδας ἴσας ἀλλήλαις καὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο pyramids equal to one another, and similar to the whole, πρίσματα ἴσα, ἔσται ὡς ἡ τῆς μιᾶς πυραμίδος βάσις πρὸς τὴν and into two equal prisms then as the base of one pyra- τῆς ἑτέρας πυραμίδος βάσιν, οὕτως τὰ ἐν τῇ μιᾷ πυραμίδι mid (is) to the base of the other pyramid, so (the sum of) πρίσματα πάντα πρὸς τὰ ἐν τῇ ἑτάρᾳ πυραμίδι πρίσματα all the prisms in one pyramid will be to (the sum of all) πάντα ἰσοπληθῇ. the equal number of prisms in the other pyramid. ῎Εστωσαν δύο πυραμίδες ὑπὸ τὸ αὐτὸ ὕψος τριγώνους Let there be two pyramids with the same height, hav- ἔχουσαι βάσεις τὰς ΑΒΓ, ΔΕΖ, κορυφὰς δὲ τὰ Η, Θ σημεῖα, ing the triangular bases ABC and DEF , (with) apexes καὶ διῃρήσθω ἑκατέρα αὐτῶν εἴς τε δύο πυραμίδας ἴσας the points G and H (respectively). And let each of them ἀλλήλαις καὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο πρίσματα ἴσα· λέγω, have been divided into two pyramids equal to one an- 477 STOIQEIWN ibþ. ELEMENTS BOOK 12 ὅτι ἐστὶν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ other, and similar to the whole, and into two equal prisms ἐν τῇ ΑΒΓΗ πυραμίδι πρίσματα πάντα πρὸς τὰ ἐν τῇ ΔΕΖΘ [Prop. 12.3]. I say that as base ABC is to base DEF , πυραμίδι πρίσματα ἰσοπληθῆ. so (the sum of) all the prisms in pyramid ABCG (is) to (the sum of) all the equal number of prisms in pyramid DEFH . N A G M L J Z E TO X H R FD S P U BK K A M L E T R D S U B C F G H O P Q V N ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ μὲν ΒΞ τῇ ΞΓ, ἡ δὲ ΑΛ τῇ ΛΓ, For since BO is equal to OC, and AL to LC, LO is παράλληλος ἄρα ἐστὶν ἡ ΛΞ τῇ ΑΒ καὶ ὅμοιον τὸ ΑΒΓ thus parallel to AB, and triangle ABC similar to triangle τρίγωνον τῷ ΛΞΓ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΔΕΖ LOC [Prop. 12.3]. So, for the same (reasons), triangle τρίγωνον τῷ ΡΦΖ τριγώνῳ ὅμοιόν ἐστιν. καὶ ἐπεὶ δι- DEF is also similar to triangle RV F . And since BC is πλασίων ἐστὶν ἡ μὲν ΒΓ τῆς ΓΞ, ἡ δὲ ΕΖ τῆς ΖΦ, ἔστιν double CO, and EF (double) FV , thus as BC (is) to ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΞ, οὕτως ἡ ΕΖ πρὸς τὴν ΖΦ. καὶ CO, so EF (is) to FV . And the similar, and similarly ἀναγέγραπται ἀπὸ μὲν τῶν ΒΓ, ΓΞ ὅμοιά τε καὶ ὁμοίως laid out, rectilinear (figures) ABC and LOC have been κείμενα εὐθύγραμμα τὰ ΑΒΓ, ΛΞΓ, ἀπὸ δὲ τῶν ΕΖ, ΖΦ described on BC and CO (respectively), and the sim- ὅμοιά τε καὶ ὁμοίως κείμενα [εὐθύγραμμα] τὰ ΔΕΖ, ΡΦΖ· ilar, and similarly laid out, [rectilinear] (figures) DEF ἔστιν ἄρα ὡς τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΛΞΓ τρίγωνον, and RV F on EF and FV (respectively). Thus, as tri- οὕτως τὸ ΔΕΖ τρίγωνον πρὸς τὸ ΡΦΖ τρίγωνον· ἐναλλὰξ angle ABC is to triangle LOC, so triangle DEF (is) to ἄρα ἐστὶν ὡς τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ [τρίγωνον], triangle RV F [Prop. 6.22]. Thus, alternately, as trian- οὕτως τὸ ΛΞΓ [τρίγωνον] πρὸς τὸ ΡΦΖ τρίγωνον. ἀλλ᾿ gle ABC is to [triangle] DEF , so [triangle] LOC (is) ὡς τὸ ΛΞΓ τρίγωνον πρὸς τὸ ΡΦΖ τρίγωνον, οὕτως τὸ to triangle RV F [Prop. 5.16]. But, as triangle LOC πρίσμα, οὗ βάσις μέν [ἐστι] τὸ ΛΞΓ τρίγωνον, ἀπεναντίον δὲ (is) to triangle RV F , so the prism whose base [is] trian- τὸ ΟΜΝ, πρὸς τὸ πρίσμα, οὗ βάσις μὲν τὸ ΡΦΖ τρίγωνον, gle LOC, and opposite (plane) PMN , (is) to the prism ἀπεναντίον δὲ τὸ ΣΤΥ· καὶ ὡς ἄρα τὸ ΑΒΓ τρίγωνον πρὸς whose base (is) triangle RV F , and opposite (plane) STU τὸ ΔΕΖ τρίγωνον, οὕτως τὸ πρίσμα, οὗ βάσις μὲν τὸ ΛΞΓ (see lemma). And, thus, as triangle ABC (is) to trian- τρίγωνον, ἀπεναντίον δὲ τὸ ΟΜΝ, πρὸς τὸ πρίσμα, οὗ βάσις gle DEF , so the prism whose base (is) triangle LOC, μὲν τὸ ΡΦΖ τρίγωνον, ἀπεναντίον δὲ τὸ ΣΤΥ. ὡς δὲ τὰ and opposite (plane) PMN , (is) to the prism whose base εἰρημένα πρίσματα πρὸς ἄλληλα, οὕτως τὸ πρίσμα, οὗ βάσις (is) triangle RV F , and opposite (plane) STU . And as μὲν τὸ ΚΒΞΛ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΟΜ the aforementioned prisms (are) to one another, so the εὐθεῖα, πρὸς τὸ πρίσμα, οὗ βάσις μὲν τὸ ΠΕΦΡ παραλ- prism whose base (is) parallelogram KBOL, and oppo- ληλόγραμμον, ἀπεναντίον δὲ ἡ ΣΤ εὐθεῖα. καὶ τὰ δύο ἄρα site (side) straight-line PM , (is) to the prism whose base πρίσματα, οὗ τε βάσις μὲν τὸ ΚΒΞΛ παραλληλόγραμμον, (is) parallelogram QEV R, and opposite (side) straight- ἀπεναντίον δὲ ἡ ΟΜ, καὶ οὗ βάσις μὲν τὸ ΛΞΓ, ἀπεναντίον line ST [Props. 11.39, 12.3]. Thus, also, (is) the (sum δὲ τὸ ΟΜΝ, πρὸς τὰ πρίσματα, οὗ τε βάσις μὲν τὸ ΠΕΦΡ, of the) two prisms—that whose base (is) parallelogram ἀπεναντίον δὲ ἡ ΣΤ εὐθεῖα, καὶ οὗ βάσις μὲν τὸ ΡΦΖ KBOL, and opposite (side) PM , and that whose base τρίγωνον, ἀπεναντίον δὲ τὸ ΣΤΥ. καὶ ὡς ἄρα ἡ ΑΒΓ βάσις (is) LOC, and opposite (plane) PMN—to (the sum of) πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ εἰρημένα δύο πρίσματα πρὸς the (two) prisms—that whose base (is) QEV R, and op- τὰ εἰρημένα δύο πρίσματα. posite (side) straight-line ST , and that whose base (is) Καὶ ὁμοίως, ἐὰν διαιρεθῶσιν αἱ ΟΜΝΗ, ΣΤΥΘ πυ- triangle RV F , and opposite (plane) STU [Prop. 5.12]. ραμίδες εἴς τε δύο πρίσματα καὶ δύο πυραμίδας, ἔσται ὡς ἡ And, thus, as base ABC (is) to base DEF , so the (sum 478 STOIQEIWN ibþ. ELEMENTS BOOK 12 ΟΜΝ βάσις πρὸς τὴν ΣΤΥ βάσιν, οὕτως τὰ ἐν τῇ ΟΜΝΗ of the first) aforementioned two prisms (is) to the (sum πυραμίδι δύο πρίσματα πρὸς τὰ ἐν τῇ ΣΤΥΘ πυραμίδι δύο of the second) aforementioned two prisms. πρίσματα. ἀλλ᾿ ὡς ἡ ΟΜΝ βάσις πρὸς τὴν ΣΤΥ βάσιν, And, similarly, if pyramids PMNG and STUH are di- οὕτως ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν· ἴσον γὰρ ἑκάτερον vided into two prisms, and two pyramids, as base PMN τῶν ΟΜΝ, ΣΤΥ τριγώνων ἑκατέρῳ τῶν ΛΞΓ, ΡΦΖ. καὶ ὡς (is) to base STU , so (the sum of) the two prisms in pyra- ἄρα ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ τέσσαρα mid PMNG will be to (the sum of) the two prisms in πρίσματα πρὸς τὰ τέσσαρα πρίσματα. ὁμοίως δὲ κἂν τὰς pyramid STUH . But, as base PMN (is) to base STU , so ὑπολειπομένας πυραμίδας διέλωμεν εἴς τε δύο πυραμίδας base ABC (is) to base DEF . For the triangles PMN καὶ εἰς δύο πρίσματα, ἔσται ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ and STU (are) equal to LOC and RV F , respectively. βάσιν, οὕτως τὰ ἐν τῇ ΑΒΓΗ πυραμίδι πρίσματα πάντα πρὸς And, thus, as base ABC (is) to base DEF , so (the sum τὰ ἐν τῇ ΔΕΖΘ πυραμίδι πρίσματα πάντα ἰσοπληθῆ· ὅπερ of) the four prisms (is) to (the sum of) the four prisms ἔδει δεῖξαι. [Prop. 5.12]. So, similarly, even if we divide the pyra- mids left behind into two pyramids and into two prisms, as base ABC (is) to base DEF , so (the sum of) all the prisms in pyramid ABCG will be to (the sum of) all the equal number of prisms in pyramid DEFH . (Which is) the very thing it was required to show.L¨mma. Lemma ῞Οτι δέ ἐστιν ὡς τὸ ΛΞΓ τρίγωνον πρὸς τὸ ΡΦΖ And one may show, as follows, that as triangle LOC τρίγωνον, οὕτως τὸ πρίσμα, οὗ βάσις τὸ ΛΞΓ τρίγωνον, is to triangle RV F , so the prism whose base (is) trian- ἀπεναντίον δὲ τὸ ΟΜΝ, πρὸς τὸ πρίσμα, οὗ βάσις μὲν τὸ gle LOC, and opposite (plane) PMN , (is) to the prism ΡΦΖ [τρίγωνον], ἀπεναντίον δὲ τὸ ΣΤΥ, οὕτω δεικτέον. whose base (is) [triangle] RV F , and opposite (plane) ᾿Επὶ γὰρ τῆς αὐτῆς καταγραφῆς νενοήσθωσαν ἀπὸ τῶν STU . Η, Θ κάθετοι ἐπὶ τὰ ΑΒΓ, ΔΕΖ ἐπίπεδα, ἴσαι δηλαδὴ For, in the same figure, let perpendiculars have been τυγχάνουσαι διὰ τὸ ἰσοϋψεῖς ὑποκεῖσθαι τὰς πυραμίδας. conceived (drawn) from (points) G and H to the planes καὶ ἐπεὶ δύο εὐθεῖαι ἥ τε ΗΓ καὶ ἡ ἀπὸ τοῦ Η κάθετος ABC and DEF (respectively). These clearly turn out to ὑπὸ παραλλήλων ἐπιπέδων τῶν ΑΒΓ, ΟΜΝ τέμνονται, εἰς be equal, on account of the pyramids being assumed (to τοὺς αὐτοὺς λόγους τμηθήσονται. καὶ τέτμηται ἡ ΗΓ δίχα be) of equal height. And since two straight-lines, GC ὑπὸ τοῦ ΟΜΝ ἐπιπέδου κατὰ τὸ Ν· καὶ ἡ ἀπὸ τοῦ Η ἄρα and the perpendicular from G, are cut by the parallel κάθετος ἐπὶ τὸ ΑΒΓ ἐπίπεδον δίχα τμηθήσεται ὑπὸ τοῦ planes ABC and PMN they will be cut in the same ra- ΟΜΝ ἐπιπέδου. διὰ τὰ αὐτὰ δὴ καὶ ἡ ἀπὸ τοῦ Θ κάθετος ἐπὶ tios [Prop. 11.17]. And GC was cut in half by the plane τὸ ΔΕΖ ἐπίπεδον δίχα τμηθήσεται ὑπὸ τοῦ ΣΤΥ ἐπιπέδου. PMN at N . Thus, the perpendicular from G to the plane καί εἰσιν ἴσαι αἱ ἀπὸ τῶν Η, Θ κάθετοι ἐπὶ τὰ ΑΒΓ, ΔΕΖ ABC will also be cut in half by the plane PMN . So, ἐπίπεδα· ἴσαι ἄρα καὶ αἱ ἀπὸ τῶν ΟΜΝ, ΣΤΥ τριγώνων for the same (reasons), the perpendicular from H to the ἐπὶ τὰ ΑΒΓ, ΔΕΖ κάθετοι. ἰσοϋψῆ ἄρα [ἐστὶ] τὰ πρίσματα, plane DEF will also be cut in half by the plane STU . And ὧν βάσεις μέν εἰσι τὰ ΛΞΓ, ΡΦΖ τρίγωνα, ἀπεναντίον δὲ the perpendiculars from G and H to the planes ABC and τὰ ΟΜΝ, ΣΤΥ. ὥστε καὶ τὰ στερεὰ παραλληλεπίπεδα τὰ DEF (respectively) are equal. Thus, the perpendiculars ἀπὸ τῶν εἰρημένων πρισμάτων ἀναγραφόμενα ἰσοϋψῆ καὶ from the triangles PMN and STU to ABC and DEF πρὸς ἄλληλά [εἰσιν] ὡς αἱ βάσεις· καὶ τὰ ἡμίση ἄρα ἐστὶν (respectively, are) also equal. Thus, the prisms whose ὡς ἡ ΛΞΓ βάσις πρὸς τὴν ΡΦΖ βάσιν, οὕτως τὰ εἰρημένα bases are triangles LOC and RV F , and opposite (sides) πρίσματα πρὸς ἄλληλα· ὅπερ ἔδει δεῖξαι. PMN and STU (respectively), [are] of equal height. And, hence, the parallelepiped solids described on the aforementioned prisms [are] of equal height and (are) to one another as their bases [Prop. 11.32]. Likewise, the halves (of the solids) [Prop. 11.28]. Thus, as base LOC is to base RV F , so the aforementioned prisms (are) to one another. (Which is) the very thing it was required to show. 479 STOIQEIWN ibþ. ELEMENTS BOOK 12eþ. Proposition 5 Αἱ ὑπὸ τὸ αὐτὸ ὕψος οὖσαι πυραμίδες καὶ τριγώνους Pyramids which are of the same height, and have tri- ἔχουσαι βάσεις πρὸς ἀλλήλας εἰσὶν ὡς αἱ βάσεις. angular bases, are to one another as their bases. Q A M L E T R FD S U B N G ZP O H J XK W A M L E T R D S Q U B C F G H O P V N K ῎Εστωσαν ὑπὸ τὸ αὐτὸ ὕψος πυραμίδες, ὧν βάσεις μὲν Let there be pyramids of the same height whose bases τὰ ΑΒΓ, ΔΕΖ τρίγωνα, κορυφαὶ δὲ τὰ Η, Θ σημεῖα· λέγω, (are) the triangles ABC and DEF , and apexes the points ὅτι ἐστὶν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως ἡ G and H (respectively). I say that as base ABC is to base ΑΒΓΗ πυραμὶς πρὸς τὴν ΔΕΖΘ πυραμίδα. DEF , so pyramid ABCG (is) to pyramid DEFH . Εἰ γὰρ μή ἐστιν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, For if base ABC is not to base DEF , as pyramid οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς τὴν ΔΕΖΘ πυραμίδα, ἔσται ὡς ABCG (is) to pyramid DEFH , then base ABC will be ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς to base DEF , as pyramid ABCG (is) to some solid ei- ἤτοι πρὸς ἔλασσόν τι τῆς ΔΕΖΘ πυραμίδος στερεὸν ἢ πρὸς ther less than, or greater than, pyramid DEFH . Let it, μεῖζον. ἔστω πρότερον πρὸς ἔλασσον τὸ Χ, καὶ διῃρὴσθω first of all, be (in this ratio) to (some) lesser (solid), W . ἡ ΔΕΖΘ πυραμὶς εἴς τε δύο πυραμίδας ἴσας ἀλλήλαις καὶ And let pyramid DEFH have been divided into two pyra- ὁμοίας τῇ ὅλῃ καὶ εἰς δύο πρίσματα ἴσα· τὰ δὴ δύο πρίσαμτα mids equal to one another, and similar to the whole, and μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος. καὶ πάλιν αἱ ἐκ into two equal prisms. So, the (sum of the) two prisms τῆς διαιρέσεως γινόμεναι πυραμίδες ὁμοίως διῃρήσθωσαν, is greater than half of the whole pyramid [Prop. 12.3]. καὶ τοῦτο ἀεὶ γινέσθω, ἕως οὗ λειφθῶσί τινες πυραμίδες ἀπὸ And, again, let the pyramids generated by the division τῆς ΔΕΖΘ πυραμίδος, αἵ εἰσιν ἐλάττονες τῆς ὑπεροχῆς, ᾗ have been similarly divided, and let this be done contin- ὑπερέχει ἡ ΔΕΖΘ πυραμὶς τοῦ Χ στερεοῦ. λελείφθωσαν καὶ ually until some pyramids are left from pyramid DEFH ἔστωσαν λόγου ἕνεκεν αἱ ΔΠΡΣ, ΣΤΥΘ· λοιπὰ ἄρα τὰ ἐν which (when added together) are less than the excess by τῇ ΔΕΖΘ πυραμίδι πρίσματα μείζονά ἐστι τοῦ Χ στερεοῦ. which pyramid DEFH exceeds the solid W [Prop. 10.1]. διῃρήσθω καὶ ἡ ΑΒΓΗ πυραμὶς ὁμοίως καὶ ἰσοπληθῶς τῇ Let them have been left, and, for the sake of argument, ΔΕΖΘ πυραμίδι· ἔστιν ἄρα ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ let them be DQRS and STUH . Thus, the (sum of the) βάσιν, οὕτως τὰ ἐν τῇ ΑΒΓΗ πυραμίδι πρίσματα πρὸς τὰ ἐν remaining prisms within pyramid DEFH is greater than τῇ ΔΕΖΘ πυραμίδι πρίσματα, ἀλλὰ καὶ ὡς ἡ ΑΒΓ βάσις πρὸς solid W . Let pyramid ABCG also have been divided sim- τὴν ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς τὸ Χ στερεόν· ilarly, and a similar number of times, as pyramid DEFH . καὶ ὡς ἄρα ἡ ΑΒΓΗ πυραμὶς πρὸς τὸ Χ στερεόν, οὕτως Thus, as base ABC is to base DEF , so the (sum of the) τὰ ἐν τῇ ΑΒΓΗ πυραμίδι πρίσματα πρὸς τὰ ἐν τῇ ΔΕΖΘ prisms within pyramid ABCG (is) to the (sum of the) πυραμίδι πρίσματα· ἐναλλὰξ ἄρα ὡς ἡ ΑΒΓΗ πυραμὶς πρὸς prisms within pyramid DEFH [Prop. 12.4]. But, also, τὰ ἐν αὐτῇ πρίσματα, οὕτως τὸ Χ στερεὸν πρὸς τὰ ἐν τῇ as base ABC (is) to base DEF , so pyramid ABCG (is) ΔΕΖΘ πυραμίδι πρίσματα. μείζων δὲ ἡ ΑΒΓΗ πυραμὶς τῶν to solid W . And, thus, as pyramid ABCG (is) to solid ἐν αὐτῇ πρισμάτων· μεῖζον ἄρα καὶ τὸ Χ στερεὸν τῶν ἐν τῇ W , so the (sum of the) prisms within pyramid ABCG ΔΕΖΘ πυραμίδι πρισμάτων. ἀλλὰ καὶ ἔλαττον· ὅπερ ἐστὶν (is) to the (sum of the) prisms within pyramid DEFH ἀδύνατον. οὐκ ἄρα ἐστὶν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ [Prop. 5.11]. Thus, alternately, as pyramid ABCG (is) to βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς ἔλασσόν τι τῆς ΔΕΖΘ the (sum of the) prisms within it, so solid W (is) to the πυραμίδος στερεόν. ὁμοίως δὴ δειχθήσεται, ὅτι οὐδὲ ὡς ἡ (sum of the) prisms within pyramid DEFH [Prop. 5.16]. ΔΕΖ βάσις πρὸς τὴν ΑΒΓ βάσιν, οὕτως ἡ ΔΕΖΘ πυραμὶς And pyramid ABCG (is) greater than the (sum of the) πρὸς ἔλαττόν τι τῆς ΑΒΓΗ πυραμίδος στερεόν. prisms within it. Thus, solid W (is) also greater than the Λέγω δή, ὅτι οὐκ ἔστιν οὐδὲ ὡς ἡ ΑΒΓ βάσις πρὸς τὴν (sum of the) prisms within pyramid DEFH [Prop. 5.14]. ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς μεῖζόν τι τῆς But, (it is) also less. This very thing is impossible. Thus, ΔΕΖΘ πυραμίδος στερεόν. as base ABC is to base DEF , so pyramid ABCG (is) 480 STOIQEIWN ibþ. ELEMENTS BOOK 12 Εἰ γὰρ δυνατόν, ἔστω πρὸς μεῖζον τὸ Χ· ἀνάπαλιν ἄρα not to some solid less than pyramid DEFH . So, simi- ἐστὶν ὡς ἡ ΔΕΖ βάσις πρὸς τὴν ΑΒΓ βάσιν, οὕτως τὸ Χ larly, we can show that base DEF is not to base ABC, στερεὸν πρὸς τὴν ΑΒΓΗ πυραμίδα. ὡς δὲ τὸ Χ στερεὸν as pyramid DEFH (is) to some solid less than pyramid πρὸς τὴν ΑΒΓΗ πυραμίδα, οὕτως ἡ ΔΕΖΘ πυραμὶς πρὸς ABCG either. ἔλασσόν τι τῆς ΑΒΓΗ πυραμίδος, ὡς ἔμπροσθεν ἐδείχθη· So, I say that neither is base ABC to base DEF , as καὶ ὡς ἄρα ἡ ΔΕΖ βάσις πρὸς τὴν ΑΒΓ βάσιν, οὕτως ἡ pyramid ABCG (is) to some solid greater than pyramid ΔΕΖΘ πυραμὶς πρὸς ἔλασσόν τι τῆς ΑΒΓΗ πυραμίδος· DEFH . ὅπερ ἄτοπον ἐδείχθη. οὐκ ἄρα ἐστὶν ὡς ἡ ΑΒΓ βάσις πρὸς For, if possible, let it be (in this ratio) to some τὴν ΔΕΖ βάσιν, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς μεῖζόν τι greater (solid), W . Thus, inversely, as base DEF τῆς ΔΕΖΘ πυραμίδος στερεόν. ἐδείχθη δέ, ὅτι οὐδὲ πρὸς (is) to base ABC, so solid W (is) to pyramid ABCG ἔλασσον. ἔστιν ἄρα ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, [Prop. 5.7. corr.]. And as solid W (is) to pyramid ABCG, οὕτως ἡ ΑΒΓΗ πυραμὶς πρὸς τὴν ΔΕΖΘ πυραμίδα· ὅπερ so pyramid DEFH (is) to some (solid) less than pyramid ἔδει δεῖξαι. ABCG, as shown before [Prop. 12.2 lem.]. And, thus, as base DEF (is) to base ABC, so pyramid DEFH (is) to some (solid) less than pyramid ABCG [Prop. 5.11]. The very thing was shown (to be) absurd. Thus, base ABC is not to base DEF , as pyramid ABCG (is) to some solid greater than pyramid DEFH . And, it was shown that neither (is it in this ratio) to a lesser (solid). Thus, as base ABC is to base DEF , so pyramid ABCG (is) to pyramid DEFH . (Which is) the very thing it was required to show.�þ. Proposition 6 Αἱ ὐπὸ τὸ αὐτὸ ὕψος οὖσαι πυραμίδες καὶ πολυγώνους Pyramids which are of the same height, and have ἔχουσαι βάσεις πρὸς ἀλλήλας εἰσὶν ὡς αἱ βάσεις. polygonal bases, are to one another as their bases. J M D B K N A LE G H Z L M D B K N A E H G F C ῎Εστωσαν ὑπὸ τὸ αὐτὸ ὕψος πυραμίδες, ὧν [αἱ] βάσεις Let there be pyramids of the same height whose bases μὲν τὰ ΑΒΓΔΕ, ΖΗΘΚΛ πολύγωνα, κορυφαὶ δὲ τὰ Μ, (are) the polygons ABCDE and FGHKL, and apexes Ν σημεῖα· λέγω, ὅτι ἐστὶν ὡς ἡ ΑΒΓΔΕ βάσις πρὸς τὴν the points M and N (respectively). I say that as base ΖΗΘΚΛ βάσιν, οὕτως ἡ ΑΒΓΔΕΜ πυραμὶς πρὸς τὴν ΖΗΘ- ABCDE is to base FGHKL, so pyramid ABCDEM (is) ΚΛΝ πυραμίδα. to pyramid FGHKLN . ᾿Επεζεύχθωσαν γὰρ αἱ ΑΓ, ΑΔ, ΖΘ, ΖΚ. ἐπεὶ οὖν For let AC, AD, FH , and FK have been joined. δύο πυραμίδες εἰσὶν αἱ ΑΒΓΜ, ΑΓΔΜ τριγώνους ἔχου- Therefore, since ABCM and ACDM are two pyramids σαι βάσεις καὶ ὕψος ἴσον, πρὸς ἀλλήλας εἰσὶν ὡς αἱ βάσεις· having triangular bases and equal height, they are to ἔστιν ἄρα ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΑΓΔ βάσιν, οὕτως ἡ one another as their bases [Prop. 12.5]. Thus, as base ΑΒΓΜ πυραμὶς πρὸς τὴν ΑΓΔΜ πυραμίδα. καὶ συνθέντι ABC is to base ACD, so pyramid ABCM (is) to pyra- ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΑΓΔ βάσιν, οὕτως ἡ ΑΒΓΔΜ mid ACDM . And, via composition, as base ABCD 481 STOIQEIWN ibþ. ELEMENTS BOOK 12 πυραμὶς πρὸς τὴν ΑΓΔΜ πυραμίδα. ἀλλὰ καὶ ὡς ἡ ΑΓΔ (is) to base ACD, so pyramid ABCDM (is) to pyra- βάσις πρὸς τὴν ΑΔΕ βάσιν, οὕτως ἡ ΑΓΔΜ πυραμὶς πρὸς mid ACDM [Prop. 5.18]. But, as base ACD (is) to τὴν ΑΔΕΜ πυραμίδα. δι᾿ ἴσου ἄρα ὡς ἡ ΑΒΓΔ βάσις πρὸς base ADE, so pyramid ACDM (is) also to pyramid τὴν ΑΔΕ βάσιν, οὕτως ἡ ΑΒΓΔΜ πυραμὶς πρὸς τὴν ΑΔΕΜ ADEM [Prop. 12.5]. Thus, via equality, as base ABCD πυραμίδα. καὶ συνθέντι πάλιν, ὡς ἡ ΑΒΓΔΕ βάσις πρὸς τὴν (is) to base ADE, so pyramid ABCDM (is) to pyramid ΑΔΕ βάσιν, οὕτως ἡ ΑΒΓΔΕΜ πυραμὶς πρὸς τὴν ΑΔΕΜ ADEM [Prop. 5.22]. And, again, via composition, as πυραμίδα. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ὡς ἡ ΖΗΘΚΛ base ABCDE (is) to base ADE, so pyramid ABCDEM βάσις πρὸς τὴν ΖΗΘ βάσιν, οὕτως καὶ ἡ ΖΗΘΚΛΝ πυραμὶς (is) to pyramid ADEM [Prop. 5.18]. So, similarly, it can πρὸς τὴν ΖΗΘΝ πυραμίδα. καὶ ἐπεὶ δύο πυραμίδες εἱσὶν αἱ also be shown that as base FGHKL (is) to base FGH , ΑΔΕΜ, ΖΗΘΝ τριγώνους ἔχουσαι βάσεις καὶ ὕψος ἴσον, so pyramid FGHKLN (is) also to pyramid FGHN . And ἔστιν ἄρα ὡς ἡ ΑΔΕ βάσις πρὸς τὴν ΖΗΘ βάσιν, οὕτως since ADEM and FGHN are two pyramids having tri- ἡ ΑΔΕΜ πυραμὶς πρὸς τὴν ΖΗΘΝ πυραμίδα. ἀλλ᾿ ὡς ἡ angular bases and equal height, thus as base ADE (is) to ΑΔΕ βάσις πρὸς τὴν ΑΒΓΔΕ βάσιν, οὕτως ἦν ἡ ΑΔΕΜ base FGH , so pyramid ADEM (is) to pyramid FGHN πυραμὶς πρὸς τὴν ΑΒΓΔΕΜ πυραμίδα. καὶ δι᾿ ἴσου ἄρα ὡς [Prop. 12.5]. But, as base ADE (is) to base ABCDE, so ἡ ΑΒΓΔΕ βάσις πρὸς τὴν ΖΗΘ βάσιν, οὕτως ἡ ΑΒΓΔΕΜ pyramid ADEM (was) to pyramid ABCDEM . Thus, via πυραμὶς πρὸς τὴν ΖΗΘΝ πυραμίδα. ἀλλὰ μὴν καὶ ὡς ἡ ΖΗΘ equality, as base ABCDE (is) to base FGH , so pyramid βάσις πρὸς τὴν ΖΗΘΚΛ βάσιν, οὕτως ἦν καὶ ἡ ΖΗΘΝ πυ- ABCDEM (is) also to pyramid FGHN [Prop. 5.22]. ραμὶς πρὸς τὴν ΖΗΘΚΛΝ πυραμίδα, καὶ δι᾿ ἴσου ἄρα ὡς ἡ But, furthermore, as base FGH (is) to base FGHKL, ΑΒΓΔΕ βάσις πρὸς τὴν ΖΗΘΚΛ βάσιν, οὕτως ἡ ΑΒΓΔΕΜ so pyramid FGHN was also to pyramid FGHKLN . πυραμὶς πρὸς τὴν ΖΗΘΚΛΝ πυραμίδα· ὅπερ ἔδει δεῖξαι. Thus, via equality, as base ABCDE (is) to base FGHKL, so pyramid ABCDEM (is) also to pyramid FGHKLN [Prop. 5.22]. (Which is) the very thing it was required to show.zþ. Proposition 7 Πᾶν πρίσμα τρίγωνον ἔχον βάσιν διαιρεῖται εἰς τρεῖς πυ- Any prism having a triangular base is divided into ραμίδας ἴσας ἀλλήλαις τριγώνους βάσεις ἐχούσας. three pyramids having triangular bases (which are) equal to one another. G B A E Z D C B A E D F ῎Εστω πρίσμα, οὗ βάσις μὲν τὸ ΑΒΓ τρίγωνον, ἀπε- Let there be a prism whose base (is) triangle ABC, ναντίον δὲ τὸ ΔΕΖ· λέγω, ὅτι τὸ ΑΒΓΔΕΖ πρίσμα διαιρεῖται and opposite (plane) DEF . I say that prism ABCDEF εἰς τρεῖς πυραμίδας ἴσας ἀλλήλαις τριγώνους ἐχούσας is divided into three pyramids having triangular bases βάσεις. (which are) equal to one another. ᾿Επεζεύχθωσαν γὰρ αἱ ΒΔ, ΕΓ, ΓΔ. ἐπεὶ παραλ- For let BD, EC, and CD have been joined. Since ληλόγραμμόν ἐστι τὸ ΑΒΕΔ, διάμετρος δὲ αὐτὸῦ ἐστιν ABED is a parallelogram, and BD is its diagonal, tri- ἡ ΒΔ, ἴσον ἄρα ἐστι τὸ ΑΒΔ τρίγωνον τῷ ΕΒΔ τρίγωνῳ· angle ABD is thus equal to triangle EBD [Prop. 1.34]. 482 STOIQEIWN ibþ. ELEMENTS BOOK 12 καὶ ἡ πυραμὶς ἄρα, ἧς βάσις μὲν τὸ ΑΒΔ τρίγωνον, κο- And, thus, the pyramid whose base (is) triangle ABD, ρυφὴ δὲ τὸ Γ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις μέν and apex the point C, is equal to the pyramid whose base ἐστι τὸ ΔΕΒ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον. ἀλλὰ is triangle DEB, and apex the point C [Prop. 12.5]. But, ἡ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΔΕΒ τρίγωνον, κορυφὴ the pyramid whose base is triangle DEB, and apex the δὲ τὸ Γ σημεῖον, ἡ αὐτή ἐστι πυραμίδι, ἧς βάσις μέν ἐστι point C, is the same as the pyramid whose base is trian- τὸ ΕΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον· ὑπὸ γὰρ τῶν gle EBC, and apex the point D. For they are contained αὐτῶν ἐπιπέδων περιέχεται. καὶ πυραμὶς ἄρα, ἧς βάσις μέν by the same planes. And, thus, the pyramid whose base ἐστι τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, ἴση ἐστὶ is ABD, and apex the point C, is equal to the pyramid πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΕΒΓ τρίγωνον, κορυφὴ whose base is EBC and apex the point D. Again, since δὲ τὸ Δ σημεῖον. πάλιν, ἐπεὶ παραλληλόγραμμόν ἐστι τὸ FCBE is a parallelogram, and CE is its diagonal, trian- ΖΓΒΕ, διάμετρος δέ ἐστιν αὐτοῦ ἡ ΓΕ, ἴσον ἐστὶ τὸ ΓΕΖ gle CEF is equal to triangle CBE [Prop. 1.34]. And, τρίγωνον τῷ ΓΒΕ τριγώνῳ. καὶ πυραμὶς ἄρα, ἧς βάσις thus, the pyramid whose base is triangle BCE, and apex μέν ἐστι τὸ ΒΓΕ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ἴση the point D, is equal to the pyramid whose base is trian- ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΕΓΖ τρίγωνον, κορυφὴ gle ECF , and apex the point D [Prop. 12.5]. And the δὲ τὸ Δ σημεῖον. ἡ δὲ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΒΓΕ pyramid whose base is triangle BCE, and apex the point τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον, ἴση ἐδείχθη πυραμίδι, ἧς D, was shown (to be) equal to the pyramid whose base is βάσις μέν ἐστι τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον· triangle ABD, and apex the point C. Thus, the pyramid καὶ πυραμὶς ἄρα, ἧς βάσις μέν ἐστι τὸ ΓΕΖ τρίγωνον, κο- whose base is triangle CEF , and apex the point D, is ρυφὴ δὲ τὸ Δ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις μέν also equal to the pyramid whose base [is] triangle ABD, [ἐστι] τὸ ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον· διῄρηται and apex the point C. Thus, the prism ABCDEF has ἄρα τὸ ΑΒΓΔΕΖ πρίσμα εἰς τρεῖς πυραμίδας ἴσας ἀλλήλαις been divided into three pyramids having triangular bases τριγώνους ἐχούσας βάσεις. (which are) equal to one another. Καὶ ἐπεὶ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΒΔ τρίγωνον, And since the pyramid whose base is triangle ABD, κορυφὴ δὲ τὸ Γ σημεῖον, ἡ αὐτή ἐστι πυραμίδι, ἧς βάσις and apex the point C, is the same as the pyramid whose τὸ ΓΑΒ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον· ὑπὸ γὰρ τῶν base is triangle CAB, and apex the point D. For they are αὐτῶν ἐπιπέδων περιέχονται· ἡ δὲ πυραμίς, ἧς βάσις τὸ contained by the same planes. And the pyramid whose ΑΒΔ τρίγωνον, κορυφὴ δὲ τὸ Γ σημεῖον, τρίτον ἐδείχθη base (is) triangle ABD, and apex the point C, was shown τοῦ πρίσματος, οὗ βάσις τὸ ΑΒΓ τρίγωνον, ἀπεναντίον δὲ (to be) a third of the prism whose base is triangle ABC, τὸ ΔΕΖ, καὶ ἡ πυραμὶς ἄρα, ἧς βάσις τὸ ΑΒΓ τρίγωνον, and opposite (plane) DEF , thus the pyramid whose base κορυφὴ δὲ τὸ Δ σημεῖον, τρίτον ἐστὶ τοῦ πρίσματος τοῦ is triangle ABC, and apex the point D, is also a third of ἔχοντος βάσις τὴν αὐτὴν τὸ ΑΒΓ τρίγωνον, ἀπεναντίον δὲ the pyramid having the same base, triangle ABC, and τὸ ΔΕΖ. opposite (plane) DEF .Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι πᾶσα πυραμὶς τρίτον μέρος And, from this, (it is) clear that any pyramid is the ἐστὶ τοῦ πρίσματος τοῦ τὴν αὐτὴν βάσιν ἔχοντος αὐτῇ καὶ third part of the prism having the same base as it, and an ὕψος ἴσον· ὅπερ ἔδει δεῖξαι. equal height. (Which is) the very thing it was required to show.hþ. Proposition 8 Αἱ ὅμοιαι πυραμίδες καὶ τριγώνους ἔχουσαι βάσεις ἐν Similar pyramids which also have triangular bases are τριπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. in the cubed ratio of their corresponding sides. ῎Εστωσαν ὅμοιαι καὶ ὁμοίως κείμεναι πυραμίδες, ὧν Let there be similar, and similarly laid out, pyramids βάσεις μέν εἰσι τὰ ΑΒΓ, ΔΕΖ τρίγωνα, κορυφαὶ δὲ τὰ Η, whose bases are triangles ABC and DEF , and apexes Θ σημεῖα· λέγω, ὅτι ἡ ΑΒΓΗ πυραμὶς πρὸς τὴν ΔΕΖΘ the points G and H (respectively). I say that pyramid πυραμίδα τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ABCG has to pyramid DEFH the cubed ratio of that BC (has) to EF . 483 STOIQEIWN ibþ. ELEMENTS BOOK 12 M Z K H N X OL B G A P E D J R M K N L B E R C O P F D Q H G A Συμπεπληρώσθω γὰρ τὰ ΒΗΜΛ, ΕΘΠΟ στερεὰ παραλ- For let the parallelepiped solids BGML and EHQP ληλεπίπεδα. καὶ ἐπεὶ ὁμοία ἐστὶν ἡ ΑΒΓΗ πυραμὶς τῇ ΔΕΖΘ have been completed. And since pyramid ABCG is simi- πυραμίδι, ἵση ἄρα ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ lar to pyramid DEFH , angle ABC is thus equal to angle γωνίᾳ, ἡ δὲ ὑπὸ ΗΒΓ τῇ ὑπὸ ΘΕΖ, ἡ δὲ ὑπὸ ΑΒΗ τῇ ὑπὸ DEF , and GBC to HEF , and ABG to DEH . And as AB ΔΕΘ, καί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΔΕ, οὕτως ἡ ΒΓ πρὸς is to DE, so BC (is) to EF , and BG to EH [Def. 11.9]. τὴν ΕΖ, καὶ ἡ ΒΗ πρὸς τὴν ΕΘ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ And since as AB is to DE, so BC (is) to EF , and (so) πρὸς τὴν ΔΕ, οὕτως ἡ ΒΓ πρὸς τὴν ΕΖ, καὶ περὶ ἴσας γωνίας the sides around equal angles are proportional, parallel- αἱ πλευραὶ ἀνάλογόν εἰσιν, ὅμοιον ἄρα ἐστὶ τὸ ΒΜ παραλ- ogram BM is thus similar to paralleleogram EQ. So, ληλόγραμμον τῷ ΕΠ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ καὶ for the same (reasons), BN is also similar to ER, and τὸ μὲν ΒΝ τῷ ΕΡ ὅμοιόν ἐστι, τὸ δὲ ΒΚ τῷ ΕΞ· τὰ τρία BK to EO. Thus, the three (parallelograms) MB, BK, ἄρα τὰ ΜΒ, ΒΚ, ΒΝ τρισὶ τοῖς ΕΠ, ΕΞ, ΕΡ ὅμοιά ἐστιν. and BN are similar to the three (parallelograms) EQ, ἀλλὰ τὰ μὲν τρία τὰ ΜΒ, ΒΚ, ΒΝ τρισὶ τοῖς ἀπεναντίον ἴσα EO, ER (respectively). But, the three (parallelograms) τε καὶ ὅμοιά ἐστιν, τὰ δὲ τρία τὰ ΕΠ, ΕΞ, ΕΡ τρισὶ τοῖς MB, BK, and BN are (both) equal and similar to the ἀπεναντίον ἴσα τε καὶ ὅμοιά ἐστιν. τὰ ΒΗΜΛ, ΕΘΠΟ ἄρα three opposite (parallelograms), and the three (parallel- στερεὰ ὑπὸ ὁμοίων ἐπιπέδων ἴσων τὸ πλῆθος περιέχεται. ograms) EQ, EO, and ER are (both) equal and simi- ὅμοιον ἄρα ἐστὶ τὸ ΒΗΜΛ στερεὸν τῷ ΕΘΠΟ στερεῷ. τὰ lar to the three opposite (parallelograms) [Prop. 11.24]. δὲ ὅμοια στερεὰ παραλληλεπίπεδα ἐν τριπλασίονι λόγῳ ἐστὶ Thus, the solids BGML and EHQP are contained by τῶν ὁμολόγων πλευρῶν. τὸ ΒΗΜΛ ἄρα στερεὸν πρὸς τὸ equal numbers of similar (and similarly laid out) planes. ΕΘΠΟ στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος Thus, solid BGML is similar to solid EHQP [Def. 11.9]. πλευρὰ ἡ ΒΓ πρὸς τὴν ὁμόλογον πλευρὰν τὴν ΕΖ. ὡς δὲ τὸ And similar parallelepiped solids are in the cubed ratio of ΒΗΜΛ στερεὸν πρὸς τὸ ΕΘΠΟ στερεόν, οὕτως ἡ ΑΒΓΗ corresponding sides [Prop. 11.33]. Thus, solid BGML πυραμὶς πρὸς τὴν ΔΕΖΘ πυραμίδα, ἐπειδήπερ ἡ πυραμὶς has to solid EHQP the cubed ratio that the correspond- ἕκτον μέρος ἐστὶ τοῦ στερεοῦ διὰ τὸ καὶ τὸ πρίσμα ἥμισυ ing side BC (has) to the corresponding side EF . And as ὂν τοῦ στερεοῦ παραλληλεπιπέδου τριπλάσιον εἶναι τῆς πυ- solid BGML (is) to solid EHQP , so pyramid ABCG (is) ραμίδος. καὶ ἡ ΑΒΓΗ ἄρα πυραμὶς πρὸς τὴν ΔΕΖΘ πυ- to pyramid DEFH , inasmuch as the pyramid is the sixth ραμίδα τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ· part of the solid, on account of the prism, being half of the ὅπερ ἔδει δεῖξαι. parallelepiped solid [Prop. 11.28], also being three times the pyramid [Prop. 12.7]. Thus, pyramid ABCG also has to pyramid DEFH the cubed ratio that BC (has) to EF . (Which is) the very thing it was required to show.Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι καὶ αἱ πολυγώνους ἔχου- So, from this, (it is) also clear that similar pyra- σαι βάσεις ὅμοιαι πυραμίδες πρὸς ἀλλήλας ἐν τριπλασίονι mids having polygonal bases (are) to one another as the λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. διαιρεθεισῶν γὰρ αὐτῶν cubed ratio of their corresponding sides. For, dividing εἰς τὰς ἐν αὐταῖς πυραμίδας τριγώνους βάσεις ἐχούσας τῷ them into the pyramids (contained) within them which καὶ τὰ ὅμοια πολύγωνα τῶν βάσεων εἰς ὅμοια τρίγωνα have triangular bases, with the similar polygons of the διαιρεῖσθαι καὶ ἴσα τῷ πλήθει καὶ ὁμόλογα τοῖς ὅλοις ἔσται bases also being divided into similar triangles (which are) 484 STOIQEIWN ibþ. ELEMENTS BOOK 12 ὡς [ἡ] ἐν τῇ ἑτέρᾳ μία πυραμὶς τρίγωνον ἔχουσα βάσιν both equal in number, and corresponding, to the wholes πρὸς τὴν ἐν τῇ ἑτέρᾳ μίαν πυραμίδα τρίγωνον ἔχουσαν [Prop. 6.20]. As one pyramid having a triangular base in βάσιν, οὕτως καὶ ἅπασαι αἱ ἐν τῇ ἑτέρᾳ πυραμίδι πυραμίδες the former (pyramid having a polygonal base is) to one τριγώνους ἔχουσαι βάσεις πρὸς τὰς ἐν τῇ ἑτέρᾳ πυραμίδι pyramid having a triangular base in the latter (pyramid πυραμίδας τριγώνους βάσεις ἐχούσας, τουτέστιν αὐτὴ ἡ having a polygonal base), so (the sum of) all the pyra- πολύγωνον βάσιν ἔχουσα πυραμὶς πρὸς τὴν πολύγωνον mids having triangular bases in the former pyramid will βάσιν ἔχουσαν πυραμίδα. ἡ δὲ τρίγωνον βάσιν ἔχουσα πυ- also be to (the sum of) all the pyramids having triangu- ραμὶς πρὸς τὴν τρίγωνον βάσιν ἔχουσαν ἐν τριπλασίονι λόγῳ lar bases in the latter pyramid [Prop. 5.12]—that is to ἐστὶ τῶν ὁμολόγον πλευρῶν· καὶ ἡ πολύγωνον ἄρα βάσιν say, the (former) pyramid itself having a polygonal base ἔχουσα πρὸς τὴν ὁμοίαν βάσιν ἔχουσαν τριπλασίονα λόγον to the (latter) pyramid having a polygonal base. And a ἔχει ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν. pyramid having a triangular base is to a (pyramid) hav- ing a triangular base in the cubed ratio of corresponding sides [Prop. 12.8]. Thus, a (pyramid) having a polygonal base also has to to a (pyramid) having a similar base the cubed ratio of a (corresponding) side to a (correspond- ing) side.jþ. Proposition 9 Τῶν ἴσων πυραμίδων καὶ τριγώνους βάσεις ἐχουσῶν The bases of equal pyramids which also have trian- ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν· καὶ ὧν πυραμίδων gular bases are reciprocally proportional to their heights. τριγώνους βάσεις ἐχουσῶν ἀντιπεπόνθασιν αἱ βάσεις τοῖς And those pyramids which have triangular bases whose ὕψεσιν, ἴσαι εἰσὶν ἐκεῖναι. bases are reciprocally proportional to their heights are equal. LXR O P B G J E DZ H MA L R B E D M A H O P QF C G ῎Εστωσαν γὰρ ἴσαι πυραμίδες τριγώνους βάσεις ἔχουσαι For let there be (two) equal pyramids having the tri- τὰς ΑΒΓ, ΔΕΖ, κορυφὰς δὲ τὰ Η, Θ σημεῖα· λέγω, ὅτι τῶν angular bases ABC and DEF , and apexes the points G ΑΒΓΗ, ΔΕΖΘ πυραμίδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς and H (respectively). I say that the bases of the pyramids ὕψεσιν, καί ἐστιν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, ABCG and DEFH are reciprocally proportional to their οὕτως τὸ τῆς ΔΕΖΘ πυραμίδος ὕψος πρὸς τὸ τῆς ΑΒΓΗ heights, and (so) that as base ABC is to base DEF , so πυραμίδος ὕψος. the height of pyramid DEFH (is) to the height of pyra- Συμπεπληρώσθω γὰρ τὰ ΒΗΜΛ, ΕΘΠΟ στερεὰ παραλ- mid ABCG. ληλεπίπεδα. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒΓΗ πυραμὶς τῇ ΔΕΖΘ For let the parallelepiped solids BGML and EHQP πυραμίδι, καί ἐστι τῆς μὲν ΑΒΓΗ πυραμίδος ἑξαπλάσιον have been completed. And since pyramid ABCG is τὸ ΒΗΜΛ στερεόν, τῆς δὲ ΔΕΖΘ πυραμίδος ἑξαπλάσιον equal to pyramid DEFH , and solid BGML is six times τὸ ΕΘΠΟ στερεόν, ἴσον ἄρα ἐστὶ τὸ ΒΗΜΛ στερεὸν τῷ pyramid ABCG (see previous proposition), and solid ΕΘΠΟ στερεῷ. τῶν δὲ ἴσων στερεῶν παραλληλεπιπώδων EHQP (is) six times pyramid DEFH , solid BGML is 485 STOIQEIWN ibþ. ELEMENTS BOOK 12 ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν· ἔστιν ἄρα ὡς ἡ ΒΜ thus equal to solid EHQP . And the bases of equal par- βάσις πρὸς τὴν ΕΠ βάσιν, οὕτως τὸ τοῦ ΕΘΠΟ στερεοῦ allelepiped solids are reciprocally proportional to their ὕψος πρὸς τὸ τοῦ ΒΗΜΛ στερεοῦ ὕψος. ἀλλ᾿ ὡς ἡ ΒΜ heights [Prop. 11.34]. Thus, as base BM is to base EQ, βάσις πρὸς τὴν ΕΠ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ so the height of solid EHQP (is) to the height of solid τρίγωνον. καὶ ὡς ἄρα τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ BGML. But, as base BM (is) to base EQ, so triangle τρίγωνον, οὕτως τὸ τοῦ ΕΘΠΟ στερεοῦ ὕψος πρὸς τὸ τοῦ ABC (is) to triangle DEF [Prop. 1.34]. And, thus, as ΒΗΜΛ στερεοῦ ὕψος. ἀλλὰ τὸ μὲν τοῦ ΕΘΠΟ στερεοῦ triangle ABC (is) to triangle DEF , so the height of solid ὕψος τὸ αὐτὸ ἐστι τῷ τῆς ΔΕΖΘ πυραμίδος ὕψει, τὸ δὲ EHQP (is) to the height of solid BGML [Prop. 5.11]. τοῦ ΒΗΜΛ στερεοῦ ὕψος τὸ αὐτό ἐστι τῷ τῆς ΑΒΓΗ πυ- But, the height of solid EHQP is the same as the height ραμίδος ὕψει· ἔστιν ἄρα ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ of pyramid DEFH , and the height of solid BGML is βάσιν, οὕτως τὸ τῆς ΔΕΖΘ πυραμίδος ὕψος πρὸς τὸ τῆς the same as the height of pyramid ABCG. Thus, as base ΑΒΓΗ πυραμίδος ὕψος. τῶν ΑΒΓΗ, ΔΕΖΘ ἄρα πυραμίδων ABC is to base DEF , so the height of pyramid DEFH ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν. (is) to the height of pyramid ABCG. Thus, the bases Ἀλλὰ δὴ τῶν ΑΒΓΗ, ΔΕΖΘ πυραμίδων ἀντιπεπονθέτ- of pyramids ABCG and DEFH are reciprocally propor- ωσαν αἱ βάσεις τοῖς ὕψεσιν, καὶ ἔστω ὡς ἡ ΑΒΓ βάσις πρὸς tional to their heights. τὴν ΔΕΖ βάσιν, οὕτως τὸ τῆς ΔΕΖΘ πυραμίδος ὕψος πρὸς And so, let the bases of pyramids ABCG and DEFH τὸ τῆς ΑΒΓΗ πυραμίδος ὕψος· λέγω, ὅτι ἴση ἐστὶν ἡ ΑΒΓΗ be reciprocally proportional to their heights, and (thus) πυραμὶς τῇ ΔΕΖΘ πυραμίδι. let base ABC be to base DEF , as the height of pyramid Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ DEFH (is) to the height of pyramid ABCG. I say that ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὸ τῆς ΔΕΖΘ πυ- pyramid ABCG is equal to pyramid DEFH . ραμίδος ὕψος πρὸς τὸ τῆς ΑΒΓΗ πυραμίδος ὕψος, ἀλλ᾿ ὡς For, with the same construction, since as base ABC ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὸ ΒΜ παραλ- is to base DEF , so the height of pyramid DEFH (is) to ληλόγραμμον πρὸς τὸ ΕΠ παραλληλόγραμμον, καὶ ὡς ἄρα the height of pyramid ABCG, but as base ABC (is) to τὸ ΒΜ παραλληλόγραμμον πρὸς τὸ ΕΠ παραλληλόγραμμον, base DEF , so parallelogram BM (is) to parallelogram οὕτως τὸ τῆς ΔΕΖΘ πυραμίδος ὕψος πρὸς τὸ τῆς ΑΒΓΗ EQ [Prop. 1.34], thus as parallelogram BM (is) to paral- πυραμίδος ὕψος. ἀλλὰ τὸ [μὲν] τῆς ΔΕΖΘ πυραμίδος ὕψος lelogram EQ, so the height of pyramid DEFH (is) also τὸ αὐτό ἐστι τῷ τοῦ ΕΘΠΟ παραλληλεπιπέδου ὕψει, τὸ δὲ to the height of pyramid ABCG [Prop. 5.11]. But, the τῆς ΑΒΓΗ πυραμίδος ὕψος τὸ αὐτό ἐστι τῷ τοῦ ΒΗΜΛ height of pyramid DEFH is the same as the height of παραλληλεπιπέδου ὕψει· ἔστιν ἄρα ὡς ἡ ΒΜ βάσις πρὸς parallelepiped EHQP , and the height of pyramid ABCG τὴν ΕΠ βάσιν, οὕτως τὸ τοῦ ΕΘΠΟ παραλληλεπιπέδου is the same as the height of parallelepiped BGML. Thus, ὕψος πρὸς τὸ τοῦ ΒΗΜΛ παραλληλεπιπέδου ὕψος. ὧν as base BM is to base EQ, so the height of parallelepiped δὲ στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις EHQP (is) to the height of parallelepiped BGML. And τοῖς ὕψεσιν, ἴσα ἐστὶν ἐκεῖνα· ἴσον ἄρα ἐστὶ τὸ ΒΗΜΛ those parallelepiped solids whose bases are reciprocally στερεὸν παραλληλεπίπεδον τῷ ΕΘΠΟ στερεῷ παραλληλε- proportional to their heights are equal [Prop. 11.34]. πιπέδῳ. καί ἐστι τοῦ μὲν ΒΗΜΛ ἕκτον μέρος ἡ ΑΒΓΗ Thus, the parallelepiped solid BGML is equal to the par- πυραμίς, τοῦ δὲ ΕΘΠΟ παραλληλεπιπέδου ἕκτον μέρος ἡ allelepiped solid EHQP . And pyramid ABCG is a sixth ΔΕΖΘ πυραμίς· ἴση ἄρα ἡ ΑΒΓΗ πυραμὶς τῇ ΔΕΖΘ πυ- part of BGML, and pyramid DEFH a sixth part of par- ραμίδι. allelepiped EHQP . Thus, pyramid ABCG is equal to Τῶν ἄρα ἴσων πυραμίδων καὶ τριγώνους βάσεις ἐχουσῶν pyramid DEFH . ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν· καὶ ὧν πυραμίδων Thus, the bases of equal pyramids which also have τριγώνους βάσεις ἐχουσῶν ἀντιπεπόνθασιν αἱ βάσεις τοῖς triangular bases are reciprocally proportional to their ὕψεσιν, ἴσαι εἰσὶν ἐκεῖναι· ὅπερ ἔδει δεῖξαι. heights. And those pyramids having triangular bases whose bases are reciprocally proportional to their heights are equal. (Which is) the very thing it was required to show.iþ. Proposition 10 Πᾶς κῶνος κυλίνδρου τρίτον μέρος ἐστὶ τοῦ τὴν αὐτὴν Every cone is the third part of the cylinder which has βάσιν ἔχοντος αὐτῷ καὶ ὕψος ἴσον. the same base as it, and an equal height. ᾿Εχέτω γὰρ κῶνος κυλίνδρῷ βάσιν τε τὴν αὐτὴν τὸν For let there be a cone (with) the same base as a cylin- 486 STOIQEIWN ibþ. ELEMENTS BOOK 12 ΑΒΓΔ κύκλον καὶ ὕψος ἴσον· λέγω, ὅτι ὁ κῶνος τοῦ der, (namely) the circle ABCD, and an equal height. I κυλίνδρου τρίτον ἐστὶ μέρος, τουτέστιν ὅτι ὁ κύλινδρος τοῦ say that the cone is the third part of the cylinder—that is κώνου τριπλασίων ἐστίν. to say, that the cylinder is three times the cone. J B Z G H D AE C B D A E F G H Εἰ γὰρ μή ἐστιν ὁ κύλινδρος τοῦ κώνου τριπλασίων, For if the cylinder is not three times the cone then the ἔσται ὁ κύλινδρος τοῦ κώνου ἤτοι μείζων ἢ τριπλασίων cylinder will be either more than three times, or less than ἢ ἐλάσσων ἢ τριπλασίων. ἔστω πρότερον μείζων ἢ τρι- three times, (the cone). Let it, first of all, be more than πλασίων, καὶ ἐγγεγράφθω εἰς τὸν ΑΒΓΔ κύκλον τετράγων- three times (the cone). And let the square ABCD have ον τὸ ΑΒΓΔ· τὸ δὴ ΑΒΓΔ τετράγωνον μείζόν ἐστιν ἢ τὸ been inscribed in circle ABCD [Prop. 4.6]. So, square ἥμισυ τοῦ ΑΒΓΔ κύκλου. καὶ ἀνεστάτω ἀπὸ τοῦ ΑΒΓΔ τε- ABCD is more than half of circle ABCD [Prop. 12.2]. τραγώνου πρίσμα ἰσοϋψὲς τῷ κυλίνδρῳ. τὸ δὴ ἀνιστάμενον And let a prism of equal height to the cylinder have been πρίσμα μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ κυλίνδου, ἐπειδήπερ set up on square ABCD. So, the prism set up is more κἂν περὶ τὸν ΑΒΓΔ κύκλον τετράγωνον περιγράψωμεν, than half of the cylinder, inasmuch as if we also circum- τὸ ἐγγεγραμμένον εἰς τὸν ΑΒΓΔ κύκλον τετράγωνον scribe a square around circle ABCD [Prop. 4.7] then the ἥμισύ ἐστι τοῦ περιγεγραμμένου· καί ἐστι τὰ ἀπ᾿ αὐτῶν square inscribed in circle ABCD is half of the circum- ἀνιστάμενα στερεὰ παραλληλεπίπεδα πρίσματα ἰσοϋψῆ· τὰ scribed (square). And the solids set up on them are par- δὲ ὑπὸ τὸ αὐτὸ ὕψος ὄντα στερεὰ παραλληλεπίπεδα πρὸς allelepiped prisms of equal height. And parallelepiped ἄλληλά ἐστιν ὡς αἱ βάσεις· καὶ τὸ ἐπὶ τοῦ ΑΒΓΔ ἄρα τε- solids having the same height are to one another as their τραγώνου ἀνασταθὲν πρίσμα ἥμισύ ἐστι τοῦ ἀνασταθέντος bases [Prop. 11.32]. And, thus, the prism set up on πρίσματος ἀπὸ τοῦ περὶ τὸν ΑΒΓΔ κύκλον περιγραφέντος square ABCD is half of the prism set up on the square τετραγώνου· καί ἐστιν ὁ κύλινδρος ἐλάττων τοῦ πρίσματος circumscribed about circle ABCD. And the cylinder is τοῦ ἀνατραθέντος ἀπὸ τοῦ περὶ τὸν ΑΒΓΔ κύκλον περι- less than the prism set up on the square circumscribed γραφέντος τετραγώνου· τὸ ἄρα πρίσμα τὸ ἀνασταθὲν ἀπὸ about circle ABCD. Thus, the prism set up on square τοῦ ΑΒΓΔ τετραγώνου ἰσοϋψὲς τῷ κυλίνδρῳ μεῖζόν ἐστι ABCD of the same height as the cylinder is more than τοῦ ἡμίσεως τοῦ κυλίνδρου. τετμήσθωσαν αἱ ΑΒ, ΒΓ, half of the cylinder. Let the circumferences AB, BC, ΓΔ, ΔΑ περιφέρειαι δίχα κατὰ τὰ Ε, Ζ, Η, Θ σημεῖα, καὶ CD, and DA have been cut in half at points E, F , G, ἐπεζεύχθωσαν αἱ ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, ΗΔ, ΔΘ, ΘΑ· and H . And let AE, EB, BF , FC, CG, GD, DH , and καὶ ἕκαστον ἄρα τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων HA have been joined. And thus each of the triangles μειζόν ἐστιν ἢ τὸ ἥμισυ τοῦ καθ᾿ ἑαυτὸ τηήματος τοῦ AEB, BFC, CGD, and DHA is more than half of the ΑΒΓΔ κύκλου, ὡς ἔμπροσθεν ἐδείκνυμεν. ἀνεστάτω ἐφ᾿ segment of circle ABCD about it, as was shown pre- ἑκάστου τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων πρίσματα viously [Prop. 12.2]. Let prisms of equal height to the ἰσοϋψῆ τῷ κυλίνδρῳ· καὶ ἕκαστον ἄρα τῶν ἀνασταθέντων cylinder have been set up on each of the triangles AEB, πρισμάτων μεῖζόν ἐστιν ἢ τὸ ἥμισυ μέρος τοῦ καθ᾿ ἑαυτὸ BFC, CGD, and DHA. And each of the prisms set up is τμήματος τοῦ κυλίνδρου, ἐπειδήπερ ἐὰν διὰ τῶν Ε, Ζ, Η, greater than the half part of the segment of the cylinder Θ σημείων παραλλήλους ταῖς ΑΒ, ΒΓ, ΓΔ, ΔΑ ἀγάγωμεν, about it—inasmuch as if we draw (straight-lines) parallel καὶ συμπληρώσωμεν τὰ ἐπὶ τῶν ΑΒ, ΒΓ, ΓΔ, ΔΑ παραλ- to AB, BC, CD, and DA through points E, F , G, and H 487 STOIQEIWN ibþ. ELEMENTS BOOK 12 ληλόγραμμα, καὶ ἀπ᾿ αὐτῶν ἀναστήσωμεν στερεὰ παραλλη- (respectively), and complete the parallelograms on AB, λεπίπεδα ἰσοϋψῆ τῷ κυλίνδρῳ, ἑκάσου τῶν ἀνασταθέντων BC, CD, and DA, and set up parallelepiped solids of ἡμίση ἐστὶ τὰ πρίσματα τὰ ἐπὶ τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ equal height to the cylinder on them, then the prisms on τριγώνων· καί ἐστι τὰ τοῦ κυλίνδρου τμήματα ἐλάττονα triangles AEB, BFC, CGD, and DHA are each half of τῶν ἀνασταθέντων στερεῶν παραλληλεπιπέδων· ὥστε καὶ the set up (parallelepipeds). And the segments of the τὰ ἐπὶ τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων πρίσματα cylinder are less than the set up parallelepiped solids. μείζονά ἐστιν ἢ τὸ ἥμισυ τῶν καθ᾿ ἑαυτὰ τοῦ κυλίνδρου Hence, the prisms on triangles AEB, BFC, CGD, and τμημάτων. τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας DHA are also greater than half of the segments of the δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ ἀνιστάντες ἐφ᾿ ἑκάσου cylinder about them. So (if) the remaining circumfer- τῶν τριγώνων πρίσματα ἰσοϋψῆ τῷ κυλίνδρῳ καὶ τοῦτο ἀεὶ ences are cut in half, and straight-lines are joined, and ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ κυλίνδρου, prisms of equal height to the cylinder are set up on each ἃ ἔσται ἐλάττονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ κυλίνδρος of the triangles, and this is done continually, then we will τοῦ τριπλασίου τοῦ κώνου. λελείφθω, καὶ ἔστω τὰ ΑΕ, (eventually) leave some segments of the cylinder whose ΕΒ, ΒΖ, ΖΓ, ΓΗ, ΗΔ, ΔΘ, ΘΑ· λοιπὸν ἄρα τὸ πρίσμα, οὗ (sum) is less than the excess by which the cylinder ex- βάσις μὲν τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, ὕψος δὲ τὸ αὐτὸ ceeds three times the cone [Prop. 10.1]. Let them have τῷ κυλίνδρῷ, μεῖζόν ἐστὶν ἢ τριπλάσιον τοῦ κώνου. ἀλλὰ been left, and let them be AE, EB, BF , FC, CG, GD, τὸ πρίσμα, οὗ βάσις μὲν ἐστὶ τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, DH , and HA. Thus, the remaining prism whose base ὕψος δὲ τὸ αὐτὸ τῷ κυλίνδρῳ, τριπλάσιόν ἐστι τῆς πυ- (is) polygon AEBFCGDH , and height the same as the ραμίδος, ἧς βάσις μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, cylinder, is greater than three times the cone. But, the κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ· καὶ ἡ πυραμὶς ἄρα, ἧς βάσις prism whose base is polygon AEBFCGDH , and height μέν [ἐστι] τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, κορυφὴ δὲ ἡ αὐτὴ the same as the cylinder, is three times the pyramid whose τῷ κώνῳ, μείζων ἐστὶ τοῦ κώνου τοῦ βάσιν ἔχοντες τὸν base is polygon AEBFCGDH , and apex the same as the ΑΒΓΔ κύκλον. ἀλλὰ καὶ ἐλάττων· ἐμπεριέχεται γὰρ ὑπ᾿ cone [Prop. 12.7 corr.]. And thus the pyramid whose αὐτοῦ· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐστὶν ὁ κύλινδρος base [is] polygon AEBFCGDH , and apex the same as τοῦ κώνου μεῖζων ἢ τριπλάσιος. the cone, is greater than the cone having (as) base circle Λέγω δή, ὅτι οὐδὲ ἐλάττων ἐστὶν ἢ τριπλάσιος ὁ ABCD. But (it is) also less. For it is encompassed by it. κύλινδρος τοῦ κώνου. The very thing (is) impossible. Thus, the cylinder is not Εἰ γὰρ δυνατόν, ἔστω ἐλάττων ἢ τριπλάσιος ὁ κύλινδρος more than three times the cone. τοῦ κώνου· ἀνάπαλιν ἄρα ὁ κῶνος τοῦ κυλίνδρου μεῖζων So, I say that neither (is) the cylinder less than three ἐστὶν ἢ τρίτον μέρος. ἐγγεγράφθω δὴ εἰς τὸν ΑΒΓΔ κύκλον times the cone. τετράγωνον τὸ ΑΒΓΔ· τὸ ΑΒΓΔ ἄρα τετράγωνον μεῖζόν For, if possible, let the cylinder be less than three times ἐστιν ἢ τὸ ἥμισυ τοῦ ΑΒΓΔ κύκλου. καὶ ἀνεστάτω ἀπὸ τοῦ the cone. Thus, inversely, the cone is greater than the ΑΒΓΔ τετραγώνου πυραμὶς τὴν αὐτὴν κορυφὴν ἔχουσα τῷ third part of the cylinder. So, let the square ABCD have κώνῳ· ἠ ἄρα ἀνασταθεῖσα πυραμὶς μείζων ἐστὶν ἢ τὸ ἥμισυ been inscribed in circle ABCD [Prop. 4.6]. Thus, square μέρος τοῦ κώνου, ἐπειδήπερ, ὡς ἕμπροσθεν ἐδείκνυμεν, ABCD is greater than half of circle ABCD. And let a ὅτι ἐὰν περὶ τὸν κύκλον τετράγωνον περιγράψωμεν, ἔσται pyramid having the same apex as the cone have been set τὸ ΑΒΓΔ τετράγωνον ἥμισυ τοῦ περὶ τὸν κύκλον περι- up on square ABCD. Thus, the pyramid set up is greater γεγραμμένου τετραγώνου· καὶ ἐὰν ἀπὸ τῶν τετραγώνων than the half part of the cone, inasmuch as we showed στερεὰ παραλληλεπίπεδα ἀναστήσωμεν ἰσοϋψῆ τῷ κώνῳ, ἂ previously that if we circumscribe a square about the cir- καὶ καλεῖται πρίσματα, ἔσται τὸ ἀνασταθὲν ἀπὸ τοῦ ΑΒΓΔ cle [Prop. 4.7] then the square ABCD will be half of the τετραγώνου ἥμισυ τοῦ ἀνασταθέντος ἀπὸ τοῦ περὶ τὸν square circumscribed about the circle [Prop. 12.2]. And κύκλον περιγραφέντος τετραγώνου· πρὸς ἄλληλα γάρ εἰσιν if we set up on the squares parallelepiped solids—which ὡς αἱ βάσεις. ὥστε καὶ τὰ τρίτα· καὶ πυραμὶς ἄρα, ἧς are also called prisms—of the same height as the cone, βάσις τὸ ΑΒΓΔ τετράγωνον, ἥμισύ ἐστι τῆς πυραμίδος τῆς then the (prism) set up on square ABCD will be half ἀνασταθείσης ἀπὸ τοῦ περὶ τὸν κύκλον περιγραφέντος τε- of the (prism) set up on the square circumscribed about τραγώνου. καί ἐστι μείζων ἡ πυραμὶς ἡ ἀνασταθεῖσα ἀπὸ the circle. For they are to one another as their bases τοῦ περὶ τὸν κύκλον τετραγώνου τοῦ κώνου· ἐμπεριέχει [Prop. 11.32]. Hence, (the same) also (goes for) the γὰρ αὐτόν. ἡ ἄρα πυραμὶς, ἧς βάσις τὸ ΑΒΓΔ τετράγωνον, thirds. Thus, the pyramid whose base is square ABCD κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ, μείζων ἐστὶν ἢ τὸ ἥμισυ τοῦ is half of the pyramid set up on the square circumscribed κώνου. τετμήσθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΑ περιφέρειαι about the circle [Prop. 12.7 corr.]. And the pyramid set δίχα κατὰ τὰ Ε, Ζ, Η, Θ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ up on the square circumscribed about the circle is greater 488 STOIQEIWN ibþ. ELEMENTS BOOK 12 ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, ΗΔ, ΔΘ, ΘΑ· καὶ ἕκαστον ἄρα than the cone. For it encompasses it. Thus, the pyramid τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων μεῖζόν ἐστιν ἢ τὸ whose base is square ABCD, and apex the same as the ἥμισυ μέρος του καθ᾿ ἑαυτὸ τμήματος τοῦ ΑΒΓΔ κύκλου. cone, is greater than half of the cone. Let the circum- καὶ ἀνεστάτωσαν ἐφ᾿ ἑκάστου τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ ferences AB, BC, CD, and DA have been cut in half τριγώνων πυραμίδες τὴν αὐτὴν κορυφὴν ἔχουσαι τῷ κώνῳ· at points E, F , G, and H (respectively). And let AE, καὶ ἑκάστη ἄρα τῶν ἀνασταθεισῶν πυραμίδων κατὰ τὸν EB, BF , FC, CG, GD, DH , and HA have been joined. αὐτὸν τρόπον μείζων ἐστὶν ἢ τὸ ἥμισυ μέρος τοῦ καθ᾿ And, thus, each of the triangles AEB, BFC, CGD, and ἑαυτὴν τμήματος τοῦ κώνου. τέμνοντες δὴ τὰς ὑπολει- DHA is greater than the half part of the segment of cir- πομένας περιφερείας δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ cle ABCD about it [Prop. 12.2]. And let pyramids having ἀνιστάντες ἐφ᾿ ἑκάστου τῶν τριγώνων πυραμίδα τὴν αὐτὴν the same apex as the cone have been set up on each of the κορυφὴν ἔχουσαν τῷ κώνῳ καὶ τοῦτο ἀεὶ ποιοῦτες κα- triangles AEB, BFC, CGD, and DHA. And, thus, in the ταλείψομέν τινα ἀποτμήματα τοῦ κώνου, ἃ ἔσται ἐλάττονα same way, each of the pyramids set up is more than the τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ κῶνος τοῦ τρίτου μέρους τοῦ half part of the segment of the cone about it. So, (if) the κυλίνδρου. λελείφθω, καὶ ἔστω τὰ ἐπὶ τῶν ΑΕ, ΕΒ, ΒΖ, ΖΓ, remaining circumferences are cut in half, and straight- ΓΗ, ΗΔ, ΔΘ, ΘΑ· λοιπὴ ἄρα ἡ πυραμίς, ἧς βάσις μέν ἐστι lines are joined, and pyramids having the same apex as τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ, the cone are set up on each of the triangles, and this is μείζων ἐστὶν ἢ τρίτον μέρος τοῦ κυλίνδρου. ἀλλ᾿ ἡ πυραμίς, done continually, then we will (eventually) leave some ἧς βάσις μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, κορυφὴ δὲ segments of the cone whose (sum) is less than the excess ἡ αυτὴ τῷ κώνῳ, τρίτον ἐστὶ μέρος τοῦ πρίσματος, οὗ βάσις by which the cone exceeds the third part of the cylinder μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ [Prop. 10.1]. Let them have been left, and let them be κυλίνδρῳ· τὸ ἄρα πρίσμα, οὗ βάσις μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ the (segments) on AE, EB, BF , FC, CG, GD, DH , and πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ κυλίνδρῳ, μεῖζόν ἐστι τοῦ HA. Thus, the remaining pyramid whose base is poly- κυλίνδρου, οὗ βάσις ἐστὶν ὁ ΑΒΓΔ κύκλος. ἀλλὰ καὶ ἔλατ- gon AEBFCGDH , and apex the same as the cone, is τον· ἐμπεριέχεται γὰρ ὑπ᾿ αὐτοῦ· ὅπερ ἐστὶν ἀδύνατον. οὐκ greater than the third part of the cylinder. But, the pyra- ἄρα ὁ κύλινδρος τοῦ κώνου ἐλάττων ἐστὶν ἢ τριπλάσιος. mid whose base is polygon AEBFCGDH , and apex the ἐδείχθη δέ, ὅτι οὐδὲ μείζων ἢ τριπλάσιος· τριπλάσιος ἄρα ὁ same as the cone, is the third part of the prism whose κύλινδρος τοῦ κώνου· ὥστε ὁ κῶνος τρίτον ἐστὶ μέρος τοῦ base is polygon AEBFCGDH , and height the same as κυλίνδρου. the cylinder [Prop. 12.7 corr.]. Thus, the prism whose Πᾶς ἄρα κῶνος κυλίνδρου τρίτον μέρος ἐστὶ τοῦ τὴν base is polygon AEBFCGDH , and height the same as αὐτὴν βάσιν ἔχοντος αὐτῷ καὶ ὕψος ἴσον· ὅπερ ἔδει δεῖξαι. the cylinder, is greater than the cylinder whose base is circle ABCD. But, (it is) also less. For it is encompassed by it. The very thing is impossible. Thus, the cylinder is not less than three times the cone. And it was shown that neither (is it) greater than three times (the cone). Thus, the cylinder (is) three times the cone. Hence, the cone is the third part of the cylinder. Thus, every cone is the third part of the cylinder which has the same base as it, and an equal height. (Which is) the very thing it was required to show.iaþ. Proposition 11 Οἱ ὑπο τὸ αὐτὸ ὕψος ὄντες κῶνοι καὶ κύλινδροι πρὸς Cones and cylinders having the same height are to one ἀλλήλους εἰσὶν ὡς αἱ βάσεις. another as their bases. ῎Εστωσαν ὑπὸ τὸ αὐτὸ ὕψος κῶνοι καὶ κύλινδροι, ὧν Let there be cones and cylinders of the same height βάσεις μὲν [εἰσιν] οἱ ΑΒΓΔ, ΕΖΗΘ κύκλοι, ἄξονες δὲ οἱ whose bases [are] the circles ABCD and EFGH , axes ΚΛ, ΜΝ, διάμετροι δὲ τῶν βάσεων αἱ ΑΓ, ΕΗ· λέγω, ὅτι KL and MN , and diameters of the bases AC and EG (re- ἐστὶν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως spectively). I say that as circle ABCD is to circle EFGH , ὁ ΑΛ κῶνος πρὸς τὸν ΕΝ κῶνον. so cone AL (is) to cone EN . 489 STOIQEIWN ibþ. ELEMENTS BOOK 12 R DT A U B L K M Z H NSJO Y XG E PF Q R D T A U B L K M N S E O P H F Q G C X V W Εἰ γὰρ μή, ἔσται ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ For if not, then as circle ABCD (is) to circle EFGH , κύκλον, οὕτως ὁ ΑΛ κῶνος ἤτοι πρὸς ἔλασσόν τι τοῦ ΕΝ so cone AL will be to some solid either less than, or κώνου στερεὸν ἢ πρὸς μεῖζον. ἔστω πρότερον πρὸς ἔλασ- greater than, cone EN . Let it, first of all, be (in this ra- σον τὸ Ξ, καὶ ᾧ ἔλασσόν ἐστι τὸ Ξ στερεὸν τοῦ ΕΝ κώνου, tio) to (some) lesser (solid), O. And let solid X be equal ἐκείνῳ ἴσον ἔστω τὸ Ψ στερεόν· ὁ ΕΝ κῶνος ἄρα ἴσος ἐστὶ to that (magnitude) by which solid O is less than cone τοῖς Ξ, Ψ στερεοῖς. ἐγγεγράφθω εἰς τὸν ΕΖΗΘ κύκλον EN . Thus, cone EN is equal to (the sum of) solids O τετράγωνον τὸ ΕΖΗΘ· τὸ ἄρα τετράγωνον μεῖζόν ἐστιν and X . Let the square EFGH have been inscribed in cir- ἢ τὸ ἥμισυ τοῦ κύκλου. ἀνεστάτω ἀπὸ τοῦ ΕΖΗΘ τε- cle EFGH [Prop. 4.6]. Thus, the square is greater than τραγώνου πυραμὶς ἰσοϋψὴς τῷ κώνῳ· ἡ ἄρα ἀνασταθεῖσα half of the circle [Prop. 12.2]. Let a pyramid of the same πυραμὶς μείζων ἐστὶν ἢ τὸ ἥμισυ τοῦ κώνου, ἐπειδήπερ ἐὰν height as the cone have been set up on square EFGH . περιγράψωμεν περὶ τὸν κύκλον τετράγωνον, καὶ ἀπ᾿ αὐτοῦ Thus, the pyramid set up is greater than half of the cone, ἀναστήσωμεν πυραμίδα ἰσοϋψῆ τῷ κώνῳ, ἡ ἐγγραφεῖσα πυ- inasmuch as, if we circumscribe a square about the cir- ραμὶς ἥμισύ ἐστι τῆς περιγραφείσης· πρὸς ἀλλήλας γάρ εἰσιν cle [Prop. 4.7], and set up on it a pyramid of the same ὡς αἱ βάσεις· ἐλάττων δὲ ὁ κῶνος τῆς περιγραφείσης πυ- height as the cone, then the inscribed pyramid is half ραμίδος. τετμήσθωσαν αἱ ΕΖ, ΖΗ, ΗΘ, ΘΕ περιφέρειαι of the circumscribed pyramid. For they are to one an- δίχα κατὰ τὰ Ο, Π, Ρ, Σ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ other as their bases [Prop. 12.6]. And the cone (is) less ΘΟ, ΟΕ, ΕΠ, ΠΖ, ΖΡ, ΡΗ, ΗΣ, ΣΘ. ἕκαστον ἄρα τῶν than the circumscribed pyramid. Let the circumferences ΘΟΕ, ΕΠΖ, ΖΡΗ, ΗΖΘ τριγώνων μεῖζόν ἐστιν ἢ τὸ ἥμισυ EF , FG, GH , and HE have been cut in half at points τοῦ καθ᾿ ἑαυτὸ τμήματος τοῦ κύκλου. ἀνεστάτω ἐφ᾿ P , Q, R, and S. And let HP , PE, EQ, QF , FR, RG, ἑκάστου τῶν ΘΟΕ, ΕΠΖ, ΖΡΗ, ΗΣΘ τριγώνων πυραμὶς GS, and SH have been joined. Thus, each of the trian- ἰσοϋψὴς τῷ κώνῳ· καὶ ἑκάστη ἄρα τῶν ἀνασταθεισῶν πυ- gles HPE, EQF , FRG, and GSH is greater than half ραμίδων μείζων ἐστὶν ἢ τὸ ἥμισυ τοῦ καθ᾿ ἑαυτὴν τμήματος of the segment of the circle about it [Prop. 12.2]. Let τοῦ κώνου. τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας pyramids of the same height as the cone have been set up δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ ἀνιστάντες ἐπὶ ἑκάστου on each of the triangles HPE, EQF , FRG, and GSH . τῶν τριγώνων πυραμίδας ἰσοϋψεῖς τῷ κώνῳ καὶ ἀεὶ τοῦτο And, thus, each of the pyramids set up is greater than ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ κώνου, ἃ half of the segment of the cone about it [Prop. 12.10]. ἔσται ἐλάσσονα τοῦ Ψ στερεοῦ. λελείφθω, καὶ ἔστω τὰ So, (if) the remaining circumferences are cut in half, and ἐπὶ τῶν ΘΟΕ, ΕΠΖ, ΖΡΗ, ΗΣΘ· λοιπὴ ἄρα ἡ πυραμίς, ἧς straight-lines are joined, and pyramids of equal height βάσις τὸ ΘΟΕΠΖΡΗΣ πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ to the cone are set up on each of the triangles, and κώνῳ, μείζων ἐστὶ τοῦ Ξ στερεοῦ. ἐγγεγράφθω καὶ εἰς τὸν this is done continually, then we will (eventually) leave ΑΒΓΔ κύκλον τῷ ΘΟΕΠΖΡΗΣ πολυγώνῳ ὅμοιόν τε καὶ some segments of the cone (the sum of) which is less ὁμοίως κείμενον πολύγωνον τὸ ΔΤΑΥΒΦΓΧ, καὶ ἀνεστάτω than solid X [Prop. 10.1]. Let them have been left, and ἐπ᾿ αὐτοῦ πυραμὶς ἰσοϋψὴς τῷ ΑΛ κώνῳ. ἐπεὶ οὖν ἐστιν ὡς let them be the (segments) on HPE, EQF , FRG, and τὸ ἀπὸ τῆς ΑΓ πρὸς τὸ ἀπὸ τῆς ΕΗ, οὕτως τὸ ΔΤΑΥΒΦΓΧ GSH . Thus, the remaining pyramid whose base is poly- πολύγωνον πρὸς τὸ ΘΟΕΠΖΡΗΣ πολύγωνον, ὡς δὲ τὸ gon HPEQFRGS, and height the same as the cone, is ἀπὸ τῆς ΑΓ πρὸς τὸ ἀπὸ τῆς ΕΗ, οὕτως ὁ ΑΒΓΔ κύκλος greater than solid O [Prop. 6.18]. And let the polygon πρὸς τὸν ΕΖΗΘ κύκλον, καὶ ὡς ἄρα ὁ ΑΒΓΔ κύκλος πρὸς DTAUBV CW , similar, and similarly laid out, to polygon τὸν ΕΖΗΘ κύκλον, οὕτως τὸ ΔΤΑΥΒΦΓΧ πολύγωνον HPEQFRGS, have been inscribed in circle ABCD. And πρὸς τὸ ΘΟΕΠΖΡΗΣ πολύγωνον. ὡς δὲ ὁ ΑΒΓΔ κύκλος on it let a pyramid of the same height as cone AL have πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΛ κῶνος πρὸς τὸ Ξ been set up. Therefore, since as the (square) on AC is στερεόν, ὡς δὲ τὸ ΔΤΑΥΒΦΓΧ πολύγωνον πρὸς τὸ ΘΟ- to the (square) on EG, so polygon DTAUBV CW (is) to ΕΠΖΡΗΣ πολύγωνον, οὕτως ἡ πυραμίς, ἧς βάσις μὲν τὸ polygon HPEQFRGS [Prop. 12.1], and as the (square) ΔΤΑΥΒΦΓΧ πολύγωνον, κορυφὴ δὲ τὸ Λ σημεῖον, πρὸς on AC (is) to the (square) on EG, so circle ABCD (is) 490 STOIQEIWN ibþ. ELEMENTS BOOK 12 τὴν πυραμίδα, ἧς βάσις μὲν τὸ ΘΟΕΠΖΡΗΣ πολύγωνον, to circle EFGH [Prop. 12.2], thus as circle ABCD (is) κορυφὴ δὲ τὸ Ν σημεῖον. καὶ ὡς ἄρα ὁ ΑΛ κῶνος πρὸς τὸ to circle EFGH , so polygon DTAUBV CW also (is) to Ξ στερεόν, οὕτως ἡ πυραμίς, ἧς βάσις μὲν τὸ ΔΤΑΥΒΦΓΧ polygon HPEQFRGS. And as circle ABCD (is) to cir- πολύγωνον, κορυφὴ δὲ τὸ Λ σημεῖον, πρὸς τὴν πυραμίδα, cle EFGH , so cone AL (is) to solid O. And as poly- ἧς βάσις μὲν τὸ ΘΟΕΠΖΡΗΣ πολύγωνον, κορυφὴ δὲ τὸ Ν gon DTAUBV CW (is) to polygon HPEQFRGS, so the σημεῖον· ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ ΑΛ κῶνος πρὸς τὴν ἐν pyramid whose base is polygon DTAUBV CW , and apex αὐτῷ πυραμίδα, οὕτως τὸ Ξ στερεὸν πρὸς τὴν ἐν τῷ ΕΝ the point L, (is) to the pyramid whose base is polygon κώνῳ πυραμίδα. μείζων δὲ ὁ ΑΛ κῶνος τῆς ἐν αὐτῷ πυ- HPEQFRGS, and apex the point N [Prop. 12.6]. And, ραμίδος· μεῖζον ἄρα καὶ τὸ Ξ στερεὸν τῆς ἐν τῷ ΕΝ κώνῳ thus, as cone AL (is) to solid O, so the pyramid whose πυραμίδος. ἀλλὰ καὶ ἔλασσον· ὅπερ ἄτοπον. οὐκ ἄρα ἐστὶν base is DTAUBV CW , and apex the point L, (is) to the ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΛ pyramid whose base is polygon HPEQFRGS, and apex κῶνος πρὸς ἔλασσόν τι τοῦ ΕΝ κώνου στερεόν. ὁμοίως the point N [Prop. 5.11]. Thus, alternately, as cone AL δὲ δείξομεν, ὅτι οὐδέ ἐστιν ὡς ὁ ΕΖΗΘ κύκλος πρὸς τὸν is to the pyramid within it, so solid O (is) to the pyramid ΑΒΓΔ κύκλον, οὕτως ὁ ΕΝ κῶνος πρὸς ἔλασσόν τι τοῦ within cone EN [Prop. 5.16]. But, cone AL (is) greater ΑΛ κώνου στερεόν. than the pyramid within it. Thus, solid O (is) also greater Λέγω δή, ὅτι οὐδέ ἐστιν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν than the pyramid within cone EN [Prop. 5.14]. But, (it ΕΖΗΘ κύκλον, οὕτως ὁ ΑΛ κῶνος πρὸς μεῖζόν τι τοῦ ΕΝ is) also less. The very thing (is) absurd. Thus, circle κώνου στερεόν. ABCD is not to circle EFGH , as cone AL (is) to some Εἰ γὰρ δυνατόν, ἕστω πρὸς μεῖζον τὸ Ξ· ἀνάπαλιν ἄρα solid less than cone EN . So, similarly, we can show that ἐστὶν ὡς ὁ ΕΖΗΘ κύκλος πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως neither is circle EFGH to circle ABCD, as cone EN (is) τὸ Ξ στερεὸν πρὸς τὸν ΑΛ κῶνον. ἀλλ᾿ ὡς τὸ Ξ στερεὸν to some solid less than cone AL. πρὸς τὸν ΑΛ κῶνον, οὕτως ὁ ΕΝ κῶνος πρὸς ἔλασσόν τι So, I say that neither is circle ABCD to circle EFGH , τοῦ ΑΛ κώνου στερεόν· καὶ ὡς ἄρα ὁ ΕΖΗΘ κύκλος πρὸς as cone AL (is) to some solid greater than cone EN . τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΝ κῶνος πρὸς ἔλασσόν τι For, if possible, let it be (in this ratio) to (some) τοῦ ΑΛ κώνου στερεόν· ὅπερ ἀδύνατον ἐδείχθη. οὐκ ἄρα greater (solid), O. Thus, inversely, as circle EFGH is to ἐστὶν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ circle ABCD, so solid O (is) to cone AL [Prop. 5.7 corr.]. ΑΛ κῶνος πρὸς μεῖζόν τι τοῦ ΕΝ κώνου στερεόν. ἐδείχθη But, as solid O (is) to cone AL, so cone EN (is) to some δέ, ὅτι οὐδὲ πρὸς ἔλασσον· ἔστιν ἄρα ὡς ὁ ΑΒΓΔ κύκλος solid less than cone AL [Prop. 12.2 lem.]. And, thus, as πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΛ κῶνος πρὸς τὸν ΕΝ circle EFGH (is) to circle ABCD, so cone EN (is) to κῶνον. some solid less than cone AL. The very thing was shown Ἀλλ᾿ ὡς ὁ κῶνος πρὸς τὸν κῶνον, ὁ κύλινδρος πρὸς (to be) impossible. Thus, circle ABCD is not to circle τὸν κύλινδρον· τριπλασίων γὰρ ἑκάτερος ἑκατέρου. καὶ ὡς EFGH , as cone AL (is) to some solid greater than cone ἄρα ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως οἱ ἐπ᾿ EN . And, it was shown that neither (is it in this ratio) to αὐτῶν ἰσοϋψεῖς. (some) lesser (solid). Thus, as circle ABCD is to circle Οἱ ἄρα ὑπὸ τὸ αὐτὸ ὕψος ὄντες κῶνοι καὶ κύλινδροι EFGH , so cone AL (is) to cone EN . πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις· ὅπερ ἔδει δεῖξαι. But, as the cone (is) to the cone, (so) the cylin- der (is) to the cylinder. For each (is) three times each [Prop. 12.10]. Thus, circle ABCD (is) also to circle EFGH , as (the ratio of the cylinders) on them (having) the same height. Thus, cones and cylinders having the same height are to one another as their bases. (Which is) the very thing it was required to show.ibþ. Proposition 12 Οἱ ὅμοιοι κῶνοι καὶ κύλινδροι πρὸς ἀλλήλους ἐν τρι- Similar cones and cylinders are to one another in the πλασίονι λόγῳ εἰσὶ τῶν ἐν ταῖς βάσεσι διαμέτρων. cubed ratio of the diameters of their bases. ῎Εστωσαν ὅμοιοι κῶνοι καὶ κύλινδροι, ὧν βάσεις μὲν Let there be similar cones and cylinders of which the οἱ ΑΒΓΔ, ΕΖΗΘ κύκλοι, διάμετροι δὲ τῶν βάσεων αἱ ΒΔ, bases (are) the circles ABCD and EFGH , the diameters ΖΘ, ἄξονες δὲ τῶν κώνων καὶ κυλίνδρων οἱ ΚΛ, ΜΝ· λέγω, of the bases (are) BD and FH , and the axes of the cones 491 STOIQEIWN ibþ. ELEMENTS BOOK 12 ὅτι ὁ κῶνος, οὗ βάσις μέν [ἐστιν] ὁ ΑΒΓΔ κύκλος, κορυφὴ and cylinders (are) KL and MN (respectively). I say δὲ τὸ Λ σημεῖον, πρὸς τὸν κῶνον, οὗ βάσις μέν [ἐστιν] that the cone whose base [is] circle ABCD, and apex the ὁ ΕΖΗΘ κύκλος, κορυφὴ δὲ τὸ Ν σημεῖον, τριπλασίονα point L, has to the cone whose base [is] circle EFGH , λόγον ἔχει ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. and apex the point N , the cubed ratio that BD (has) to FH . J L B U G D Q K AT ES R H M N X F P O Z R L U DK A T E S M N C V W H G Q OF B P Εἰ γὰρ μὴ ἔχει ὁ ΑΒΓΔΛ κῶνος πρὸς τὸν ΕΖΗΘΝ For if cone ABCDL does not have to cone EFGHN κῶνον πριπλασίονα λόγον ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ, ἕξει the cubed ratio that BD (has) to FH then cone ABCDL ὁ ΑΒΓΔΛ κῶνος ἢ πρὸς ἔλασσόν τι τοῦ ΕΖΗΘΝ κώνου will have the cubed ratio to some solid either less than, or στερεὸν τριπλασίονα λόγον ἢ πρὸς μεῖζον. ἐχέτω πρότερον greater than, cone EFGHN . Let it, first of all, have (such πρὸς ἔλασσον τὸ Ξ, καὶ ἐγγεγράφθω εἰς τὸν ΕΖΗΘ κύκλον a ratio) to (some) lesser (solid), O. And let the square τετράγωνον τὸ ΕΖΗΘ· τὸ ἄρα ΕΖΗΘ τετράγωνον μεῖζόν EFGH have been inscribed in circle EFGH [Prop. 4.6]. ἐστιν ἢ τὸ ἥμισυ τοῦ ΕΖΗΘ κύκλου. καὶ ἀνεστάτω ἐπὶ Thus, square EFGH is greater than half of circle EFGH τοῦ ΕΖΗΘ τετραγώνου πυραμὶς τὴν αὐτὴν κορυφὴν ἔχουσα [Prop. 12.2]. And let a pyramid having the same apex τῷ κώνῳ· ἡ ἄρα ἀνασταθεῖσα πυραμὶς μείζων ἐστὶν ἢ τὸ as the cone have been set up on square EFGH . Thus, ἥμισυ μέρος τοῦ κώνου. τετμήσθωσαν δὴ αἱ ΕΖ, ΖΗ, the pyramid set up is greater than the half part of the ΗΘ, ΘΕ περιφέρειαι δίχα κατὰ τὰ Ο, Π, Ρ, Σ σημεῖα, καὶ cone [Prop. 12.10]. So, let the circumferences EF , FG, ἐπεζεύχθωσαν αἱ ΕΟ, ΟΖ, ΖΠ, ΠΗ, ΗΡ, ΡΘ, ΘΣ, ΣΕ. καὶ GH , and HE have been cut in half at points P , Q, R, ἕκαστον ἄρα τῶν ΕΟΖ, ΖΠΗ, ΗΡΘ, ΘΣΕ τριγώνων μεῖζόν and S (respectively). And let EP , PF , FQ, QG, GR, ἐστιν ἢ τὸ ἥμισυ μέρος τοῦ καθ᾿ ἑαυτὸ τμήματος τοῦ ΕΖΗΘ RH , HS, and SE have been joined. And, thus, each κύκλου. καὶ ἀνεστάτω ἐφ᾿ ἑκάστου τῶν ΕΟΖ, ΖΠΗ, ΗΡΘ, of the triangles EPF , FQG, GRH , and HSE is greater ΘΣΕ τριγώνων πυραμὶς τὴν αὐτὴν κορυφὴν ἔχουσα τῷ than the half part of the segment of circle EFGH about it κώνῳ· καὶ ἑκάστη ἄρα τῶν ἀνασταθεισῶν πυραμίδων μείζων [Prop. 12.2]. And let a pyramid having the same apex as ἐστὶν ἢ τὸ ἥμισυ μέρος τοῦ καθ᾿ ἑαυτὴν τμήματος τοῦ the cone have been set up on each of the triangles EPF , κώνου. τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας δίχα FQG, GRH , and HSE. And thus each of the pyramids καὶ ἐπιζευγνύντες εὐθείας καὶ ἀνιστάντες ἐφ᾿ ἑκάστου τῶν set up is greater than the half part of the segment of the τριγώνων πυραμίδας τὴν αὐτὴν κορυφὴν ἐχούσας τῷ κώνῳ cone about it [Prop. 12.10]. So, (if) the the remaining cir- καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ cumferences are cut in half, and straight-lines are joined, κώνου, ἃ ἔσται ἐλάσσονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ and pyramids having the same apex as the cone are set ΕΖΗΘΝ κῶνος τοῦ Ξ στερεοῦ. λελείφθω, καὶ ἔστω τὰ up on each of the triangles, and this is done continu- ἐπὶ τῶν ΕΟ, ΟΖ, ΖΠ, ΠΗ, ΗΡ, ΡΘ, ΘΣ, ΣΕ· λοιπὴ ἄρα ἡ ally, then we will (eventually) leave some segments of the πυραμίς, ἧς βάσις μέν ἐστι τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, cone whose (sum) is less than the excess by which cone κορυφὴ δὲ τὸ Ν σημεῖον, μείζων ἐστὶ τοῦ Ξ στερεοῦ. EFGHN exceeds solid O [Prop. 10.1]. Let them have ἐγγεγράφθω καὶ εἰς τὸν ΑΒΓΔ κύκλον τῷ ΕΟΖΠΗΡΘΣ been left, and let them be the (segments) on EP , PF , πολυγώνῳ ὅμοιόν τε καὶ ὁμοίως κείμενον πολύγωνον τὸ FQ, QG, GR, RH , HS, and SE. Thus, the remaining ΑΤΒΥΓΦΔΧ, καὶ ἀνεστάτω ἐπὶ τοῦ ΑΤΒΥΓΦΔΧ πο- pyramid whose base is polygon EPFQGRHS, and apex λυγώνου πυραμὶς τὴν αὐτὴν κορυφὴν ἔχουσα τῷ κώνῳ, the point N , is greater than solid O. And let the polygon καὶ τῶν μὲν περιεχόντων τὴν πυραμίδα, ἧς βάσις μέν ἐστι ATBUCV DW , similar, and similarly laid out, to poly- τὸ ΑΤΒΥΓΦΔΧ πολύγωνον, κορυφὴ δὲ τὸ Λ σημεῖον, gon EPFQGRHS, have been inscribed in circle ABCD ἓν τρίγωνον ἔστω τὸ ΛΒΤ, τῶν δὲ περειχόντων τὴν πυ- [Prop. 6.18]. And let a pyramid having the same apex ραμίδα, ἧς βάσις μέν ἐστι τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, as the cone have been set up on polygon ATBUCV DW . 492 STOIQEIWN ibþ. ELEMENTS BOOK 12 κορυφὴ δὲ τὸ Ν σημεῖον, ἓν τρίγωνον ἔστω τὸ ΝΖΟ, καὶ And let LBT be one of the triangles containing the pyra- ἐπεζεύχθωσαν αἱ ΚΤ, ΜΟ. καὶ ἐπεὶ ὅμοιός ἐστιν ὁ ΑΒΓΔΛ mid whose base is polygon ATBUCV DW , and apex the κῶνος τῷ ΕΖΗΘΝ κώνῳ, ἔστιν ἄρα ὡς ἡ ΒΔ πρὸς τὴν point L. And let NFP be one of the triangles containing ΖΘ, οὕτως ὁ ΚΛ ἄξων πρὸς τὸν ΜΝ ἄξονα. ὡς δὲ ἡ ΒΔ the pyramid whose base is triangle EPFQGRHS, and πρὸς τὴν ΖΘ, οὕτως ἡ ΒΚ πρὸς τὴν ΖΜ· καὶ ὡς ἄρα ἡ ΒΚ apex the point N . And let KT and MP have been joined. πρὸς τὴν ΖΜ, οὕτως ἡ ΚΛ πρὸς τὴν ΜΝ. καὶ ἐναλλὰξ ὡς And since cone ABCDL is similar to cone EFGHN , thus ἡ ΒΚ πρὸς τὴν ΚΛ, οὕτως ἡ ΖΜ πρὸς τὴν ΜΝ. καὶ περὶ as BD is to FH , so axis KL (is) to axis MN [Def. 11.24]. ἴσας γωνίας τὰς ὑπὸ ΒΚΛ, ΖΜΝ αἱ πλευραὶ ἀνάλογόν εἰσιν· And as BD (is) to FH , so BK (is) to FM . And, thus, as ὅμοιον ἄρα ἐστὶ τὸ ΒΚΛ τρίγωνον τῷ ΖΜΝ τριγώνῳ. πάλιν, BK (is) to FM , so KL (is) to MN . And, alternately, as ἐπεί ἐστιν ὡς ἡ ΒΚ πρὸς τὴν ΚΤ, οὕτως ἡ ΖΜ πρὸς τὴν BK (is) to KL, so FM (is) to MN [Prop. 5.16]. And ΜΟ, καὶ περὶ ἴσας γωνίας τὰς ὑπὸ ΒΚΤ, ΖΜΟ, ἐπειδήπερ, the sides around the equal angles BKL and FMN are ὃ μέρος ἐστὶν ἡ ὑπὸ ΒΚΤ γωνία τῶν πρὸς τῷ Κ κέντρῳ proportional. Thus, triangle BKL is similar to triangle τεσσάρων ὀρθῶν, τὸ αὐτὸ μέρος ἐστὶ καὶ ἡ ὑπὸ ΖΜΟ γωνία FMN [Prop. 6.6]. Again, since as BK (is) to KT , so τῶν πρὸς τῷ Μ κέντρῳ τεσσάρων ὀρθῶν· ἐπεὶ οὖν περὶ ἴσας FM (is) to MP , and (they are) about the equal angles γωνίας αἱ πλευραὶ ἀνάλογόν εἰσιν, ὅμοιον ἄρα ἐστι τὸ ΒΚΤ BKT and FMP , inasmuch as whatever part angle BKT τρίγωνον τῷ ΖΜΟ τριγώνῳ. πάλιν, ἐπεὶ ἐδείχθη ὡς ἡ ΒΚ is of the four right-angles at the center K, angle FMP is πρὸς τὴν ΚΛ, οὕτως ἡ ΖΜ πρὸς τὴν ΜΝ, ἴση δὲ ἡ μὲν also the same part of the four right-angles at the cen- ΒΚ τῇ ΚΤ, ἡ δὲ ΖΜ τῇ ΟΜ, ἔστιν ἄρα ὡς ἡ ΤΚ πρὸς ter M . Therefore, since the sides about equal angles τὴν ΚΛ, οὕτως ἡ ΟΜ πρὸς τὴν ΜΝ. καὶ περὶ ἴσας γωνίας are proportional, triangle BKT is thus similar to train- τὰς ὑπὸ ΤΚΛ, ΟΜΝ· ὀρθαὶ γάρ· αἱ πλευραὶ ἀνάλογόν εἰσιν· gle FMP [Prop. 6.6]. Again, since it was shown that ὅμοιον ἄρα ἐστὶ τὸ ΛΚΤ τρίγωνον τῷ ΝΜΟ τριγώνῳ. καὶ as BK (is) to KL, so FM (is) to MN , and BK (is) ἐπεὶ διὰ τὴν ὁμοιότητα τῶν ΛΚΒ, ΝΜΖ τριγώνων ἐστὶν equal to KT , and FM to PM , thus as TK (is) to KL, ὡς ἡ ΛΒ πρὸς τὴν ΒΚ, οὕτως ἡ ΝΖ πρὸς τὴν ΖΜ, διὰ δὲ so PM (is) to MN . And the sides about the equal angles τὴν ὁμοιότητα τῶν ΒΚΤ, ΖΜΟ τριγώνων ἐστὶν ὡς ἡ ΚΒ TKL and PMN—for (they are both) right-angles—are πρὸς τὴν ΒΤ, οὕτως ἡ ΜΖ πρὸς τὴν ΖΟ, δι᾿ ἴσου ἄρα ὡς proportional. Thus, triangle LKT (is) similar to triangle ἡ ΛΒ πρὸς τὴν ΒΤ, οὕτως ἡ ΝΖ πρὸς τὴν ΖΟ. πάλιν, ἐπεὶ NMP [Prop. 6.6]. And since, on account of the similarity διὰ τὴν ομοιότητα τῶν ΛΤΚ, ΝΟΜ τριγώνων ἐστὶν ὡς ἡ of triangles LKB and NMF , as LB (is) to BK, so NF ΛΤ πρὸς τὴν ΤΚ, οὕτως ἡ ΝΟ πρὸς τὴν ΟΜ, διὰ δὲ τὴν (is) to FM , and, on account of the similarity of triangles ὁμοιότητα τῶν ΤΚΒ, ΟΜΖ τριγώνων ἐστὶν ὡς ἡ ΚΤ πρὸς BKT and FMP , as KB (is) to BT , so MF (is) to FP τὴν ΤΒ, οὕτως ἡ ΜΟ πρὸς τὴν ΟΖ, δι᾿ ἴσου ἄρα ὡς ἡ ΛΤ [Def. 6.1], thus, via equality, as LB (is) to BT , so NF πρὸς τὴν ΤΒ, οὕτως ἡ ΝΟ πρὸς τὴν ΟΖ. ἐδείχθη δὲ καὶ (is) to FP [Prop. 5.22]. Again, since, on account of the ὡς ἡ ΤΒ πρὸς τὴν ΒΛ, οὕτως ἡ ΟΖ πρὸς τὴν ΖΝ. δι᾿ ἴσου similarity of triangles LTK and NPM , as LT (is) to TK, ἄρα ὡς ἡ ΤΛ πρὸς τὴν ΛΒ, οὕτως ἡ ΟΝ πρὸς τὴν ΝΖ. so NP (is) to PM , and, on account of the similarity of τῶν ΛΤΒ, ΝΟΖ ἄρα τριγώνων ἀνάλογόν εἰσιν αἱ πλευραί· triangles TKB and PMF , as KT (is) to TB, so MP (is) ἰσογώνια ἄρα ἐστὶ τὰ ΛΤΒ, ΝΟΖ τρίγωνα· ὥστε καὶ ὅμοια. to PF , thus, via equality, as LT (is) to TB, so NP (is) καὶ πυραμὶς ἄρα, ἧς βάσις μὲν τὸ ΒΚΤ τρίγωνον, κορυφὴ to PF [Prop. 5.22]. And it was shown that as TB (is) δὲ τὸ Λ σημεῖον, ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μὲν τὸ to BL, so PF (is) to FN . Thus, via equality, as TL (is) ΖΜΟ τρίγωνον, κορυφὴ δὲ τὸ Ν σημεῖον· ὑπὸ γὰρ ὅμοίων to LB, so PN (is) to NF [Prop. 5.22]. Thus, the sides ἐπιπέδων περιέχονται ἴσων τὸ πλῆθος. αἱ δὲ ὅμοιαι πυ- of triangles LTB and NPF are proportional. Thus, tri- ραμίδες καὶ τριγώνους ἔχουσαι βάσεις ἐν τριπλασίονι λόγῳ angles LTB and NPF are equiangular [Prop. 6.5]. And, εἰσὶ τῶν ὁμολόγων πλευρῶν. ἡ ἄρα ΒΚΤΛ πυραμὶς πρὸς τὴν hence, (they are) similar [Def. 6.1]. And, thus, the pyra- ΖΜΟΝ πυραμίδα τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΚ πρὸς mid whose base is triangle BKT , and apex the point L, τὴν ΖΜ. ὁμοίως δὴ ἐπιζευγνύντες ἀπὸ τῶν Α, Χ, Δ, Φ, Γ, Υ is similar to the pyramid whose base is triangle FMP , ἐπὶ τὸ Κ εὐθείας καὶ ἀπὸ τῶν Ε, Σ, Θ, Ρ, Η, Π ἐπὶ τὸ Μ καὶ and apex the point N . For they are contained by equal ἀνιστάντες ἐφ᾿ ἑκάστου τῶν τριγώνων πυραμίδας τὴν αὐτὴν numbers of similar planes [Def. 11.9]. And similar pyra- κορυφὴν ἐχούσας τοῖς κώνοις δείξομεν, ὅτι καὶ ἑκάστη τῶν mids which also have triangular bases are in the cubed ὁμοταγῶν πυραμίδων πρὸς ἑκάστην ὁμοταγῆ πυραμίδα τρι- ratio of corresponding sides [Prop. 12.8]. Thus, pyramid πλασίονα λόγον ἕξει ἤπερ ἡ ΒΚ ὁμόλογος πλευρὰ πρὸς τὴν BKTL has to pyramid FMPN the cubed ratio that BK ΖΜ ὁμόλογον πλευράν, τουτέστιν ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. (has) to FM . So, similarly, joining straight-lines from καὶ ὡς ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως (points) A, W , D, V , C, and U to (center) K, and from ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα· ἔστιν ἄρα (points) E, S, H , R, G, and Q to (center) M , and set- 493 STOIQEIWN ibþ. ELEMENTS BOOK 12 καὶ ὡς ἡ ΒΚΤΛ πυραμὶς πρὸς τὴν ΖΜΟΝ πυραμίδα, οὕτως ting up pyramids having the same apexes as the cones ἡ ὅλη πυραμίς, ἧς βάσις τὸ ΑΤΒΥΓΦΔΧ πολύγωνον, κο- on each of the triangles (so formed), we can also show ρυφὴ δὲ τὸ Λ σημεῖον, πρὸς τὴν ὅλην πυραμίδα, ἧς βάσις that each of the pyramids (on base ABCD taken) in or- μὲν τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, κορυφὴ δὲ τὸ Ν σημεῖον· der will have to each of the pyramids (on base EFGH ὥστε καὶ πυραμίς, ἧς βάσις μὲν τὸ ΑΤΒΥΓΦΔΧ, κορυφὴ δὲ taken) in order the cubed ratio that the corresponding τὸ Λ, πρὸς τὴν πυραμίδα, ἧς βάσις [μὲν] τὸ ΕΟΖΠΗΡΘΣ side BK (has) to the corresponding side FM—that is to πολύγωνον, κορυφὴ δὲ τὸ Ν σημεῖον, τριπλασίονα λόγον say, that BD (has) to FH . And (for two sets of propor- ἔχει ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. ὑπόκειται δὲ καὶ ὁ κῶνος, οὗ tional magnitudes) as one of the leading (magnitudes is) βάσις [μὲν] ὁ ΑΒΓΔ κύκλος, κορυφὴ δὲ τὸ Λ σημεῖον, πρὸς to one of the following, so (the sum of) all of the leading τὸ Ξ στερεὸν τριπλασίονα λόγον ἔχων ἤπερ ἡ ΒΔ πρὸς τὴν (magnitudes is) to (the sum of) all of the following (mag- ΖΘ· ἔστιν ἄρα ὡς ὁ κῶνος, οὗ βάσις μέν ἐστιν ὁ ΑΒΓΔ nitudes) [Prop. 5.12]. And, thus, as pyramid BKTL (is) κύκλος, κορυφὴ δὲ τὸ Λ, πρὸς τὸ Ξ στερεόν, οὕτως ἡ πυ- to pyramid FMPN , so the whole pyramid whose base ραμίς, ἧς βάσις μὲν τὸ ΑΤΒΥΓΦΔΧ [πολύγωνον], κορυφὴ is polygon ATBUCV DW , and apex the point L, (is) to δὲ τὸ Λ, πρὸς τὴν πυραμίδα, ἧς βάσις μέν ἐστι τὸ ΕΟΖ- the whole pyramid whose base is polygon EPFQGRHS, ΠΗΡΘΣ πολύγωνον, κορυφὴ δὲ τὸ Ν· ἐναλλὰξ ἄρα, ὡς ὁ and apex the point N . And, hence, the pyramid whose κῶνος, οὗ βάσις μὲν ὁ ΑΒΓΔ κύκλος, κορυφὴ δὲ τὸ Λ, base is polygon ATBUCV DW , and apex the point L, πρὸς τὴν ἐν αὐτῷ πυραμίδα, ἧς βάσις μὲν τὸ ΑΤΒΥΓΦΔΧ has to the pyramid whose base is polygon EPFQGRHS, πολύγωνον, κορυφὴ δὲ τὸ Λ, οὕτως τὸ Ξ [στερεὸν] πρὸς τὴν and apex the point N , the cubed ratio that BD (has) πυραμίδα, ἧς βάσις μέν ἐστι τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, to FH . And it was also assumed that the cone whose κορυφὴ δὲ τὸ Ν. μείζων δὲ ὁ εἰρημένος κῶνος τῆς ἐν αὐτῷ base is circle ABCD, and apex the point L, has to solid πυραμίδος· ἐμπεριέχει γὰρ αὐτὴν. μεῖζον ἄρα καὶ τὸ Ξ O the cubed ratio that BD (has) to FH . Thus, as the στερεὸν τῆς πυραμίδος, ἧς βάσις μέν ἐστι τὸ ΕΟΖΠΗΡΘΣ cone whose base is circle ABCD, and apex the point L, πολύγωνον, κορυφὴ δὲ τὸ Ν. ἀλλὰ καὶ ἔλαττον· ὅπερ ἐστὶν is to solid O, so the pyramid whose base (is) [polygon] ἀδύνατον. οὐκ ἄρα ὁ κῶνος, οὗ βάσις ὁ ΑΒΓΔ κύκλος, κο- ATBUCV DW , and apex the point L, (is) to the pyramid ρυφὴ δὲ τὸ Λ [σημεῖον], πρὸς ἔλαττόν τι τοῦ κώνου στερεόν, whose base is polygon EPFQGRHS, and apex the point οὗ βάσις μὲν ὁ ΕΖΗΘ κύκλος, κορυφὴ δὲ τὸ Ν σημεῖον, τρι- N . Thus, alternately, as the cone whose base (is) circle πλασίονα λόγον ἔχει ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. ὁμοίως δὴ ABCD, and apex the point L, (is) to the pyramid within δείξομεν, ὅτι οὐδὲ ὁ ΕΖΗΘΝ κῶνος πρὸς ἔλαττόν τι τοῦ it whose base (is) the polygon ATBUCV DW , and apex ΑΒΓΔΛ κώνου στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΖΘ the point L, so the [solid] O (is) to the pyramid whose πρὸς τὴν ΒΔ. base is polygon EPFQGRHS, and apex the point N Λέγω δή, ὅτι οὐδὲ ὁ ΑΒΓΔΛ κῶνος πρὸς μεῖζόν τι τοῦ [Prop. 5.16]. And the aforementioned cone (is) greater ΕΖΗΘΝ κώνου στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΔ than the pyramid within it. For it encompasses it. Thus, πρὸς τὴν ΖΘ. solid O (is) also greater than the pyramid whose base is Εἰ γὰρ δυνατόν, ἐχέτω πρὸς μεῖζον τὸ Ξ. ἀνάπαλιν ἄρα polygon EPFQGRHS, and apex the point N . But, (it τὸ Ξ στερεὸν πρὸς τὸν ΑΒΓΔΛ κῶνον τριπλασίονα λόγον is) also less. The very thing is impossible. Thus, the cone ἔχει ἤπερ ἡ ΖΘ πρὸς τὴν ΒΔ. ὡς δὲ τὸ Ξ στερεὸν πρὸς whose base (is) circle ABCD, and apex the [point] L, τὸν ΑΒΓΔΛ κῶνον, οὕτως ὁ ΕΖΗΘΝ κῶνος πρὸς ἔλαττόν does not have to some solid less than the cone whose τι τοῦ ΑΒΓΔΛ κώνου στερεόν. καὶ ὁ ΕΖΗΘΝ ἄρα κῶνος base (is) circle EFGH , and apex the point N , the cubed πρὸς ἔλαττόν τι τοῦ ΑΒΓΔΛ κώνου στερεὸν τριπλασίονα ratio that BD (has) to EH . So, similarly, we can show λόγον ἔχει ἤπερ ἡ ΖΘ πρὸς τὴν ΒΔ· ὅπερ ἀδύνατον ἐδείχθη. that neither does cone EFGHN have to some solid less οὐκ ἄρα ὁ ΑΒΓΔΛ κῶνος πρὸς μεῖζόν τι τοῦ ΕΖΗΘΝ than cone ABCDL the cubed ratio that FH (has) to BD. κώνου στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΔ πρὸς So, I say that neither does cone ABCDL have to some τὴν ΖΘ. ἐδείχθη δέ, ὅτι οὐδὲ πρὸς ἔλαττον. ὁ ΑΒΓΔΛ ἄρα solid greater than cone EFGHN the cubed ratio that BD κῶνος πρὸς τὸν ΕΖΗΘΝ κῶνον τριπλασίονα λόγον ἔχει (has) to FH . ἤπερ ἡ ΒΔ πρὸς τὴν ΖΘ. For, if possible, let it have (such a ratio) to a greater ῾Ως δὲ ὁ κῶνος πρὸς τὸν κῶνον, ὁ κύλινδρος πρὸς τὸν (solid), O. Thus, inversely, solid O has to cone ABCDL κύλινδρον· τριπλάσιος γὰρ ὁ κύλινδρος τοῦ κώνου ὁ ἐπὶ τῆς the cubed ratio that FH (has) to BD [Prop. 5.7 corr.]. αὐτῆς βάσεως τῷ κώνῳ καὶ ἰσοϋψὴς αὐτῷ. καὶ ὁ κύλινδρος And as solid O (is) to cone ABCDL, so cone EFGHN ἄρα πρὸς τὸν κύλινδρον τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΔ (is) to some solid less than cone ABCDL [12.2 lem.]. πρὸς τὴν ΖΘ. Thus, cone EFGHN also has to some solid less than cone Οἱ ἄρα ὅμοιοι κῶνοι καὶ κύλινδροι πρὸς ἀλλήλους ἐν ABCDL the cubed ratio that FH (has) to BD. The very 494 STOIQEIWN ibþ. ELEMENTS BOOK 12 τριπλασίονι λόγῳ εἰσὶ τῶν ἐν ταῖς βάσεσι διαμέτρων· ὅπερ thing was shown (to be) impossible. Thus, cone ABCDL ἔδει δεῖξαι. does not have to some solid greater than cone EFGHN the cubed ratio than BD (has) to FH . And it was shown that neither (does it have such a ratio) to a lesser (solid). Thus, cone ABCDL has to cone EFGHN the cubed ra- tio that BD (has) to FG. And as the cone (is) to the cone, so the cylinder (is) to the cylinder. For a cylinder is three times a cone on the same base as the cone, and of the same height as it [Prop. 12.10]. Thus, the cylinder also has to the cylinder the cubed ratio that BD (has) to FH . Thus, similar cones and cylinders are in the cubed ra- tio of the diameters of their bases. (Which is) the very thing it was required to show.igþ. Proposition 13 ᾿Εὰν κύλινδρος ἐπιπέδῳ τμηθῇ παραλλήλῳ ὄντι τοῖς ἀπε- If a cylinder is cut by a plane which is parallel to the ναντίον ἐπιπέδοις, ἔσται ὡς ὁ κύλινδρος πρὸς τὸν κύλινδρον, opposite planes (of the cylinder) then as the cylinder (is) οὕτως ὁ ἄξων πρὸς τὸν ἄξονα. to the cylinder, so the axis will be to the axis. Q O R A H G T F L N E K Z X M P S B J D U F R A T L N E K M S B D U P G C V WQ H O Κύλινδρος γὰρ ὁ ΑΔ ἐπιπέδῳ τῷ ΗΘ τετμήσθω πα- For let the cylinder AD have been cut by the plane ραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις τοῖς ΑΒ, ΓΔ, καὶ GH which is parallel to the opposite planes (of the cylin- συμβαλλέτω τῷ ἄξονι τὸ ΗΘ ἐπίπεδον κατὰ τὸ Κ σημεῖον· der), AB and CD. And let the plane GH have met the λέγω, ὅτι ἐστὶν ὡς ὁ ΒΗ κύλινδρος πρὸς τὸν ΗΔ κύλινδρον, axis at point K. I say that as cylinder BG is to cylinder οὕτως ὁ ΕΚ ἄξων πρὸς τὸν ΚΖ ἄξονα. GD, so axis EK (is) to axis KF . ᾿Εκβεβλήσθω γὰρ ὁ ΕΖ ἄξων ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ For let axis EF have been produced in each direction τὰ Λ, Μ σημεῖα, καὶ ἐκκείσθωσαν τῷ ΕΚ ἄξονι ἴσοι ὁσοι- to points L and M . And let any number whatsoever (of δηποτοῦν οἱ ΕΝ, ΝΛ, τῷ δὲ ΖΚ ἴσοι ὁσοιδηποτοῦν οἱ ΖΞ, lengths), EN and NL, equal to axis EK, be set out (on ΞΜ, καὶ νοείσθω ὁ ἐπὶ τοῦ ΛΜ ἄξονος κύλινδρος ὁ ΟΧ, the axis EL), and any number whatsoever (of lengths), οὗ βάσεις οἱ ΟΠ, ΦΧ κύκλοι. καὶ ἐκβεβλήσθω διὰ τῶν FO and OM , equal to (axis) FK, (on the axis KM). Ν, Ξ σημείων ἐπίπεδα παράλληλα τοῖς ΑΒ, ΓΔ καὶ ταῖς And let the cylinder PW , whose bases (are) the circles βάσεσι τοῦ ΟΧ κυλίνδρου καὶ ποιείτωσαν τοὺς ΡΣ, ΤΥ PQ and V W , have been conceived on axis LM . And κύκλους περὶ τὰ Ν, Ξ κέντρα. καὶ ἐπεὶ οἱ ΛΝ, ΝΕ, ΕΚ let planes parallel to AB, CD, and the bases of cylinder ἄξονες ἴσοι εἰσὶν ἀλλήλοις, οἱ ἄρα ΠΡ, ΡΒ, ΒΗ κύλινδροι PW , have been produced through points N and O, and πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις. ἴσαι δέ εἰσιν αἱ βάσεις· let them have made the circles RS and TU around the ἴσοι ἄρα καὶ οἱ ΠΡ, ΡΒ, ΒΗ κύλινδροι ἀλλήλοις. επεὶ οὖν centers N and O (respectively). And since axes LN , NE, οἱ ΛΝ, ΝΕ, ΕΚ ἄξονες ἴσοι εἰσὶν ἀλλήλοις, εἰσὶ δὲ καὶ οἱ and EK are equal to one another, the cylinders QR, RB, ΠΡ, ΡΒ, ΒΗ κύλινδροι ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ and BG are to one another as their bases [Prop. 12.11]. πλῆθος τῷ πλήθει, ὁσαπλασίων ἄρα ὁ ΚΛ ἄξων τοῦ ΕΚ But the bases are equal. Thus, the cylinders QR, RB, ἄξονος, τοσαυταπλασίων ἔσται καὶ ὁ ΠΗ κύλινδρος τοῦ ΗΒ and BG (are) also equal to one another. Therefore, since κυλίνδρου. διὰ τὰ αὐτὰ δὴ καὶ ὁσαπλασίων ἐστὶν ὁ ΜΚ ἄξων the axes LN , NE, and EK are equal to one another, τοῦ ΚΖ ἄξονος, τοσαυταπλασίων ἐστὶ καὶ ὁ ΧΗ κύλινδρος and the cylinders QR, RB, and BG are also equal to one τοῦ ΗΔ κυλίνδρου. καὶ εἰ μὲν ἴσος ἐστὶν ὁ ΚΛ ἄξων τῷ another, and the number (of the former) is equal to the ΚΜ ἄξονι, ἴσος ἔσται καὶ ὁ ΠΗ κύλινδρος τῷ ΗΧ κυλίνδρῳ, number (of the latter), thus as many multiples as axis KL 495 STOIQEIWN ibþ. ELEMENTS BOOK 12 εἰ δὲ μείζων ὁ ἄξων τοῦ ἄξονος, μείζων καὶ ὁ κύλινδρος is of axis EK, so many multiples is cylinder QG also of τοῦ κυλίνδρου, καὶ εἰ ἐλάσσων, ἐλάσσων. τεσσάρων δὴ με- cylinder GB. And so, for the same (reasons), as many γεθῶν ὄντων, ἀξόνων μὲν τῶν ΕΚ, ΚΖ, κυλίνδρων δὲ τῶν multiples as axis MK is of axis KF , so many multiples ΒΗ, ΗΔ, εἴληπται ἰσάκις πολλαπλάσια, τοῦ μὲν ΕΚ ἄξονος is cylinder WG also of cylinder GD. And if axis KL is καὶ τοῦ ΒΗ κυλίνδρου ὅ τε ΛΚ ἄξων καὶ ὁ ΠΗ κύλινδρος, equal to axis KM then cylinder QG will also be equal τοῦ δὲ ΚΖ ἄξονες καὶ τοῦ ΗΔ κυλίνδρου ὅ τε ΚΜ ἄξων to cylinder GW , and if the axis (is) greater than the axis καὶ ὁ ΗΧ κύλινδρος, καὶ δέδεικται, ὅτι εἰ ὑπερέχει ὁ ΚΛ then the cylinder (will also be) greater than the cylinder, ἄξων τοῦ ΚΜ ἄξονος, ὑπερέχει καὶ ὁ ΠΗ κύλινδρος τοῦ and if (the axis is) less then (the cylinder will also be) ΗΧ κυλίνδρου, καὶ εἰ ἴσος, ἴσος, καὶ εἰ ἐλάσσων, ἐλάσσων. less. So, there are four magnitudes—the axes EK and ἔστιν ἄρα ὡς ὁ ΕΚ ἄξων πρὸς τὸν ΚΖ ἄξονα, οὕτως ὁ ΒΗ KF , and the cylinders BG and GD—and equal multiples κύλινδρος πρὸς τὸν ΗΔ κύλινδρον· ὅπερ ἔδει δεῖξαι. have been taken of axis EK and cylinder BG—(namely), axis LK and cylinder QG—and of axis KF and cylinder GD—(namely), axis KM and cylinder GW . And it has been shown that if axis KL exceeds axis KM then cylin- der QG also exceeds cylinder GW , and if (the axes are) equal then (the cylinders are) equal, and if (KL is) less then (QG is) less. Thus, as axis EK is to axis KF , so cylinder BG (is) to cylinder GD [Def. 5.5]. (Which is) the very thing it was required to show.idþ. Proposition 14 Οἱ ἐπὶ ἴσων βάσεων ὄντες κῶνοι καὶ κύλινδροι πρὸς Cones and cylinders which are on equal bases are to αλλήλους εἰσὶν ὡς τὰ ὕψη. one another as their heights. LE A B K D N M Z G J H L E A B K G H C D N M F ῎Εστωσαν γὰρ ἐπὶ ἴσων βάσεων τῶν ΑΒ, ΓΔ κύκλων For let EB and FD be cylinders on equal bases, κύλινδροι οἱ ΕΒ, ΖΔ· λέγω, ὅτι ἐστὶν ὡς ὁ ΕΒ κύλινδρος (namely) the circles AB and CD (respectively). I say πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΗΘ ἄξων πρὸς τὸν ΚΛ that as cylinder EB is to cylinder FD, so axis GH (is) to ἄξονα. axis KL. ᾿Εκβεβλήσθω γὰρ ὁ ΚΛ ἄξων ἐπὶ τὸ Ν σημεῖον, καὶ For let the axis KL have been produced to point N . κείσθω τῷ ΗΘ ἄξονι ἴσος ὁ ΛΝ, καὶ περὶ ἄξονα τὸν ΛΝ And let LN be made equal to axis GH . And let the cylin- κύλινδρος νενοήσθω ὁ ΓΜ. ἐπεὶ οὖν οἱ ΕΒ, ΓΜ κύλινδροι der CM have been conceived about axis LN . Therefore, ὑπὸ τὸ αὐτὸ ὕψος εἰσίν, πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις. since cylinders EB and CM have the same height they ἴσαι δέ εἰσίν αἱ βάσεις ἀλλήλαις· ἴσοι ἄρα εἰσὶ καὶ οἱ ΕΒ, ΓΜ are to one another as their bases [Prop. 12.11]. And the κύλινδροι. καὶ ἐπεὶ κύλινδρος ὁ ΖΜ ἐπιπέδῳ τέτμηται τῷ bases are equal to one another. Thus, cylinders EB and ΓΔ παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔστιν ἄρα ὡς CM are also equal to one another. And since cylinder ὁ ΓΜ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΛΝ ἄξων FM has been cut by the plane CD, which is parallel to πρὸς τὸν ΚΛ ἄξονα. ἴσος δέ ἐστιν ὁ μὲν ΓΜ κύλινδρος τῷ its opposite planes, thus as cylinder CM is to cylinder ΕΒ κυλίνδρῳ, ὁ δὲ ΛΝ ἄξων τῷ ΗΘ ἄξονι· ἔστιν ἄρα ὡς ὁ FD, so axis LN (is) to axis KL [Prop. 12.13]. And cylin- ΕΒ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΗΘ ἄξων der CM is equal to cylinder EB, and axis LN to axis GH . πρὸς τὸν ΚΛ ἄξονα. ὡς δὲ ὁ ΕΒ κύλινδρος πρὸς τὸν ΖΔ Thus, as cylinder EB is to cylinder FD, so axis GH (is) 496 STOIQEIWN ibþ. ELEMENTS BOOK 12 κύλινδρον, οὑτως ὁ ΑΒΗ κῶνος πρὸς τὸν ΓΔΚ κῶνον. καὶ to axis KL. And as cylinder EB (is) to cylinder FD, so ὡς ἄρα ὁ ΗΘ ἄξων πρὸς τὸν ΚΛ ἄξονα, οὕτως ὁ ΑΒΗ cone ABG (is) to cone CDK [Prop. 12.10]. Thus, also, κῶνος πρὸς τὸν ΓΔΚ κῶνον καὶ ὁ ΕΒ κύλινδρος πρὸς τὸν as axis GH (is) to axis KL, so cone ABG (is) to cone ΖΔ κύλινδρον· ὅπερ ἔδει δεῖξαι. CDK, and cylinder EB to cylinder FD. (Which is) the very thing it was required to show.ieþ. Proposition 15 Τῶν ἴσων κώνων καὶ κυλίνδρων ἀντιπεπόνθασιν αἱ The bases of equal cones and cylinders are recipro- βάσεις τοῖς ὕψεσιν· καὶ ὧν κώνων καὶ κυλίνδρων ἀντι- cally proportional to their heights. And, those cones and πεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, ἴσοι εἰσὶν ἐκεῖνοι. cylinders whose bases (are) reciprocally proportional to their heights are equal. UA L DK B G X H E ZSM R J NP O T T A L D K B E S M R N C O F G H Q P U ῎Εστωσαν ἴσοι κῶνοι καὶ κύλινδροι, ὧν βάσεις μὲν οἱ Let there be equal cones and cylinders whose bases ΑΒΓΔ, ΕΖΗΘ κύκλοι, διάμετροι δὲ αὐτῶν αἱ ΑΓ, ΕΗ, are the circles ABCD and EFGH , and the diameters ἅξονες δὲ οἱ ΚΛ, ΜΝ, οἵτινες καὶ ὕψη εἰσὶ τῶν κώνων ἢ of (the bases) AC and EG, and (whose) axes (are) KL κυλίνδρων, καὶ συμπεπληρώσθωσαν οἱ ΑΞ, ΕΟ κύλινδροι. and MN , which are also the heights of the cones and λέγω, ὅτι τῶν ΑΞ, ΕΟ κυλίνδρων ἀντιπεπόνθασιν αἱ βάσεις cylinders (respectively). And let the cylinders AO and τοῖς ὕψεσιν, καί ἐστιν ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ EP have been completed. I say that the bases of cylinders βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος. AO and EP are reciprocally proportional to their heights, Τὸ γὰρ ΛΚ ὕψος τῷ ΜΝ ὕψει ἤτοι ἴσον ἐστὶν ἢ οὔ. and (so) as base ABCD is to base EFGH , so height MN ἔστω πρότερον ἴσον. ἔστι δὲ καὶ ὁ ΑΞ κύλινδρος τῷ ΕΟ (is) to height KL. κυλίνδρῳ ἴσος. οἱ δὲ ὑπὸ τὸ αὐτὸ ὕψος ὄντες κῶνοι καὶ For height LK is either equal to height MN , or not. κύλινδροι πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις· ἴση ἄρα καὶ Let it, first of all, be equal. And cylinder AO is also equal ἡ ΑΒΓΔ βάσις τῇ ΕΖΗΘ βάσει. ὥστε καὶ ἀντιπέπονθεν, to cylinder EP . And cones and cylinders having the same ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ height are to one another as their bases [Prop. 12.11]. ὕψος πρὸς τὸ ΚΛ ὕψος. ἀλλὰ δὴ μὴ ἔστω τὸ ΛΚ ὕψος Thus, base ABCD (is) also equal to base EFGH . And, τῷ ΜΝ ἴσον, ἀλλ᾿ ἔστω μεῖζον τὸ ΜΝ, καὶ ἀφῃρήσθω ἀπὸ hence, reciprocally, as base ABCD (is) to base EFGH , τοῦ ΜΝ ὕψους τῷ ΚΛ ἴσον τὸ ΠΝ, καὶ διὰ τοῦ Π σημείου so height MN (is) to height KL. And so, let height LK τετμήσθω ὁ ΕΟ κύλινδρος ἐπιπέδῳ τῷ ΤΥΣ παραλλήλῳ not be equal to MN , but let MN be greater. And let QN , τοῖς τῶν ΕΖΗΘ, ΡΟ κύκλων ἐπιπέδοις, καὶ ἀπὸ βάσεως μὲν equal to KL, have been cut off from height MN . And τοῦ ΕΖΗΘ κύκλου, ὕψους δὲ τοῦ ΝΠ κύλινδρος νενοήσθω let the cylinder EP have been cut, through point Q, by ὁ ΕΣ. καί ἐπεὶ ἴσος ἐστὶν ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ, the plane TUS (which is) parallel to the planes of the ἔστιν ἄρα ὡς ὁ ΑΞ κύλινδρος πρὸς τὸν ΕΣ κυλίνδρον, οὕτως circles EFGH and RP . And let cylinder ES have been ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον. ἀλλ᾿ ὡς μὲν ὁ ΑΞ conceived, with base the circle EFGH , and height NQ. κύλινδρος πρὸς τὸν ΕΣ κύλινδρον, οὕτως ἡ ΑΒΓΔ βάσις And since cylinder AO is equal to cylinder EP , thus, as πρὸς τὴν ΕΖΗΘ· ὑπὸ γὰρ τὸ αὐτὸ ὕψος εἰσὶν οἱ ΑΞ, ΕΣ cylinder AO (is) to cylinder ES, so cylinder EP (is) to κύλινδροι· ὡς δὲ ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ, οὕτως τὸ cylinder ES [Prop. 5.7]. But, as cylinder AO (is) to cylin- ΜΝ ὕψος πρὸς τὸ ΠΝ ὕψος· ὁ γὰρ ΕΟ κύλινδρος ἐπιπέδῳ der ES, so base ABCD (is) to base EFGH . For cylinders τέτμηται παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις. ἔστιν AO and ES (have) the same height [Prop. 12.11]. And ἄρα καὶ ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ as cylinder EP (is) to (cylinder) ES, so height MN (is) ΜΝ ὕψος πρὸς τὸ ΠΝ ὕψος. ἴσον δὲ τὸ ΠΝ ὕψος τῷ ΚΛ to height QN . For cylinder EP has been cut by a plane ὕψει· ἔστιν ἄρα ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, which is parallel to its opposite planes [Prop. 12.13]. οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος. τῶν ἄρα ΑΞ, ΕΟ And, thus, as base ABCD is to base EFGH , so height κυλίνδρων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν. MN (is) to height QN [Prop. 5.11]. And height QN 497 STOIQEIWN ibþ. ELEMENTS BOOK 12 Ἀλλὰ δὴ τῶν ΑΞ, ΕΟ κυλίνδρων ἀντιπεπονθέτωσαν αἱ (is) equal to height KL. Thus, as base ABCD is to base βάσεις τοῖς ὕψεσιν, καὶ ἔστω ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν EFGH , so height MN (is) to height KL. Thus, the bases ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος· λέγω, of cylinders AO and EP are reciprocally proportional to ὅτι ἴσος ἐστὶν ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ. their heights. Τῶν γὰρ αὐτῶν κατασκευασθέντων ἐπεί ἐστιν ὡς ἡ And, so, let the bases of cylinders AO and EP be ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος reciprocally proportional to their heights, and (thus) let πρὸς τὸ ΚΛ ὕψος, ἴσον δὲ τὸ ΚΛ ὕψος τῷ ΠΝ ὕψει, ἔσται base ABCD be to base EFGH , as height MN (is) to ἄρα ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ height KL. I say that cylinder AO is equal to cylinder ΜΝ ὕψος πρὸς τὸ ΠΝ ὕψος. ἀλλ᾿ ὡς μὲν ἡ ΑΒΓΔ βάσις EP . πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως ὁ ΑΞ κύλινδρος πρὸς τὸν For, with the same construction, since as base ABCD ΕΣ κύλινδρον· ὑπὸ γὰρ τὸ αὐτὸ ὕψος εἰσίν· ὡς δὲ τὸ ΜΝ is to base EFGH , so height MN (is) to height KL, and ὕψος πρὸς τὸ ΠΝ [ὕψος], οὕτως ὁ ΕΟ κύλινδρος πρὸς τὸν height KL (is) equal to height QN , thus, as base ABCD ΕΣ κύλινδρον· ἔστιν ἄρα ὡς ὁ ΑΞ κύλινδρος πρὸς τὸν ΕΣ (is) to base EFGH , so height MN will be to height κύλινδρον, οὕτως ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ. ἴσος ἄρα QN . But, as base ABCD (is) to base EFGH , so cylin- ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ. ὡσαύτως δὲ καὶ ἐπὶ τῶν der AO (is) to cylinder ES. For they are the same height κώνων· ὅπερ ἔδει δεῖξαι. [Prop. 12.11]. And as height MN (is) to [height] QN , so cylinder EP (is) to cylinder ES [Prop. 12.13]. Thus, as cylinder AO is to cylinder ES, so cylinder EP (is) to (cylinder) ES [Prop. 5.11]. Thus, cylinder AO (is) equal to cylinder EP [Prop. 5.9]. In the same manner, (the proposition can) also (be demonstrated) for the cones. (Which is) the very thing it was required to show.i�þ. Proposition 16 Δύο κύκλων περὶ τὸ αὐτὸ κέντρον ὄντων εἰς τὸν There being two circles about the same center, to μείζονα κύκλον πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον inscribe an equilateral and even-sided polygon in the ἐγγράψαι μὴ ψαῦον τοῦ ἐλάσσονος κύκλου. greater circle, not touching the lesser circle. H D Z KE L N M J B A G M D KE L N B AH F C G ῎Εστωσαν οἱ δοθέντες δύο κύκλοι οἱ ΑΒΓΔ, ΕΖΗΘ Let ABCD and EFGH be the given two circles, about περὶ τὸ αὐτὸ κέντρον τὸ Κ· δεῖ δὴ εἰς τὸν μείζονα κύκλον the same center, K. So, it is necessary to inscribe an τὸν ΑΒΓΔ πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον equilateral and even-sided polygon in the greater circle ἐγγράψαι μὴ ψαῦον τοῦ ΕΖΗΘ κύκλου. ABCD, not touching circle EFGH . ῎Ηχθω γὰρ διὰ τοῦ Κ κέντρου εὐθεῖα ἡ ΒΚΔ, καὶ Let the straight-line BKD have been drawn through ἀπὸ τοῦ Η σημείου τῇ ΒΔ εὐθείᾳ πρὸς ὀρθὰς ἤχθω ἡ the center K. And let GA have been drawn, at right- ΗΑ καὶ διήχθω ἐπὶ τὸ Γ· ἡ ΑΓ ἄρα ἐφάπτεται τοῦ ΕΖΗΘ angles to the straight-line BD, through point G, and let it κύκλου. τέμνοντες δὴ τὴν ΒΑΔ περιφέρειαν δίχα καὶ τὴν have been drawn through to C. Thus, AC touches circle ἡμίσειαν αὐτῆς δίχα καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομεν EFGH [Prop. 3.16 corr.]. So, (by) cutting circumference περιφέρειαν ἐλάσσονα τῆς ΑΔ. λελείφθω, καὶ ἔστω ἡ ΛΔ, BAD in half, and the half of it in half, and doing this con- καὶ ἀπὸ τοῦ Λ ἐπὶ τὴν ΒΔ κάθετος ἤχθω ἡ ΛΜ καὶ διήχθω tinually, we will (eventually) leave a circumference less 498 STOIQEIWN ibþ. ELEMENTS BOOK 12 ἐπὶ τὸ Ν, καὶ ἐπεζεύχθωσαν αἱ ΛΔ, ΔΝ· ἴση ἄρα ἐστὶν ἡ than AD [Prop. 10.1]. Let it have been left, and let it be ΛΔ τῇ ΔΝ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΛΝ τῇ ΑΓ, ἡ δὲ ΑΓ LD. And let LM have been drawn, from L, perpendicu- ἐφάπτεται τοῦ ΕΖΗΘ κύκλου, ἡ ΛΝ ἄρα οὐκ ἐφάπτεται τοῦ lar to BD, and let it have been drawn through to N . And ΕΖΗΘ κύκλου· πολλῷ ἄρα αἱ ΛΔ, ΔΝ οὐκ ἐφάπτονται τοῦ let LD and DN have been joined. Thus, LD is equal to ΕΖΗΘ κύκλου. ἐὰν δὴ τῇ ΛΔ εὐθείᾳ ἴσας κατὰ τὸ συνεχὲς DN [Props. 3.3, 1.4]. And since LN is parallel to AC ἐναρμόσωμεν εἰς τὸν ΑΒΓΔ κύκλον, ἐγγραφήσεται εἰς τὸν [Prop. 1.28], and AC touches circle EFGH , LN thus ΑΒΓΔ κύκλον πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον does not touch circle EFGH . Thus, even more so, LD μὴ ψαῦον τοῦ ἐλάσσονος κύκλου τοῦ ΕΖΗΘ· ὅπερ ἔδει and DN do not touch circle EFGH . And if we continu- ποιῆσαι. ously insert (straight-lines) equal to straight-line LD into circle ABCD [Prop. 4.1] then an equilateral and even- sided polygon, not touching the lesser circle EFGH , will have been inscribed in circle ABCD.† (Which is) the very thing it was required to do. † Note that the chord of the polygon, LN , does not touch the inner circle either.izþ. Proposition 17 Δύο σφαιρῶν περὶ τὸ αὐτὸ κέντρον οὐσῶν εἰς τὴν There being two spheres about the same center, to in- μείζονα σφαῖραν στερεὸν πολύεδρον ἐγγράψαι μὴ ψαῦον τῆς scribe a polyhedral solid in the greater sphere, not touch- ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν. ing the lesser sphere on its surface.E U D N G ML K B W A ZQ J RS TP XOY F H U D N M L K B A R S T F H Y W Q G C O P X E V Νενοήσθωσαν δύο σφαῖραι περὶ τὸ αὐτὸ κέντρον τὸ Α· Let two spheres have been conceived about the same δεῖ δὴ εἰς τὴν μείζονα σφαῖραν στερεὸν πολύεδρον ἐγγράψαι center, A. So, it is necessary to inscribe a polyhedral solid μὴ ψαῦον τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν. in the greater sphere, not touching the lesser sphere on Τετμήσθωσαν αἱ σφαῖραι ἐπιπέδῳ τινὶ διὰ τοῦ κέντρου· its surface. ἔσονται δὴ αἱ τομαὶ κύκλοι, ἐπειδήπερ μενούσης τῆς Let the spheres have been cut by some plane through διαμέτρου καὶ περιφερομένου τοῦ ἡμικυκλίου ἐγιγνετο ἡ the center. So, the sections will be circles, inasmuch σφαῖρα· ὥστε καὶ καθ᾿ οἵας ἂν θέσεως ἐπινοήσωμεν τὸ as a sphere is generated by the diameter remaining be- ἡμικύκλιον, τὸ δι᾿ αὐτοῦ ἐκβαλλόμενον ἐπίπεδον ποιήσει hind, and a semi-circle being carried around [Def. 11.14]. ἐπὶ τῆς ἐπιφανείας τῆς σφαίρας κύκλον. καὶ φανερόν, And, hence, whatever position we conceive (of for) the ὅτι καὶ μέγιστον, ἐπειδήπερ ἡ διάμετρος τῆς σφαίρας, ἥτις semi-circle, the plane produced through it will make a 499 STOIQEIWN ibþ. ELEMENTS BOOK 12 ἐστὶ καὶ τοῦ ἡμικυκλίου διάμετρος δηλαδὴ καὶ τοῦ κύκλου, circle on the surface of the sphere. And (it is) clear μείζων ἐστὶ πασῶν τῶν εἰς τὸν κύκλον ἢ τὴν σφαῖραν δια- that (it is) also a great (circle), inasmuch as the diam- γομένων [εὐθειῶν]. ἔστω οὖν ἐν μὲν τῇ μείζονι σφαίρᾳ eter of the sphere, which is also manifestly the diame- κύκλος ὁ ΒΓΔΕ, ἐν δὲ τῇ ἐλάσσονι σφαίρᾳ κύκλος ὁ ΖΗΘ, ter of the semi-circle and the circle, is greater than all καὶ ἤχθωσαν αὐτῶν δύο διάμετροι πρὸς ὀρθὰς ἀλλήλαις αἱ of the (other) [straight-lines] drawn across in the cir- ΒΔ, ΓΕ, καὶ δύο κύκλων περὶ τὸ αὐτὸ κέντρον ὄντων τῶν cle or the sphere [Prop. 3.15]. Therefore, let BCDE be ΒΓΔΕ, ΖΗΘ εἰς τὸν μείζονα κύκλον τὸν ΒΓΔΕ πολύγωνον the circle in the greater sphere, and FGH the circle in ἰσόπλευρον καὶ ἀρτιόπλευρον ἐγγεγράφθω μὴ ψαῦον τοῦ the lesser sphere. And let two diameters of them have ἐλάσσονος κύκλου τοῦ ΖΗΘ, οὗ πλευραὶ ἔστωσαν ἐν τῷ ΒΕ been drawn at right-angles to one another, (namely), τεταρτημορίῳ αἱ ΒΚ, ΚΛ, ΛΜ, ΜΕ, καὶ ἐπιζευχθεῖσα ἡ ΚΑ BD and CE. And there being two circles about the διήχθω ἐπὶ τὸ Ν, καὶ ἀνεστάτω ἀπὸ τοῦ Α σημείου τῷ τοῦ same center—(namely), BCDE and FGH—let an equi- ΒΓΔΕ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΑΞ καὶ συμβαλλέτω lateral and even-sided polygon have been inscribed in τῇ ἐπιφανείᾳ τῆς σφαίρας κατὰ τὸ Ξ, καὶ διὰ τῆς ΑΞ καὶ the greater circle, BCDE, not touching the lesser circle, ἑκατέρας τῶν ΒΔ, ΚΝ ἐπίπεδα ἐκβεβλήσθω· ποιήσουσι δὴ FGH [Prop. 12.16], of which let the sides in the quad- διὰ τὰ εἰρημένα ἐπὶ τῆς ἐπιφανείας τῆς σφαίρας μεγίστους rant BE be BK, KL, LM , and ME. And, KA being κύκλους. ποιείτωσαν, ὧν ἡμικύκλια ἔστω ἐπὶ τῶν ΒΔ, ΚΝ joined, let it have been drawn across to N . And let AO διαμέτρων τὰ ΒΞΔ, ΚΞΝ. καὶ ἐπεὶ ἡ ΞΑ ὀρθή ἐστι πρὸς τὸ have been set up at point A, at right-angles to the plane of τοῦ ΒΓΔΕ κύκλου ἐπίπεδον, καὶ πάντα ἄρα τὰ διὰ τῆς ΞΑ circle BCDE. And let it meet the surface of the (greater) ἐπίπεδά ἐστιν ὀρθὰ πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον· sphere at O. And let planes have been produced through ὥστε καὶ τὰ ΒΞΔ, ΚΞΝ ἡμικύκλια ὀρθά ἐστι πρὸς τὸ τοῦ AO and each of BD and KN . So, according to the afore- ΒΓΔΕ κύκλου ἐπίπεδον. καὶ ἐπεὶ ἴσα ἐστὶ τὰ ΒΕΔ, ΒΞΔ, mentioned (discussion), they will make great circles on ΚΞΝ ἡμικύκλια· ἐπὶ γὰρ ἴσων εἰσὶ διαμέτρων τῶν ΒΔ, ΚΝ· the surface of the (greater) sphere. Let them make (great ἴσα ἐστὶ καὶ τὰ ΒΕ, ΒΞ, ΚΞ τεταρτημόρια ἀλλήλοις. ὄσαι circles), of which let BOD and KON be semi-circles on ἄρα εἰσὶν ἐν τῷ ΒΕ τεταρτημορίῳ πλευραὶ τοῦ πολυγώνου, the diameters BD and KN (respectively). And since OA τοσαῦταί εἰσι καὶ ἐν τοῖς ΒΞ, ΚΞ τεταρτημορίοις ἴσαι ταῖς is at right-angles to the plane of circle BCDE, all of the ΒΚ, ΚΛ, ΛΜ, ΜΕ εὐθείαις. ἐγγεγράφθωσαν καὶ ἔστωσαν planes through OA are thus also at right-angles to the αἱ ΒΟ, ΟΠ, ΠΡ, ΡΞ, ΚΣ, ΣΤ, ΤΥ, ΥΞ, καὶ ἐπεζεύχθωσαν plane of circle BCDE [Prop. 11.18]. And, hence, the αἱ ΣΟ, ΤΠ, ΥΡ, καὶ ἀπὸ τῶν Ο, Σ ἐπὶ τὸ τοῦ ΒΓΔΕ semi-circles BOD and KON are also at right-angles to κύκλου ἐπίπεδον κάθετοι ἤχθωσαν· πεσοῦνται δὴ ἐπὶ τὰς the plane of circle BCDE. And since semi-circles BED, κοινὰς τομὰς τῶν ἐπιπέδων τὰς ΒΔ, ΚΝ, ἐπειδήπερ καὶ BOD, and KON are equal—for (they are) on the equal τὰ τῶν ΒΞΔ, ΚΞΝ ἐπίπεδα ὀρθά ἐστι πρὸς τὸ τοῦ ΒΓΔΕ diameters BD and KN [Def. 3.1]—the quadrants BE, κύκλου ἐπίπεδον. πιπτέτωσαν, καὶ ἔστωσαν αἱ ΟΦ, ΣΧ, BO, and KO are also equal to one another. Thus, as καὶ ἐπεζεύχθω ἡ ΧΦ. καὶ ἐπεὶ ἐν ἴσοις ἡμικυκλίοις τοῖς many sides of the polygon as are in quadrant BE, so ΒΞΔ, ΚΞΝ ἴσαι ἀπειλημμέναι εἰσὶν αἱ ΒΟ, ΚΣ, καὶ κάθετοι many are also in quadrants BO and KO equal to the ἠγμέναι εἰσὶν αἱ ΟΦ, ΣΧ, ἴση [ἄρα] ἐστὶν ἡ μὲν ΟΦ τῇ ΣΧ, straight-lines BK, KL, LM , and ME. Let them have ἡ δὲ ΒΦ τῇ ΚΧ. ἔστι δὲ καὶ ὅλη ἡ ΒΑ ὅλῃ τῇ ΚΑ ἴση· καὶ been inscribed, and let them be BP , PQ, QR, RO, KS, λοιπὴ ἄρα ἡ ΦΑ λοιπῇ τῇ ΧΑ ἐστιν ἴση· ἔστιν ἄρα ὡς ἡ ΒΦ ST , TU , and UO. And let SP , TQ, and UR have been πρὸς τὴν ΦΑ, οὕτως ἡ ΚΧ πρὸς τὴν ΧΑ· παράλληλος ἄρα joined. And let perpendiculars have been drawn from P ἐστὶν ἡ ΧΦ τῇ ΚΒ. καὶ ἐπεὶ ἑκατέρα τῶν ΟΦ, ΣΧ ὀρθή and S to the plane of circle BCDE [Prop. 11.11]. So, ἐστι πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον, παράλληλος ἄρα they will fall on the common sections of the planes BD ἐστὶν ἡ ΟΦ τῇ ΣΧ. ἐδείχθη δὲ αὐτῇ καὶ ἴση· καὶ αἱ ΧΦ, ΣΟ and KN (with BCDE), inasmuch as the planes of BOD ἄρα ἴσαι εἰσὶ καὶ παράλληλοι. καὶ ἐπεὶ παράλληλός ἐστιν and KON are also at right-angles to the plane of circle ἡ ΧΦ τῇ ΣΟ, ἀλλὰ ἡ ΧΦ τῇ ΚΒ ἐστι παράλληλος, καὶ BCDE [Def. 11.4]. Let them have fallen, and let them be ἡ ΣΟ ἄρα τῇ ΚΒ ἐστι παράλληλος. καὶ ἐπιζευγνύουσιν PV and SW . And let WV have been joined. And since αὐτὰς αἱ ΒΟ, ΚΣ· τὸ ΚΒΟΣ ἄρα τετράπλευρον ἐν ἑνί BP and KS are equal (circumferences) having been cut ἐστιν ἐπιπέδῳ, ἐπειδήπερ, ἐὰν ὦσι δύο εὐθεῖαι παράλληλοι, off in the equal semi-circles BOD and KON [Def. 3.28], καὶ ἐφ᾿ ἑκατέρας αὐτῶν ληφθῇ τυχόντα σημεῖα, ἡ ἐπὶ τὰ and PV and SW are perpendiculars having been drawn σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐν τῷ αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς (from them), PV is [thus] equal to SW , and BV to KW παραλλήλοις. διὰ τὰ αὐτὰ δὴ καὶ ἑκάτερον τῶν ΣΟΠΤ, [Props. 3.27, 1.26]. And the whole of BA is also equal ΤΠΡΥ τετραπλεύρων ἐν ἑνί ἐστιν ἐπιπέδῳ. ἔστι δὲ καὶ to the whole of KA. And, thus, as BV is to V A, so KW τὸ ΥΡΞ τρίγωνον ἐν ἑνὶ ἑπιπέδῳ. ἐὰν δὴ νοήσωμεν ἀπὸ (is) to WA. WV is thus parallel to KB [Prop. 6.2]. And 500 STOIQEIWN ibþ. ELEMENTS BOOK 12 τῶν Ο, Σ, Π, Τ, Ρ, Υ σημείων ἐπὶ τὸ Α ἐπιζευγνυμένας since PV and SW are each at right-angles to the plane εὐθείας, συσταθήσεταί τι σχῆμα στερεὸν πολύεδρον ματαξὺ of circle BCDE, PV is thus parallel to SW [Prop. 11.6]. τῶν ΒΞ, ΚΞ περιφερειῶν ἐκ πυραμίδων συγκείμενον, ὧν And it was also shown (to be) equal to it. And, thus, βάσεις μὲν τὰ ΚΒΟΣ, ΣΟΠΤ, ΤΠΡΥ τετράπλευρα καὶ τὸ WV and SP are equal and parallel [Prop. 1.33]. And ΥΡΞ τρίγωνον, κορυφὴ δὲ τὸ Α σημεῖον. ἐὰν δὲ καὶ ἐπὶ since WV is parallel to SP , but WV is parallel to KB, ἑκάστης τῶν ΚΛ, ΛΜ, ΜΕ πλευρῶν καθάπερ ἐπὶ τῆς ΒΚ τὰ SP is thus also parallel to KB [Prop. 11.1]. And BP αὐτὰ κατασκευάσωμεν καὶ ἔτι τῶν λοιπῶν τριῶν τεταρτη- and KS join them. Thus, the quadrilateral KBPS is in μορίων, συσταθήσεταί τι σχῆμα πολύεδρον ἐγγεγραμμένον one plane, inasmuch as if there are two parallel straight- εἰς τὴν σφαῖραν πυραμίσι περιεχόμενον, ὧν βάσιες [μὲν] τὰ lines, and a random point is taken on each of them, then εἰρημένα τετράπλευρα καὶ τὸ ΥΡΞ τρίγωνον καὶ τὰ ὁμοταγῆ the straight-line joining the points is in the same plane αὐτοῖς, κορυφὴ δὲ τὸ Α σημεῖον. as the parallel (straight-lines) [Prop. 11.7]. So, for the Λέγω ὅτι τὸ εἰρημένον πολύεδρον οὐκ ἐφάψεται τῆς same (reasons), each of the quadrilaterals SPQT and ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν, ἐφ᾿ ἧς ἐστιν ὁ TQRU is also in one plane. And triangle URO is also ΖΗΘ κύκλος. in one plane [Prop. 11.2]. So, if we conceive straight- ῎Ηχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὸ τοῦ ΚΒΟΣ τε- lines joining points P , S, Q, T , R, and U to A then τραπλεύρου ἐπίπεδον κάθετος ἡ ΑΨ καὶ συμβαλλέτω τῷ some solid polyhedral figure will have been constructed ἐπιπέδῳ κατὰ τὸ Ψ σημεῖον, καὶ ἐπεζεύχθωσαν αἱ ΨΒ, ΨΚ. between the circumferences BO and KO, being com- καὶ ἐπεὶ ἡ ΑΨ ὀρθή ἐστι πρὸς τὸ τοῦ ΚΒΟΣ τετραπλεύρου posed of pyramids whose bases (are) the quadrilaterals ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας KBPS, SPQT , TQRU , and the triangle URO, and apex καὶ οὔσας ἐν τῷ τοῦ τετραπλεύρου ἐπιπέδῳ ὀρθή ἐστιν. ἡ the point A. And if we also make the same construction ΑΨ ἄρα ὀρθή ἐστι πρὸς ἑκατέραν τῶν ΒΨ, ΨΚ. καὶ ἐπεὶ on each of the sides KL, LM , and ME, just as on BK, ἴση ἐστὶν ἡ ΑΒ τῇ ΑΚ, ἵσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΒ τῷ and, further, (repeat the construction) in the remaining ἀπὸ τῆς ΑΚ. καί ἐστι τῷ μὲν ἀπὸ τῆς ΑΒ ἴσα τὰ ἀπὸ τῶν three quadrants, then some polyhedral figure which has ΑΨ, ΨΒ· ὀρθὴ γὰρ ἡ πρὸς τῷ Ψ· τῷ δὲ ἀπὸ τῆς ΑΚ ἴσα τὰ been inscribed in the sphere will have been constructed, ἀπὸ τῶν ΑΨ, ΨΚ. τὰ ἄρα ἀπὸ τῶν ΑΨ, ΨΒ ἴσα ἐστὶ τοῖς being contained by pyramids whose bases (are) the afore- ἀπὸ τῶν ΑΨ, ΨΚ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΑΨ· λοιπὸν mentioned quadrilaterals, and triangle URO, and the ἄρα τὸ ἀπὸ τῆς ΒΨ λοιπῷ τῷ ἀπὸ τῆς ΨΚ ἴσον ἐστίν· ἴση (quadrilaterals and triangles) similarly arranged to them, ἄρα ἡ ΒΨ τῇ ΨΚ. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ἀπὸ τοῦ Ψ and apex the point A. ἐπὶ τὰ Ο, Σ ἐπιζευγνύμεναι εὐθεῖαι ἴσαι εἰσὶν ἑκατέρᾳ τῶν So, I say that the aforementioned polyhedron will not ΒΨ, ΨΚ. ὁ ἄρα κέντρῳ τῷ Ψ καὶ διαστήματι ἑνὶ τῶν ΨΒ, touch the lesser sphere on the surface on which the circle ΨΚ γραφόμενος κύκλος ἥξει καὶ διὰ τῶν Ο, Σ, καὶ ἔσται FGH is (situated). ἐν κύκλῳ τὸ ΚΒΟΣ τετράπλευρον. Let the perpendicular (straight-line) AX have been Καὶ ἐπεὶ μείζων ἐστὶν ἡ ΚΒ τῆς ΧΦ, ἴση δὲ ἡ ΧΦ τῇ drawn from point A to the plane KBPS, and let it meet ΣΟ, μείζων ἄρα ἡ ΚΒ τῆς ΣΟ. ἴση δὲ ἡ ΚΒ ἑκατέρᾳ τῶν the plane at point X [Prop. 11.11]. And let XB and ΚΣ, ΒΟ· καὶ ἑκατέρα ἄρα τῶν ΚΣ, ΒΟ τῆς ΣΟ μείζων XK have been joined. And since AX is at right-angles to ἐστίν. καὶ ἐπεὶ ἐν κύκλῳ τετράπλευρόν ἐστι τὸ ΚΒΟΣ, καὶ the plane of quadrilateral KBPS, it is thus also at right- ἴσαι αἱ ΚΒ, ΒΟ, ΚΣ, καὶ ἐλάττων ἡ ΟΣ, καὶ ἐκ τοῦ κέντρου angles to all of the straight-lines joined to it which are τοῦ κύκλου ἐστὶν ἡ ΒΨ, τὸ ἄρα ἀπὸ τῆς ΚΒ τοῦ ἀπὸ τῆς also in the plane of the quadrilateral [Def. 11.3]. Thus, ΒΨ μεῖζόν ἐστιν ἢ διπλάσιον. ἤχθω ἀπὸ τοῦ Κ ἐπὶ τὴν ΒΦ AX is at right-angles to each of BX and XK. And since κάθετος ἡ ΚΩ. καὶ ἐπεὶ ἡ ΒΔ τῆς ΔΩ ἐλάττων ἐστὶν ἢ AB is equal to AK, the (square) on AB is also equal to διπλῆ, καί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΔΩ, οὕτως τὸ ὑπὸ τῶν the (square) on AK. And the (sum of the squares) on AX ΔΒ, ΒΩ πρὸς τὸ ὑπὸ [τῶν] ΔΩ, ΩΒ, ἀναγραφομένου ἀπὸ and XB is equal to the (square) on AB. For the angle τῆς ΒΩ τετραγώνου καὶ συμπληρουμένου τοῦ ἐπὶ τῆς ΩΔ at X (is) a right-angle [Prop. 1.47]. And the (sum of παραλληλογράμμου καὶ τὸ ὑπὸ ΔΒ, ΒΩ ἄρα τοῦ ὑπὸ ΔΩ, the squares) on AX and XK is equal to the (square) on ΩΒ ἔλαττόν ἐστιν ἢ διπλάσιον. καί ἐστι τῆς ΚΔ ἐπιζευ- AK [Prop. 1.47]. Thus, the (sum of the squares) on AX γνυμένης τὸ μὲν ὑπὸ ΔΒ, ΒΩ ἴσον τῷ ἀπὸ τῆς ΒΚ, τὸ δὲ and XB is equal to the (sum of the squares) on AX and ὑπὸ τῶν ΔΩ, ΩΒ ἴσον τῷ ἀπὸ τῆς ΚΩ· τὸ ἄρα ἀπὸ τῆς ΚΒ XK. Let the (square) on AX have been subtracted from τοῦ ἀπὸ τῆς ΚΩ ἔλασσόν ἐστιν ἢ διπλάσιον. ἀλλὰ τὸ ἀπὸ both. Thus, the remaining (square) on BX is equal to the τῆς ΚΒ τοῦ ἀπὸ τῆς ΒΨ μεῖζόν ἐστιν ἢ διπλάσιον· μεῖζον remaining (square) on XK. Thus, BX (is) equal to XK. ἄρα τὸ ἀπὸ τῆς ΚΩ τοῦ ἀπὸ τῆς ΒΨ. καὶ ἐπεὶ ἴση ἐστὶν ἡ So, similarly, we can show that the straight-lines joined ΒΑ τῇ ΚΑ, ἴσον ἐστὶ τὸ ἀπὸ τῆς ΒΑ τῷ ἀπὸ τῆς ΑΚ. καί from X to P and S are equal to each of BX and XK. 501 STOIQEIWN ibþ. ELEMENTS BOOK 12 ἐστι τῷ μὲν ἀπὸ τῆς ΒΑ ἴσα τὰ ἀπὸ τῶν ΒΨ, ΨΑ, τῷ δὲ ἀπὸ Thus, a circle drawn (in the plane of the quadrilateral) τῆς ΚΑ ἴσα τὰ ἀπὸ τῶν ΚΩ, ΩΑ· τὰ ἄρα ἀπὸ τῶν ΒΨ, ΨΑ with center X , and radius one of XB or XK, will also ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΚΩ, ΩΑ, ὧν τὸ ἀπὸ τῆς ΚΩ μεῖζον pass through P and S, and the quadrilateral KBPS will τοῦ ἀπὸ τῆς ΒΨ· λοιπὸν ἄρα τὸ ἀπὸ τῆς ΩΑ ἔλασσόν ἐστι be inside the circle. τοῦ ἀπὸ τῆς ΨΑ. μείζων ἄρα ἡ ΑΨ τῆς ΑΩ· πολλῷ ἄρα ἡ And since KB is greater than WV , and WV (is) equal ΑΨ μείζων ἐστὶ τῆς ΑΗ. καί ἐστιν ἡ μὲν ΑΨ ἐπὶ μίαν τοῦ to SP , KB (is) thus greater than SP . And KB (is) πολυέδρου βάσιν, ἡ δὲ ΑΗ ἐπὶ τὴν τῆς ἐλάσσονος σφαίρας equal to each of KS and BP . Thus, KS and BP are ἐπιφάνειαν· ὥστε τὸ πολύεδρον οὐ ψαύσει τῆς ἐλάσσονος each greater than SP . And since quadrilateral KBPS σφαίρας κατὰ τὴν ἐπιφάνειαν. is in a circle, and KB, BP , and KS are equal (to one Δύο ἄρα σφαιρῶν περὶ τὸ αὐτὸ κέντρον οὐσῶν εἰς τὴν another), and PS (is) less (than them), and BX is the μείζονα σφαῖραν στερεὸν πολύεδρον ἐγγέγραπται μὴ ψαῦον radius of the circle, the (square) on KB is thus greater τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν· ὅπερ ἔδει than double the (square) on BX .† Let the perpendicular ποιῆσαι. KY have been drawn from K to BV .‡ And since BD is less than double DY , and as BD is to DY , so the (rect- angle contained) by DB and BY (is) to the (rectangle contained) by DY and Y B—a square being described on BY , and a (rectangular) parallelogram (with short side equal to BY ) completed on Y D—the (rectangle con- tained) by DB and BY is thus also less than double the (rectangle contained) by DY and Y B. And, KD being joined, the (rectangle contained) by DB and BY is equal to the (square) on BK, and the (rectangle contained) by DY and Y B equal to the (square) on KY [Props. 3.31, 6.8 corr.]. Thus, the (square) on KB is less than double the (square) on KY . But, the (square) on KB is greater than double the (square) on BX . Thus, the (square) on KY (is) greater than the (square) on BX . And since BA is equal to KA, the (square) on BA is equal to the (square) on AK. And the (sum of the squares) on BX and XA is equal to the (square) on BA, and the (sum of the squares) on KY and Y A (is) equal to the (square) on KA [Prop. 1.47]. Thus, the (sum of the squares) on BX and XA is equal to the (sum of the squares) on KY and Y A, of which the (square) on KY (is) greater than the (square) on BX . Thus, the remaining (square) on Y A is less than the (square) on XA. Thus, AX (is) greater than AY . Thus, AX is much greater than AG.§ And AX is (a perpendicular) on one of the bases of the polyhe- dron, and AG (is a perpendicular) on the surface of the lesser sphere. Hence, the polyhedron will not touch the lesser sphere on its surface. Thus, there being two spheres about the same cen- ter, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere on its sur- face. (Which is) the very thing it was required to do. † Since KB, BP , and KS are greater than the sides of an inscribed square, which are each of length √ 2 BX. ‡ Note that points Y and V are actually identical. § This conclusion depends on the fact that the chord of the polygon in proposition 12.16 does not touch the inner circle. 502 STOIQEIWN ibþ. ELEMENTS BOOK 12Pìrisma. Corollary ᾿Εὰν δὲ καὶ εἰς ἑτάραν σφαῖραν τῷ ἐν τῇ ΒΓΔΕ σφαίρᾳ And, also, if a similar polyhedral solid to that in στερεῷ πολυέδρῳ ὅμοιον στερεὸν πολύεδρον ἐγγραφῇ, τὸ sphere BCDE is inscribed in another sphere then the ἐν τῇ ΒΓΔΕ σφαίρᾳ στερεὸν πολύεδρον πρὸς τὸ ἐν τῇ polyhedral solid in sphere BCDE has to the polyhedral ἑτέρᾳ σφαίρᾳ στερεὸν πολύεδρον τριπλασίονα λόγον ἔχει, solid in the other sphere the cubed ratio that the diameter ἤπερ ἡ τῆς ΒΓΔΕ σφαίρας διάμετρος πρὸς τὴν τῆς ἑτέρας of sphere BCDE has to the diameter of the other sphere. σφαίρας διάμετρον. διαιρεθέντων γὰρ τῶν στερεῶν εἰς For if the solids are divided into similarly numbered, and τὰς ὁμοιοπληθεῖς καὶ ὁμοιοταγεῖς πυραμίδας ἔσονται αἱ πυ- similarly situated, pyramids, then the pyramids will be ραμίδες ὅμοιαι. αἱ δὲ ὅμοιαι πυραμίδες πρὸς ἀλλήλας ἐν similar. And similar pyramids are in the cubed ratio of τριπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν· ἡ ἄρα πυ- corresponding sides [Prop. 12.8 corr.]. Thus, the pyra- ραμίς, ἧς βάσις μέν ἐστι τὸ ΚΒΟΣ τετράπλευρον, κορυφὴ mid whose base is quadrilateral KBPS, and apex the δὲ τὸ Α σημεῖον, πρὸς τὴν ἐν τῇ ἑτέρᾳ σφαίρᾳ ὁμοιοταγῆ point A, will have to the similarly situated pyramid in the πυραμίδα τριπλασίονα λόγον ἔχει, ἤπερ ἡ ὁμόλογος πλευρὰ other sphere the cubed ratio that a corresponding side πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἤπερ ἡ ΑΒ ἐκ τοῦ (has) to a corresponding side. That is to say, that of ra- κέντρου τῆς σφαίρας τῆς περὶ κέντρον τὸ Α πρὸς τὴν ἐκ τοῦ dius AB of the sphere about center A to the radius of the κέντρου τῆς ἑτέρας σφαίρας. ὁμοίως καὶ ἑκάστη πυραμὶς other sphere. And, similarly, each pyramid in the sphere τῶν ἐν τῇ περὶ κέντρον τὸ Α σφαίρᾳ πρὸς ἑκάστην ὁμοταγῆ about center A will have to each similarly situated pyra- πυραμίδα τῶν ἐν τῇ ἑτέρᾳ σφαίρᾳ τριπλασίονα λόγον ἕξει, mid in the other sphere the cubed ratio that AB (has) to ἤπερ ἡ ΑΒ πρὸς τὴν ἐκ τοῦ κέντρου τῆς ἑτέρας σφαίρας. the radius of the other sphere. And as one of the leading καὶ ὡς ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως (magnitudes is) to one of the following (in two sets of ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα· ὥστε ὅλον proportional magnitudes), so (the sum of) all the lead- τὸ ἐν τῇ περὶ κέντρον τὸ Α σφαίρᾳ στερεὸν πολύεδρον ing (magnitudes is) to (the sum of) all of the following πρὸς ὅλον τὸ ἐν τῇ ἑτέρᾳ [σφαίρᾳ] στερεὸν πολύεδρον τρι- (magnitudes) [Prop. 5.12]. Hence, the whole polyhedral πλασίονα λόγον ἕξει, ἤπερ ἡ ΑΒ πρὸς τὴν ἐκ τοῦ κέντρου solid in the sphere about center A will have to the whole τῆς ἑτέρας σφαίρας, τουτέστιν ἤπερ ἡ ΒΔ διάμετρος πρὸς polyhedral solid in the other [sphere] the cubed ratio that τὴν τῆς ἑτέρας σφαίρας διάμετρον· ὅπερ ἔδει δεῖξαι. (radius) AB (has) to the radius of the other sphere. That is to say, that diameter BD (has) to the diameter of the other sphere. (Which is) the very thing it was required to show.ihþ. Proposition 18 Αἱ σφαῖραι πρὸς ἀλλήλας ἐν τριπλασίονι λόγῳ εἰσὶ τῶν Spheres are to one another in the cubed ratio of their ἰδίων διαμέτρων. respective diameters. D L E J K H G N A B Z M G L E K N A B M D C H F 503 STOIQEIWN ibþ. ELEMENTS BOOK 12 Νενοήσθωσαν σφαῖραι αἱ ΑΒΓ, ΔΕΖ, διάμετροι δὲ Let the spheres ABC and DEF have been conceived, αὐτῶν αἱ ΒΓ, ΕΖ· λέγω, ὅτι ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΔΕΖ and (let) their diameters (be) BC and EF (respectively). σφαῖραν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. I say that sphere ABC has to sphere DEF the cubed ratio Εἰ γὰρ μὴ ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΔΕΖ σφαῖραν τρι- that BC (has) to EF . πλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ, ἕξει ἄρα ἡ For if sphere ABC does not have to sphere DEF the ΑΒΓ σφαῖρα πρὸς ἐλάσσονά τινα τῆς ΔΕΖ σφαίρας τρι- cubed ratio that BC (has) to EF then sphere ABC will πλασίονα λόγον ἢ πρὸς μείζονα ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. have to some (sphere) either less than, or greater than, ἐχέτω πρότερον πρὸς ἐλάσσονα τὴν ΗΘΚ, καὶ νενοήσθω ἡ sphere DEF the cubed ratio that BC (has) to EF . Let ΔΕΖ τῇ ΗΘΚ περὶ τὸ αὐτὸ κέντρον, καὶ ἐγγεγράφθω εἰς it, first of all, have (such a ratio) to a lesser (sphere), τὴν μείζονα σφαῖραν τὴν ΔΕΖ στερεὸν πολύεδρον μὴ ψαῦον GHK. And let DEF have been conceived about the τῆς ἐλάσσονος σφαίρας τῆς ΗΘΚ κατὰ τὴν ἐπιφάνειαν, same center as GHK. And let a polyhedral solid have ἐγγεγράφθω δὲ καὶ εἰς τὴν ΑΒΓ σφαῖραν τῷ ἐν τῇ ΔΕΖ been inscribed in the greater sphere DEF , not touching σφαίρᾳ στερεῷ πολυέδρῳ ὅμοιον στερεὸν πολύεδρον· τὸ the lesser sphere GHK on its surface [Prop. 12.17]. And ἄρα ἐν τῇ ΑΒΓ στερεὸν πολύεδρον πρὸς τὸ ἐν τῇ ΔΕΖ let a polyhedral solid, similar to the polyhedral solid in στερεὸν πολύεδρον τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς sphere DEF , have also been inscribed in sphere ABC. τὴν ΕΖ. ἔχει δὲ καὶ ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΗΘΚ σφαῖραν Thus, the polyhedral solid in sphere ABC has to the τριπλασίονα λόγον ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ· ἔστιν ἄρα ὡς polyhedral solid in sphere DEF the cubed ratio that BC ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΗΘΚ σφαῖραν, οὕτως τὸ ἐν τῇ (has) to EF [Prop. 12.17 corr.]. And sphere ABC also ΑΒΓ σφαίρᾳ στερεὸν πολύεδρον πρὸς τὸ ἐν τῇ ΔΕΖ σφαίρᾳ has to sphere GHK the cubed ratio that BC (has) to EF . στερεὸν πολύεδρον· ἐναλλὰξ [ἄρα] ὡς ἡ ΑΒΓ σφαῖρα πρὸς Thus, as sphere ABC is to sphere GHK, so the polyhe- τὸ ἐν αὐτῇ πολύεδρον, οὕτως ἡ ΗΘΚ σφαῖρα πρὸς τὸ ἐν τῇ dral solid in sphere ABC (is) to the polyhedral solid is ΔΕΖ σφαίρᾳ στερεὸν πολύεδρον. μείζων δὲ ἡ ΑΒΓ σφαῖρα sphere DEF . [Thus], alternately, as sphere ABC (is) to τοῦ ἐν αὐτῇ πολυέδρου· μείζων ἄρα καὶ ἡ ΗΘΚ σφαῖρα the polygon within it, so sphere GHK (is) to the polyhe- τοῦ ἐν τῇ ΔΕΖ σφαίρᾳ πολυέδρου. ἀλλὰ καὶ ἐλάττων· dral solid within sphere DEF [Prop. 5.16]. And sphere ἐμπεριέχεται γὰρ ὑπ᾿ αὐτοῦ. οὐκ ἄρα ἡ ΑΒΓ σφαῖρα πρὸς ABC (is) greater than the polyhedron within it. Thus, ἐλάσσονα τῆς ΔΕΖ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ sphere GHK (is) also greater than the polyhedron within ΒΓ διάμετρος πρὸς τὴν ΕΖ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἡ sphere DEF [Prop. 5.14]. But, (it is) also less. For it is ΔΕΖ σφαῖρα πρὸς ἐλάσσονα τῆς ΑΒΓ σφαίρας τριπλασίονα encompassed by it. Thus, sphere ABC does not have to λόγον ἔχει ἤπερ ἡ ΕΖ πρὸς τὴν ΒΓ. (a sphere) less than sphere DEF the cubed ratio that di- Λέγω δή, ὅτι οὐδὲ ἡ ΑΒΓ σφαῖρα πρὸς μείζονά τινα τῆς ameter BC (has) to EF . So, similarly, we can show that ΔΕΖ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν sphere DEF does not have to (a sphere) less than sphere ΕΖ. ABC the cubed ratio that EF (has) to BC either. Εἰ γὰρ δυνατόν, ἐχέτω πρὸς μείζονα τὴν ΛΜΝ· ἀνάπαλιν So, I say that sphere ABC does not have to some ἄρα ἡ ΛΜΝ σφαῖρα πρὸς τὴν ΑΒΓ σφαῖραν τριπλασίονα (sphere) greater than sphere DEF the cubed ratio that λόγον ἔχει ἤπερ ἡ ΕΖ διάμετρος πρὸς τὴν ΒΓ διάμετρον. BC (has) to EF either. ὡς δὲ ἡ ΛΜΝ σφαῖρα πρὸς τὴν ΑΒΓ σφαῖραν, οὕτως ἡ ΔΕΖ For, if possible, let it have (the cubed ratio) to a σφαῖρα πρὸς ἐλάσσονά τινα τῆς ΑΒΓ σφαίρας, ἐπειδήπερ greater (sphere), LMN . Thus, inversely, sphere LMN μείζων ἐστὶν ἡ ΛΜΝ τῆς ΔΕΖ, ὡς ἔμπροσθεν ἐδείχθη. καὶ (has) to sphere ABC the cubed ratio that diameter ἡ ΔΕΖ ἄρα σφαῖρα πρὸς ἐλάσσονά τινα τῆς ΑΒΓ σφαίρας EF (has) to diameter BC [Prop. 5.7 corr.]. And as τριπλασίονα λόγον ἔχει ἤπερ ἡ ΕΖ πρὸς τὴν ΒΓ· ὅπερ sphere LMN (is) to sphere ABC, so sphere DEF ἀδύνατον ἐδείχθη. οὐκ ἄρα ἡ ΑΒΓ σφαῖρα πρὸς μείζονά (is) to some (sphere) less than sphere ABC, inasmuch τινα τῆς ΔΕΖ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ as LMN is greater than DEF , as was shown before πρὸς τὴν ΕΖ. ἐδείχθη δέ, ὅτι οὐδὲ πρὸς ἐλάσσονα. ἡ ἄρα [Prop. 12.2 lem.]. And, thus, sphere DEF has to some ΑΒΓ σφαῖρα πρὸς τὴν ΔΕΖ σφαῖραν τριπλασίονα λόγον ἔχει (sphere) less than sphere ABC the cubed ratio that EF ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ· ὅπερ ἔδει δεῖξαι. (has) to BC. The very thing was shown (to be) impossi- ble. Thus, sphere ABC does not have to some (sphere) greater than sphere DEF the cubed ratio that BC (has) to EF . And it was shown that neither (does it have such a ratio) to a lesser (sphere). Thus, sphere ABC has to sphere DEF the cubed ratio that BC (has) to EF . (Which is) the very thing it was required to show. 504 ELEMENTS BOOK 13 The Platonic Solids† †The five regular solids—the cube, tetrahedron (i.e., pyramid), octahedron, icosahedron, and dodecahedron—were problably discovered by the school of Pythagoras. They are generally termed “Platonic” solids because they feature prominently in Plato’s famous dialogue Timaeus. Many of the theorems contained in this book—particularly those which pertain to the last two solids—are ascribed to Theaetetus of Athens. 505 STOIQEIWN igþ. ELEMENTS BOOK 13aþ. Proposition 1 ᾿Εὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, If a straight-line is cut in extreme and mean ratio τὸ μεῖζον τμῆμα προσλαβὸν τὴν ἡμίσειαν τῆς ὅλης πεν- then the square on the greater piece, added to half of ταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τετραγώνου. the whole, is five times the square on the half. G O D H E B ZL N AX K JM C D E B L N A K F H G P M O Εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ ἄκρον καὶ μέσον λόγον For let the straight-line AB have been cut in extreme τετμήσθω κατὰ τὸ Γ σημεῖον, καὶ ἔστω μεῖζον τμῆμα τὸ and mean ratio at point C, and let AC be the greater ΑΓ, καὶ ἐκβεβλήσθω ἐπ᾿ εὐθείας τῇ ΓΑ εὐθεῖα ἡ ΑΔ, καὶ piece. And let the straight-line AD have been produced κείσθω τῆς ΑΒ ἡμίσεια ἡ ΑΔ· λέγω, ὅτι πενταπλάσιόν ἐστι in a straight-line with CA. And let AD be made (equal τὸ ἀπὸ τῆς ΓΔ τοῦ ἀπὸ τῆς ΔΑ. to) half of AB. I say that the (square) on CD is five times Ἀναγεγράφθωσαν γὰρ ἀπὸ τῶν ΑΒ, ΔΓ τετράγωνα the (square) on DA. τὰ ΑΕ, ΔΖ, καὶ καταγεγράφθω ἐν τῷ ΔΖ τὸ σχῆμα, καὶ For let the squares AE and DF have been described διήχθω ἡ ΖΓ ἐπὶ τὸ Η. καὶ ἐπεὶ ἡ ΑΒ ἄκρον καὶ μέσον on AB and DC (respectively). And let the figure in DF λόγον τέτμηται κατὰ τὸ Γ, τὸ ἄρα ὑπὸ τῶν ΑΒΓ ἴσον ἐστὶ have been drawn. And let FC have been drawn across to τῷ ἀπὸ τῆς ΑΓ. καί ἐστι τὸ μὲν ὑπὸ τῶν ΑΒΓ τὸ ΓΕ, τὸ G. And since AB has been cut in extreme and mean ratio δὲ ἀπὸ τῆς ΑΓ τὸ ΖΘ· ἴσον ἄρα τὸ ΓΕ τῷ ΖΘ. καὶ ἐπεὶ at C, the (rectangle contained) by ABC is thus equal to διπλῆ ἐστιν ἡ ΒΑ τῆς ΑΔ, ἴση δὲ ἡ μὲν ΒΑ τῇ ΚΑ, ἡ δὲ the (square) on AC [Def. 6.3, Prop. 6.17]. And CE is ΑΔ τῇ ΑΘ, διπλῆ ἄρα καὶ ἡ ΚΑ τῆς ΑΘ. ὡς δὲ ἡ ΚΑ πρὸς the (rectangle contained) by ABC, and FH the (square) τὴν ΑΘ, οὕτως τὸ ΓΚ πρὸς τὸ ΓΘ· διπλάσιον ἄρα τὸ ΓΚ on AC. Thus, CE (is) equal to FH . And since BA is τοῦ ΓΘ. εἰσὶ δὲ καὶ τὰ ΛΘ, ΘΓ διπλάσια τοῦ ΓΘ. ἴσον ἄρα double AD, and BA (is) equal to KA, and AD to AH , τὸ ΚΓ τοῖς ΛΘ, ΘΓ. ἐδείχθη δὲ καὶ τὸ ΓΕ τῷ ΘΖ ἴσον· KA (is) thus also double AH . And as KA (is) to AH , so ὅλον ἄρα τὸ ΑΕ τετράγωνον ἴσον ἐστὶ τῷ ΜΝΞ γνώμονι. CK (is) to CH [Prop. 6.1]. Thus, CK (is) double CH . καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΒΑ τῆς ΑΔ, τετραπλάσιόν ἐστι τὸ And LH plus HC is also double CH [Prop. 1.43]. Thus, ἀπὸ τῆς ΒΑ τοῦ ἀπὸ τῆς ΑΔ, τουτέστι τὸ ΑΕ τοῦ ΔΘ. KC (is) equal to LH plus HC. And CE was also shown ἴσον δὲ τὸ ΑΕ τῷ ΜΝΞ γνώμονι· καὶ ὁ ΜΝΞ ἄρα γνώμων (to be) equal to HF . Thus, the whole square AE is equal τετραπλάσιός ἐστι τοῦ ΑΟ· ὅλον ἄρα τὸ ΔΖ πενταπλάσιόν to the gnomon MNO. And since BA is double AD, the ἐστι τοῦ ΑΟ. καί ἐστι τὸ μὲν ΔΖ τὸ ἀπὸ τῆς ΔΓ, τὸ δὲ ΑΟ (square) on BA is four times the (square) on AD—that τὸ ἀπὸ τῆς ΔΑ· τὸ ἄρα ἀπὸ τῆς ΓΔ πενταπλάσιόν ἐστι τοῦ is to say, AE (is four times) DH . And AE (is) equal to ἀπὸ τῆς ΔΑ. gnomon MNO. And, thus, gnomon MNO is also four ᾿Εὰν ἄρα εὐθεῖα ἄκρον καὶ μέσον λόγον τμηθῇ, τὸ μεῖζον times AP . Thus, the whole of DF is five times AP . And τμῆμα προσλαβὸν τὴν ἡμίσειαν τῆς ὅλης πενταπλάσιον DF is the (square) on DC, and AP the (square) on DA. δύναται τοῦ ἀπὸ τῆς ἡμισείας τετραγώνου· ὅπερ ἔδει δεῖξαι. Thus, the (square) on CD is five times the (square) on DA. Thus, if a straight-line is cut in extreme and mean ra- tio then the square on the greater piece, added to half of 506 STOIQEIWN igþ. ELEMENTS BOOK 13 the whole, is five times the square on the half. (Which is) the very thing it was required to show.bþ. Proposition 2 ᾿Εὰν εὐθεῖα γραμμὴ τμήματος ἑαυτῆς πενταπλάσιον If the square on a straight-line is five times the δύνηται, τῆς διπλασίας τοῦ εἰρημένου τμήματος ἄκρον καὶ (square) on a piece of it, and double the aforementioned μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμα τὸ λοιπὸν μέρος piece is cut in extreme and mean ratio, then the greater ἐστὶ τῆς ἐξ ἀρχῆς εὐθείας. piece is the remaining part of the original straight-line. G ZL N X K JM A B D E H O L N K M A B D E C H G F Εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ τμήματος ἑαυτῆς τοῦ ΑΓ πεν- For let the square on the straight-line AB be five times ταπλάσιον δυνάσθω, τῆς δὲ ΑΓ διπλῆ ἔστω ἡ ΓΔ. λέγω, the (square) on the piece of it, AC. And let CD be double ὅτι τῆς ΓΔ ἄκρον καὶ μέσον λόγον τεμνομένος τὸ μεῖζον AC. I say that if CD is cut in extreme and mean ratio τμῆμά ἐστιν ἡ ΓΒ. then the greater piece is CB. Ἀναγεγράφθω γὰρ ἀφ᾿ ἑκατέρας τῶν ΑΒ, ΓΔ τετράγωνα For let the squares AF and CG have been described τὰ ΑΖ, ΓΗ, καὶ καταγεγράφθω ἐν τῷ ΑΖ τὸ σχῆμα, καὶ on each of AB and CD (respectively). And let the figure διήχθω ἡ ΒΕ. καὶ ἐπεὶ πενταπλάσιόν ἐστι τὸ ἀπό τῆς ΒΑ in AF have been drawn. And let BE have been drawn τοῦ ἀπὸ τῆς ΑΓ, πενταπλάσιόν ἐστι τὸ ΑΖ τοῦ ΑΘ. τε- across. And since the (square) on BA is five times the τραπλάσιος ἄρα ὁ ΜΝΞ γνώμων τοῦ ΑΘ. καὶ ἐπεὶ διπλῆ (square) on AC, AF is five times AH . Thus, gnomon ἐστιν ἡ ΔΓ τῆς ΓΑ, τετραπλάσιον ἄρα ἐστὶ τὸ ἀπὸ ΔΓ τοῦ MNO (is) four times AH . And since DC is double CA, ἀπὸ ΓΑ, τουτέστι τὸ ΓΗ τοῦ ΑΘ. ἐδείχθη δὲ καὶ ὁ ΜΝΞ the (square) on DC is thus four times the (square) on γνώμων τετραπλάσιος τοῦ ΑΘ· ἴσος ἄρα ὁ ΜΝΞ γνώμων CA—that is to say, CG (is four times) AH . And the τῷ ΓΗ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΔΓ τῆς ΓΑ, ἴση δὲ ἡ μὲν gnomon MNO was also shown (to be) four times AH . ΔΓ τῇ ΓΚ, ἡ δὲ ΑΓ τῇ ΓΘ, [διπλῆ ἄρα καὶ ἡ ΚΓ τῆς ΓΘ], Thus, gnomon MNO (is) equal to CG. And since DC is διπλάσιον ἄρα καὶ τὸ ΚΒ τοῦ ΒΘ. εἰσὶ δὲ καὶ τὰ ΛΘ, ΘΒ double CA, and DC (is) equal to CK, and AC to CH , τοῦ ΘΒ διπλάσια· ἴσον ἄρα τὸ ΚΒ τοῖς ΛΘ, ΘΒ. ἐδείχθη [KC (is) thus also double CH], (and) KB (is) also dou- δὲ καὶ ὅλος ὁ ΜΝΞ γνώμων ὅλῳ τῷ ΓΗ ἴσος· καὶ λοιπὸν ble BH [Prop. 6.1]. And LH plus HB is also double HB ἄρα τὸ ΘΖ τῷ ΒΗ ἐστιν ἴσον. καί ἐστι τὸ μὲν ΒΗ τὸ ὑπὸ [Prop. 1.43]. Thus, KB (is) equal to LH plus HB. And τῶν ΓΔΒ· ἴση γὰρ ἡ ΓΔ τῇ ΔΗ· τὸ δὲ ΘΖ τὸ ἀπὸ τῆς ΓΒ· the whole gnomon MNO was also shown (to be) equal τὸ ἄρα ὑπὸ τῶν ΓΔΒ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΒ. ἔστιν ἄρα to the whole of CG. Thus, the remainder HF is also ὡς ἡ ΔΓ πρὸς τὴν ΓΒ, οὕτως ἡ ΓΒ πρὸς τὴν ΒΔ. μείζων equal to (the remainder) BG. And BG is the (rectangle δὲ ἡ ΔΓ τῆς ΓΒ· μείζων ἄρα καὶ ἡ ΓΒ τῆς ΒΔ. τῆς ΓΔ contained) by CDB. For CD (is) equal to DG. And HF ἄρα εὐθείας ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον (is) the square on CB. Thus, the (rectangle contained) τμῆμά ἐστιν ἡ ΓΒ. by CDB is equal to the (square) on CB. Thus, as DC ᾿Εὰν ἄρα εὐθεῖα γραμμὴ τμήματος ἑαυτῆς πενταπλάσιον is to CB, so CB (is) to BD [Prop. 6.17]. And DC (is) δύνηται, τῆς διπλασίας τοῦ εἰρημένου τμήματος ἄκρον καὶ greater than CB (see lemma). Thus, CB (is) also greater μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμα τὸ λοιπὸν μέρος than BD [Prop. 5.14]. Thus, if the straight-line CD is cut 507 STOIQEIWN igþ. ELEMENTS BOOK 13 ἐστὶ τῆς ἐξ ἀρχῆς εὐθείας· ὅπερ ἔδει δεῖξαι. in extreme and mean ratio then the greater piece is CB. Thus, if the square on a straight-line is five times the (square) on a piece of itself, and double the afore- mentioned piece is cut in extreme and mean ratio, then the greater piece is the remaining part of the original straight-line. (Which is) the very thing it was required to show.L¨mma. Lemma ῞Οτι δὲ ἡ διπλῆ τῆς ΑΓ μείζων ἐστὶ τῆς ΒΓ, οὕτως And it can be shown that double AC (i.e., DC) is δεικτέον. greater than BC, as follows. Εἰ γὰρ μή, ἔστω, εἰ δυνατόν, ἡ ΒΓ διπλῆ τῆς ΓΑ. τε- For if (double AC is) not (greater than BC), if possi- τραπλάσιον ἄρα τὸ ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς ΓΑ· πενταπλάσια ble, let BC be double CA. Thus, the (square) on BC (is) ἄρα τὰ ἀπὸ τῶν ΒΓ, ΓΑ τοῦ ἀπὸ τῆς ΓΑ. ὑπόκειται δὲ καὶ four times the (square) on CA. Thus, the (sum of) the τὸ ἀπὸ τῆς ΒΑ πενταπλάσιον τοῦ ἀπὸ τῆς ΓΑ· τὸ ἄρα ἀπὸ (squares) on BC and CA (is) five times the (square) on τῆς ΒΑ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΓ, ΓΑ· ὅπερ ἀδύνατον. CA. And the (square) on BA was assumed (to be) five οὐκ ἄρα ἡ ΓΒ διπλασία ἐστὶ τῆς ΑΓ. ὁμοίως δὴ δείξομεν, times the (square) on CA. Thus, the (square) on BA is ὅτι οὐδὲ ἡ ἐλάττων τῆς ΓΒ διπλασίων ἐστὶ τῆς ΓΑ· πολλῷ equal to the (sum of) the (squares) on BC and CA. The γὰρ [μεῖζον] τὸ ἄτοπον. very thing (is) impossible [Prop. 2.4]. Thus, CB is not ῾Η ἄρα τῆς ΑΓ διπλῆ μείζων ἐστὶ τῆς ΓΒ· ὅπερ ἔδει double AC. So, similarly, we can show that a (straight- δεῖξαι. line) less than CB is not double AC either. For (in this case) the absurdity is much [greater]. Thus, double AC is greater than CB. (Which is) the very thing it was required to show.gþ. Proposition 3 ᾿Εὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, If a straight-line is cut in extreme and mean ratio then τὸ ἔλασσον τμῆμα προσλαβὸν τὴν ἡμίσειαν τοῦ μείζονος the square on the lesser piece added to half of the greater τμήματος πενταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τοῦ piece is five times the square on half of the greater piece. μείζονος τμήματος τετραγώνου. S R G M N E K Z BA J OH D L PX QR M N E K BA D L S F O G C H P Εὐθεῖα γάρ τις ἡ ΑΒ ἄκρον καὶ μέσον λόγον τετμήσθω For let some straight-line AB have been cut in ex- κατὰ τὸ Γ σημεῖον, καὶ ἔστω μεῖζον τμῆμα τὸ ΑΓ, καὶ treme and mean ratio at point C. And let AC be the τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Δ· λέγω, ὅτι πενταπλάσιόν greater piece. And let AC have been cut in half at D. I ἐστι τὸ ἀπὸ τῆς ΒΔ τοῦ ἀπὸ τῆς ΔΓ. say that the (square) on BD is five times the (square) on Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΕ, καὶ DC. 508 STOIQEIWN igþ. ELEMENTS BOOK 13 καταγεγράφθω διπλοῦν τὸ σχῆμα. ἐπεὶ διπλῆ ἐστιν ἡ ΑΓ For let the square AE have been described on AB. τῆς ΔΓ, τετραπλάσιον ἄρα τὸ ἀπὸ τῆς ΑΓ τοῦ ἀπὸ τῆς ΔΓ, And let the figure have been drawn double. Since AC is τουτέστι τὸ ΡΣ τοῦ ΖΗ. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΒΓ ἴσον double DC, the (square) on AC (is) thus four times the ἐστὶ τῷ ἀπὸ τῆς ΑΓ, καί ἐστι τὸ ὑπὸ τῶν ΑΒΓ τὸ ΓΕ, τὸ (square) on DC—that is to say, RS (is four times) FG. ἄρα ΓΕ ἴσον ἐστὶ τῷ ΡΣ. τετραπλάσιον δὲ τὸ ΡΣ τοῦ ΖΗ· And since the (rectangle contained) by ABC is equal to τετραπλάσιον ἄρα καὶ τὸ ΓΕ τοῦ ΖΗ. πάλιν ἐπεὶ ἴση ἐστὶν the (square) on AC [Def. 6.3, Prop. 6.17], and CE is the ἡ ΑΔ τῇ ΔΓ, ἴση ἐστὶ καὶ ἡ ΘΚ τῇ ΚΖ. ὥστε καὶ τὸ ΗΖ (rectangle contained) by ABC, CE is thus equal to RS. τετράγωνον ἴσον ἐστὶ τῷ ΘΛ τετραγώνῳ. ἴση ἄρα ἡ ΗΚ And RS (is) four times FG. Thus, CE (is) also four times τῇ ΚΛ, τουτέστιν ἡ ΜΝ τῇ ΝΕ· ὥστε καὶ τὸ ΜΖ τῷ ΖΕ FG. Again, since AD is equal to DC, HK is also equal to ἐστιν ἴσον. ἀλλὰ τὸ ΜΖ τῷ ΓΗ ἐστιν ἴσον· καὶ τὸ ΓΗ ἄρα KF . Hence, square GF is also equal to square HL. Thus, τῷ ΖΕ ἐστιν ἴσον. κοινὸν προσκείσθω τὸ ΓΝ· ὁ ἄρα ΞΟΠ GK (is) equal to KL—that is to say, MN to NE. Hence, γνώμων ἴσος ἐστὶ τῷ ΓΕ. ἀλλὰ τὸ ΓΕ τετραπλάσιον ἐδείχθῃ MF is also equal to FE. But, MF is equal to CG. Thus, τοῦ ΗΖ· καὶ ὁ ΞΟΠ ἄρα γνώμων τετραπλάσιός ἐστι τοῦ ΖΗ CG is also equal to FE. Let CN have been added to τετραγώνου. ὁ ΞΟΠ ἄρα γνώμων καὶ τὸ ΖΗ τετράγωνον both. Thus, gnomon OPQ is equal to CE. But, CE was πενταπλάσιός ἐστι τοῦ ΖΗ. ἀλλὰ ὁ ΞΟΠ γνώμων καὶ τὸ shown (to be) equal to four times GF . Thus, gnomon ΖΗ τετράγωνόν ἐστι τὸ ΔΝ. καί ἐστι τὸ μὲν ΔΝ τὸ ἀπὸ OPQ is also four times square FG. Thus, gnomon OPQ τῆς ΔΒ, τὸ δὲ ΗΖ τὸ ἀπὸ τῆς ΔΓ. τὸ ἄρα ἀπὸ τῆς ΔΒ plus square FG is five times FG. But, gnomon OPQ plus πενταπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΔΓ· ὅπερ ἔδει δεῖξαι. square FG is (square) DN . And DN is the (square) on DB, and GF the (square) on DC. Thus, the (square) on DB is five times the (square) on DC. (Which is) the very thing it was required to show.dþ. Proposition 4 ᾿Εὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, τὸ If a straight-line is cut in extreme and mean ratio then ἀπὸ τῆς ὅλης καὶ τοῦ ἐλάσσονος τμήματος, τὰ συναμφότερα the sum of the squares on the whole and the lesser piece τετράγωνα, τριπλάσιά ἐστι τοῦ ἀπὸ τοῦ μείζονος τμήματος is three times the square on the greater piece. τετραγώνου.A G B KJ D H E ML NZ A B K D E M L N H C G F ῎Εστω εὐθεῖα ἡ ΑΒ, καὶ τετμήσθω ἄκρον καὶ μέσον Let AB be a straight-line, and let it have been cut in λόγον κατὰ τὸ Γ, καὶ ἔστω μεῖζον τμῆμα τὸ ΑΓ· λέγω, extreme and mean ratio at C, and let AC be the greater ὅτι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΓΑ. piece. I say that the (sum of the squares) on AB and BC Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ, is three times the (square) on CA. καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν ἡ ΑΒ ἄκρον καὶ For let the square ADEB have been described on AB, μέσον λόγον τέτμηται κατὰ τὸ Γ, καὶ τὸ μεῖζον τμῆμά ἐστιν and let the (remainder of the) figure have been drawn. ἡ ΑΓ, τὸ ἄρα ὑπὸ τῶν ΑΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ. καί Therefore, since AB has been cut in extreme and mean ἐστι τὸ μὲν ὑπὸ τῶν ΑΒΓ τὸ ΑΚ, τὸ δὲ ἀπὸ τῆς ΑΓ τὸ ΘΗ· ratio at C, and AC is the greater piece, the (rectangle 509 STOIQEIWN igþ. ELEMENTS BOOK 13 ἴσον ἄρα ἐστὶ τὸ ΑΚ τῷ ΘΗ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΑΖ τῷ contained) by ABC is thus equal to the (square) on AC ΖΕ, κοινὸν προσκείσθω τὸ ΓΚ· ὅλον ἄρα τὸ ΑΚ ὅλῳ τῷ [Def. 6.3, Prop. 6.17]. And AK is the (rectangle con- ΓΕ ἐστιν ἴσον· τὰ ἄρα ΑΚ, ΓΕ τοῦ ΑΚ ἐστι διπλάσια. ἀλλὰ tained) by ABC, and HG the (square) on AC. Thus, τὰ ΑΚ, ΓΕ ὁ ΛΜΝ γνώμων ἐστὶ καὶ τὸ ΓΚ τετράγωνον· ὁ AK is equal to HG. And since AF is equal to FE ἄρα ΛΜΝ γνώμων καὶ τὸ ΓΚ τετράγωνον διπλάσιά ἐστι τοῦ [Prop. 1.43], let CK have been added to both. Thus, ΑΚ. ἀλλὰ μὴν καὶ τὸ ΑΚ τῷ ΘΗ ἐδείχθη ἴσον· ὁ ἄρα ΛΜΝ the whole of AK is equal to the whole of CE. Thus, AK γνώμων καὶ [τὸ ΓΚ τετράγωνον διπλάσιά ἐστι τοῦ ΘΗ· plus CE is double AK. But, AK plus CE is the gnomon ὥστε ὁ ΛΜΝ γνώμων καὶ] τὰ ΓΚ, ΘΗ τετράγωνα τριπλάσιά LMN plus the square CK. Thus, gnomon LMN plus ἐστι τοῦ ΘΗ τετραγώνου. καί ἐστιν ὁ [μὲν] ΛΜΝ γνώμων square CK is double AK. But, indeed, AK was also καὶ τὰ ΓΚ, ΘΗ τετράγωνα ὅλον τὸ ΑΕ καὶ τὸ ΓΚ, ἅπερ ἐστὶ shown (to be) equal to HG. Thus, gnomon LMN plus τὰ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα, τὸ δὲ ΗΘ τὸ ἀπὸ τῆς ΑΓ [square CK is double HG. Hence, gnomon LMN plus] τετράγωνον. τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα τριπλάσιά the squares CK and HG is three times the square HG. ἐστι τοῦ ἀπὸ τῆς ΑΓ τετραγώνου· ὅπερ ἔδει δεῖξαι. And gnomon LMN plus the squares CK and HG is the whole of AE plus CK—which are the squares on AB and BC (respectively)—and GH (is) the square on AC. Thus, the (sum of the) squares on AB and BC is three times the square on AC. (Which is) the very thing it was required to show.eþ. Proposition 5 ᾿Εὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, καὶ If a straight-line is cut in extreme and mean ratio, and προστεθῇ αὐτῇ ἴση τῷ μείζονι τμήματι, ἡ ὅλη εὐθεῖα ἄκρον a (straight-line) equal to the greater piece is added to it, καὶ μέσον λόγον τέτμηται, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ἐξ then the whole straight-line has been cut in extreme and ἀρχῆς εὐθεῖα. mean ratio, and the original straight-line is the greater piece. J D A G B K E L CD A B K E L H Εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ ἄκρον καὶ μέσον λόγον For let the straight-line AB have been cut in extreme τετμήσθω κατὰ τὸ Γ σημεῖον, καὶ ἔστω μεῖζον τμῆμα ἡ and mean ratio at point C. And let AC be the greater ΑΓ, καὶ τῇ ΑΓ ἴση [κείσθω] ἡ ΑΔ. λέγω, ὅτι ἡ ΔΒ εὐθεῖα piece. And let AD be [made] equal to AC. I say that the ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Α, καὶ τὸ μεῖζον straight-line DB has been cut in extreme and mean ratio τμῆμά ἐστιν ἡ ἐξ ἀρχῆς εὐθεῖα ἡ ΑΒ. at A, and that the original straight-line AB is the greater Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΕ, καὶ piece. καταγεγράφθω τὸ σχῆμα. ἐπεὶ ἡ ΑΒ ἄκρον καὶ μέσον λόγον For let the square AE have been described on AB, τέτμηται κατὰ τὸ Γ, τὸ ἄρα ὑπὸ ΑΒΓ ἴσον ἐστὶ τῷ ἀπὸ ΑΓ. and let the (remainder of the) figure have been drawn. καί ἐστι τὸ μὲν ὑπὸ ΑΒΓ τὸ ΓΕ, τὸ δὲ ἀπὸ τῆς ΑΓ τὸ ΓΘ· And since AB has been cut in extreme and mean ratio at ἴσον ἄρα τὸ ΓΕ τῷ ΘΓ. ἀλλὰ τῷ μὲν ΓΕ ἴσον ἐστὶ τὸ ΘΕ, C, the (rectangle contained) by ABC is thus equal to the τῷ δὲ ΘΓ ἴσον τὸ ΔΘ· καὶ τὸ ΔΘ ἄρα ἴσον ἐστὶ τῷ ΘΕ (square) on AC [Def. 6.3, Prop. 6.17]. And CE is the [κοινὸν προσκείσθω τὸ ΘΒ]. ὅλον ἄρα τὸ ΔΚ ὅλῳ τῷ ΑΕ (rectangle contained) by ABC, and CH the (square) on ἐστιν ἴσον. καί ἐστι τὸ μὲν ΔΚ τὸ ὑπὸ τῶν ΒΔ, ΔΑ· ἴση AC. But, HE is equal to CE [Prop. 1.43], and DH equal 510 STOIQEIWN igþ. ELEMENTS BOOK 13 γὰρ ἡ ΑΔ τῇ ΔΛ· τὸ δὲ ΑΕ τὸ ἀπὸ τῆς ΑΒ· τὸ ἄρα ὑπὸ to HC. Thus, DH is also equal to HE. [Let HB have τῶν ΒΔΑ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ. ἔστιν ἄρα ὡς ἡ ΔΒ been added to both.] Thus, the whole of DK is equal to πρὸς τὴν ΒΑ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΔ. μείζων δὲ ἡ ΔΒ the whole of AE. And DK is the (rectangle contained) τῆς ΒΑ· μείζων ἄρα καὶ ἡ ΒΑ τῆς ΑΔ. by BD and DA. For AD (is) equal to DL. And AE (is) ῾Η ἄρα ΔΒ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Α, the (square) on AB. Thus, the (rectangle contained) by καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΑΒ· ὅπερ ἔδει δεῖξαι. BDA is equal to the (square) on AB. Thus, as DB (is) to BA, so BA (is) to AD [Prop. 6.17]. And DB (is) greater than BA. Thus, BA (is) also greater than AD [Prop. 5.14]. Thus, DB has been cut in extreme and mean ratio at A, and the greater piece is AB. (Which is) the very thing it was required to show.�þ. Proposition 6 ᾿Εὰν εὐθεῖα ῥητη ἄκρον καὶ μέσον λόγον τμηθῇ, If a rational straight-line is cut in extreme and mean ἑκάτερον τῶν τμημάτων ἄλογός ἐστιν ἡ καλουμένη ἀπο- ratio then each of the pieces is that irrational (straight- τομή. line) called an apotome.GD A B CD A B ῎Εστω εὐθεῖα ῥητὴ ἡ ΑΒ καὶ τετμήσθω ἄκρον καὶ μέσον Let AB be a rational straight-line cut in extreme and λόγον κατὰ τὸ Γ, καὶ ἔστω μεῖζον τμῆμα ἡ ΑΓ· λέγω, ὅτι mean ratio at C, and let AC be the greater piece. I say ἑκατέρα τῶν ΑΓ, ΓΒ ἄλογός ἐστιν ἡ καλουμένη ἀποτομή. that AC and CB is each that irrational (straight-line) ᾿Εκβεβλήσθω γὰρ ἡ ΒΑ, καὶ κείσθω τῆς ΒΑ ἡμίσεια called an apotome. ἡ ΑΔ. ἐπεὶ οὖν εὐθεῖα ἡ ΑΒ τέτμηται ἄκρον καὶ μέσον For let BA have been produced, and let AD be made λόγον κατὰ τὸ Γ, καὶ τῷ μείζονι τμήματι τῷ ΑΓ πρόσκειται (equal) to half of BA. Therefore, since the straight- ἡ ΑΔ ἡμίσεια οὖσα τῆς ΑΒ, τὸ ἄρα ἀπὸ ΓΔ τοῦ ἀπὸ ΔΑ line AB has been cut in extreme and mean ratio at C, πενταπλάσιόν ἐστιν. τὸ ἄρα ἀπὸ ΓΔ πρὸς τὸ ἀπὸ ΔΑ λόγον and AD, which is half of AB, has been added to the ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν· σύμμετρον ἄρα τὸ ἀπὸ ΓΔ greater piece AC, the (square) on CD is thus five times τῷ ἀπὸ ΔΑ. ῥητὸν δὲ τὸ ἀπὸ ΔΑ· ῥητὴ γάρ [ἐστιν] ἡ ΔΑ the (square) on DA [Prop. 13.1]. Thus, the (square) on ἡμίσεια οὖσα τῆς ΑΒ ῥητῆς οὔσης· ῥητὸν ἄρα καὶ τὸ ἀπὸ CD has to the (square) on DA the ratio which a number ΓΔ· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΓΔ. καὶ ἐπεὶ τὸ ἀπὸ ΓΔ πρὸς (has) to a number. The (square) on CD (is) thus com- τὸ ἀπὸ ΔΑ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς mensurable with the (square) on DA [Prop. 10.6]. And τετράγωνον ἀριθμόν, ἀσύμμετρος ἄρα μήκει ἡ ΓΔ τῇ ΔΑ· αἱ the (square) on DA (is) rational. For DA [is] rational, ΓΔ, ΔΑ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἀποτομὴ being half of AB, which is rational. Thus, the (square) ἄρα ἐστὶν ἡ ΑΓ. πάλιν, ἐπεὶ ἡ ΑΒ ἄκρον καὶ μέσον λόγον on CD (is) also rational [Def. 10.4]. Thus, CD is also τέτμηται, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΑΓ, τὸ ἄρα ὑπὸ ΑΒ, rational. And since the (square) on CD does not have ΒΓ τῷ ἀπὸ ΑΓ ἴσον ἐστίν. τὸ ἄρα ἀπὸ τῆς ΑΓ ἀποτομῆς to the (square) on DA the ratio which a square num- παρὰ τὴν ΑΒ ῥητὴν παραβληθὲν πλάτος ποιεῖ τὴν ΒΓ. τὸ ber (has) to a square number, CD (is) thus incommensu- δὲ ἀπὸ ἀποτομῆς παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ rable in length with DA [Prop. 10.9]. Thus, CD and DA ἀποτομὴν πρώτην· ἀποτομὴ ἄρα πρώτη ἐστὶν ἡ ΓΒ. ἐδείχθη are rational (straight-lines which are) commensurable in δὲ καὶ ἡ ΓΑ ἀποτομή. square only. Thus, AC is an apotome [Prop. 10.73]. ᾿Εὰν ἄρα εὐθεῖα ῥητὴ ἄκρον καὶ μέσον λόγον τμηθῇ, Again, since AB has been cut in extreme and mean ratio, ἑκάτερον τῶν τμημάτων ἄλογός ἐστιν ἡ καλουμένη ἀπο- and AC is the greater piece, the (rectangle contained) by τομή· ὅπερ ἔδει δεῖξαι. AB and BC is thus equal to the (square) on AC [Def. 6.3, Prop. 6.17]. Thus, the (square) on the apotome AC, ap- plied to the rational (straight-line) AB, makes BC as width. And the (square) on an apotome, applied to a rational (straight-line), makes a first apotome as width [Prop. 10.97]. Thus, CB is a first apotome. And CA was also shown (to be) an apotome. 511 STOIQEIWN igþ. ELEMENTS BOOK 13 Thus, if a rational straight-line is cut in extreme and mean ratio then each of the pieces is that irrational (straight-line) called an apotome.zþ. Proposition 7 ᾿Εὰν πενταγώνου ἰσοπλεύρου αἱ τρεῖς γωνίαι ἤτοι αἱ If three angles, either consecutive or not consecutive, κατὰ τὸ ἑξῆς ἢ αἱ μὴ κατὰ τὸ ἑξῆς ἴσαι ὦσιν, ἰσογώνιον of an equilateral pentagon are equal then the pentagon ἔσται τὸ πεντάγωνον. will be equiangular. D B Z G E A C B E A D F Πενταγώνου γὰρ ἰσοπλεύρον τοῦ ΑΒΓΔΕ αἱ τρεῖς For let three angles of the equilateral pentagon γωνίαι πρότερον αἱ κατὰ τὸ ἑξῆς αἱ πρὸς τοῖς Α, Β, Γ ἴσαι ABCDE—first of all, the consecutive (angles) at A, B, ἀλλήλαις ἔστωσαν· λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓΔΕ and C—-be equal to one another. I say that pentagon πεντάγωνον. ABCDE is equiangular. ᾿Επεζεύχθωσαν γὰρ αἱ ΑΓ, ΒΕ, ΖΔ. καὶ ἐπεὶ δύο αἱ For let AC, BE, and FD have been joined. And since ΓΒ, ΒΑ δυσὶ ταῖς ΒΑ, ΑΕ ἴσαι ἐισὶν ἑκατέρα ἑκατέρᾳ, καὶ the two (straight-lines) CB and BA are equal to the two γωνία ἡ ὑπὸ ΓΒΑ γωνίᾳ τῇ ὑπὸ ΒΑΕ ἐστιν ἴση, βάσις ἄρα ἡ (straight-lines) BA and AE, respectively, and angle CBA ΑΓ βάσει τῇ ΒΕ ἐστιν ἴση, καὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΒΕ is equal to angle BAE, base AC is thus equal to base τριγώνῳ ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι BE, and triangle ABC equal to triangle ABE, and the ἔσονται, ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν, ἡ μὲν ὑπὸ ΒΓΑ remaining angles will be equal to the remaining angles τῇ ὑπὸ ΒΕΑ, ἡ δὲ ὑπὸ ΑΒΕ τῇ ὑπὸ ΓΑΒ· ὥστε καὶ πλευρὰ which the equal sides subtend [Prop. 1.4], (that is), BCA ἡ ΑΖ πλευρᾷ τῇ ΒΖ ἐστιν ἴση. ἐδείχθη δὲ καὶ ὅλη ἡ ΑΓ (equal) to BEA, and ABE to CAB. And hence side AF ὅλῃ τῇ ΒΕ ἴση· καὶ λοιπὴ ἄρα ἡ ΖΓ λοιπῇ τῇ ΖΕ ἐστιν ἴση. is also equal to side BF [Prop. 1.6]. And the whole of AC ἔστι δὲ καὶ ἡ ΓΔ τῇ ΔΕ ἴση. δύο δὴ αἱ ΖΓ, ΓΔ δυσὶ ταῖς was also shown (to be) equal to the whole of BE. Thus, ΖΕ, ΕΔ ἴσαι εἰσίν· καὶ βάσις αὐτῶν κοινὴ ἡ ΖΔ· γωνία ἄρα the remainder FC is also equal to the remainder FE. ἡ ὑπὸ ΖΓΔ γωνίᾳ τῇ ὑπὸ ΖΕΔ ἐστιν ἴση. ἐδείχθη δὲ καὶ And CD is also equal to DE. So, the two (straight-lines) ἡ ὑπὸ ΒΓΑ τῇ ὑπὸ ΑΕΒ ἴση· καὶ ὅλη ἄρα ἡ ὑπὸ ΒΓΔ ὅλῃ FC and CD are equal to the two (straight-lines) FE and τῇ ὑπὸ ΑΕΔ ἴση. ἀλλ᾿ ἡ ὑπὸ ΒΓΔ ἴση ὑπόκειται ταῖς πρὸς ED (respectively). And FD is their common base. Thus, τοῖς Α, Β γωνίαις· καὶ ἡ ὑπὸ ΑΕΔ ἄρα ταῖς πρὸς τοῖς Α, Β angle FCD is equal to angle FED [Prop. 1.8]. And BCA γωνίαις ἴση ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ὑπὸ ΓΔΕ was also shown (to be) equal to AEB. And thus the γωνία ἴση ἐστὶ ταῖς πρὸς τοῖς Α, Β, Γ γωνίαις· ἰσογώνιον whole of BCD (is) equal to the whole of AED. But, ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον. (angle) BCD was assumed (to be) equal to the angles at Ἀλλὰ δὴ μὴ ἔστωσαν ἴσαι αἱ κατὰ τὸ ἑξῆς γωνίαι, ἀλλ᾿ A and B. Thus, (angle) AED is also equal to the angles ἔστωσαν ἴσαι αἱ πρὸς τοῖς Α, Γ, Δ σημείοις· λέγω, ὅτι καὶ at A and B. So, similarly, we can show that angle CDE οὕτως ἰσογώνιόν ἐστι τὸ ΑΒΓΔΕ πεντάγωνον. is also equal to the angles at A, B, C. Thus, pentagon ᾿Επεζεύχθω γὰρ ἡ ΒΔ. καὶ ἐπεὶ δύο αἱ ΒΑ, ΑΕ δυσὶ ABCDE is equiangular. ταῖς ΒΓ, ΓΔ ἴσαι εἰσὶ καὶ γωνίας ἴσας περιέχουσιν, βάσις And so let consecutive angles not be equal, but let ἄρα ἡ ΒΕ βάσει τῇ ΒΔ ἴση ἐστίν, καὶ τὸ ΑΒΕ τρίγωνον τῷ the (angles) at points A, C, and D be equal. I say that ΒΓΔ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς pentagon ABCDE is also equiangular in this case. γωνίαις ἴσαι ἔσονται, ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν· For let BD have been joined. And since the two 512 STOIQEIWN igþ. ELEMENTS BOOK 13 ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΕΒ γωνία τῇ ὑπὸ ΓΔΒ. ἔστι δὲ καὶ (straight-lines) BA and AE are equal to the (straight- ἡ ὑπὸ ΒΕΔ γωνία τῇ ὑπὸ ΒΔΕ ἴση, ἐπεὶ καὶ πλευρὰ ἡ ΒΕ lines) BC and CD, and they contain equal angles, base πλευρᾷ τῇ ΒΔ ἐστιν ἴση. καὶ ὅλη ἄρα ἡ ὑπὸ ΑΕΔ γωνία BE is thus equal to base BD, and triangle ABE is equal ὅλῃ τῇ ὑπὸ ΓΔΕ ἐστιν ἴση. ἀλλὰ ἡ ὑπὸ ΓΔΕ ταῖς πρὸς τοῖς to triangle BCD, and the remaining angles will be equal Α, Γ γωνίαις ὑπόκειται ἴση· καὶ ἡ ὑπὸ ΑΕΔ ἄρα γωνία ταῖς to the remaining angles which the equal sides subtend πρὸς τοῖς Α, Γ ἴση ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΑΒΓ ἴση [Prop. 1.4]. Thus, angle AEB is equal to (angle) CDB. ἐστὶ ταῖς πρὸς τοῖς Α, Γ, Δ γωνίαις. ἰσογώνιον ἄρα ἐστὶ τὸ And angle BED is also equal to (angle) BDE, since side ΑΒΓΔΕ πεντάγωνον· ὅπερ ἔδει δεῖξαι. BE is also equal to side BD [Prop. 1.5]. Thus, the whole angle AED is also equal to the whole (angle) CDE. But, (angle) CDE was assumed (to be) equal to the angles at A and C. Thus, angle AED is also equal to the (angles) at A and C. So, for the same (reasons), (angle) ABC is also equal to the angles at A, C, and D. Thus, pentagon ABCDE is equiangular. (Which is) the very thing it was required to show.hþ. Proposition 8 ᾿Εὰν πενταγώνου ἰσοπλεύρου καὶ ἰσογωνίου τὰς κατὰ If straight-lines subtend two consecutive angles of an τὸ ἑξῆς δύο γωνίας ὑποτείνωσιν εὐθεῖαι, ἄκρον καὶ μέσον equilateral and equiangular pentagon then they cut one λόγον τέμνουσιν ἀλλήλας, καὶ τὰ μείζονα αὐτῶν τμήματα another in extreme and mean ratio, and their greater ἴσα ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ. pieces are equal to the sides of the pentagon. J B A D G E H B A D E C Πενταγώνου γὰρ ἰσοπλεύρον καὶ ἰσογωνίου τοῦ ΑΒΓΔΕ For let the two straight-lines, AC and BE, cutting one δύο γωνίας τὰς κατὰ τὸ ἑξῆς τὰς πρὸς τοῖς Α, Β ὑπο- another at point H , have subtended two consecutive an- τεινέτωσαν εὐθεῖαι αἱ ΑΓ, ΒΕ τέμνουσαι ἀλλήλας κατὰ τὸ gles, at A and B (respectively), of the equilateral and Θ σημεὶον· λέγω, ὅτι ἑκατέρα αὐτῶν ἄκρον καὶ μέσον λόγον equiangular pentagon ABCDE. I say that each of them τέτμηται κατὰ τὸ Θ σημεῖον, καὶ τὰ μείζονα αὐτῶν τμήματα has been cut in extreme and mean ratio at point H , and ἴσα ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ. that their greater pieces are equal to the sides of the pen- Περιγεγράφθω γὰρ περὶ τὸ ΑΒΓΔΕ πεντάγωνον κύκλος tagon. ὁ ΑΒΓΔΕ. καὶ ἐπεὶ δύο εὐθεῖαι αἱ ΕΑ, ΑΒ δυσὶ ταῖς For let the circle ABCDE have been circumscribed ΑΒ, ΒΓ ἴσαι εἰσὶ καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα about pentagon ABCDE [Prop. 4.14]. And since the two ἡ ΒΕ βάσει τῇ ΑΓ ἴση ἐστίν, καὶ τὸ ΑΒΕ τρίγωνον τῷ straight-lines EA and AB are equal to the two (straight- ΑΒΓ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς lines) AB and BC (respectively), and they contain equal γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ angles, the base BE is thus equal to the base AC, and tri- ὑποτείνουσιν. ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΑΒΕ· angle ABE is equal to triangle ABC, and the remaining διπλῆ ἄρα ἡ ὑπὸ ΑΘΕ τῆς ὑπὸ ΒΑΘ. ἔστι δὲ καὶ ἡ ὑπὸ ΕΑΓ angles will be equal to the remaining angles, respectively, τῆς ὑπὸ ΒΑΓ διπλῆ, ἐπειδήπερ καὶ περιφέρεια ἡ ΕΔΓ περι- which the equal sides subtend [Prop. 1.4]. Thus, angle φερείας τῆς ΓΒ ἐστι διπλῆ· ἴση ἄρα ἡ ὑπὸ ΘΑΕ γωνία τῇ BAC is equal to (angle) ABE. Thus, (angle) AHE (is) ὑπὸ ΑΘΕ· ὥστε καὶ ἡ ΘΕ εὐθεῖα τῇ ΕΑ, τουτέστι τῇ ΑΒ double (angle) BAH [Prop. 1.32]. And EAC is also dou- 513 STOIQEIWN igþ. ELEMENTS BOOK 13 ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΑ εὐθεῖα τῇ ΑΕ, ἴση ἐστὶ ble BAC, inasmuch as circumference EDC is also dou- καὶ γωνία ἡ ὑπὸ ΑΒΕ τῇ ὑπὸ ΑΕΒ. ἀλλὰ ἡ ὑπὸ ΑΒΕ τῇ ble circumference CB [Props. 3.28, 6.33]. Thus, angle ὑπὸ ΒΑΘ ἐδείχθη ἴση· καὶ ἡ ὑπὸ ΒΕΑ ἄρα τῇ ὑπὸ ΒΑΘ HAE (is) equal to (angle) AHE. Hence, straight-line ἐστιν ἴση. καὶ κοινὴ τῶν δύο τριγώνων τοῦ τε ΑΒΕ καὶ HE is also equal to (straight-line) EA—that is to say, τοῦ ΑΒΘ ἐστιν ἡ ὑπὸ ΑΒΕ· λοιπὴ ἄρα ἡ ὑπὸ ΒΑΕ γωνία to (straight-line) AB [Prop. 1.6]. And since straight-line λοιπῇ τῇ ὑπὸ ΑΘΒ ἐστιν ἴση· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΕ BA is equal to AE, angle ABE is also equal to AEB τρίγωνον τῷ ΑΒΘ τριγώνῳ· ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΕΒ [Prop. 1.5]. But, ABE was shown (to be) equal to BAH . πρὸς τὴν ΒΑ, οὕτως ἡ ΑΒ πρὸς τὴν ΒΘ. ἴση δὲ ἡ ΒΑ τῇ Thus, BEA is also equal to BAH . And (angle) ABE is ΕΘ· ὡς ἄρα ἡ ΒΕ πρὸς τὴν ΕΘ, οὕτως ἡ ΕΘ πρὸς τὴν ΘΒ. common to the two triangles ABE and ABH . Thus, the μείζων δὲ ἡ ΒΕ τῆς ΕΘ· μείζων ἄρα καὶ ἡ ΕΘ τῆς ΘΒ. ἡ remaining angle BAE is equal to the remaining (angle) ΒΕ ἅρα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Θ, καὶ τὸ AHB [Prop. 1.32]. Thus, triangle ABE is equiangular to μεῖζον τμῆμα τὸ ΘΕ ἴσον ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ. triangle ABH . Thus, proportionally, as EB is to BA, so ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΑΓ ἄκρον καὶ μέσον λόγον AB (is) to BH [Prop. 6.4]. And BA (is) equal to EH . τέτμηται κατὰ τὸ Θ, καὶ τὸ μεῖζον αὐτῆς τμῆμα ἡ ΓΘ ἴσον Thus, as BE (is) to EH , so EH (is) to HB. And BE ἐστὶ τῇ τοῦ πενταγώνου πλευρᾷ· ὅπερ ἔδει δεῖξαι. (is) greater than EH . EH (is) thus also greater than HB [Prop. 5.14]. Thus, BE has been cut in extreme and mean ratio at H , and the greater piece HE is equal to the side of the pentagon. So, similarly, we can show that AC has also been cut in extreme and mean ratio at H , and that its greater piece CH is equal to the side of the pentagon. (Which is) the very thing it was required to show.jþ. Proposition 9 ᾿Εὰν ἡ τοῦ ἑξαγώνου πλευρὰ καὶ ἡ τοῦ δεκαγώνου τῶν If the side of a hexagon and of a decagon inscribed εἰς τὸν αὐτὸν κύκλον ἐγγραφομένων συντεθῶσιν, ἡ ὅλη in the same circle are added together then the whole εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται, καὶ τὸ μεῖξον αὐτῆς straight-line has been cut in extreme and mean ratio (at τμῆμά ἐστιν ἡ τοῦ ἑξαγώνου πλευρά. the junction point), and its greater piece is the side of the hexagon.† AE G ZD B F E D B A C ῎Εστω κύκλος ὁ ΑΒΓ, καὶ τῶν εἰς τὸν ΑΒΓ κύκλον Let ABC be a circle. And of the figures inscribed in ἐγγραφομένων σχημάτων, δεκαγώνου μὲν ἔστω πλευρὰ ἡ circle ABC, let BC be the side of a decagon, and CD (the ΒΓ, ἑξαγώνου δὲ ἡ ΓΔ, καὶ ἔστωσαν ἐπ᾿ εὐθείας· λέγω, ὅτι side) of a hexagon. And let them be (laid down) straight- ἡ ὅλη εὐθεῖα ἡ ΒΔ ἄκρον καὶ μέσον λόγον τέτμηται, καὶ τὸ on (to one another). I say that the whole straight-line μεῖζον αὐτῆς τμῆμά ἐστιν ἡ ΓΔ. BD has been cut in extreme and mean ratio (at C), and Εἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου τὸ Ε σημεῖον, καὶ that CD is its greater piece. ἐπεζεύχθωσαν αἱ ΕΒ, ΕΓ, ΕΔ, καὶ διήχθω ἡ ΒΕ ἐπὶ τὸ For let the center of the circle, point E, have been 514 STOIQEIWN igþ. ELEMENTS BOOK 13 Α. ἐπεὶ δεκαγώνου ἰσοπλεύρον πλευρά ἐστιν ἡ ΒΓ, πεντα- found [Prop. 3.1], and let EB, EC, and ED have been πλασίων ἄρα ἡ ΑΓΒ περιφέρεια τῆς ΒΓ περιφερείας· τε- joined, and let BE have been drawn across to A. Since τραπλασίων ἄρα ἡ ΑΓ περιφέρεια τῆς ΓΒ. ὡς δὲ ἡ ΑΓ πε- BC is a side on an equilateral decagon, circumference ριφέρεια πρὸς τὴν ΓΒ, οὕτως ἡ ὑπὸ ΑΕΓ γωνία πρὸς τὴν ACB (is) thus five times circumference BC. Thus, cir- ὑπὸ ΓΕΒ· τετραπλασίων ἄρα ἡ ὑπὸ ΑΕΓ τῆς ὑπὸ ΓΕΒ. καὶ cumference AC (is) four times CB. And as circumference ἐπεὶ ἴση ἡ ὑπὸ ΕΒΓ γωνία τῇ ὑπὸ ΕΓΒ, ἡ ἄρα ὑπὸ ΑΕΓ AC (is) to CB, so angle AEC (is) to CEB [Prop. 6.33]. γωνία διπλασία ἐστὶ τῆς ὑπὸ ΕΓΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΕΓ Thus, (angle) AEC (is) four times CEB. And since angle εὐθεῖα τῇ ΓΔ· ἑκατέρα γὰρ αὐτῶν ἴση ἐστὶ τῇ τοῦ ἑξαγώνου EBC (is) equal to ECB [Prop. 1.5], angle AEC is thus πλευρᾷ τοῦ εἰς τὸν ΑΒΓ κύκλον [ἐγγραφομένου]· ἴση ἐστὶ double ECB [Prop. 1.32]. And since straight-line EC is καὶ ἡ ὑπὸ ΓΕΔ γωνία τῇ ὑπὸ ΓΔΕ γωνίᾳ· διπλασία ἄρα equal to CD—for each of them is equal to the side of the ἡ ὑπὸ ΕΓΒ γωνία τῆς ὑπὸ ΕΔΓ. ἀλλὰ τῆς ὑπὸ ΕΓΒ δι- hexagon [inscribed] in circle ABC [Prop. 4.15 corr.]— πλασία ἐδείχθη ἡ ὑπὸ ΑΕΓ· τετραπλασία ἄρα ἡ ὑπὸ ΑΕΓ angle CED is also equal to angle CDE [Prop. 1.5]. Thus, τῆς ὑπὸ ΕΔΓ. ἐδείχθη δὲ καὶ τῆς ὑπὸ ΒΕΓ τετραπλασία angle ECB (is) double EDC [Prop. 1.32]. But, AEC ἡ ὑπὸ ΑΕΓ· ἴση ἄρα ἡ ὑπὸ ΕΔΓ τῇ ὑπὸ ΒΕΓ. κοινὴ δὲ was shown (to be) double ECB. Thus, AEC (is) four τῶν δύο τριγώνων, τοῦ τε ΒΕΓ καὶ τοῦ ΒΕΔ, ἡ ὑπὸ ΕΒΔ times EDC. And AEC was also shown (to be) four times γωνία· καὶ λοιπὴ ἄρα ἡ ὑπὸ ΒΕΔ τῇ ὑπὸ ΕΓΒ ἐστιν ἴση· BEC. Thus, EDC (is) equal to BEC. And angle EBD ἰσογώνιον ἄρα ἐστὶ τὸ ΕΒΔ τρίγωνον τῷ ΕΒΓ τριγώνῳ. (is) common to the two triangles BEC and BED. Thus, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΕΒ the remaining (angle) BED is equal to the (remaining πρὸς τὴν ΒΓ. ἴση δὲ ἡ ΕΒ τῇ ΓΔ. ἔστιν ἄρα ὡς ἡ ΒΔ πρὸς angle) ECB [Prop. 1.32]. Thus, triangle EBD is equian- τὴν ΔΓ, οὕτως ἡ ΔΓ πρὸς τὴν ΓΒ. μείζων δὲ ἡ ΒΔ τῆς gular to triangle EBC. Thus, proportionally, as DB is to ΔΓ· μείζων ἄρα καὶ ἡ ΔΓ τῆς ΓΒ. ἡ ΒΔ ἄρα εὐθεῖα ἄκρον BE, so EB (is) to BC [Prop. 6.4]. And EB (is) equal καὶ μέσον λόγον τέτμηται [κατὰ τὸ Γ], καὶ τὸ μεῖζον τμῆμα to CD. Thus, as BD is to DC, so DC (is) to CB. And αὐτῆς ἐστιν ἡ ΔΓ· ὅπερ ἔδει δεῖξαι. BD (is) greater than DC. Thus, DC (is) also greater than CB [Prop. 5.14]. Thus, the straight-line BD has been cut in extreme and mean ratio [at C], and DC is its greater piece. (Which is), the very thing it was required to show. † If the circle is of unit radius then the side of the hexagon is 1, whereas the side of the decagon is (1/2) ( √ 5 − 1).iþ. Proposition 10 ᾿Εὰν εἰς κύκλον πεντάγωνον ἰσόπλευρον ἐγγραφῇ, ἡ τοῦ If an equilateral pentagon is inscribed in a circle then πενταγώνου πλευρὰ δύναται τήν τε τοῦ ἑξαγώνου καὶ τὴν the square on the side of the pentagon is (equal to) the τοῦ δεκαγώνου τῶν εἰς τὸν αὐτὸν κύκλον ἐγγραφομένων. (sum of the squares) on the (sides) of the hexagon and of the decagon inscribed in the same circle.† L H Z K A E G D B N M J E K A B N M G C F H L D 515 STOIQEIWN igþ. ELEMENTS BOOK 13 ῎Εστω κύκλος ὁ ΑΒΓΔΕ, καὶ εἰς τὸ ΑΒΓΔΕ κύκλον Let ABCDE be a circle. And let the equilateral pen- πεντάγωνον ἰσόπλευρον ἐγγεγράφθω τὸ ΑΒΓΔΕ. λέγω, tagon ABCDE have been inscribed in circle ABCDE. I ὅτι ἡ τοῦ ΑΒΓΔΕ πενταγώνου πλευρὰ δύναται τήν τε τοῦ say that the square on the side of pentagon ABCDE is ἑξαγώνου καὶ τὴν τοῦ δεκαγώνου πλευρὰν τῶν εἰς τὸν the (sum of the squares) on the sides of the hexagon and ΑΒΓΔΕ κύκλον ἐγγραφομένων. of the decagon inscribed in circle ABCDE. Εἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου τὸ Ζ σημεὶον, καὶ For let the center of the circle, point F , have been ἐπιζευχθεῖσα ἡ ΑΖ διήχθω ἐπὶ τὸ Η σημεῖον, καὶ ἐπεζεύχθω found [Prop. 3.1]. And, AF being joined, let it have been ἡ ΖΒ, καὶ ἀπὸ τοῦ Ζ ἐπὶ τὴν ΑΒ κάθετος ἤχθω ἡ ΖΘ, καὶ drawn across to point G. And let FB have been joined. διήχθω ἐπὶ τὸ Κ, καὶ ἐπεζεύχθωσαν αἱ ΑΚ, ΚΒ, καὶ πάλιν And let FH have been drawn from F perpendicular to ἀπὸ τοῦ Ζ ἐπὶ τὴν ΑΚ κάθετος ἤχθω ἡ ΖΛ, καὶ διήχθω ἐπὶ AB. And let it have been drawn across to K. And let AK τὸ Μ, καὶ ἐπεζεύχθω ἡ ΚΝ. and KB have been joined. And, again, let FL have been ᾿Επεὶ ἴση ἐστὶν ἡ ΑΒΓΗ περιφέρεια τῇ ΑΕΔΗ περι- drawn from F perpendicular to AK. And let it have been φερείᾳ, ὧν ἡ ΑΒΓ τῇ ΑΕΔ ἐστιν ἴση, λοιπὴ ἄρα ἡ ΓΗ drawn across to M . And let KN have been joined. περιφέρεια λοιπῇ τῇ ΗΔ ἐστιν ἴση. πενταγώνου δὲ ἡ ΓΔ· Since circumference ABCG is equal to circumference δεκαγώνου ἄρα ἡ ΓΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΖΑ τῇ ΖΒ, καὶ AEDG, of which ABC is equal to AED, the remain- κάθετος ἡ ΖΘ, ἴση ἄρα καὶ ἡ ὑπὸ ΑΖΚ γωνία τῇ ὑπὸ ΚΖΒ. ing circumference CG is thus equal to the remaining ὥστε καὶ περιφέρεια ἡ ΑΚ τῇ ΚΒ ἐστιν ἴση· διπλῆ ἄρα ἡ (circumference) GD. And CD (is the side) of the pen- ΑΒ περιφέρεια τῆς ΒΚ περιφερείας· δεκαγώνου ἄρα πλευρά tagon. CG (is) thus (the side) of the decagon. And since ἐστιν ἡ ΑΚ εὐθεῖα. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΑΚ τῆς ΚΜ ἐστι FA is equal to FB, and FH is perpendicular (to AB), διπλῆ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΑΒ περιφέρεια τῆς ΒΚ περι- angle AFK (is) thus also equal to KFB [Props. 1.5, φερείας, ἴση δὲ ἡ ΓΔ περιφέρεια τῇ ΑΒ περιφερείᾳ, διπλῆ 1.26]. Hence, circumference AK is also equal to KB ἄρα καὶ ἡ ΓΔ περιφέρεια τῆς ΒΚ περιφερείας. ἔστι δὲ ἡ ΓΔ [Prop. 3.26]. Thus, circumference AB (is) double cir- περιφέρεια καὶ τῆς ΓΗ διπλῆ· ἴση ἄρα ἡ ΓΗ περιφέρεια τῇ cumference BK. Thus, straight-line AK is the side of ΒΚ περιφερείᾳ. ἀλλὰ ἡ ΒΚ τῆς ΚΜ ἐστι διπλῆ, ἐπεὶ καὶ the decagon. So, for the same (reasons, circumference) ἡ ΚΑ· καὶ ἡ ΓΗ ἄρα τῆς ΚΜ ἐστι διπλῆ. ἀλλὰ μὴν καὶ ἡ AK is also double KM . And since circumference AB ΓΒ περιφέρεια τῆς ΒΚ περιφερείας ἐστὶ διπλῆ· ἴση γὰρ ἡ is double circumference BK, and circumference CD (is) ΓΒ περιφέρεια τῇ ΒΑ. καὶ ὅλη ἄρα ἡ ΗΒ περιφέρεια τῆς equal to circumference AB, circumference CD (is) thus ΒΜ ἐστι διπλῆ· ὥστε καὶ γωνία ἡ ὑπὸ ΗΖΒ γωνίας τῆς ὑπὸ also double circumference BK. And circumference CD ΒΖΜ [ἐστι] διπλῆ. ἔστι δὲ ἡ ὑπὸ ΗΖΒ καὶ τῆς ὑπὸ ΖΑΒ is also double CG. Thus, circumference CG (is) equal διπλῆ· ἴση γὰρ ἡ ὑπὸ ΖΑΒ τῇ ὑπὸ ΑΒΖ. καὶ ἡ ὑπὸ ΒΖΝ ἄρα to circumference BK. But, BK is double KM , since τῇ ὑπὸ ΖΑΒ ἐστιν ἴση. κοινὴ δὲ τῶν δύο τριγώνων, τοῦ KA (is) also (double KM). Thus, (circumference) CG τε ΑΒΖ καὶ τοῦ ΒΖΝ, ἡ ὑπὸ ΑΒΖ γωνία· λοιπὴ ἄρα ἡ ὑπὸ is also double KM . But, indeed, circumference CB is ΑΖΒ λοιπῇ τῇ ὑπὸ ΒΝΖ ἐστιν ἴση· ἴσογώνιον ἄρα ἐστὶ τὸ also double circumference BK. For circumference CB ΑΒΖ τρίγωνον τῷ ΒΖΝ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ (is) equal to BA. Thus, the whole circumference GB ΑΒ εὐθεῖα πρὸς τὴν ΒΖ, οὕτως ἡ ΖΒ πρὸς τὴν ΒΝ· τὸ ἄρα is also double BM . Hence, angle GFB [is] also dou- ὑπὸ τῶν ΑΒΝ ἴσον ἐστὶ τῷ ἀπὸ ΒΖ. πάλιν ἐπεὶ ἴση ἐστὶν ἡ ble angle BFM [Prop. 6.33]. And GFB (is) also dou- ΑΛ τῇ ΛΚ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΛΝ, βάσις ἄρα ἡ ΚΝ ble FAB. For FAB (is) equal to ABF . Thus, BFN βάσει τῇ ΑΝ ἐστιν ἴση· καὶ γωνία ἄρα ἡ ὑπὸ ΛΚΝ γωνίᾳ τῇ is also equal to FAB. And angle ABF (is) common to ὑπὸ ΛΑΝ ἐστιν ἴση. ἀλλὰ ἡ ὑπὸ ΛΑΝ τῇ ὑπὸ ΚΒΝ ἐστιν the two triangles ABF and BFN . Thus, the remain- ἴση· καὶ ἡ ὑπὸ ΛΚΝ ἄρα τῇ ὑπὸ ΚΒΝ ἐστιν ἴση. καὶ κοινὴ ing (angle) AFB is equal to the remaining (angle) BNF τῶν δύο τριγώνων τοῦ τε ΑΚΒ καὶ τοῦ ΑΚΝ ἡ πρὸς τῷ [Prop. 1.32]. Thus, triangle ABF is equiangular to trian- Α. λοιπὴ ἄρα ἡ ὑπὸ ΑΚΒ λοιπῇ τῇ ὑπὸ ΚΝΑ ἐστιν ἴση· gle BFN . Thus, proportionally, as straight-line AB (is) ἰσογώνιον ἄρα ἐστὶ τὸ ΚΒΑ τρίγωνον τῷ ΚΝΑ τριγώνῳ. to BF , so FB (is) to BN [Prop. 6.4]. Thus, the (rectan- ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΑ εὐθεῖα πρὸς τὴν ΑΚ, οὕτως gle contained) by ABN is equal to the (square) on BF ἡ ΚΑ πρὸς τὴν ΑΝ· τὸ ἄρα ὑπὸ τῶν ΒΑΝ ἴσον ἐστὶ τῷ [Prop. 6.17]. Again, since AL is equal to LK, and LN ἀπὸ τῆς ΑΚ. ἐδείχθη δὲ καὶ τὸ ὑπὸ τῶν ΑΒΝ ἴσον τῷ ἀπὸ is common and at right-angles (to KA), base KN is thus τῆς ΒΖ· τὸ ἄρα ὑπὸ τῶν ΑΒΝ μετὰ τοῦ ὑπὸ ΒΑΝ, ὅπερ equal to base AN [Prop. 1.4]. And, thus, angle LKN ἐστὶ τὸ ἀπὸ τῆς ΒΑ, ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΖ μετὰ τοῦ ἀπὸ is equal to angle LAN . But, LAN is equal to KBN τῆς ΑΚ. καί ἐστιν ἡ μὲν ΒΑ πενταγώνου πλευρά, ἡ δὲ ΒΖ [Props. 3.29, 1.5]. Thus, LKN is also equal to KBN . ἑξαγώνου, ἡ δὲ ΑΚ δεκαγώνου. And the (angle) at A (is) common to the two triangles ῾Η ἄρα τοῦ πενταγώνου πλευρὰ δύναται τήν τε τοῦ AKB and AKN . Thus, the remaining (angle) AKB is 516 STOIQEIWN igþ. ELEMENTS BOOK 13 ἑξαγώνου καὶ τὴν τοῦ δεκαγώνου τῶν εἰς τὸν αὐτὸν κύκλον equal to the remaining (angle) KNA [Prop. 1.32]. Thus, ἐγγραφομένων· ὅπερ ἔδει δεῖξαι. triangle KBA is equiangular to triangle KNA. Thus, proportionally, as straight-line BA is to AK, so KA (is) to AN [Prop. 6.4]. Thus, the (rectangle contained) by BAN is equal to the (square) on AK [Prop. 6.17]. And the (rectangle contained) by ABN was also shown (to be) equal to the (square) on BF . Thus, the (rectangle con- tained) by ABN plus the (rectangle contained) by BAN , which is the (square) on BA [Prop. 2.2], is equal to the (square) on BF plus the (square) on AK. And BA is the side of the pentagon, and BF (the side) of the hexagon [Prop. 4.15 corr.], and AK (the side) of the decagon. Thus, the square on the side of the pentagon (in- scribed in a circle) is (equal to) the (sum of the squares) on the (sides) of the hexagon and of the decagon in- scribed in the same circle. † If the circle is of unit radius then the side of the pentagon is (1/2) p 10 − 2 √ 5.iaþ. Proposition 11 ᾿Εὰν εἰς κύκλον ῥητὴν ἔχοντα τὴν διάμετρον πεντάγω- If an equilateral pentagon is inscribed in a circle which νον ἰσόπλευρον ἐγγραφῇ, ἡ τοῦ πενταγώνου πλευρὰ ἄλογός has a rational diameter then the side of the pentagon is ἐστιν ἡ καλουμένη ἐλάσσων. that irrational (straight-line) called minor. D A H EM N B G L Z K J F A E M N B L K DC G H Εἰς γὰρ κύκλον τὸν ΑΒΓΔΕ ῥητὴν ἔχοντα τὴν δίαμετρον For let the equilateral pentagon ABCDE have been πεντάγωνον ἰσόπλευρον ἐγγεγράφθω τὸ ΑΒΓΔΕ· λέγω, inscribed in the circle ABCDE which has a rational di- ὅτι ἡ τοῦ [ΑΒΓΔΕ] πενταγώνου πλευρὰ ἄλογός ἐστιν ἡ ameter. I say that the side of pentagon [ABCDE] is that καλουμένη ἐλάσσων. irrational (straight-line) called minor. Εἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου τὸ Ζ σημεῖον, For let the center of the circle, point F , have been καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΖΒ καὶ διήχθωσαν ἐπὶ τὰ Η, Θ found [Prop. 3.1]. And let AF and FB have been joined. σημεῖα, καὶ ἐπεζεύχθω ἡ ΑΓ, καὶ κείσθω τῆς ΑΖ τέταρτον And let them have been drawn across to points G and H μέρος ἡ ΖΚ. ῥητὴ δὲ ἡ ΑΖ· ῥητὴ ἄρα καὶ ἡ ΖΚ. ἔστι δὲ (respectively). And let AC have been joined. And let FK καὶ ἡ ΒΖ ῥητή· ὅλη ἄρα ἡ ΒΚ ῥητή ἐστιν. καὶ ἐπεὶ ἴση made (equal) to the fourth part of AF . And AF (is) ratio- ἐστὶν ἡ ΑΓΗ περιφέρεια τῇ ΑΔΗ περιφερείᾳ, ὧν ἡ ΑΒΓ nal. FK (is) thus also rational. And BF is also rational. τῇ ΑΕΔ ἐστιν ἴση, λοιπὴ ἄρα ἡ ΓΗ λοιπῇ τῇ ΗΔ ἐστιν Thus, the whole of BK is rational. And since circum- ἴση. καὶ ἐὰν ἐπιζεύξωμεν τὴν ΑΔ, συνάγονται ὀρθαὶ αἱ ference ACG is equal to circumference ADG, of which 517 STOIQEIWN igþ. ELEMENTS BOOK 13 πρὸς τῷ Λ γωνίαι, καὶ διπλῆ ἡ ΓΔ τῆς ΓΛ. διὰ τὰ αὐτὰ ABC is equal to AED, the remainder CG is thus equal δὴ καὶ αἱ πρὸς τῷ Μ ὀρθαί εἰσιν, καὶ διπλῆ ἡ ΑΓ τῆς ΓΜ. to the remainder GD. And if we join AD then the angles ἐπεὶ οὖν ἴση ἐστὶν ἡ ὑπὸ ΑΛΓ γωνία τῇ ὑπὸ ΑΜΖ, κοινὴ at L are inferred (to be) right-angles, and CD (is inferred δὲ τῶν δύο τριγώνων τοῦ τε ΑΓΛ καὶ τοῦ ΑΜΖ ἡ ὑπὸ to be) double CL [Prop. 1.4]. So, for the same (reasons), ΛΑΓ, λοιπὴ ἄρα ἡ ὑπὸ ΑΓΛ λοιπῇ τῇ ὑπὸ ΜΖΑ ἐστιν ἴση· the (angles) at M are also right-angles, and AC (is) dou- ἰσογώνιον ἄρα ἐστὶ τὸ ΑΓΛ τρίγωνον τῷ ΑΜΖ τριγώνῳ· ble CM . Therefore, since angle ALC (is) equal to AMF , ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΛΓ πρὸς ΓΑ, οὕτως ἡ ΜΖ πρὸς and (angle) LAC (is) common to the two triangles ACL ΖΑ· καὶ τῶν ἡγουμένων τὰ διπλάσια· ὡς ἄρα ἡ τῆς ΛΓ and AMF , the remaining (angle) ACL is thus equal to διπλῆ πρὸς τὴν ΓΑ, οὕτως ἡ τῆς ΜΖ διπλῆ πρὸς τὴν ΖΑ. the remaining (angle) MFA [Prop. 1.32]. Thus, triangle ὡς δὲ ἡ τῆς ΜΖ διπλῆ πρὸς τὴν ΖΑ, οὕτως ἡ ΜΖ πρὸς τὴν ACL is equiangular to triangle AMF . Thus, proportion- ἡμίσειαν τῆς ΖΑ· καὶ ὡς ἄρα ἡ τῆς ΛΓ διπλῆ πρὸς τὴν ΓΑ, ally, as LC (is) to CA, so MF (is) to FA [Prop. 6.4]. And οὕτως ἡ ΜΖ πρὸς τὴν ἡμίσειαν τῆς ΖΑ· καὶ τῶν ἑπομένων (we can take) the doubles of the leading (magnitudes). τὰ ἡμίσεα· ὡς ἄρα ἡ τῆς ΛΓ διπλῆ πρὸς τὴν ἡμίσειαν τῆς Thus, as double LC (is) to CA, so double MF (is) to ΓΑ, οὕτως ἡ ΜΖ πρὸς τὸ τέτατρον τῆς ΖΑ. καί ἐστι τῆς FA. And as double MF (is) to FA, so MF (is) to half of μὲν ΛΓ διπλῆ ἡ ΔΓ, τῆς δὲ ΓΑ ἡμίσεια ἡ ΓΜ, τῆς δὲ ΖΑ FA. And, thus, as double LC (is) to CA, so MF (is) to τέτατρον μέρος ἡ ΖΚ· ἔστιν ἄρα ὡς ἡ ΔΓ πρὸς τὴν ΓΜ, half of FA. And (we can take) the halves of the following οὕτως ἡ ΜΖ πρὸς τὴν ΖΚ. συνθέντι καὶ ὡς συναμφότερος (magnitudes). Thus, as double LC (is) to half of CA, so ἡ ΔΓΜ πρὸς τὴν ΓΜ, οὕτως ἡ ΜΚ πρὸς ΚΖ· καὶ ὡς ἄρα τὸ MF (is) to the fourth of FA. And DC is double LC, and ἀπὸ συναμφοτέρου τῆς ΔΓΜ πρὸς τὸ ἀπὸ ΓΜ, οὕτως τὸ CM half of CA, and FK the fourth part of FA. Thus, ἀπὸ ΜΚ πρὸς τὸ ἀπὸ ΚΖ. καὶ ἐπεὶ τῆς ὑπὸ δύο πλευρὰς τοῦ as DC is to CM , so MF (is) to FK. Via composition, as πενταγώνου ὑποτεινούσης, οἷον τῆς ΑΓ, ἄκρον καὶ μέσον the sum of DCM (i.e., DC and CM) (is) to CM , so MK λόγον τεμνομένης τὸ μεῖζον τμῆμα ἴσον ἐστὶ τῇ τοῦ πεν- (is) to KF [Prop. 5.18]. And, thus, as the (square) on the ταγώνου πλευρᾷ, τουτέστι τῇ ΔΓ, τὸ δὲ μεῖζον τμῆμα προ- sum of DCM (is) to the (square) on CM , so the (square) σλαβὸν τὴν ἡμίσειαν τῆς ὅλῆς πενταπλάσιον δύναται τοῦ on MK (is) to the (square) on KF . And since the greater ἀπὸ τῆς ἡμισείας τῆς ὅλης, καί ἐστιν ὅλης τῆς ΑΓ ἡμίσεια piece of a (straight-line) subtending two sides of a pen- ἡ ΓΜ, τὸ ἄρα ἀπὸ τῆς ΔΓΜ ὡς μιᾶς πενταπλάσιόν ἐστι tagon, such as AC, (which is) cut in extreme and mean τοῦ ἀπὸ τῆς ΓΜ. ὡς δὲ τὸ ἀπὸ τῆς ΔΓΜ ὡς μιᾶς πρὸς τὸ ratio is equal to the side of the pentagon [Prop. 13.8]— ἀπὸ τῆς ΓΜ, οὕτως ἐδείχθη τὸ ἀπὸ τῆς ΜΚ πρὸς τὸ ἀπὸ that is to say, to DC—-and the square on the greater piece τῆς ΚΖ· πενταπλάσιον ἄρα τὸ ἀπὸ τῆς ΜΚ τοῦ ἀπὸ τῆς added to half of the whole is five times the (square) on ΚΖ. ῥητὸν δὲ τὸ ἀπὸ τῆς ΚΖ· ῥητὴ γὰρ ἡ διάμετρος· ῥητὸν half of the whole [Prop. 13.1], and CM (is) half of the ἄρα καὶ τὸ ἀπὸ τῆς ΜΚ· ῥητὴ ἄρα ἐστὶν ἡ ΜΚ [δυνάμει whole, AC, thus the (square) on DCM , (taken) as one, μόνον]. καὶ ἐπεὶ τετραπλασία ἐστὶν ἡ ΒΖ τῆς ΖΚ, πεντα- is five times the (square) on CM . And the (square) on πλασία ἄρα ἐστὶν ἡ ΒΚ τῆς ΚΖ· εἰκοσιπενταπλάσιον ἄρα τὸ DCM , (taken) as one, (is) to the (square) on CM , so ἀπὸ τῆς ΒΚ τοῦ ἀπὸ τῆς ΚΖ. πενταπλάσιον δὲ τὸ ἀπὸ τῆς the (square) on MK was shown (to be) to the (square) ΜΚ τοῦ ἀπὸ τῆς ΚΖ· πενταπλάσιον ἄρα τὸ ἀπὸ τῆς ΒΚ on KF . Thus, the (square) on MK (is) five times the τοῦ ἀπὸ τῆς ΚΜ· τὸ ἄρα ἀπὸ τῆς ΒΚ πρὸς τὸ ἀπὸ ΚΜ (square) on KF . And the square on KF (is) rational. λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον For the diameter (is) rational. Thus, the (square) on ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΚ τῇ ΚΜ μήκει. καί ἐστι MK (is) also rational. Thus, MK is rational [in square ῥητὴ ἑκατέρα αὐτῶν. αἱ ΒΚ, ΚΜ ἄρα ῥηταί εἰσι δυνάμει only]. And since BF is four times FK, BK is thus five μόνον σύμμετροι. ἐὰν δὲ ἀπὸ ῥητῆς ῥητὴ ἀφαιρεθῇ δυνάμει times KF . Thus, the (square) on BK (is) twenty-five μόνον σύμμετρος οὖσα τῇ ὅλῃ, ἡ λοιπὴ ἄλογός ἐστιν ἀπο- times the (square) on KF . And the (square) on MK τομή· ἀποτομὴ ἄρα ἐστὶν ἡ ΜΒ, προσαρμόζουσα δὲ αὐτῇ ἡ (is) five times the square on KF . Thus, the (square) ΜΚ. λέγω δή, ὅτι καὶ τετάρτη. ᾧ δὴ μεῖζόν ἐστι τὸ ἀπὸ on BK (is) five times the (square) on KM . Thus, the τῆς ΒΚ τοῦ ἀπὸ τῆς ΚΜ, ἐκείνῳ ἴσον ἔστω τὸ ἀπὸ τῆς Ν· (square) on BK does not have to the (square) on KM ἡ ΒΚ ἄρα τῆς ΚΜ μεῖζον δύναται τῇ Ν. καὶ ἐπεὶ σύμμετρός the ratio which a square number (has) to a square num- ἐστιν ἡ ΚΖ τῇ ΖΒ, καὶ συνθέντι σύμμετρός ἐστι ἡ ΚΒ τῇ ber. Thus, BK is incommensurable in length with KM ΖΒ. ἀλλὰ ἡ ΒΖ τῇ ΒΘ σύμμετρός ἐστιν· καὶ ἡ ΒΚ ἄρα τῇ [Prop. 10.9]. And each of them is a rational (straight- ΒΘ σύμμετρός ἐστιν. καὶ ἐπεὶ πενταπλάσιόν ἐστι τὸ ἀπὸ line). Thus, BK and KM are rational (straight-lines τῆς ΒΚ τοῦ ἀπὸ τῆς ΚΜ, τὸ ἄρα ἀπὸ τῆς ΒΚ πρὸς τὸ which are) commensurable in square only. And if from ἀπὸ τῆς ΚΜ λόγον ἔχει, ὃν ε πρὸς ἕν. ἀναστρέψαντι ἄρα a rational (straight-line) a rational (straight-line) is sub- τὸ ἀπὸ τῆς ΒΚ πρὸς τὸ ἀπὸ τῆς Ν λόγον ἔχει, ὃν ε πρὸς tracted, which is commensurable in square only with the 518 STOIQEIWN igþ. ELEMENTS BOOK 13 δ, οὐχ ὃν τετράγωνος πρὸς τετράγωνον· ἀσύμμετρος ἄρα whole, then the remainder is that irrational (straight-line ἐστὶν ἡ ΒΚ τῇ Ν· ἡ ΒΚ ἄρα τῆς ΚΜ μεῖζον δύναται τῷ ἀπὸ called) an apotome [Prop. 10.73]. Thus, MB is an apo- ἀσυμμέτρου ἑαυτῇ. ἐπεὶ οὖν ὅλη ἡ ΒΚ τῆς προσαρμοζούσης tome, and MK its attachment. So, I say that (it is) also τῆς ΚΜ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ὅλη a fourth (apotome). So, let the (square) on N be (made) ἡ ΒΚ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΒΘ, ἀποτομὴ equal to that (magnitude) by which the (square) on BK ἄρα τετάρτη ἐστὶν ἡ ΜΒ. τὸ δὲ ὑπὸ ῥητῆς καὶ ἀποτομῆς is greater than the (square) on KM . Thus, the square on τετάρτης περιεχόμενον ὀρθογώνιον ἄλογόν ἐστιν, καὶ ἡ δυ- BK is greater than the (square) on KM by the (square) ναμένη αὐτὸ ἄλογός ἐστιν, καλεῖται δὲ ἐλάττων. δύναται on N . And since KF is commensurable (in length) with δὲ τὸ ὑπὸ τῶν ΘΒΜ ἡ ΑΒ διὰ τὸ ἐπιζευγνυμένης τῆς ΑΘ FB then, via composition, KB is also commensurable (in ἰσογώνιον γίνεσθαι τὸ ΑΒΘ τρίγωνον τῷ ΑΒΜ τριγώνῳ length) with FB [Prop. 10.15]. But, BF is commensu- καὶ εἶναι ὡς τὴν ΘΒ πρὸς τὴν ΒΑ, οὕτως τὴν ΑΒ πρὸς τὴν rable (in length) with BH . Thus, BK is also commen- ΒΜ. surable (in length) with BH [Prop. 10.12]. And since ῾Η ἄρα ΑΒ τοῦ πενταγώνου πλευρὰ ἄλογός ἐστιν ἡ κα- the (square) on BK is five times the (square) on KM , λουμένη ἐλάττων· ὅπερ ἔδει δεῖξαι. the (square) on BK thus has to the (square) on KM the ratio which 5 (has) to one. Thus, via conversion, the (square) on BK has to the (square) on N the ratio which 5 (has) to 4 [Prop. 5.19 corr.], which is not (that) of a square (number) to a square (number). BK is thus in- commensurable (in length) with N [Prop. 10.9]. Thus, the square on BK is greater than the (square) on KM by the (square) on (some straight-line which is) incom- mensurable (in length) with (BK). Therefore, since the square on the whole, BK, is greater than the (square) on the attachment, KM , by the (square) on (some straight- line which is) incommensurable (in length) with (BK), and the whole, BK, is commensurable (in length) with the (previously) laid down rational (straight-line) BH , MB is thus a fourth apotome [Def. 10.14]. And the rectangle contained by a rational (straight-line) and a fourth apotome is irrational, and its square-root is that irrational (straight-line) called minor [Prop. 10.94]. And the square on AB is the rectangle contained by HBM , on account of joining AH , (so that) triangle ABH be- comes equiangular with triangle ABM [Prop. 6.8], and (proportionally) as HB is to BA, so AB (is) to BM . Thus, the side AB of the pentagon is that irrational (straight-line) called minor.† (Which is) the very thing it was required to show. † If the circle has unit radius then the side of the pentagon is (1/2) p 10 − 2 √ 5. However, this length can be written in the “minor” form (see Prop. 10.94) (ρ/ √ 2) q 1 + k/ √ 1 + k2 − (ρ/ √ 2) q 1 − k/ √ 1 + k2, with ρ = p 5/2 and k = 2.ibþ. Proposition 12 ᾿Εὰν εἰς κύκλον τρίγωνον ἰσόπλευρον ἐγγραφῇ, ἡ τοῦ If an equilateral triangle is inscribed in a circle then τριγώνου πλευρὰ δυνάμει τριπλασίων ἐστὶ τῆς ἐκ τοῦ the square on the side of the triangle is three times the κέντρου τοῦ κύκλου. (square) on the radius of the circle. ᾿Εστω κύκλος ὁ ΑΒΓ, καὶ εἰς αὐτὸν τρίγωνον ἰσόπλευρ- Let there be a circle ABC, and let the equilateral tri- ον ἐγγεγράφθω τὸ ΑΒΓ· λέγω, ὅτι τοῦ ΑΒΓ τριγώνου μία angle ABC have been inscribed in it [Prop. 4.2]. I say πλευρὰ δυνάμει τριπλασίων ἐστὶ τῆς ἐκ τοῦ κέντρου τοῦ that the square on one side of triangle ABC is three times ΑΒΓ κύκλου. the (square) on the radius of circle ABC. 519 STOIQEIWN igþ. ELEMENTS BOOK 13 B G E D A C E D A B Εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓ κύκλου τὸ Δ, καὶ For let the center, D, of circle ABC have been found ἐπιζευχθεῖσα ἡ ΑΔ διήχθω ἐπὶ τὸ Ε, καὶ ἐπεζεύχθω ἡ ΒΕ. [Prop. 3.1]. And AD (being) joined, let it have been Καὶ ἐπεὶ ἰσόπλευρόν ἐστι τὸ ΑΒΓ τρίγωνον, ἡ ΒΕΓ ἄρα drawn across to E. And let BE have been joined. περιφέρεια τρίτον μέρος ἐστὶ τῆς τοῦ ΑΒΓ κύκλου περι- And since triangle ABC is equilateral, circumference φερείας. ἡ ἄρα ΒΕ περιφέρεια ἕκτον ἐστὶ μέρος τῆς τοῦ BEC is thus the third part of the circumference of cir- κύκλου περιφερείας· ἑξαγώνου ἄρα ἐστὶν ἡ ΒΕ εὐθεῖα· ἴση cle ABC. Thus, circumference BE is the sixth part of ἄρα ἐστὶ τῇ ἐκ τοῦ κέντρου τῇ ΔΕ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ the circumference of the circle. Thus, straight-line BE is ΑΕ τῆς ΔΕ, τετραπλάσιον ἐστι τὸ ἀπὸ τῆς ΑΕ τοῦ ἀπὸ τῆς (the side) of a hexagon. Thus, it is equal to the radius ΕΔ, τουτέστι τοῦ ἀπὸ τῆς ΒΕ. ἴσον δὲ τὸ ἀπὸ τῆς ΑΕ τοῖς DE [Prop. 4.15 corr.]. And since AE is double DE, the ἀπὸ τῶν ΑΒ, ΒΕ· τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΕ τετραπλάσιά ἐστι (square) on AE is four times the (square) on ED—that τοῦ ἀπὸ τῆς ΒΕ. διελόντι ἄρα τὸ ἀπὸ τῆς ΑΒ τριπλάσιόν is to say, of the (square) on BE. And the (square) on ἐστι τοῦ ἀπὸ ΒΕ. ἴση δὲ ἡ ΒΕ τῇ ΔΕ· τὸ ἄρα ἀπὸ τῆς ΑΒ AE (is) equal to the (sum of the squares) on AB and BE τριπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΔΕ. [Props. 3.31, 1.47]. Thus, the (sum of the squares) on ῾Η ἄρα τοῦ τριγώνου πλευρὰ δυνάμει τριπλασία ἐστὶ τῆς AB and BE is four times the (square) on BE. Thus, ἐκ τοῦ κέντρου [τοῦ κύκλου]· ὅπερ ἔδει δεῖξαι. via separation, the (square) on AB is three times the (square) on BE. And BE (is) equal to DE. Thus, the (square) on AB is three times the (square) on DE. Thus, the square on the side of the triangle is three times the (square) on the radius [of the circle]. (Which is) the very thing it was required to show.igþ. Proposition 13 Πυραμίδα συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ To construct a (regular) pyramid (i.e., a tetrahedron), καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει ἡμιολία ἐστὶ and to enclose (it) in a given sphere, and to show that τῆς πλευρᾶς τῆς πυραμίδος. the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid. 520 STOIQEIWN igþ. ELEMENTS BOOK 13 D J K E GA B HZ L H K E A B L D C F G ᾿Εκκείσθω ἡ τῆς δοθείσης σφαίρας δίαμετρος ἡ ΑΒ, καὶ Let the diameter AB of the given sphere be laid out, τετμήσθω κατὰ τὸ Γ σημεῖον, ὥστε διπλασίαν εἶναι τὴν ΑΓ and let it have been cut at point C such that AC is double τῆς ΓΒ· καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, CB [Prop. 6.10]. And let the semi-circle ADB have been καὶ ἤχθω ἀπὸ τοῦ Γ σημείου τῇ ΑΒ πρὸς ὀρθὰς ἡ ΓΔ, drawn on AB. And let CD have been drawn from point C καὶ ἐπεζεύχθω ἡ ΔΑ· καὶ ἐκκείσθω κύκλος ὁ ΕΖΗ ἴσην at right-angles to AB. And let DA have been joined. And ἔχων τὴν ἐκ τοῦ κέντρου τῇ ΔΓ, καὶ ἐγγεγράφθω εἰς τὸν let the circle EFG be laid down having a radius equal ΕΖΗ κύκλον τρίγωνον ἰσόπλευρον τὸ ΕΖΗ· καὶ εἰλήφθω to DC, and let the equilateral triangle EFG have been τὸ κέντρον τοῦ κύκλου τὸ Θ σημεῖον, καὶ ἐπεζεύχθωσαν inscribed in circle EFG [Prop. 4.2]. And let the center αἱ ΕΘ, ΘΖ, ΘΗ· καὶ ἀνεστάτω ἀπὸ τοῦ Θ σημείου τῷ τοῦ of the circle, point H , have been found [Prop. 3.1]. And ΕΖΗ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΘΚ, καὶ ἀφῃρήσθω ἀπὸ let EH , HF , and HG have been joined. And let HK τῆς ΘΚ τῇ ΑΓ εὐθείᾳ ἴση ἡ ΘΚ, καὶ ἐπεζεύχθωσαν αἱ ΚΕ, have been set up, at point H , at right-angles to the plane ΚΖ, ΚΗ. καὶ ἐπεὶ ἡ ΚΘ ὀρθή ἐστι πρὸς τὸ τοῦ ΕΖΗ κύκλου of circle EFG [Prop. 11.12]. And let HK, equal to the ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας straight-line AC, have been cut off from HK. And let καὶ οὔσας ἐν τῷ τοῦ ΕΖΗ κύκλου ἐπιπέδῳ ὀρθὰς ποιήσει KE, KF , and KG have been joined. And since KH is at γωνίας. ἅπτεται δὲ αὐτῆς ἑκάστη τῶν ΘΕ, ΘΖ, ΘΗ· ἡ ΘΚ right-angles to the plane of circle EFG, it will thus also ἄρα πρὸς ἑκάστη τῶν ΘΕ, ΘΖ, ΘΗ ὀρθή ἐστιν. καὶ ἐπεὶ ἴση make right-angles with all of the straight-lines joining it ἐστὶν ἡ μὲν ΑΓ τῇ ΘΚ, ἡ δὲ ΓΔ τῇ ΘΕ, καὶ ὀρθὰς γωνίας (which are) also in the plane of circle EFG [Def. 11.3]. περιέχουσιν, βάσις ἄρα ἡ ΔΑ βάσει τῇ ΚΕ ἐστιν ἴση. διὰ And HE, HF , and HG each join it. Thus, HK is at τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ΚΖ, ΚΗ τῇ ΔΑ ἐστιν ἴση· αἱ right-angles to each of HE, HF , and HG. And since τρεῖς ἄρα αἱ ΚΕ, ΚΖ, ΚΗ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ διπλῆ AC is equal to HK, and CD to HE, and they contain ἐστιν ἡ ΑΓ τῆς ΓΒ, τριπλῆ ἄρα ἡ ΑΒ τῆς ΒΓ. ὡς δὲ ἡ ΑΒ right-angles, the base DA is thus equal to the base KE πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΔ πρὸς τὸ ἀπὸ τῆς ΔΓ, [Prop. 1.4]. So, for the same (reasons), KF and KG is ὡς ἑξῆς δειχθήσεται. τριπλάσιον ἄρα τὸ ἀπὸ τῆς ΑΔ τοῦ each equal to DA. Thus, the three (straight-lines) KE, ἀπὸ τῆς ΔΓ. ἔστι δὲ καὶ τὸ ἀπὸ τῆς ΖΕ τοῦ ἀπὸ τῆς ΕΘ KF , and KG are equal to one another. And since AC is τριπλάσιον, καί ἐστιν ἴση ἡ ΔΓ τῇ ΕΘ· ἴση ἄρα καὶ ἡ ΔΑ double CB, AB (is) thus triple BC. And as AB (is) to τῇ ΕΖ. ἀλλὰ ἡ ΔΑ ἑκάστῃ τῶν ΚΕ, ΚΖ, ΚΗ ἐδείχθη ἴση· BC, so the (square) on AD (is) to the (square) on DC, καὶ ἑκάστη ἄρα τῶν ΕΖ, ΖΗ, ΗΕ ἑκάστῃ τῶν ΚΕ, ΚΖ, ΚΗ as will be shown later [see lemma]. Thus, the (square) ἐστιν ἴση· ἰσόπλευρα ἄρα ἐστὶ τὰ τέσσαρα τρίγωνα τὰ ΕΖΗ, on AD (is) three times the (square) on DC. And the ΚΕΖ, ΚΖΗ, ΚΕΗ. πυραμὶς ἄρα συνέσταται ἐκ τεσσάρων (square) on FE is also three times the (square) on EH τριγώνων ἰσοπλέυρων, ἧς βάσις μέν ἐστι τὸ ΕΖΗ τρίγωνον, [Prop. 13.12], and DC is equal to EH . Thus, DA (is) 521 STOIQEIWN igþ. ELEMENTS BOOK 13 κορυφὴ δὲ τὸ Κ σημεῖον. also equal to EF . But, DA was shown (to be) equal to Δεῖ δὴ αὐτὴν καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ each of KE, KF , and KG. Thus, EF , FG, and GE are δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος ἡμιολία ἐστὶ δυνάμει equal to KE, KF , and KG, respectively. Thus, the four τῆς πλευρᾶς τῆς πυραμίδος. triangles EFG, KEF , KFG, and KEG are equilateral. ᾿Εκβεβλήσθω γὰρ ἐπ᾿ εὐθείας τῇ ΚΘ εὐθεῖα ἡ ΘΛ, καὶ Thus, a pyramid, whose base is triangle EFG, and apex κείσθω τῇ ΓΒ ἴση ἡ ΘΛ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΓ πρὸς τὴν the point K, has been constructed from four equilateral ΓΔ, οὕτως ἡ ΓΔ πρὸς τὴν ΓΒ, ἴση δὲ ἡ μὲν ΑΓ τῇ ΚΘ, ἡ δὲ triangles. ΓΔ τῇ ΘΕ, ἡ δὲ ΓΒ τῇ ΘΛ, ἔστιν ἄρα ὡς ἡ ΚΘ πρὸς τὴν ΘΕ, So, it is also necessary to enclose it in the given οὕτως ἡ ΕΘ πρὸς τὴν ΘΛ· τὸ ἄρα ὑπὸ τῶν ΚΘ, ΘΛ ἴσον sphere, and to show that the square on the diameter of ἐστὶ τῷ ἀπὸ τῆς ΕΘ. καί ἐστιν ὀρθὴ ἑκατέρα τῶν ὑπὸ ΚΘΕ, the sphere is one and a half times the (square) on the side ΕΘΛ γωνιῶν· τὸ ἄρα ἐπὶ τῆς ΚΛ γραφόμενον ἡμικύκλιον of the pyramid. ἥξει καὶ διὰ τοῦ Ε [ἐπειδήπερ ἐὰν ἐπιζεύξωμεν τὴν ΕΛ, ὀρθὴ For let the straight-line HL have been produced in γίνεται ἡ ὑπὸ ΛΕΚ γωνία διὰ τὸ ἰσογώνιον γίνεσθαι τὸ a straight-line with KH , and let HL be made equal to ΕΛΚ τρίγωνον ἑκατέρῳ τῶν ΕΛΘ, ΕΘΚ τριγώνων]. ἐὰν CB. And since as AC (is) to CD, so CD (is) to CB δὴ μενούσης τῆς ΚΛ περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ [Prop. 6.8 corr.], and AC (is) equal to KH , and CD to πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, ἥξει καὶ διὰ HE, and CB to HL, thus as KH is to HE, so EH (is) τῶν Ζ, Η σημείων ἐπιζευγνυμένων τῶν ΖΛ, ΛΗ καὶ ὀρθῶν to HL. Thus, the (rectangle contained) by KH and HL ὁμοίως γινομένων τῶν πρὸς τοῖς Ζ, Η γωνιῶν· καὶ ἔσται is equal to the (square) on EH [Prop. 6.17]. And each ἡ πυραμὶς σφαίρᾳ περιειλημμένη τῇ δοθείσῇ. ἡ γὰρ ΚΛ of the angles KHE and EHL is a right-angle. Thus, τῆς σφαίρας διάμετρος ἴση ἐστὶ τῇ τῆς δοθείσης σφαίρας the semi-circle drawn on KL will also pass through E διαμετρῳ τῇ ΑΒ, ἐπειδήπερ τῇ μὲν ΑΓ ἴση κεῖται ἡ ΚΘ, τῇ [inasmuch as if we join EL then the angle LEK be- δὲ ΓΒ ἡ ΘΛ. comes a right-angle, on account of triangle ELK becom- Λέγω δή, ὅτι ἡ τῆς σφαίρας διάμετρος ἡμιολία ἐστὶ ing equiangular to each of the triangles ELH and EHK δυνάμει τῆς πλευρᾶς τῆς πυραμίδος. [Props. 6.8, 3.31] ]. So, if KL remains (fixed), and the ᾿Επεὶ γὰρ διπλῆ ἐστιν ἡ ΑΓ τῆς ΓΒ, τριπλῆ ἄρα ἐστὶν semi-circle is carried around, and again established at the ἡ ΑΒ τῆς ΒΓ· ἀναστρέψαντι ἡμιολία ἄρα ἐστὶν ἡ ΒΑ τῆς same (position) from which it began to be moved, it will ΑΓ. ὡς δὲ ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΒΑ πρὸς also pass through points F and G, (because) if FL and τὸ ἀπὸ τῆς ΑΔ [ἐπειδήπερ ἐπιζευγνμένης τῆς ΔΒ ἐστιν LG are joined, the angles at F and G will similarly be- ὡς ἡ ΒΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΔΑ πρὸς τὴν ΑΓ διὰ come right-angles. And the pyramid will have been en- τὴν ὁμοιότητα τῶν ΔΑΒ, ΔΑΓ τριγώνων, καὶ εἶναι ὡς τὴν closed by the given sphere. For the diameter, KL, of the πρώτην πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς τρώτης πρὸς τὸ sphere is equal to the diameter, AB, of the given sphere— ἀπὸ τῆς δευτέρας]. ἡμιόλιον ἄρα καὶ τὸ ἀπὸ τῆς ΒΑ τοῦ inasmuch as KH was made equal to AC, and HL to CB. ἀπὸ τῆς ΑΔ. καί ἐστιν ἡ μὲν ΒΑ ἡ τῆς δοθείσης σφαίρας So, I say that the square on the diameter of the sphere διάμετρος, ἡ δὲ ΑΔ ἴση τῇ πλευρᾷ τῆς πυραμίδος. is one and a half times the (square) on the side of the ῾Η ἄρα τῆς σφαίρας διάμετρος ἡμιολία ἐστὶ τῆς πλευρᾶς pyramid. τῆς πυραμίδος· ὅπερ ἔδει δεῖξαι. For since AC is double CB, AB is thus triple BC. Thus, via conversion, BA is one and a half times AC. And as BA (is) to AC, so the (square) on BA (is) to the (square) on AD [inasmuch as if DB is joined then as BA is to AD, so DA (is) to AC, on account of the similarity of triangles DAB and DAC. And as the first is to the third (of four proportional magnitudes), so the (square) on the first (is) to the (square) on the second.] Thus, the (square) on BA (is) also one and a half times the (square) on AD. And BA is the diameter of the given sphere, and AD (is) equal to the side of the pyramid. Thus, the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid.† (Which is) the very thing it was required to show. † If the radius of the sphere is unity then the side of the pyramid (i.e., tetrahedron) is p 8/3. 522 STOIQEIWN igþ. ELEMENTS BOOK 13 G D E Z BA C D E BA FL¨mma. Lemma Δεικτέον, ὅτι ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ It must be shown that as AB is to BC, so the (square) ἀπὸ τῆς ΑΔ πρὸς τὸ ἀπὸ τῆς ΔΓ. on AD (is) to the (square) on DC. ᾿Εκκείσθω γὰρ ἡ τοῦ ἡμικυκλίου καταγραφή, καὶ For, let the figure of the semi-circle have been set ἐπεζεύχθω ἡ ΔΒ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΑΓ τετράγωνον out, and let DB have been joined. And let the square τὸ ΕΓ, καὶ συμπεπληρώσθω τὸ ΖΒ παραλληλόγραμμον. EC have been described on AC. And let the parallel- ἐπεὶ οὖν διὰ τὸ ἰσογώνιον εἶναι τὸ ΔΑΒ τρίγωνον τῷ ΔΑΓ ogram FB have been completed. Therefore, since, on τριγώνῳ ἐστὶν ὡς ἡ ΒΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΔΑ πρὸς account of triangle DAB being equiangular to triangle τὴν ΑΓ, τὸ ἄρα ὑπὸ τῶν ΒΑ, ΑΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς DAC [Props. 6.8, 6.4], (proportionally) as BA is to AD, ΑΔ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ΕΒ so DA (is) to AC, the (rectangle contained) by BA and πρὸς τὸ ΒΖ, καί ἐστι τὸ μὲν ΕΒ τὸ ὑπὸ τῶν ΒΑ, ΑΓ· ἴση AC is thus equal to the (square) on AD [Prop. 6.17]. γὰρ ἡ ΕΑ τῇ ΑΓ· τὸ δὲ ΒΖ τὸ ὑπὸ τῶν ΑΓ, ΓΒ, ὡς ἄρα ἡ And since as AB is to BC, so EB (is) to BF [Prop. 6.1]. ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ὑπὸ τῶν ΒΑ, ΑΓ πρὸv τὸ ὑπὸ And EB is the (rectangle contained) by BA and AC—for τῶν ΑΓ, ΓΒ. καί ἐστι τὸ μὲν ὑπὸ τῶν ΒΑ, ΑΓ ἴσον τῷ ἀπὸ EA (is) equal to AC. And BF the (rectangle contained) τῆς ΑΔ, τὸ δὲ ὑπὸ τῶν ΑΓΒ ἴσον τῷ ἀπὸ τῆς ΔΓ· ἡ γὰρ by AC and CB. Thus, as AB (is) to BC, so the (rectan- ΔΓ κάθετος τῶν τῆς βάσεως τμημάτων τῶν ΑΓ, ΓΒ μέση gle contained) by BA and AC (is) to the (rectangle con- ἀνάλογόν ἐστι διὰ τὸ ὀρθὴν εἶναι τὴν ὑπὸ ΑΔΒ. ὡς ἄρα ἡ tained) by AC and CB. And the (rectangle contained) ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΔ πρὸς τὸ ἀπὸ τῆς by BA and AC is equal to the (square) on AD, and the ΔΓ· ὅπερ ἔδει δεῖξαι. (rectangle contained) by ACB (is) equal to the (square) on DC. For the perpendicular DC is the mean propor- tional to the pieces of the base, AC and CB, on account of ADB being a right-angle [Prop. 6.8 corr.]. Thus, as AB (is) to BC, so the (square) on AD (is) to the (square) on DC. (Which is) the very thing it was required to show.idþ. Proposition 14 ᾿Οκτάεδρον συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν, ᾗ καὶ To construct an octahedron, and to enclose (it) in a τὰ πρότερα, καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει (given) sphere, like in the preceding (proposition), and διπλασία ἐστὶ τῆς πλευρᾶς τοῦ ὀκταέδρου. to show that the square on the diameter of the sphere is ᾿Εκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ, double the (square) on the side of the octahedron. καὶ τετμήσθω δίχα κατὰ τὸ Γ, καὶ γεγράφθω ἐπὶ τῆς ΑΒ Let the diameter AB of the given sphere be laid out, ἡμικύκλιον τὸ ΑΔΒ, καὶ ἤχθω ἀπὸ τοῦ Γ τῇ ΑΒ πρὸς ὀρθὰς and let it have been cut in half at C. And let the semi- ἡ ΓΔ, καὶ ἐπεζεύχθω ἡ ΔΒ, καὶ ἐκκείσθω τετράγωνον circle ADB have been drawn on AB. And let CD be τὸ ΕΖΗΘ ἴσην ἔχον ἑκάστην τῶν πλευρῶν τῇ ΔΒ, καὶ drawn from C at right-angles to AB. And let DB have 523 STOIQEIWN igþ. ELEMENTS BOOK 13 ἐπεζεύχθωσαν αἱ ΘΖ, ΕΗ, καὶ ἀνεστάτω ἀπὸ τοῦ Κ σημείου been joined. And let the square EFGH , having each of τῷ τοῦ ΕΖΗΘ τετραγώνου ἐπιπέδῳ πρὸς ὀρθὰς εὐθεῖα ἡ its sides equal to DB, be laid out. And let HF and EG ΚΛ καὶ διήχθω ἐπὶ τὰ ἕτερα μέρη τοῦ ἐπιπέδου ὡς ἡ ΚΜ, have been joined. And let the straight-line KL have been καὶ ἀφῃρήσθω ἀφ᾿ ἑκατέρας τῶν ΚΛ, ΚΜ μιᾷ τῶν ΕΚ, ΖΚ, set up, at point K, at right-angles to the plane of square ΗΚ, ΘΚ ἴση ἑκατέρα τῶν ΚΛ, ΚΜ, καὶ ἐπεζεύχθωσαν αἱ EFGH [Prop. 11.12]. And let it have been drawn across ΛΕ, ΛΖ, ΛΗ, ΛΘ, ΜΕ, ΜΖ, ΜΗ, ΜΘ. on the other side of the plane, like KM . And let KL and KM , equal to one of EK, FK, GK, and HK, have been cut off from KL and KM , respectively. And let LE, LF , LG, LH , ME, MF , MG, and MH have been joined. G D A B M E L K F C H H D A B E L K F C G M Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΚΕ τῇ ΚΘ, καί ἐστιν ὀρθὴ ἡ ὑπὸ And since KE is equal to KH , and angle EKH is a ΕΚΘ γωνία, τὸ ἄρα ἀπὸ τῆς ΘΕ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς right-angle, the (square) on the HE is thus double the ΕΚ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΛΚ τῇ ΚΕ, καί ἐστιν ὀρθὴ ἡ (square) on EK [Prop. 1.47]. Again, since LK is equal ὑπὸ ΛΚΕ γωνία, τὸ ἄρα ἀπὸ τῆς ΕΛ διπλάσιόν ἐστι τοῦ ἀπὸ to KE, and angle LKE is a right-angle, the (square) on ΕΚ. ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς ΘΕ διπλάσιον τοῦ ἀπὸ τῆς EL is thus double the (square) on EK [Prop. 1.47]. And ΕΚ· τὸ ἄρα ἀπὸ τῆς ΛΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ· ἴση ἄρα the (square) on HE was also shown (to be) double the ἐστὶν ἡ ΛΕ τῇ ΕΘ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΛΘ τῇ ΘΕ ἐστιν (square) on EK. Thus, the (square) on LE is equal to ἴση· ἰσόπλευρον ἄρα ἐστὶ τὸ ΛΕΘ τρίγωνον. ὁμοίως δὴ the (square) on EH . Thus, LE is equal to EH . So, for δείξομεν, ὅτι καὶ ἕκαστον τῶν λοιπῶν τριγώνων, ὧν βάσεις the same (reasons), LH is also equal to HE. Triangle μέν εἰσιν αἱ τοῦ ΕΖΗΘ τετραγώνου πλευραί, κορυφαὶ δὲ τὰ LEH is thus equilateral. So, similarly, we can show that Λ, Μ σημεῖα, ἰσόπλευρόν ἐστιν· ὀκτάεδρον ἄρα συνέσταται each of the remaining triangles, whose bases are the sides ὑπὸ ὀκτὼ τριγώνων ἰσοπλεύρων περιεχόμενον. of the square EFGH , and apexes the points L and M , Δεῖ δὴ αὐτὸ καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, are equilateral. Thus, an octahedron contained by eight ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει διπλασίων ἐστὶ τῆς τοῦ equilateral triangles has been constructed. ὀκταέδρου πλευρᾶς. So, it is also necessary to enclose it by the given ᾿Επεὶ γὰρ αἱ τρεῖς αἱ ΛΚ, ΚΜ, ΚΕ ἴσαι ἀλλήλαις εἰσίν, sphere, and to show that the square on the diameter of τὸ ἄρα ἐπὶ τῆς ΛΜ γραφόμενον ἡμικύκλιον ἥξει καὶ διὰ the sphere is double the (square) on the side of the octa- τοῦ Ε. καὶ διὰ τὰ αὐτά, ἐὰν μενούσης τῆς ΛΜ περιενεχθὲν hedron. τὸ ἡμικύκλιον εἰς τὸ αὐτὸ ἀποκατασταθῇ, ὅθεν ἤρξατο For since the three (straight-lines) LK, KM , and KE φέρεσθαι, ἥξει καὶ διὰ τῶν Ζ, Η, Θ σημείων, καὶ ἔσται are equal to one another, the semi-circle drawn on LM σφαίρᾳ περιειλημμένον τὸ ὀκτάεδρον. λέγω δή, ὅτι καὶ τῇ will thus also pass through E. And, for the same (rea- δοθείσῃ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΛΚ τῇ ΚΜ, κοινὴ δὲ ἡ ΚΕ, sons), if LM remains (fixed), and the semi-circle is car- 524 STOIQEIWN igþ. ELEMENTS BOOK 13 καὶ γωνίας ὀρθὰς περιέχουσιν, βάσις ἄρα ἡ ΛΕ βάσει τῇ ried around, and again established at the same (position) ΕΜ ἐστιν ἴση. καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΛΕΜ γωνία· ἐν from which it began to be moved, then it will also pass ἡμικυκλίῳ γάρ· τὸ ἄρα ἀπὸ τῆς ΛΜ διπλάσιόν ἐστι τοῦ ἀπὸ through points F , G, and H , and the octahedron will τῆς ΛΕ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, διπλασία ἐστὶν have been enclosed by a sphere. So, I say that (it is) ἡ ΑΒ τῆς ΒΓ. ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ also (enclosed) by the given (sphere). For since LK is τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΔ· διπλάσιον ἄρα ἐστὶ τὸ ἀπὸ equal to KM , and KE (is) common, and they contain τῆς ΑΒ τοῦ ἀπὸ τῆς ΒΔ. ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς ΛΜ right-angles, the base LE is thus equal to the base EM διπλάσιον τοῦ ἀπὸ τῆς ΛΕ. καί ἐστιν ἴσον τὸ ἀπὸ τῆς ΔΒ [Prop. 1.4]. And since angle LEM is a right-angle—for τῷ ἀπὸ τῆς ΛΕ· ἴση γὰρ κεῖται ἡ ΕΘ τῇ ΔΒ. ἴσον ἄρα καὶ (it is) in a semi-circle [Prop. 3.31]—the (square) on LM τὸ ἀπὸ τῆς ΑΒ τῷ ἀπὸ τῆς ΛΜ· ἴση ἄρα ἡ ΑΒ τῇ ΛΜ. καί is thus double the (square) on LE [Prop. 1.47]. Again, ἐστιν ἡ ΑΒ ἡ τῆς δοθείσης σφαίρας διάμετρος· ἡ ΛΜ ἄρα since AC is equal to CB, AB is double BC. And as AB ἴση ἐστὶ τῇ τῆς δοθείσης σφαίρας διαμέτρῳ. (is) to BC, so the (square) on AB (is) to the (square) Περιείληπται ἄρα τὸ ὀκτάεδρον τῇ δοθείσῃ σφαίρᾳ. καὶ on BD [Prop. 6.8, Def. 5.9]. Thus, the (square) on AB is συναποδέδεικται, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει δι- double the (square) on BD. And the (square) on LM was πλασίων ἐστὶ τῆς τοῦ ὀκταέδρου πλευρᾶς· ὅπερ ἔδει δεῖξαι. also shown (to be) double the (square) on LE. And the (square) on DB is equal to the (square) on LE. For EH was made equal to DB. Thus, the (square) on AB (is) also equal to the (square) on LM . Thus, AB (is) equal to LM . And AB is the diameter of the given sphere. Thus, LM is equal to the diameter of the given sphere. Thus, the octahedron has been enclosed by the given sphere, and it has been simultaneously proved that the square on the diameter of the sphere is double the (square) on the side of the octahedron.† (Which is) the very thing it was required to show. † If the radius of the sphere is unity then the side of octahedron is √ 2.ieþ. Proposition 15 Κύβον συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν, ᾗ καὶ τὴν To construct a cube, and to enclose (it) in a sphere, πυραμίδα, καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει like in the (case of the) pyramid, and to show that the τριπλασίων ἐστὶ τῆς τοῦ κύβου πλευρᾶς. square on the diameter of the sphere is three times the ᾿Εκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ καὶ (square) on the side of the cube. τετμήσθω κατὰ τὸ Γ ὥστε διπλῆν εἶναι τὴν ΑΓ τῆς ΓΒ, καὶ Let the diameter AB of the given sphere be laid out, γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ ἀπὸ τοῦ Γ and let it have been cut at C such that AC is double τῇ ΑΒ πρὸς ὀρθὰς ἤχθω ἡ ΓΔ, καὶ ἐπεζεύχθω ἡ ΔΒ, καὶ CB. And let the semi-circle ADB have been drawn on ἐκκείσθω τετράγωνον τὸ ΕΖΗΘ ἴσην ἔχον τὴν πλευρὰν τῇ AB. And let CD have been drawn from C at right- ΔΒ, καὶ ἀπὸ τῶν Ε, Ζ, Η, Θ τῷ τοῦ ΕΖΗΘ τετραγώνου angles to AB. And let DB have been joined. And let the ἐπιπέδῳ πρὸς ὀρθὰς ἤχθωσαν αἱ ΕΚ, ΖΛ, ΗΜ, ΘΝ, καὶ square EFGH , having (its) side equal to DB, be laid out. ἀφῃρήσθω ἀπὸ ἑκάστης τῶν ΕΚ, ΖΛ, ΗΜ, ΘΝ μιᾷ τῶν And let EK, FL, GM , and HN have been drawn from ΕΖ, ΖΗ, ΗΘ, ΘΕ ἴση ἑκάστη τῶν ΕΚ, ΖΛ, ΗΜ, ΘΝ, καὶ (points) E, F , G, and H , (respectively), at right-angles to ἐπεζεύχθωσαν αἱ ΚΛ, ΛΜ, ΜΝ, ΝΚ· κύβος ἄρα συνέσταται the plane of square EFGH . And let EK, FL, GM , and ὁ ΖΝ ὑπὸ ἓξ τετραγώνων ἴσων περιεχόμενος. HN , equal to one of EF , FG, GH , and HE, have been Δεῖ δὴ αὐτὸν καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ cut off from EK, FL, GM , and HN , respectively. And let δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει τριπλασία ἐστὶ KL, LM , MN , and NK have been joined. Thus, a cube τῆς πλευρᾶς τοῦ κύβου. contained by six equal squares has been constructed. So, it is also necessary to enclose it by the given sphere, and to show that the square on the diameter of the sphere is three times the (square) on the side of the cube. 525 STOIQEIWN igþ. ELEMENTS BOOK 13 K GA B L MD HZ E N J H A B L M D F G C K E N ᾿Επεζεύχθωσαν γὰρ αἱ ΚΗ, ΕΗ. καὶ ἐπεὶ ὀρθή ἐστιν For let KG and EG have been joined. And since an- ἡ ὑπὸ ΚΕΗ γωνία διὰ τὸ καὶ τὴν ΚΕ ὀρθὴν εἶναι πρὸς gle KEG is a right-angle—on account of KE also being τὸ ΕΗ ἐπίπεδον δηλαδὴ καὶ πρὸς τὴν ΕΗ εὐθεῖαν, τὸ ἄρα at right-angles to the plane EG, and manifestly also to ἐπὶ τῆς ΚΗ γραφόμενον ἡμικύκλιον ἥξει καὶ διὰ τοῦ Ε the straight-line EG [Def. 11.3]—the semi-circle drawn σημείου. πάλιν, ἐπεὶ ἡ ΗΖ ὀρθή ἐστι πρὸς ἑκατέραν τῶν ΖΛ, on KG will thus also pass through point E. Again, since ΖΕ, καὶ πρὸς τὸ ΖΚ ἄρα ἐπίπεδον ὀρθή ἐστιν ἡ ΗΖ· ὥστε GF is at right-angles to each of FL and FE, GF is thus καὶ ἐὰν ἐπιζεύξωμεν τὴν ΖΚ, ἡ ΗΖ ὀρθὴ ἔσται καὶ πρὸς also at right-angles to the plane FK [Prop. 11.4]. Hence, τὴν ΖΚ· καὶ δὶα τοῦτο πάλιν τὸ ἐπὶ τῆς ΗΚ γραφόμενον if we also join FK then GF will also be at right-angles ἡμικύκλιον ἥξει καὶ διὰ τοῦ Ζ. ὁμοίως καὶ δὶα τῶν λοιπῶν to FK. And, again, on account of this, the semi-circle τοῦ κύβου σημείων ἥξει. ἐὰν δὴ μενούσης τῆς ΚΗ πε- drawn on GK will also pass through point F . Similarly, ριενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ ἀποκατασταθῇ, ὅθεν it will also pass through the remaining (angular) points of ἤρξατο φέρεσθαι, ἔσται σφαίρᾳ περιειλημμένος ὁ κύβος. the cube. So, if KG remains (fixed), and the semi-circle is λέγω δή, ὅτι καὶ τῇ δοθείσῃ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΗΖ τῇ carried around, and again established at the same (posi- ΖΕ, καί ἐστιν ὀρθὴ ἡ πρὸς τῷ Ζ γωνία, τὸ ἄρα ἀπὸ τῆς ΕΗ tion) from which it began to be moved, then the cube will διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΕΖ. ἴση δὲ ἡ ΕΖ τῇ ΕΚ· τὸ ἄρα have been enclosed by a sphere. So, I say that (it is) also ἀπὸ τῆς ΕΗ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΕΚ· ὥστε τὰ ἀπὸ (enclosed) by the given (sphere). For since GF is equal τῶν ΗΕ, ΕΚ, τουτέστι τὸ ἀπὸ τῆς ΗΚ, τριπλάσιόν ἐστι τοῦ to FE, and the angle at F is a right-angle, the (square) ἀπὸ τῆς ΕΚ. καὶ ἐπεὶ τριπλασίων ἐστὶν ἡ ΑΒ τῆς ΒΓ, ὡς on EG is thus double the (square) on EF [Prop. 1.47]. δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ And EF (is) equal to EK. Thus, the (square) on EG τῆς ΒΔ, τριπλάσιον ἄρα τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΒΔ. is double the (square) on EK. Hence, the (sum of the ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς ΗΚ τοῦ ἀπὸ τῆς ΚΕ τριπλάσιον. squares) on GE and EK—that is to say, the (square) on καὶ κεῖται ἴση ἡ ΚΕ τῇ ΔΒ· ἴση ἄρα καὶ ἡ ΚΗ τῇ ΑΒ. καί GK [Prop. 1.47]—is three times the (square) on EK. ἐστιν ἡ ΑΒ τῆς δοθείσης σφαίρας διάμετρος· καὶ ἡ ΚΗ ἄρα And since AB is three times BC, and as AB (is) to ἴση ἐστὶ τῇ τῆς δοθείσης σφαίρας διαμέτρῳ. BC, so the (square) on AB (is) to the (square) on BD Τῇ δοθείσῃ ἄρα σφαίρα περιείληπται ὁ κύβος· καὶ συ- [Prop. 6.8, Def. 5.9], the (square) on AB (is) thus three ναποδέδεικται, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει τρι- times the (square) on BD. And the (square) on GK was πλασίων ἐστὶ τῆς τοῦ κύβου πλευρᾶς· ὅπερ ἔδει δεῖξαι. also shown (to be) three times the (square) on KE. And KE was made equal to DB. Thus, KG (is) also equal to AB. And AB is the radius of the given sphere. Thus, KG is also equal to the diameter of the given sphere. Thus, the cube has been enclosed by the given sphere. And it has simultaneously been shown that the square on the diameter of the sphere is three times the (square) on 526 STOIQEIWN igþ. ELEMENTS BOOK 13 the side of the cube.† (Which is) the very thing it was required to show. † If the radius of the sphere is unity then the side of the cube is p 4/3.i�þ. Proposition 16 Εἰκοσάεδρον συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν, ᾗ καὶ To construct an icosahedron, and to enclose (it) in a τὰ προειρημένα σχήματα, καὶ δεῖξαι, ὅτι ἡ τοῦ εἰκοσαέδρου sphere, like the aforementioned figures, and to show that πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάττων. the side of the icosahedron is that irrational (straight- line) called minor.D A G B CA B D ᾿Εκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ καὶ Let the diameter AB of the given sphere be laid out, τετμήσθω κατὰ τὸ Γ ὥστε τετραπλῆν εἶναι τὴν ΑΓ τῆς ΓΒ, and let it have been cut at C such that AC is four times καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ ἤχθω CB [Prop. 6.10]. And let the semi-circle ADB have been ἀπὸ τοῦ Γ τῇ ΑΒ πρὸς ορθὰς γωνίας εὐθεῖα γραμμὴ ἡ ΓΔ, drawn on AB. And let the straight-line CD have been καί ἐπεζεύχθω ἡ ΔΒ, καὶ ἐκκείσθω κύκλος ὁ ΕΖΗΘΚ, drawn from C at right-angles to AB. And let DB have οὗ ἡ ἐν τοῦ κέντρου ἴση ἔστω τῇ ΔΒ, καὶ ἐγγεγράφθω been joined. And let the circle EFGHK be set down, εἰς τὸν ΕΖΗΘΚ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ and let its radius be equal to DB. And let the equilat- ἰσογώνιον τὸ ΕΖΗΘΚ, καὶ τετμήσθωσαν αἱ ΕΖ, ΖΗ, ΗΘ, eral and equiangular pentagon EFGHK have been in- ΘΚ, ΚΕ περιφέρειαι δίχα κατὰ τὸ Λ, Μ, Ν, Ξ, Ο σημεῖα, καὶ scribed in circle EFGHK [Prop. 4.11]. And let the cir- ἐπεζεύχθωσαν αἱ ΛΜ, ΜΝ, ΝΞ, ΞΟ, ΟΛ, ΕΟ. ἴσόπλευρον cumferences EF , FG, GH , HK, and KE have been cut ἄρα ἐστὶ καὶ τὸ ΛΜΝΞΟ πεντάγωνον, καὶ δεκαγώνου ἡ in half at points L, M , N , O, and P (respectively). And ΕΟ εὐθεῖα. καὶ ἀνεστάτωσαν ἄπὸ τῶν Ε, Ζ, Η, Θ, Κ let LM , MN , NO, OP , PL, and EP have been joined. σημείων τῷ τοῦ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς γωνίας εὐθεῖαι Thus, pentagon LMNOP is also equilateral, and EP (is) αἱ ΕΠ, ΖΡ, ΗΣ, ΘΤ, ΚΥ ἴσαι οὖσαι τῇ ἐκ τοῦ κέντρου τοῦ the side of the decagon (inscribed in the circle). And let ΕΖΗΘΚ κύκλου, καὶ ἐπεζεύχθωσαν αἱ ΠΡ, ΡΣ, ΣΤ, ΤΥ, the straight-lines EQ, FR, GS, HT , and KU , which are ΥΠ, ΠΛ, ΛΡ, ΡΜ, ΜΣ, ΣΝ, ΝΤ, ΤΞ, ΞΥ, ΥΟ, ΟΠ. equal to the radius of circle EFGHK, have been set up Καὶ ἐπεὶ ἑκατέρα τῶν ΕΠ, ΚΥ τῷ αὐτῷ ἐπιπέδῳ πρὸς at right-angles to the plane of the circle, at points E, F , ὀρθάς ἐστιν, παράλληλος ἄρα ἐστὶν ἡ ΕΠ τῇ ΚΥ. ἔστι G, H , and K (respectively). And let QR, RS, ST , TU , δὲ αὐτῇ καὶ ἴση· αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπι- UQ, QL, LR, RM , MS, SN , NT , TO, OU , UP , and PQ ζευγνύουσαι ἐπὶ τὰ αὐτὰ μέρη εὐθεῖαι ἴσαι τε καὶ παράλληλοί have been joined. εἰσιν. ἡ ΠΥ ἄρα τῇ ΕΚ ἴση τε καὶ παράλληλός ἐστιν. πεν- And since EQ and KU are each at right-angles to the ταγώνου δὲ ἰσοπλεύρου ἡ ΕΚ· πενταγώνου ἄρα ἰσοπλεύρου same plane, EQ is thus parallel to KU [Prop. 11.6]. And καὶ ἡ ΠΥ τοῦ εἰς τὸν ΕΖΗΘΚ κύκλον ἐγγραφομένου. it is also equal to it. And straight-lines joining equal and διὰ τὰ αὐτὰ δὴ καὶ ἑκάστη τῶν ΠΡ, ΡΣ, ΣΤ, ΤΥ πεν- parallel (straight-lines) on the same side are (themselves) ταγώνου ἐστὶν ἰσοπλεύρου τοῦ εἰς τὸν ΕΖΗΘΚ κύκλον equal and parallel [Prop. 1.33]. Thus, QU is equal and ἐγγραφομένου· ἰσόπλευρον ἄρα τὸ ΠΡΣΤΥ πεντάγωνον. parallel to EK. And EK (is the side) of an equilateral καὶ ἐπεὶ ἑξαγώνου μέν ἐστιν ἡ ΠΕ, δεκαγώνου δὲ ἡ ΕΟ, pentagon (inscribed in circle EFGHK). Thus, QU (is) καί ἐστιν ὀρθὴ ἡ ὑπὸ ΠΕΟ, πενταγώνου ἄρα ἐστὶν ἡ ΠΟ· ἡ also the side of an equilateral pentagon inscribed in circle γὰρ τοῦ πενταγώνου πλευρὰ δύναται τήν τε τοῦ ἑξαγώνου EFGHK. So, for the same (reasons), QR, RS, ST , and καὶ τὴν τοῦ δεκαγώνου τῶν εἰς τὸν αὐτὸν κύκλον ἐγγρα- TU are also the sides of an equilateral pentagon inscribed φομένων. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΟΥ πενταγώνου ἐστὶ in circle EFGHK. Pentagon QRSTU (is) thus equilat- 527 STOIQEIWN igþ. ELEMENTS BOOK 13 πλευρά. ἔστι δὲ καὶ ἡ ΠΥ πενταγώνου· ἰσόπλευρον ἄρα eral. And side QE is (the side) of a hexagon (inscribed ἐστὶ τὸ ΠΟΥ τρίγωνον. διὰ τὰ αὐτὰ δὴ καὶ ἕκαστον τῶν in circle EFGHK), and EP (the side) of a decagon, and ΠΛΡ, ΡΜΣ, ΣΝΤ, ΤΞΥ ἰσόπλευρόν ἐστιν. καὶ ἐπεὶ πεν- (angle) QEP is a right-angle, thus QP is (the side) of a ταγώνου ἐδείχθη ἑκατέρα τῶν ΠΛ, ΠΟ, ἔστι δὲ καὶ ἡ ΛΟ pentagon (inscribed in the same circle). For the square πενταγώνου, ἰσόπλευρον ἄρα ἐστὶ τὸ ΠΛΟ τρίγωνον. διὰ on the side of a pentagon is (equal to the sum of) the τὰ αὐτὰ δὴ καὶ ἕκαστον τῶν ΛΡΜ, ΜΣΝ, ΝΤΞ, ΞΥΟ (squares) on (the sides of) a hexagon and a decagon in- τριγώνων ἰσόπλευρόν ἐστιν. scribed in the same circle [Prop. 13.10]. So, for the same (reasons), PU is also the side of a pentagon. And QU is also (the side) of a pentagon. Thus, triangle QPU is equilateral. So, for the same (reasons), (triangles) QLR, RMS, SNT , and TOU are each also equilateral. And since QL and QP were each shown (to be the sides) of a pentagon, and LP is also (the side) of a pentagon, trian- gle QLP is thus equilateral. So, for the same (reasons), triangles LRM , MSN , NTO, and OUP are each also equilateral. a P S Z R T N U X O E M K J H L Q WFY a S T N U E K L F G O Q H Z W P R M X V Εἰλήφθω τὸ κέντρον τοῦ ΕΖΗΘΚ κύκλου τὸ Φ σημεῖον· Let the center, point V , of circle EFGHK have been καὶ ἀπὸ τοῦ Φ τῷ τοῦ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς ἀνεστάτω found [Prop. 3.1]. And let V Z have been set up, at ἡ ΦΩ, καὶ ἐκβεβλήσθω ἐπὶ τὰ ἕτερα μέρη ὡς ἡ ΦΨ, καὶ (point) V , at right-angles to the plane of the circle. And ἀφῃρήσθω ἑξαγώνου μὲν ἡ ΦΧ, δεκαγώνου δὲ ἑκατέρα τῶν let it have been produced on the other side (of the cir- ΦΨ, ΧΩ, καὶ ἐπεζεύχθωσαν αἱ ΠΩ, ΠΧ, ΥΩ, ΕΦ, ΛΦ, ΛΨ, cle), like V X . And let V W have been cut off (from XZ ΨΜ. so as to be equal to the side) of a hexagon, and each of Καὶ ἐπεὶ ἑκατέρα τῶν ΦΧ, ΠΕ τῷ τοῦ κύκλου ἐπιπέδῳ V X and WZ (so as to be equal to the side) of a decagon. πρὸς ὀρθάς ἐστιν, παράλληλος ἄρα ἐστὶν ἡ ΦΧ τῇ ΠΕ. εἰσὶ And let QZ, QW , UZ, EV , LV , LX , and XM have been δὲ καὶ ἴσαι· καὶ αἱ ΕΦ, ΠΧ ἄρα ἴσαι τε καὶ παράλληλοί joined. εἰσιν. ἑξαγώνου δὲ ἡ ΕΦ· ἑξαγώνου ἄρα καὶ ἡ ΠΧ. καὶ And since V W and QE are each at right-angles ἐπεὶ ἑξαγώνου μέν ἐστιν ἡ ΠΧ, δεκαγώνου δὲ ἡ ΧΩ, καὶ to the plane of the circle, V W is thus parallel to QE ὀρθή ἐστιν ἡ ὑπὸ ΠΧΩ γωνία, πενταγώνου ἄρα ἐστὶν ἡ [Prop. 11.6]. And they are also equal. EV and QW are ΠΩ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΥΩ πενταγώνου ἐστίν, ἐπειδήπερ, thus equal and parallel (to one another) [Prop. 1.33]. 528 STOIQEIWN igþ. ELEMENTS BOOK 13 ἐὰν ἐπιζεύξωμεν τὰς ΦΚ, ΧΥ, ἴσαι καὶ ἀπεναντίον ἔσον- And EV (is the side) of a hexagon. Thus, QW (is) also ται, καί ἐστιν ἡ ΦΚ ἐκ τοῦ κέντρου οὖσα ἑξαγώνου. (the side) of a hexagon. And since QW is (the side) of ἑξαγώνου ἄρα καὶ ἡ ΧΥ. δεκαγώνου δὲ ἡ ΧΩ, καὶ ὀρθὴ a hexagon, and WZ (the side) of a decagon, and angle ἡ ὑπὸ ΥΧΩ· πενταγώνου ἄρα ἡ ΥΩ. ἔστι δὲ καὶ ἡ ΠΥ QWZ is a right-angle [Def. 11.3, Prop. 1.29], QZ is thus πενταγώνου· ἰσόπλευρον ἄρα ἐστὶ τὸ ΠΥΩ τρίγωνον. διὰ (the side) of a pentagon [Prop. 13.10]. So, for the same τὰ αὐτὰ δὴ καὶ ἕκαστον τῶν λοιπῶν τριγώνων, ὧν βάσεις (reasons), UZ is also (the side) of a pentagon—inasmuch μέν εἰσιν αἱ ΠΡ, ΡΣ, ΣΤ, ΤΥ εὐθεῖαι, κορυφὴ δὲ τὸ Ω as, if we join V K and WU then they will be equal and σημεῖον, ἰσόπλευρόν ἐστιν. πάλιν, ἐπεὶ ἑξαγώνου μὲν ἡ opposite. And V K, being (equal) to the radius (of the cir- ΦΛ, δεκαγώνου δὲ ἡ ΦΨ, καὶ ὀρθή ἐστιν ἡ ὑπὸ ΛΦΨ cle), is (the side) of a hexagon [Prop. 4.15 corr.]. Thus, γωνία, πενταγώνου ἄρα ἐστὶν ἡ ΛΨ. διὰ τὰ αὐτὰ δὴ ἐὰν WU (is) also the side of a hexagon. And WZ (is the side) ἐπιζεύξωμεν τὴν ΜΦ οὖσαν ἑξαγώνου, συνάγεται καὶ ἡ ΜΨ of a decagon, and (angle) UWZ (is) a right-angle. Thus, πενταγώνου. ἔστι δὲ καὶ ἡ ΛΜ πενταγώνου· ἰσόπλευρον UZ (is the side) of a pentagon [Prop. 13.10]. And QU ἄρα ἐστὶ τὸ ΛΜΨ τρίγωνον. ὁμοίως δὴ δειχθήσεται, ὅτι is also (the side) of a pentagon. Triangle QUZ is thus καὶ ἕκαστον τῶν λοιπῶν τριγώνων, ὧν βάσεις μέν εἰσιν αἱ equilateral. So, for the same (reasons), each of the re- ΜΝ, ΝΞ, ΞΟ, ΟΛ, κορυφὴ δὲ τὸ Ψ σημεὶον, ἰσόπλευρόν maining triangles, whose bases are the straight-lines QR, ἐστιν. συνέσταται ἄρα εἰκοσάεδρον ὑπὸ εἴκοσι τριγώνων RS, ST , and TU , and apexes the point Z, are also equi- ἰσοπλεύρων περιεχόμενον. lateral. Again, since V L (is the side) of a hexagon, and Δεῖ δὴ αὐτὸ καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, V X (the side) of a decagon, and angle LV X is a right- ὅτι ἡ τοῦ εἰκοσαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη angle, LX is thus (the side) of a pentagon [Prop. 13.10]. ἐλάσσων. So, for the same (reasons), if we join MV , which is (the ᾿Επεὶ γὰρ ἑξαγώνου ἐστὶν ἡ ΦΧ, δεκαγώνου δὲ ἡ ΧΩ, ἡ side) of a hexagon, MX is also inferred (to be the side) ΦΩ ἄρα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Χ, καὶ τὸ of a pentagon. And LM is also (the side) of a pentagon. μεῖζον αὐτῆς τμῆμά ἐστιν ἡ ΦΧ· ἔστιν ἄρα ὡς ἡ ΩΦ πρὸς Thus, triangle LMX is equilateral. So, similarly, it can τὴν ΦΧ, οὕτως ἡ ΦΧ πρὸς τὴν ΧΩ. ἴση δὲ ἡ μὲν ΦΧ τῇ be shown that each of the remaining triangles, whose ΦΕ, ἡ δὲ ΧΩ τῇ ΦΨ· ἔστιν ἄρα ὡς ἡ ΩΦ πρὸς τὴν ΦΕ, bases are the (straight-lines) MN , NO, OP , and PL, οὕτως ἡ ΕΦ πρὸς τὴν ΦΨ. καί εἰσιν ὀρθαὶ αἱ ὑπὸ ΩΦΕ, and apexes the point X , are also equilateral. Thus, an ΕΦΨ γωνίαι· ἐὰν ἄρα ἐπιζεύξωμεν τὴν ΕΩ εὐθεὶαν, ὀρθὴ icosahedron contained by twenty equilateral triangles has ἔσται ἡ ὑπὸ ΨΕΩ γωνία διὰ τὴν ὁμοιότητα τῶν ΨΕΩ, ΦΕΩ been constructed. τριγώνων. διὰ τὰ αὐτὰ δὴ ἐπεί ἐστιν ὡς ἡ ΩΦ πρὸς τὴν So, it is also necessary to enclose it in the given ΦΧ, οὕτως ἡ ΦΧ πρὸς τὴν ΧΩ, ἴση δὲ ἡ μὲν ΩΦ τῇ ΨΧ, sphere, and to show that the side of the icosahedron is ἡ δὲ ΦΧ τῇ ΧΠ, ἔστιν ἄρα ὡς ἡ ΨΧ πρὸς τὴν ΧΠ, οὕτως that irrational (straight-line) called minor. ἡ ΠΧ πρὸς τὴν ΧΩ. καὶ διὰ τοῦτο πάλιν ἐὰν ἐπιζεύξωμεν For, since V W is (the side) of a hexagon, and WZ τὴν ΠΨ, ὀρθὴ ἔσται ἡ πρὸς τῷ Π γωνία· τὸ ἄρα ἐπὶ τῆς (the side) of a decagon, V Z has thus been cut in ex- ΨΩ γραφόμενον ἡμικύκλιον ἥξει καὶ δὶα τοῦ Π. καὶ ἐὰν treme and mean ratio at W , and V W is its greater piece μενούσης τῆς ΨΩ περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ [Prop. 13.9]. Thus, as ZV is to V W , so V W (is) to WZ. πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, ἥξει καὶ διὰ And V W (is) equal to V E, and WZ to V X . Thus, as τοῦ Π καὶ τῶν λοιπῶν σημείων τοῦ εἰκοσαέδρου, καὶ ἔσται ZV is to V E, so EV (is) to V X . And angles ZV E and σφαίρᾳ περιειλημμένον τὸ εἰκοσάεδρον. λέγω δή, ὅτι καὶ EV X are right-angles. Thus, if we join straight-line EZ τῇ δοθείσῃ. τετμήσθω γὰρ ἡ ΦΧ δίχα κατὰ τὸ α. καὶ ἐπεὶ then angle XEZ will be a right-angle, on account of the εὐθεῖα γραμμὴ ἡ ΦΩ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ similarity of triangles XEZ and V EZ. [Prop. 6.8]. So, τὸ Χ, καὶ τὸ ἔλασσον αὐτῆς τμῆμά ἐστιν ἡ ΩΧ, ἡ ἄρα ΩΧ for the same (reasons), since as ZV is to V W , so V W προσλαβοῦσα τὴν ἡμίσειαν τοῦ μείζονος τμήματος τὴν Χα (is) to WZ, and ZV (is) equal to XW , and V W to WQ, πενταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τοῦ μείζονος thus as XW is to WQ, so QW (is) to WZ. And, again, τμήματος· πενταπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς Ωα τοῦ ἀπὸ on account of this, if we join QX then the angle at Q will τῆς αΧ. καί ἐστι τῆς μὲν Ωα διπλῆ ἡ ΩΨ, τὴς δὲ αΧ διπλῆ be a right-angle [Prop. 6.8]. Thus, the semi-circle drawn ἡ ΦΧ· πενταπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΩΨ τοῦ ἀπὸ τῆς on XZ will also pass through Q [Prop. 3.31]. And if XZ ΧΦ. καὶ ἐπεὶ τετραπλῆ ἐστιν ἡ ΑΓ τῆς ΓΒ, πενταπλῆ ἄρα remains fixed, and the semi-circle is carried around, and ἐστὶν ἡ ΑΒ τῆς ΒΓ. ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ again established at the same (position) from which it ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΔ· πενταπλάσιον ἄρα ἐστὶ began to be moved, then it will also pass through (point) τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΒΔ. ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς Q, and (through) the remaining (angular) points of the ΩΨ πενταπλάσιον τοῦ ἀπὸ τῆς ΦΧ. καί ἐστιν ἴση ἡ ΔΒ τῇ icosahedron. And the icosahedron will have been en- 529 STOIQEIWN igþ. ELEMENTS BOOK 13 ΦΧ· ἑκατέρα γὰρ αὐτῶν ἴση ἐστὶ τῇ ἐκ τοῦ κέντρου τοῦ closed by a sphere. So, I say that (it is) also (enclosed) ΕΖΗΘΚ κύκλου· ἴση ἄρα καὶ ἡ ΑΒ τῇ ΨΩ. καί ἐστιν ἡ ΑΒ by the given (sphere). For let V W have been cut in half ἡ τῆς δοθείσης σφαίρας διάμετρος· καὶ ἡ ΨΩ ἄρα ἴση ἐστὶ at a. And since the straight-line V Z has been cut in ex- τῇ τῆς δοθείσης σφαίρας διαμέτρῳ· τῇ ἄρα δοθείσῃ σφαίρᾳ treme and mean ratio at W , and ZW is its lesser piece, περιείληπται τὸ εἰκοσάεδρον. then the square on ZW added to half of the greater piece, Λέγω δή, ὅτι ἡ τοῦ εἰκοσαέδρου πλευρὰ ἄλογός ἐστιν ἡ Wa, is five times the (square) on half of the greater piece καλουμένη ἐλάττων. ἐπεὶ γὰρ ῥητή ἐστιν ἡ τῆς σφαίρας [Prop. 13.3]. Thus, the (square) on Za is five times the διάμετρος, καί ἐστι δυνάμει πενταπλασίων τῆς ἐκ τοῦ (square) on aW . And ZX is double Za, and V W double κέντρου τοῦ ΕΖΗΘΚ κύκλου, ῥητὴ ἄρα ἐστὶ καὶ ἡ ἑκ aW . Thus, the (square) on ZX is five times the (square) τοῦ κέντρου τοῦ ΕΖΗΘΚ κύκλου· ὥστε καὶ ἡ διάμετρος on WV . And since AC is four times CB, AB is thus αὐτοῦ ῥητή ἐστιν. ἐὰν δὲ εἰς κύκλον ῥητὴν ἔχοντα τὴν five times BC. And as AB (is) to BC, so the (square) διάμετρον πεντάγωνον ἰσόπλευρον ἐγγραφῃ, ἡ τοῦ πεν- on AB (is) to the (square) on BD [Prop. 6.8, Def. 5.9]. ταγώνου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάττων. ἡ δὲ Thus, the (square) on AB is five times the (square) on τοῦ ΕΖΗΘΚ πενταγώνου πλευρὰ ἡ τοῦ εἰκοσαέδρου ἐστίν. BD. And the (square) on ZX was also shown (to be) ἡ ἄρα τοῦ είκοσαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη five times the (square) on V W . And DB is equal to V W . ἐλάττων. For each of them is equal to the radius of circle EFGHK. Thus, AB (is) also equal to XZ. And AB is the diameter of the given sphere. Thus, XZ is equal to the diameter of the given sphere. Thus, the icosahedron has been en- closed by the given sphere. So, I say that the side of the icosahedron is that irra- tional (straight-line) called minor. For since the diameter of the sphere is rational, and the square on it is five times the (square) on the radius of circle EFGHK, the radius of circle EFGHK is thus also rational. Hence, its di- ameter is also rational. And if an equilateral pentagon is inscribed in a circle having a rational diameter then the side of the pentagon is that irrational (straight-line) called minor [Prop. 13.11]. And the side of pentagon EFGHK is (the side) of the icosahedron. Thus, the side of the icosahedron is that irrational (straight-line) called minor.Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι ἡ τῆς σφαίρας διάμετρος So, (it is) clear, from this, that the square on the di- δυνάμει πενταπλασίων ἐστὶ τῆς ἐκ τοῦ κέντρου τοῦ κύκλου, ameter of the sphere is five times the square on the ra- ἀφ᾿ οὗ τὸ εἰκοσάεδρον ἀναγέγραπται, καὶ ὅτι ἡ τῆς σφαίρας dius of the circle from which the icosahedron has been διάμετρος σύγκειται ἔκ τε τῆς τοῦ ἑξαγώνου καὶ δύο τῶν described, and that the the diameter of the sphere is the τοῦ δεκαγώνου τῶν εἰς τὸν αὐτὸν κύκλον ἐγγραφομένων. sum of (the side) of the hexagon, and two of (the sides) ὅπερ ἔδει δεῖξαι. of the decagon, inscribed in the same circle.† † If the radius of the sphere is unity then the radius of the circle is 2/ √ 5, and the sides of the hexagon, decagon, and pentagon/icosahedron are 2/ √ 5, 1 − 1/ √ 5, and (1/ √ 5) p 10 − 2 √ 5, respectively.izþ. Proposition 17 Δωδεκάεδρον συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν, ᾗ καὶ To construct a dodecahedron, and to enclose (it) in a τὰ προειρημένα σχήματα, καὶ δεῖξαι, ὅτι ἡ τοῦ δωδεκαέδρου sphere, like the aforementioned figures, and to show that πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἀποτομή. the side of the dodecahedron is that irrational (straight- line) called an apotome. 530 STOIQEIWN igþ. ELEMENTS BOOK 13 T L P K G N E M Z A R B S F X H J U D Q W Y O T L K N E M A D G F O CB H S R Q U V W Z P X ᾿Εκκείσθωσαν τοῦ προειρημένου κύβου δύο ἐπίπεδα Let two planes of the aforementioned cube [Prop. πρὸς ὀρθὰς ἀλλήλοις τὰ ΑΒΓΔ, ΓΒΕΖ, καὶ τετμήσθω 13.15], ABCD and CBEF , (which are) at right-angles ἑκάστη τῶν ΑΒ, ΒΓ, ΓΔ, ΔΑ, ΕΖ, ΕΒ, ΖΓ πλευρῶν δίχα to one another, be laid out. And let the sides AB, BC, κατὰ τὰ Η, Θ, Κ, Λ, Μ, Ν, Ξ, καὶ ἐπεζεύχθωσαν αἱ ΗΚ, CD, DA, EF , EB, and FC have each been cut in half at ΘΛ, ΜΘ, ΝΞ, καὶ τετηήσθω ἑκάστη τῶν ΝΟ, ΟΞ, ΘΠ points G, H , K, L, M , N , and O (respectively). And let ἄκρον καὶ μέσον λόγον κατὰ τὰ Ρ, Σ, Τ σημεῖα, καὶ ἔστω GK, HL, MH , and NO have been joined. And let NP , αὐτῶν μείζονα τμήματα τὰ ΡΟ, ΟΣ, ΤΠ, καὶ ἀνεστάτωσαν PO, and HQ have each been cut in extreme and mean ἀπὸ τῶν Ρ, Σ, Τ σημείων τοῖς τοῦ κύβου ἐπιπέδοις πρὸς ratio at points R, S, and T (respectively). And let their ὀρθὰς ἐπὶ τὰ ἐκτὸς μέρη τοῦ κύβου αἱ ΡΥ, ΣΦ, ΤΧ, καὶ greater pieces be RP , PS, and TQ (respectively). And κείσθωσαν ἴσαι ταῖς ΡΟ, ΟΣ, ΤΠ, καὶ ἐπεζεύχθωσαν αἱ let RU , SV , and TW have been set up on the exterior ΥΒ, ΒΧ, ΧΓ, ΓΦ, ΦΥ. side of the cube, at points R, S, and T (respectively), at Λέγω, ὅτι τὸ ΥΒΧΓΦ πεντάγωνον ἰσόπλευρόν τε καὶ ἐν right-angles to the planes of the cube. And let them be ἑνὶ ἐπιπέδῳ καὶ ἔτι ἰσογώνιόν ἐστιν. ἐπεζεύχθωσαν γὰρ αἱ made equal to RP , PS, and TQ. And let UB, BW , WC, ΡΒ, ΣΒ, ΦΒ. καὶ ἐπεὶ εὐθεῖα ἡ ΝΟ ἄκρον καὶ μέσον λόγον CV , and V U have been joined. τέτμηται κατὰ τὸ Ρ, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΡΟ, τὰ ἄρα I say that the pentagon UBWCV is equilateral, and ἀπὸ τῶν ΟΝ, ΝΡ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΡΟ. ἴση δὲ ἡ in one plane, and, further, equiangular. For let RB, SB, μὲν ΟΝ τῇ ΝΒ, ἡ δὲ ΟΡ τῇ ΡΥ· τὰ ἄρα ἀπὸ τῶν ΒΝ, ΝΡ and V B have been joined. And since the straight-line NP τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΡΥ. τοῖς δὲ ἀπὸ τῶν ΒΝ, ΝΡ τὸ has been cut in extreme and mean ratio at R, and RP is ἀπὸ τῆς ΒΡ ἐστιν ἴσον· τὸ ἄρα ἀπὸ τῆς ΒΡ τριπλάσιόν ἐστι the greater piece, the (sum of the squares) on PN and τοῦ ἀπὸ τῆς ΡΥ· ὥστε τὰ ἀπὸ τῶν ΒΡ, ΡΥ τετραπλάσιά NR is thus three times the (square) on RP [Prop. 13.4]. ἐστι τοῦ ἀπὸ τῆς ΡΥ. τοῖς δὲ ἀπὸ τῶν ΒΡ, ΡΥ ἴσον ἐστι τὸ And PN (is) equal to NB, and PR to RU . Thus, the ἀπὸ τῆς ΒΥ· τὸ ἄρα ἄπὸ τῆς ΒΥ τετραπλάσιόν ἐστι τοῦ ἀπὸ (sum of the squares) on BN and NR is three times the τῆς ΥΡ· διπλῆ ἄρα ἐστὶν ἡ ΒΥ τῆς ΡΥ. ἔστι δὲ καὶ ἡ ΦΥ τῆς (square) on RU . And the (square) on BR is equal to ΥΡ διπλῆ, ἐπειδήπερ καὶ ἡ ΣΡ τῆς ΟΡ, τουτέστι τῆς ΡΥ, the (sum of the squares) on BN and NR [Prop. 1.47]. ἐστι διπλῆ· ἴση ἄρα ἡ ΒΥ τῇ ΥΦ. ὁμοίως δὴ δειχθήσεται, Thus, the (square) on BR is three times the (square) on ὅτι καὶ ἑκάστη τῶν ΒΧ, ΧΓ, ΓΦ ἑκατέρᾳ τῶν ΒΥ, ΥΦ RU . Hence, the (sum of the squares) on BR and RU ἐστιν ἴση. ἰσόπλευρον ἄρα ἐστὶ τὸ ΒΥΦΓΧ πεντάγωνον. is four times the (square) on RU . And the (square) on λέγω δή, ὅτι καὶ ἐν ἑνί ἐστιν ἐπιπέδῳ. ἤχθω γὰρ ἀπὸ τοῦ Ο BU is equal to the (sum of the squares) on BR and RU ἑκατέρᾳ τῶν ΡΥ, ΣΦ παράλληλος ἐπὶ τὰ ἐκτὸς τοῦ κύβου [Prop. 1.47]. Thus, the (square) on BU is four times the μέρη ἡ ΟΨ, καὶ ἐπεζεύχθωσαν αἱ ΨΘ, ΘΧ· λέγω, ὅτι ἡ (square) on UR. Thus, BU is double RU . And V U is also ΨΘΧ εὐθεῖά ἐστιν. ἐπεὶ γὰρ ἡ ΘΠ ἄκρον καὶ μέσον λόγον double UR, inasmuch as SR is also double PR—that is τέτμηται κατὰ τὸ Τ, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστιν ἡ ΠΤ, to say, RU . Thus, BU (is) equal to UV . So, similarly, it ἔστιν ἄρα ὡς ἡ ΘΠ πρὸς τὴν ΠΤ, οὕτως ἡ ΠΤ πρὸς τὴν can be shown that each of BW , WC, CV is equal to each 531 STOIQEIWN igþ. ELEMENTS BOOK 13 ΤΘ. ἴση δὲ ἡ μὲν ΘΠ τῇ ΘΟ, ἡ δὲ ΠΤ ἑκατέρᾳ τῶν ΤΧ, of BU and UV . Thus, pentagon BUV CW is equilateral. ΟΨ· ἔστιν ἄρα ὡς ἡ ΘΟ πρὸς τὴν ΟΨ, οὕτως ἡ ΧΤ πρὸς So, I say that it is also in one plane. For let PX have τὴν ΤΘ. καί ἐστι παράλληλος ἡ μὲν ΘΟ τῇ ΤΧ· ἑκατέρα been drawn from P , parallel to each of RU and SV , on γὰρ αὐτῶν τῷ ΒΔ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν· ἡ δὲ ΤΘ τῇ the exterior side of the cube. And let XH and HW have ΟΨ· ἑκατέρα γὰρ αὐτῶν τῷ ΒΖ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. been joined. I say that XHW is a straight-line. For since ἐὰν δὲ δύο τρίγωνα συντεθῇ κατὰ μίαν γωνίαν, ὡς τὰ ΨΟΘ, HQ has been cut in extreme and mean ratio at T , and ΘΤΧ, τὰς δύο πλευρὰς ταῖς δυνὶν ἀνάλογον ἔχοντα, ὥστε QT is its greater piece, thus as HQ is to QT , so QT (is) τὰς ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι, αἱ to TH . And HQ (is) equal to HP , and QT to each of λοιπαὶ εὐθεῖαι ἐπ᾿ εὐθείας ἔσονται· ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ TW and PX . Thus, as HP is to PX , so WT (is) to ΨΘ τῇ ΘΧ. πᾶσα δὲ εὐθεῖα ἐν ἑνί ἐστιν ἐπιπέδῳ· ἐν ἑνὶ ἄρα TH . And HP is parallel to TW . For of each of them is ἐπιπέδῳ ἐστὶ τὸ ΥΒΧΓΦ πεντάγωνον. at right-angles to the plane BD [Prop. 11.6]. And TH Λέγω δή, ὅτι καὶ ἰσογώνιόν ἐστιν. (is parallel) to PX . For each of them is at right-angles ᾿Επεὶ γὰρ εὐθεῖα γραμμὴ ἡ ΝΟ ἄκρον καὶ μέσον λόγον to the plane BF [Prop. 11.6]. And if two triangles, like τέτμηται κατὰ τὸ Ρ, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΟΡ [ἔστιν XPH and HTW , having two sides proportional to two ἄρα ὡς συναμφότερος ἡ ΝΟ, ΟΡ πρὸς τὴν ΟΝ, οὕτως ἡ sides, are placed together at a single angle such that their ΝΟ πρὸς τὴν ΟΡ], ἴση δὲ ἡ ΟΡ τῇ ΟΣ [ἔστιν ἄρα ὡς ἡ ΣΝ corresponding sides are also parallel then the remaining πρὸς τὴν ΝΟ, οὕτως ἡ ΝΟ πρὸς τὴν ΟΣ], ἡ ΝΣ ἄρα ἄκρον sides will be straight-on (to one another) [Prop. 6.32]. καὶ μέσον λόγον τέτμηται κατὰ τὸ Ο, καὶ τὸ μεῖζον τμῆμά Thus, XH is straight-on to HW . And every straight-line ἐστιν ἡ ΝΟ· τὰ ἄρα ἀπὸ τῶν ΝΣ, ΣΟ τριπλάσιά ἐστι τοῦ is in one plane [Prop. 11.1]. Thus, pentagon UBWCV is ἀπὸ τῆς ΝΟ. ἴση δὲ ἡ μὲν ΝΟ τῇ ΝΒ, ἡ δὲ ΟΣ τῇ ΣΦ· τὰ in one plane. ἄρα ἀπὸ τῶν ΝΣ, ΣΦ τετράγωνα τριπλάσιά ἐστι τοῦ ἀπὸ So, I say that it is also equiangular. τῆς ΝΒ· ὥστε τὰ ἀπὸ τῶν ΦΣ, ΣΝ, ΝΒ τετραπλάσιά ἐστι For since the straight-line NP has been cut in extreme τοῦ ἀπὸ τῆς ΝΒ. τοῖς δὲ ἀπὸ τῶν ΣΝ, ΝΒ ἴσον ἐστὶ τὸ ἀπὸ and mean ratio at R, and PR is the greater piece [thus as τῆς ΣΒ· τὰ ἄρα ἀπὸ τῶν ΒΣ, ΣΦ, τουτέστι τὸ ἀπὸ τῆς ΒΦ the sum of NP and PR is to PN , so NP (is) to PR], and [ὀρθὴ γὰρ ἡ ὑπὸ ΦΣΒ γωνία], τετραπλάσιόν ἐστι τοῦ ἀπὸ PR (is) equal to PS [thus as SN is to NP , so NP (is) to τῆς ΝΒ· διπλῆ ἄρα ἐστὶν ἡ ΦΒ τῆς ΒΝ. ἔστι δὲ καὶ ἡ ΒΓ PS], NS has thus also been cut in extreme and mean τῆς ΒΝ διπλῆ· ἴση ἄρα ἐστὶν ἡ ΒΦ τῇ ΒΓ. καὶ ἐπεὶ δύο αἱ ratio at P , and NP is the greater piece [Prop. 13.5]. ΒΥ, ΥΦ δυσὶ ταῖς ΒΧ, ΧΓ ἴσαι εἰσίν, καὶ βάσις ἡ ΒΦ βάσει Thus, the (sum of the squares) on NS and SP is three τῇ ΒΓ ἴση, γωνία ἄρα ἡ ὑπὸ ΒΥΦ γωνίᾳ τῇ ὑπὸ ΒΧΓ ἐστιν times the (square) on NP [Prop. 13.4]. And NP (is) ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ὑπὸ ΥΦΓ γωνία ἴση ἐστὶ equal to NB, and PS to SV . Thus, the (sum of the) τῇ ὑπὸ ΒΧΓ· αἱ ἄρα ὑπὸ ΒΧΓ, ΒΥΦ, ΥΦΓ τρεῖς γωνίαι squares on NS and SV is three times the (square) on ἴσαι ἀλλήλαις εἰσίν. ἐὰν δὲ πενταγώνου ἰσοπλεύρου αἱ τρεῖς NB. Hence, the (sum of the squares) on V S, SN , and γωνίαι ἴσαι ἀλλήλαις ὦσιν, ἰσογώνιον ἔσται τὸ πεντάγωνον· NB is four times the (square) on NB. And the (square) ἰσογώνιον ἄρα ἐστὶ τὸ ΒΥΦΓΧ πεντάγωνον. ἐδείχθη δὲ καὶ on SB is equal to the (sum of the squares) on SN and ἰσόπλευρον· τὸ ἄρα ΒΥΦΓΧ πεντάγωνον ἰσόπλευρόν ἐστι NB [Prop. 1.47]. Thus, the (sum of the squares) on BS καὶ ἰσογώνιον, καί ἐστιν ἐπὶ μιᾶς τοῦ κύβου πλευρᾶς τῆς and SV —that is to say, the (square) on BV [for angle ΒΓ. ἐὰν ἄρα ἐφ᾿ ἑκάστης τῶν τοῦ κύβου δώδεκα πλευρῶν V SB (is) a right-angle]—is four times the (square) on τὰ αὐτὰ κατασκευάσωμεν, συσταθήσεταί τι σχῆμα στερεὸν NB [Def. 11.3, Prop. 1.47]. Thus, V B is double BN . ὑπὸ δώδεκα πενταγώνων ἰσοπλεύρων τε καὶ ἰσογωνίων πε- And BC (is) also double BN . Thus, BV is equal to BC. ριεχόμενον, ὃ καλεῖται δωδεκάεδρον. And since the two (straight-lines) BU and UV are equal Δεῖ δὴ αὐτὸ καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, to the two (straight-lines) BW and WC (respectively), ὅτι ἡ τοῦ δωδεκαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη and the base BV (is) equal to the base BC, angle BUV ἀποτομή. is thus equal to angle BWC [Prop. 1.8]. So, similarly, we ᾿Εκβεβλήσθω γὰρ ἡ ΨΟ, καὶ ἔστω ἡ ΨΩ· συμβάλλει ἄρα can show that angle UV C is equal to angle BWC. Thus, ἡ ΟΩ τῇ τοῦ κύβου διαμέτρῳ, καὶ δίχα τέμνουσιν ἀλλήλας· the three angles BWC, BUV , and UV C are equal to one τοῦτο γὰρ δέδεικται ἐν τῷ παρατελεύτῳ θεωρήματι τοῦ another. And if three angles of an equilateral pentagon ἑνδεκάτου βιβλίου. τεμνέτωσαν κατὰ τὸ Ω· τὸ Ω ἄρα are equal to one another then the pentagon is equiangu- κέντρον ἐστὶ τῆς σφαίρας τῆς περιλαμβανούσης τὸν κύβον, lar [Prop. 13.7]. Thus, pentagon BUV CW is equiangu- καὶ ἡ ΩΟ ἡμίσεια τῆς πλευρᾶς τοῦ κύβου. ἐπεζεύχθω δὴ ἡ lar. And it was also shown (to be) equilateral. Thus, pen- ΥΩ. καὶ ἐπεὶ εὐθεῖα γραμμὴ ἡ ΝΣ ἄκρον καὶ μέσον λόγον tagon BUV CW is equilateral and equiangular, and it is τέτμηται κατὰ τὸ Ο, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστιν ἡ ΝΟ, on one of the sides, BC, of the cube. Thus, if we make the 532 STOIQEIWN igþ. ELEMENTS BOOK 13 τὰ ἄρα ἀπὸ τῶν ΝΣ, ΣΟ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΝΟ. same construction on each of the twelve sides of the cube ἴση δὲ ἡ μὲν ΝΣ τῇ ΨΩ, ἐπειδήπερ καὶ ἡ μὲν ΝΟ τῇ ΟΩ then some solid figure contained by twelve equilateral ἐστιν ἴση, ἡ δὲ ΨΟ τῇ ΟΣ. ἀλλὰ μὴν καὶ ἡ ΟΣ τῇ ΨΥ, ἐπεὶ and equiangular pentagons will have been constructed, καὶ τῇ ΡΟ· τὰ ἄρα ἀπὸ τῶν ΩΨ, ΨΥ τριπλάσιά ἐστι τοῦ ἀπὸ which is called a dodecahedron. τῆς ΝΟ. τοῖς δὲ ἀπὸ τῶν ΩΨ, ΨΥ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΥΩ· So, it is necessary to enclose it in the given sphere, τὸ ἄρα ἀπὸ τῆς ΥΩ τριπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΝΟ. ἔστι and to show that the side of the dodecahedron is that δὲ καὶ ἡ ἐκ τοῦ κέντρου τῆς σφαίρας τῆς περιλαμβανούσης irrational (straight-line) called an apotome. τὸν κύβον δυνάμει τριπλασίων τῆς ἡμισείας τῆς τοῦ κύβου For let XP have been produced, and let (the produced πλευρᾶς· προδέδεικται γὰρ κύβον συστήσασθαι καὶ σφαίρᾳ straight-line) be XZ. Thus, PZ meets the diameter of the περιλαβεῖν καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει cube, and they cut one another in half. For, this has been τριπλασίων ἐστὶ τῆς πλευρᾶς τοῦ κύβου. εἰ δὲ ὅλη τῆς ὅλης, proved in the penultimate theorem of the eleventh book καὶ [ἡ] ἡμίσεια τῆς ἡμισείας· καί ἐστιν ἡ ΝΟ ἡμίσεια τῆς τοῦ [Prop. 11.38]. Let them cut (one another) at Z. Thus, κύβου πλευρᾶς· ἡ ἄρα ΥΩ ἴση ἐστὶ τῇ ἐκ τοῦ κέντρου τῆς Z is the center of the sphere enclosing the cube, and ZP σφαίρας τῆς περιλαμβανούσης τὸν κύβον. καί ἐστι τὸ Ω (is) half the side of the cube. So, let UZ have been joined. κέντρον τῆς σφαίρας τῆς περιλαμβανούσης τὸν κύβον· τὸ And since the straight-line NS has been cut in extreme Υ ἄρα σημεῖον πρὸς τῇ ἐπιφανείᾳ ἐστι τῆς σφαίρας. ὁμοίως and mean ratio at P , and its greater piece is NP , the δὴ δείξομεν, ὅτι καὶ ἑκάστη τῶν λοιπῶν γωνιῶν τοῦ δω- (sum of the squares) on NS and SP is thus three times δεκαέδρου πρὸς τῇ ἐπιφανείᾳ ἐστὶ τῆς σφαίρας· περιείληπται the (square) on NP [Prop. 13.4]. And NS (is) equal to ἄρα τὸ δωδεκαέδρον τῇ δοθείσῃ σφαίρᾳ. XZ, inasmuch as NP is also equal to PZ, and XP to Λέγω δή, ὅτι ἡ τοῦ δωδεκαέδρου πλευρὰ ἄλογός ἐστιν PS. But, indeed, PS (is) also (equal) to XU , since (it ἡ καλουμένη ἀποτομή. is) also (equal) to RP . Thus, the (sum of the squares) ᾿Επεὶ γὰρ τῆς ΝΟ ἄκρον καὶ μέσον λόγον τετμημένης τὸ on ZX and XU is three times the (square) on NP . And μεῖζον τμῆμά ἐστιν ὁ ΡΟ, τῆς δὲ ΟΞ ἄκρον καὶ μέσον λόγον the (square) on UZ is equal to the (sum of the squares) τετμημένης τὸ μεῖζον τμῆμά ἐστιν ἡ ΟΣ, ὅλης ἄρα τῆς ΝΞ on ZX and XU [Prop. 1.47]. Thus, the (square) on UZ ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμά ἐστιν ἡ is three times the (square) on NP . And the square on ΡΣ. [οἷον ἐπεί ἐστιν ὡς ἡ ΝΟ πρὸς τὴν ΟΡ, ἡ ΟΡ πρὸς τὴν the radius of the sphere enclosing the cube is also three ΡΝ, καὶ τὰ διπλάσια· τὰ γὰρ μέρη τοῖς ἰσάκις πολλαπλασίοις times the (square) on half the side of the cube. For it τὸν αὐτὸν ἔχει λόγον· ὡς ἄρα ἡ ΝΞ πρὸς τὴν ΡΣ, οὕτως ἡ has previously been demonstrated (how to) construct the ΡΣ πρὸς συναμφότερον τὴν ΝΡ, ΣΞ. μείζων δὲ ἡ ΝΞ τῆς cube, and to enclose (it) in a sphere, and to show that ΡΣ· μείζων ἄρα καὶ ἡ ΡΣ συναμφοτέρου τῆς ΝΡ, ΣΞ· ἡ ΝΞ the square on the diameter of the sphere is three times ἄρα ἄκρον καὶ μέσον λόγον τέτμηται, καὶ τὸ μεῖζον αὐτῆς the (square) on the side of the cube [Prop. 13.15]. And τμῆμά ἐστιν ἡ ΡΣ.] ἴση δὲ ἡ ΡΣ τῇ ΥΦ· τῆς ἄρα ΝΞ ἄκρον if the (square on the) whole (is three times) the (square καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμά ἐστιν ἡ ΥΦ. on the) whole, then the (square on the) half (is) also καὶ ἐπεὶ ῥητή ἐστιν τῆς σφαίρας διάμετρος καί ἐστι δυνάμει (three times) the (square on the) half. And NP is half τριπλασίων τῆς τοῦ κύβου πλευρᾶς, ῥητὴ ἄρα ἐστὶν ἡ ΝΞ of the side of the cube. Thus, UZ is equal to the radius πλευρὰ οὖσα τοῦ κύβου. ἐὰν δὲ ῥητὴ γραμμὴ ἄκρον καὶ of the sphere enclosing the cube. And Z is the center of μέσον λόγον τμηθῇ, ἑκάτερον τῶν τμημάτων ἄλογός ἐστιν the sphere enclosing the cube. Thus, point U is on the ἀποτομή. surface of the sphere. So, similarly, we can show that ῾Η ΥΦ ἄρα πλευρὰ οὖσα τοῦ δωδεκαέδρου ἄλογός ἐστιν each of the remaining angles of the dodecahedron is also ἀποτομή. on the surface of the sphere. Thus, the dodecahedron has been enclosed by the given sphere. So, I say that the side of the dodecahedron is that irrational straight-line called an apotome. For since RP is the greater piece of NP , which has been cut in extreme and mean ratio, and PS is the greater piece of PO, which has been cut in extreme and mean ratio, RS is thus the greater piece of the whole of NO, which has been cut in extreme and mean ratio. [Thus, since as NP is to PR, (so) PR (is) to RN , and (the same is also true) of the doubles. For parts have the same ratio as similar multiples (taken in corresponding 533 STOIQEIWN igþ. ELEMENTS BOOK 13 order) [Prop. 5.15]. Thus, as NO (is) to RS, so RS (is) to the sum of NR and SO. And NO (is) greater than RS. Thus, RS (is) also greater than the sum of NR and SO [Prop. 5.14]. Thus, NO has been cut in extreme and mean ratio, and RS is its greater piece.] And RS (is) equal to UV . Thus, UV is the greater piece of NO, which has been cut in extreme and mean ratio. And since the diameter of the sphere is rational, and the square on it is three times the (square) on the side of the cube, NO, which is the side of the cube, is thus rational. And if a rational (straight)-line is cut in extreme and mean ra- tio then each of the pieces is the irrational (straight-line called) an apotome. Thus, UV , which is the side of the dodecahedron, is the irrational (straight-line called) an apotome [Prop. 13.6].Pìrisma. Corollary ᾿Εκ δὴ τούτου φανερόν, ὅτι τῆς τοῦ κύβου πλευρᾶς So, (it is) clear, from this, that the side of the dodeca- ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμά ἐστιν hedron is the greater piece of the side of the cube, when ἡ τοῦ δωδεκαέδρου πλευρά. ὅπερ ἔδει δεῖξαι. it is cut in extreme and mean ratio.† (Which is) the very thing it was required to show. † If the radius of the circumscribed sphere is unity then the side of the cube is p 4/3, and the side of the dodecahedron is (1/3) ( √ 15 − √ 3).ihþ. Proposition 18 Τὰς πλευρὰς τῶν πέντε σχημάτων ἐκθέσθαι καὶ συγκρῖν- To set out the sides of the five (aforementioned) fig- αι πρὸς ἀλλήλας. ures, and to compare (them) with one another.† N A G E Z M D L BK H J N A E M D L BC F G K H ᾿Εκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ, καὶ Let the diameter, AB, of the given sphere be laid out. τετμήσθω κατὰ τὸ Γ ὥστε ἴσην εἶναι τὴν ΑΓ τῇ ΓΒ, κατὰ δὲ And let it have been cut at C, such that AC is equal to τὸ Δ ὥστε διπλασίονα εἶναι τὴν ΑΔ τῆς ΔΒ, καὶ γεγράφθω CB, and at D, such that AD is double DB. And let the ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΕΒ, καὶ ἀπὸ τῶν Γ, Δ τῇ ΑΒ semi-circle AEB have been drawn on AB. And let CE πρὸς ὀρθὰς ἤχθωσαν αἱ ΓΕ, ΔΖ, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, and DF have been drawn from C and D (respectively), ΖΒ, ΕΒ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΑΔ τῆς ΔΒ, τριπλῆ ἄρα at right-angles to AB. And let AF , FB, and EB have ἐστὶν ἡ ΑΒ τῆς ΒΔ. ἀναστρέψαντι ἡμιολία ἄρα ἐστὶν ἡ ΒΑ been joined. And since AD is double DB, AB is thus τῆς ΑΔ. ὡς δὲ ἡ ΒΑ πρὸς τὴν ΑΔ, οὕτως τὸ ἀπὸ τῆς ΒΑ triple BD. Thus, via conversion, BA is one and a half 534 STOIQEIWN igþ. ELEMENTS BOOK 13 πρὸς τὸ ἀπὸ τῆς ΑΖ· ἰσογώνιον γάρ ἐστι τὸ ΑΖΒ τρίγωνον times AD. And as BA (is) to AD, so the (square) on τῷ ΑΖΔ τριγώνῳ· ἡμιόλιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΑ τοῦ ἀπὸ BA (is) to the (square) on AF [Def. 5.9]. For triangle τῆς ΑΖ. ἔστι δὲ καὶ ἡ τῆς σφαίρας διάμετρος δυνάμει ἡμιολία AFB is equiangular to triangle AFD [Prop. 6.8]. Thus, τῆς πλευρᾶς τῆς πυραμίδος. καί ἐστιν ἡ ΑΒ ἡ τῆς σφαίρας the (square) on BA is one and a half times the (square) διάμετρος· ἡ ΑΖ ἄρα ἴση ἐστὶ τῇ πλευρᾷ τῆς πυραμίδος. on AF . And the square on the diameter of the sphere is Πάλιν, ἐπεὶ διπλασίων ἐστὶν ἡ ΑΔ τῆς ΔΒ, τριπλῆ ἄρα also one and a half times the (square) on the side of the ἐστὶν ἡ ΑΒ τῆς ΒΔ. ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΔ, οὕτως τὸ pyramid [Prop. 13.13]. And AB is the diameter of the ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΖ· τριπλάσιον ἄρα ἐστὶ τὸ sphere. Thus, AF is equal to the side of the pyramid. ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΒΖ. ἔστι δὲ καὶ ἡ τῆς σφαίρας Again, since AD is double DB, AB is thus triple BD. διάμετρος δυνάμει τριπλασίων τῆς τοῦ κύβου πλευρᾶς. καί And as AB (is) to BD, so the (square) on AB (is) to the ἐστιν ἡ ΑΒ ἡ τῆς σφαίρας διάμετρος· ἡ ΒΖ ἄρα τοῦ κύβου (square) on BF [Prop. 6.8, Def. 5.9]. Thus, the (square) ἐστὶ πλευρά. on AB is three times the (square) on BF . And the square Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, διπλῆ ἄρα ἐστὶν ἡ ΑΒ on the diameter of the sphere is also three times the τῆς ΒΓ. ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΒ (square) on the side of the cube [Prop. 13.15]. And AB πρὸς τὸ ἀπὸ τῆς ΒΕ· διπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΒ τοῦ is the diameter of the sphere. Thus, BF is the side of the ἀπὸ τῆς ΒΕ. ἔστι δὲ καὶ ἡ τῆς σφαίρας διάμετρος δυνάμει cube. διπλασίων τῆς τοῦ ὀκταέδρου πλευρᾶς. καὶ ἐστιν ἡ ΑΒ ἡ And since AC is equal to CB, AB is thus double BC. τῆς δοθείσης σφαίρας διάμετρος· ἡ ΒΕ ἄρα τοῦ ὀκταέδρου And as AB (is) to BC, so the (square) on AB (is) to the ἐστὶ πλευρά. (square) on BE [Prop. 6.8, Def. 5.9]. Thus, the (square) ῎Ηχθω δὴ ἀπὸ τοῦ Α σημείου τῇ ΑΒ εὐθείᾳ πρὸς ὀρθὰς on AB is double the (square) on BE. And the square ἡ ΑΗ, καὶ κείσθω ἡ ΑΗ ἴση τῇ ΑΒ, καὶ ἐπεζεύχθω ἡ ΗΓ, on the diameter of the sphere is also double the (square) καὶ ἀπὸ τοῦ Θ ἐπὶ τὴν ΑΒ κάθετος ἤχθω ἡ ΘΚ. καὶ ἐπεὶ on the side of the octagon [Prop. 13.14]. And AB is the διπλῆ ἐστιν ἡ ΗΑ τῆς ΑΓ· ἴση γὰρ ἡ ΗΑ τῇ ΑΒ· ὡς δὲ ἡ diameter of the given sphere. Thus, BE is the side of the ΗΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΘΚ πρὸς τὴν ΚΓ, διπλῆ ἄρα octagon. καὶ ἡ ΘΚ τῆς ΚΓ. τετραπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΘΚ So let AG have been drawn from point A at right- τοῦ ἀπὸ τῆς ΚΓ· τὰ ἄρα ἀπὸ τῶν ΘΚ, ΚΓ, ὅπερ ἐστὶ τὸ angles to the straight-line AB. And let AG be made equal ἀπὸ τῆς ΘΓ, πενταπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΚΓ. ἴση δὲ to AB. And let GC have been joined. And let HK have ἡ ΘΓ τῇ ΓΒ· πενταπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΓ τοῦ been drawn from H , perpendicular to AB. And since GA ἀπὸ τῆς ΓΚ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΑΒ τῆς ΓΒ, ὧν ἡ is double AC. For GA (is) equal to AB. And as GA (is) ΑΔ τῆς ΔΒ ἐστι διπλῆ, λοιπὴ ἄρα ἡ ΒΔ λοιπῆς τῆς ΔΓ to AC, so HK (is) to KC [Prop. 6.4]. HK (is) thus also ἐστι διπλῆ. τριπλῆ ἄρα ἡ ΒΓ τῆς ΓΔ· ἐνναπλάσιον ἄρα double KC. Thus, the (square) on HK is four times the τὸ ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς ΓΔ. πενταπλάσιον δὲ τὸ ἀπὸ (square) on KC. Thus, the (sum of the squares) on HK τῆς ΒΓ τοῦ ἀπὸ τῆς ΓΚ· μεῖζον ἄρα τὸ ἀπὸ τῆς ΓΚ τοῦ and KC, which is the (square) on HC [Prop. 1.47], is ἀπὸ τῆς ΓΔ. μείζων ἄρα ἐστὶν ἡ ΓΚ τῆς ΓΔ. κείσθω τῇ five times the (square) on KC. And HC (is) equal to CB. ΓΚ ἴση ἡ ΓΛ, καὶ ἀπὸ τοῦ Λ τῇ ΑΒ πρὸς ὀρθὰς ἤχθω ἡ Thus, the (square) on BC (is) five times the (square) on ΛΜ, καὶ ἐπεζεύχθω ἡ ΜΒ. καὶ ἐπεὶ πενταπλάσιόν ἐστι τὸ CK. And since AB is double CB, of which AD is double ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς ΓΚ, καί ἐστι τῆς μὲν ΒΓ διπλῆ DB, the remainder BD is thus double the remainder DC. ἡ ΑΒ, τῆς δὲ ΓΚ διπλῆ ἡ ΚΛ, πενταπλάσιον ἄρα ἐστὶ τὸ BC (is) thus triple CD. The (square) on BC (is) thus ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΚΛ. ἔστι δὲ καὶ ἡ τῆς σφαίρας nine times the (square) on CD. And the (square) on BC διάμετρος δυνάμει πενταπλασίων τῆς ἐκ τοῦ κέντρου τοῦ (is) five times the (square) on CK. Thus, the (square) κύκλου, ἀφ᾿ οὗ τὸ εἰκοσάεδρον ἀναγέγραπται. καί ἐστιν ἡ on CK (is) greater than the (square) on CD. CK is thus ΑΒ ἡ τῆς σφαίρας διάμετρος· ἡ ΚΛ ἄρα ἐκ τοῦ κέντρου greater than CD. Let CL be made equal to CK. And ἐστὶ τοῦ κύκλου, ἀφ᾿ οὗ τὸ εἰκοσάεδρον ἀναγέγραπται· let LM have been drawn from L at right-angles to AB. ἡ ΚΛ ἄρα ἑξαγώνου ἐστὶ πλευρὰ τοῦ εἰρημένου κύκλου. And let MB have been joined. And since the (square) on καὶ ἐπεὶ ἡ τῆς σφαίρας διάμετρος σύγκειται ἔκ τε τῆς τοῦ BC is five times the (square) on CK, and AB is double ἑξαγώνου καὶ δύο τῶν τοῦ δεκαγώνου τῶν εἰς τὸν εἰρημένον BC, and KL double CK, the (square) on AB is thus five κύκλον ἐγγραφομένων, καί ἐστιν ἡ μὲν ΑΒ ἡ τῆς σφαίρας times the (square) on KL. And the square on the diam- διάμετρος, ἡ δὲ ΚΛ ἑξαγώνου πλευρά, καὶ ἴση ἡ ΑΚ τῇ eter of the sphere is also five times the (square) on the ΛΒ, ἑκατέρα ἄρα τῶν ΑΚ, ΛΒ δεκαγώνου ἐστὶ πλευρὰ radius of the circle from which the icosahedron has been τοῦ ἐγγραφομένου εἰς τὸν κύκλον, ἀφ᾿ οὗ τὸ εἰκοσάεδρον described [Prop. 13.16 corr.]. And AB is the diameter ἀναγέγραπται. καὶ ἐπεὶ δεκαγώνου μὲν ἡ ΛΒ, ἑξαγώνου of the sphere. Thus, KL is the radius of the circle from 535 STOIQEIWN igþ. ELEMENTS BOOK 13 δὲ ἡ ΜΛ· ἴση γάρ ἐστι τῇ ΚΛ, ἐπεὶ καὶ τῇ ΘΚ· ἴσον γὰρ which the icosahedron has been described. Thus, KL is ἀπέχουσιν ἀπὸ τοῦ κέντρου· καί ἐστιν ἑκατέρα τῶν ΘΚ, (the side) of the hexagon (inscribed) in the aforemen- ΚΛ διπλασίων τῆς ΚΓ· πενταγώνου ἄρα ἐστὶν ἡ ΜΒ. ἡ δὲ tioned circle [Prop. 4.15 corr.]. And since the diameter of τοῦ πενταγώνου ἐστὶν ἡ τοῦ εἰκοσαέδρου· εἰκοσαέδρου ἄρα the sphere is composed of (the side) of the hexagon, and ἐστὶν ἡ ΜΒ. two of (the sides) of the decagon, inscribed in the afore- Καὶ ἐπεὶ ἡ ΖΒ κύβου ἐστὶ πλευρά, τετμήσθω ἄκρον καὶ mentioned circle, and AB is the diameter of the sphere, μέσον λόγον κατὰ τὸ Ν, καὶ ἔστω μεῖζον τμῆμα τὸ ΝΒ· ἡ and KL the side of the hexagon, and AK (is) equal to ΝΒ ἄρα δωδεκαέδρου ἐστὶ πλευρά. LB, thus AK and LB are each sides of the decagon in- Καὶ ἐπεὶ ἡ τῆς σφαίρας διάμετρος ἐδείχθη τῆς μὲν scribed in the circle from which the icosahedron has been ΑΖ πλευρᾶς τῆς πυραμίδος δυνάμει ἡμιολία, τῆς δὲ τοῦ described. And since LB is (the side) of the decagon. ὀκταέδρου τῆς ΒΕ δυνάμει διπλασίων, τῆς δὲ τοῦ κύβου τῆς And ML (is the side) of the hexagon—for (it is) equal to ΖΒ δυνάμει τριπλασίων, οἵων ἄρα ἡ τῆς σφαίρας διάμετρος KL, since (it is) also (equal) to HK, for they are equally δυνάμει ἕξ, τοιούτων ἡ μὲν τῆς πυραμίδος τεσσάρων, ἡ δὲ far from the center. And HK and KL are each double τοῦ ὀκταέδρου τριῶν, ἡ δὲ τοῦ κύβου δύο. ἡ μὲν ἄρα τῆς KC. MB is thus (the side) of the pentagon (inscribed πυραμίδος πλευρὰ τῆς μὲν τοῦ ὀκταέδρου πλευρᾶς δυνάμει in the circle) [Props. 13.10, 1.47]. And (the side) of the ἐστὶν ἐπίτριτος, τῆς δὲ τοῦ κύβου δυνάμει διπλῆ, ἡ δὲ τοῦ pentagon is (the side) of the icosahedron [Prop. 13.16]. ὀκταέδρου τῆς τοῦ κύβου δυνάμει ἡμιολία. αἱ μὲν οὖν Thus, MB is (the side) of the icosahedron. εἰρημέναι τῶν τριῶν σχημάτων πλευραί, λέγω δὴ πυραμίδος And since FB is the side of the cube, let it have been καὶ ὀκταέδρου καὶ κύβου, πρὸς ἀλλήλας εἰσὶν ἐν λόγοις cut in extreme and mean ratio at N , and let NB be the ῥητοῖς. αἱ δὲ λοιπαὶ δύο, λέγω δὴ ἥ τε τοῦ εἰκοσαέδρου greater piece. Thus, NB is the side of the dodecahedron καὶ ἡ τοῦ δωδεκαέδρου, οὔτε πρὸς ἀλλήλας οὔτε πρὸς τὰς [Prop. 13.17 corr.]. προειρημένας εἰσὶν ἐν λόγοις ῥητοῖς· ἄλογοι γάρ εἰσιν, ἡ μὲν And since the (square) on the diameter of the sphere ἐλάττων, ἡ δὲ ἀποτομή. was shown (to be) one and a half times the square on the ῞Οτι μείζων ἐστὶν ἡ τοῦ εἰκοσαέδρου πλευρὰ ἡ ΜΒ τῆς side, AF , of the pyramid, and twice the square on (the τοῦ δωδεκαέδρου τῆς ΝΒ, δείξομεν οὕτως. side), BE, of the octagon, and three times the square ᾿Επεὶ γὰρ ἰσογώνιόν ἐστι τὸ ΖΔΒ τρίγωνον τῷ ΖΑΒ on (the side), FB, of the cube, thus, of whatever (parts) τριγώνῳ, ἀνάλογόν ἐστιν ὡς ἡ ΔΒ πρὸς τὴν ΒΖ, οὕτως the (square) on the diameter of the sphere (makes) six, ἡ ΒΖ πρὸς τὴν ΒΑ. καὶ ἐπεὶ τρεῖς εὐθεῖαι ἀνάλογόν εἰσιν, of such (parts) the (square) on (the side) of the pyramid ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης (makes) four, and (the square) on (the side) of the oc- πρὸς τὸ ἀπὸ τῆς δευτέρας· ἔστιν ἄρα ὡς ἡ ΔΒ πρὸς τὴν ΒΑ, tagon three, and (the square) on (the side) of the cube οὕτως τὸ ἀπὸ τῆς ΔΒ πρὸς τὸ ἀπὸ τῆς ΒΖ· ἀνάπαλιν ἄρα two. Thus, the (square) on the side of the pyramid is one ὡς ἡ ΑΒ πρὸς τὴν ΒΔ, οὕτως τὸ ἀπὸ τῆς ΖΒ πρὸς τὸ ἀπὸ and a third times the square on the side of the octagon, τῆς ΒΔ. τριπλῆ δὲ ἡ ΑΒ τῆς ΒΔ· τριπλάσιον ἄρα τὸ ἀπὸ and double the square on (the side) of the cube. And the τῆς ΖΒ τοῦ ἀπὸ τῆς ΒΔ. ἔστι δὲ καὶ τὸ ἀπὸ τῆς ΑΔ τοῦ (square) on (the side) of the octahedron is one and a half ἀπὸ τῆς ΔΒ τετραπλάσιον· διπλῆ γὰρ ἡ ΑΔ τῆς ΔΒ· μεῖζον times the square on (the side) of the cube. Therefore, ἄρα τὸ ἀπὸ τῆς ΑΔ τοῦ ἀπὸ τῆς ΖΒ· μείζων ἄρα ἡ ΑΔ τῆς the aforementioned sides of the three figures—I mean, of ΖΒ· πολλῷ ἄρα ἡ ΑΛ τῆς ΖΒ μείζων ἐστίν. καὶ τῆς μὲν the pyramid, and of the octahedron, and of the cube— ΑΛ ἄκρον καὶ μέσον λόγον τεμνομένης τὸ μεῖζον τμῆμά are in rational ratios to one another. And (the sides ἐστιν ἡ ΚΛ, ἐπειδήπερ ἡ μὲν ΛΚ ἑξαγώνου ἐστίν, ἡ δὲ ΚΑ of) the remaining two (figures)—I mean, of the icosahe- δεκαγώνου· τῆς δὲ ΖΒ ἄκρον καὶ μέσον λόγον τεμνομένης dron, and of the dodecahedron—are neither in rational τὸ μεῖζον τμῆμά ἐστιν ἡ ΝΒ· μείζων ἄρα ἡ ΚΛ τῆς ΝΒ. ratios to one another, nor to the (sides) of the aforemen- ἴση δὲ ἡ ΚΛ τῇ ΛΜ· μείζων ἄρα ἡ ΛΜ τῆς ΝΒ [τῆς δὲ tioned (three figures). For they are irrational (straight- ΛΜ μείζων ἐστὶν ἡ ΜΒ]. πολλῷ ἄρα ἡ ΜΒ πλευρὰ οὖσα lines): (namely), a minor [Prop. 13.16], and an apotome τοῦ εἰκοσαέδρου μείζων ἐστὶ τῆς ΝΒ πλευρᾶς οὔσης τοῦ [Prop. 13.17]. δωδεκαέδρου· ὅπερ ἔδει δεῖξαι. (And), we can show that the side, MB, of the icosahe- dron is greater that the (side), NB, or the dodecahedron, as follows. For, since triangle FDB is equiangular to triangle FAB [Prop. 6.8], proportionally, as DB is to BF , so BF (is) to BA [Prop. 6.4]. And since three straight-lines are (continually) proportional, as the first (is) to the third, 536 STOIQEIWN igþ. ELEMENTS BOOK 13 so the (square) on the first (is) to the (square) on the second [Def. 5.9, Prop. 6.20 corr.]. Thus, as DB is to BA, so the (square) on DB (is) to the (square) on BF . Thus, inversely, as AB (is) to BD, so the (square) on FB (is) to the (square) on BD. And AB (is) triple BD. Thus, the (square) on FB (is) three times the (square) on BD. And the (square) on AD is also four times the (square) on DB. For AD (is) double DB. Thus, the (square) on AD (is) greater than the (square) on FB. Thus, AD (is) greater than FB. Thus, AL is much greater than FB. And KL is the greater piece of AL, which is cut in extreme and mean ratio—inasmuch as LK is (the side) of the hexagon, and KA (the side) of the decagon [Prop. 13.9]. And NB is the greater piece of FB, which is cut in extreme and mean ratio. Thus, KL (is) greater than NB. And KL (is) equal to LM . Thus, LM (is) greater than NB [and MB is greater than LM]. Thus, MB, which is (the side) of the icosahedron, is much greater than NB, which is (the side) of the dodecahe- dron. (Which is) the very thing it was required to show. † If the radius of the given sphere is unity then the sides of the pyramid (i.e., tetrahedron), octahedron, cube, icosahedron, and dodecahedron, respectively, satisfy the following inequality: p 8/3 >
√

2 >
p

4/3 > (1/
√

5)
p

10 − 2
√

5 > (1/3) (
√

15 −
√

3).

Λέγω δή, ὅτι παρὰ τὰ εἰρημένα πέντε σχήματα οὐ συ- So, I say that, beside the five aforementioned figures,
σταθήσεται ἕτερον σχῆμα περιεχόμενον ὑπὸ ἰσοπλεύρων τε no other (solid) figure can be constructed (which is) con-
καὶ ἰσογωνίων ἴσων ἀλλήλοις. tained by equilateral and equiangular (planes), equal to
῾Υπὸ μὲν γὰρ δύο τριγώνων ἢ ὅλως ἐπιπέδων στερεὰ one another.

γωνία οὐ συνίσταται. ὑπὸ δὲ τριῶν τριγώνων ἡ τῆς πυ- For a solid angle cannot be constructed from two tri-
ραμίδος, ὑπὸ δὲ τεσσάρων ἡ τοῦ ὀκταέδρου, ὑπὸ δὲ πέντε angles, or indeed (two) planes (of any sort) [Def. 11.11].
ἡ τοῦ εἰκοσαέδρου· ὑπὸ δὲ ἓξ τριγώνων ἰσοπλεύρων τε And (the solid angle) of the pyramid (is constructed)
καὶ ἰσογωνίων πρὸς ἑνὶ σημείῳ συνισταμένων οὐκ ἔσται from three (equiangular) triangles, and (that) of the oc-
στερεὰ γωνία· οὔσης γὰρ τῆς τοῦ ἰσοπλεύρου τριγώνου tahedron from four (triangles), and (that) of the icosahe-
γωνίας διμοίρου ὀρθῆς ἔσονται αἱ ἓξ τέσσαρσιν ὀρθαῖς ἴσαι· dron from (five) triangles. And a solid angle cannot be
ὅπερ ἀδύνατον· ἅπασα γὰρ στερεὰ γωνία ὑπὸ ἐλασσόνων (made) from six equilateral and equiangular triangles set
ἢ τεσσάρων ὀρθῶν περέχεται. διὰ τὰ αὐτὰ δὴ οὐδὲ ὑπὸ up together at one point. For, since the angles of a equi-
πλειόνων ἢ ἓξ γωνιῶν ἐπιπέδων στερεὰ γωνία συνίσταται. lateral triangle are (each) two-thirds of a right-angle, the
ὑπὸ δὲ τετραγώνων τριῶν ἡ τοῦ κύβου γωνία περιέχεται· (sum of the) six (plane) angles (containing the solid an-
ὑπὸ δὲ τεσσάρων ἀδύνατον· ἔσονται γὰρ πάλιν τέσσαρες gle) will be four right-angles. The very thing (is) impos-
ὀρθαί. ὑπὸ δὲ πενταγώνων ἰσοπλεύρων καὶ ἰσογωνίων, ὑπὸ sible. For every solid angle is contained by (plane angles
μὲν τριῶν ἡ τοῦ δωδεκαέδρου· ὑπὸ δὲ τεσσάρων ἀδύνατον· whose sum is) less than four right-angles [Prop. 11.21].
οὔσης γὰρ τῆς τοῦ πενταγώνου ἰσοπλεύρου γωνίας ὀρθῆς So, for the same (reasons), a solid angle cannot be con-
καὶ πέμπτου, ἔσονται αἱ τέσσαρες γωνίαι τεσσάρων ὀρθῶν structed from more than six plane angles (equal to two-
μείζους· ὅπερ ἀδύνατον. οὐδὲ μὴν ὑπὸ πολυγώνων ἑτέρων thirds of a right-angle) either. And the (solid) angle of
σχημάτων περισχεθήσεται στερεὰ γωνία διὰ τὸ αὐτὸ ἄτο- a cube is contained by three squares. And (a solid angle
πον. contained) by four (squares is) impossible. For, again, the
Οὐκ ἄρα παρὰ τὰ εἰρημένα πέντε σχήματα ἕτερον σχῆμα (sum of the plane angles containing the solid angle) will

στερεὸν συσταθήσεται ὑπὸ ἰσοπλεύρων τε καὶ ἰσογωνίων be four right-angles. And (the solid angle) of a dodec-
περιεχόμενον· ὅπερ ἔδει δεῖξαι. ahedron (is contained) by three equilateral and equian-

gular pentagons. And (a solid angle contained) by four

537

STOIQEIWN igþ. ELEMENTS BOOK 13
(equiangular pentagons is) impossible. For, the angle of

an equilateral pentagon being one and one-fifth of right-

angle, four (such) angles will be greater (in sum) than
four right-angles. The very thing (is) impossible. And,

on account of the same absurdity, a solid angle cannot
be constructed from any other (equiangular) polygonal

figures either.

Thus, beside the five aforementioned figures, no other
solid figure can be constructed (which is) contained by

equilateral and equiangular (planes). (Which is) the very

thing it was required to show.

Z
D

A
E B

G CD

E B

L¨mma. Lemma
῞Οτι δὲ ἡ τοῦ ἰσοπλεύρου καὶ ἰσογωνίου πενταγώνου It can be shown that the angle of an equilateral and

γωνία ὀρθή ἐστι καὶ πέμπτου, οὕτω δεικτέον. equiangular pentagon is one and one-fifth of a right-
῎Εστω γὰρ πεντάγωνον ἰσόπλευρον καὶ ἰσογώνιον τὸ angle, as follows.

ΑΒΓΔΕ, καὶ περιγεγράφθω περὶ αὐτὸ κύκλος ὁ ΑΒΓΔΕ, For let ABCDE be an equilateral and equiangular
καὶ εἰλήφθω αὐτοῦ τὸ κέντρον τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ pentagon, and let the circle ABCDE have been circum-
ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ. δίχα ἄρα τέμνουσι τὰς πρὸς τοῖς Α, scribed about it [Prop. 4.14]. And let its center, F , have
Β, Γ, Δ, Ε τοῦ πενταγώνου γωνίας. καὶ ἐπεὶ αἱ πρὸς τῷ Ζ been found [Prop. 3.1]. And let FA, FB, FC, FD,
πέντε γωνίαι τέσσαρσιν ὀρθαῖς ἴσαι εἰσὶ καί εἰσιν ἴσαι, μία and FE have been joined. Thus, they cut the angles
ἄρα αὐτῶν, ὡς ἡ ὑπὸ ΑΖΒ, μιᾶς ὀρθῆς ἐστι παρὰ πέμπτον· of the pentagon in half at (points) A, B, C, D, and E
λοιπαὶ ἄρα αἱ ὑπὸ ΖΑΒ, ΑΒΖ μιᾶς εἰσιν ὀρθῆς καὶ πέμπτου. [Prop. 1.4]. And since the five angles at F are equal (in
ἴση δὲ ἡ ὑπὸ ΖΑΒ τῇ ὑπὸ ΖΒΓ· καὶ ὅλη ἄρα ἡ ὑπὸ ΑΒΓ τοῦ sum) to four right-angles, and are also equal (to one an-
πενταγώνου γωνία μιᾶς ἐστιν ὀρθῆς καὶ πέμπτου· ὅπερ ἔδει other), (any) one of them, like AFB, is thus one less a
δεῖξαι. fifth of a right-angle. Thus, the (sum of the) remaining

(angles in triangle ABF ), FAB and ABF , is one plus a

fifth of a right-angle [Prop. 1.32]. And FAB (is) equal

to FBC. Thus, the whole angle, ABC, of the pentagon
is also one and one-fifth of a right-angle. (Which is) the

very thing it was required to show.

538

GREEK-ENGLISH LEXICON

539

STOIQEIWN GREEK–ENGLISH LEXICON
ABBREVIATIONS: act – active; adj – adjective; adv – adverb; conj
– conjunction; fut – future; gen – genitive; imperat – imperative;
impf – imperfect; ind – indeclinable; indic – indicative; intr – in-
transitive; mid – middle; neut – neuter; no – noun; par – particle;
part – participle; pass – passive; perf – perfect; pre – preposition;
pres – present; pro – pronoun; sg – singular; tr – transitive; vb –
verb.�gw, �xw, ¢gagon, -ªqa, ªgmai, ¢qjhn : vb, lead, draw (a line).�dÔnato
-on : adj, impossible.�e� : adv, always, for ever.aÉrèw, aÉr sw, e[Ù℄lon, ¥rhka, ¥rhmai, �rèjhn : vb, grasp.�itèw, aÊt sw, ¢thsa, �thka, �thmai, �t jh : vb, postulate.aÒthma -ato
, tì : no, postulate.�kìloujo
-on : adj, analogous, consequent on, in conformity

with.�kro
-a -on : adj, outermost, end, extreme.�ll� : conj, but, otherwise.�logo
-on : adj, irrational.�ma : adv, at once, at the same time, together.�mblug¸nio
-on : adj, obtuse-angled; tä �mblug¸nion, no, ob-
tuse angle.�mblÔ
-eØa -Ô : adj, obtuse.�mfìtero
-a -on : pro, both.�nagr�fw : vb, describe (a figure); see gr�fw.�nalog�a, � : no, proportion, (geometric) progression.�n�logo
-on : adj, proportional.�n�palin : adv, inverse(ly).anaplhrìw : vb, fill up.�nastrèfw : vb, turn upside down, convert (ratio); see strèfw.�nastrof , � : no, turning upside down, conversion (of ratio).�njufairèw : vb, take away in turn; see aÉrèw.�n�sthmi : vb, set up; see Òsthmi.�niso
-on : adj, unequal, uneven.�ntip�sqw : vb, be reciprocally proportional; see p�sqw.�xwn -ono
, å : vb, axis.�pax : adv, once.�pa
, �pasa, �pan : adj, quite all, the whole.�peiro
-on : adj, infinite.�penant�on : ind, opposite.�pèqw : vb, be far from, be away from; see êqw.�plat
-è
: adj, without breadth.�pìdeixi
-ew
, � : no, proof.�pokaj�sthmi : vb, re-establish, restore; see Òsthmi.�polamb�nw : vb, take from, subtract from, cut off from; seelamb�nw.�potèmnw : vb, cut off, subtend.�pìtmhma -ato
, tì : no, piece cut off, segment.�potom , � : vb, piece cut off, apotome.

�ptw, �yw, ©ya, —, ©mmai, — : vb, touch, join, meet.�p¸tero
-a -on : adj, further off.�ra : par, thus, as it seems (inferential).�rijmì
, å : no, number.�rti�ki
: adv, an even number of times.artiìpleuro
-on : adj, having a even number of sides.�rqw, �rxw, ªrxa, ªrqa, ªrgmai, ªrqjhn : vb, rule; mid., be-
gin.�sÔmmetro
-on : adj, incommensurable.�sÔmptwto
-on : adj, not touching, not meeting.�rtio
-a -on : adj, even, perfect.�tmhto
-on : adj, uncut.�tìpo
-on : adj, absurd, paradoxical.aÎtìjen : adv, immediately, obviously.�fa�rew : vb, take from, subtract from, cut off from; see aÉrèw.�f , � : no, point of contact.b�jo
-eo
, tì : no, depth, height.ba�nw, -b somai, -èbhn, bèbhka, —, — : vb, walk; perf, stand
(of angle).b�llw, balÀ, êbalon, bèblhka, bèblhmai, âbl jhn : vb, throw.b�si
-ew
, � : no, base (of a triangle).g�r : conj, for (explanatory).g�[g℄nomai, gen somai, âgenìmhn, gègona, gegènhmai, — : vb,
happen, become.gn¸mwn -ono
, � : no, gnomon.gramm , � : no, line.gr�fw, gr�yw, êgra[y/f℄a, gègrafa, gègrammai, âray�mhn : vb,
draw (a figure).gwn�a, � : no, angle.deØ : vb, be necessary; deØ, it is necessary; êdei, it was necassary;dèon, being necessary.de�knumi, de�xw, êdeixa, dèdeiqa, dèdeigmai, âde�qjhn : vb, show,
demonstrate.deiktèon : ind, one must show.deØxi
-ew
, � : no, proof.dekag¸no
-on : adj, ten-sided; tä dekag¸non, no, decagon.dèqomai, dèxomai, âdex�mhn, —, dèdegmai, âdèqjhn : vb, receive,
accept.d : conj, so (explanatory).dhlad : ind, quite clear, manifest.d¨lo
-h -on : adj, clear.dhlonìti : adv, manifestly.di�gw : vb, carry over, draw through, draw across; see �gw.diag¸nio
-on : adj, diagonal.diale�pw : vb, leave an interval between.di�metro
-on : adj, diametrical; � di�metro
, no, diameter, di-
agonal.dia�resi
-ew
, � : no, division, separation.

540

STOIQEIWN GREEK–ENGLISH LEXICONdiairèw : vb, divide (in two); diarejènto
-h -on, adj, separated
(ratio); see aÉrèw.di�sthma -ato
, tì : no, radius.diafèrw : vb, differ; see fèrw.d�dwmi, d¸sw, êdwka, dèdwka, dèdomai, âdìjhn : vb, give.dimo�ro
-on : adj, two-thirds.diplasi�zw : vb, double.dipl�sio
-a -on : adj, double, twofold.diplas�wn -on : adj, double, twofold.diploÜ
-¨ -oÜn : adj, double.d�
: adv, twice.d�qa : adv, in two, in half.diqorom�a, � : no, point of bisection.du�
-�do
, � : no, the number two, dyad.dÔnamai : vb, be able, be capable, generate, square, be when
squared; dunamènh, �, no, square-root (of area)—i.e., strai-
ght-line whose square is equal to a given area.dÔnami
-ew
, � : no, power (usually 2nd power when used in
mathematical sence, hence), square.dunatì
– -ìn : adj, possible.dwdek�edro
-on : adj, twelve-sided.áautoÜ -¨
-oÜ : adj, of him/her/it/self, his/her/its/own.âgg�wn -on : adj, nearer, nearest.âggr�fw : vb, inscribe; see gr�fw.eÚdo
-eo
, tì : no, figure, form, shape.eÊkos�edro
-on : adj, twenty-sided.eÒrw/lègw, ârÀ/erèw, eÚpon, eÒrhka, eÒrhmai, ârr jhn : vb,
say, speak; per pass part, eirhmèno
-h -on, adj, said, afore-
mentioned.eÒte . . . eÒte : ind, either . . . or.ékasto
-h -on : pro, each, every one.ákatèro
-a -on : pro, each (of two).âkb�llw : vb, produce (a line); see b�llw.âkjèw : vb, set out.êkkeimai : vb, be set out, be taken; see keØmai.âkt�jhmi : vb, set out; see t�jhmi.âktì
: pre + gen, outside, external.âl�sswn/âl�ttwn -on : adj, less, lesser.âl�qisto
-h -on : adj, least.âlle�pw : vb, be less than, fall short of.âmp�ptw : vb, meet (of lines), fall on; see p�ptw.êmprosjen : adv, in front.ânall�x : adv, alternate(ly).ânarmìzw : vb, insert; perf indic pass 3rd sg, ân rmostai.ândèqomai : vb, admit, allow.éneken : ind, on account of, for the sake of.ânnapl�sio
-a -on : adj, nine-fold, nine-times.ênnoia, � : no, notion.

enperièqw : vb, encompass; see êqw.ânp�ptw : see âmp�ptw.ântì
: pre + gen, inside, interior, within, internal.áx�gwno
-on : adj, hexagonal; tä áx�gwnon, no, hexagon.áxapl�sio
-a -on : adj, sixfold.áx¨
: adv, in order, successively, consecutively.êxwjen : adv, outside, extrinsic.âp�nw : adv, above.âpaf , � : no, point of contact.âpe� : conj, since (causal).âpeid per : ind, inasmuch as, seeing that.âpizeÔgnÜmi, âpizeÔxw, âpèzeuxa, —, âpèzeugmai, âpèzeÔqjhn :
vb, join (by a line).âpilog�zomai : vb, conclude.âpinoèw : vb, think of, contrive.âpipèdo
-on : adj, level, flat, plane; tä âpipèdon, no, plane.âpiskèptomai : vb, investigate.âp�skeyi
-ew
, � : no, inspection, investigation.âpit�ssw : vb, put upon, enjoin; tä âpitaqjèn, no, the (thing)
prescribed; see t�ssw.âp�trito
-on : adj, one and a third times.âpif�neia, � : no, surface.êpomai : vb, follow.êrqomai, âleÔsomai, ªljon, âl luja, —, — : vb, come, go.êsqato
-h -on : adj, outermost, uttermost, last.âterìmhkh
-e
: adj, oblong; tä âterìmhke
, no, rectangle.étero
-a -on : adj, other (of two).êti : par, yet, still, besides.eÎjÔgrammo
-on : adj, rectilinear; tä eÎjÔgrammon, no, recti-
linear figure.eÎjÔ
-eØa -Ô : adj, straight; � eÎjeØa, no, straight-line; âp>eÎjeØa
, in a straight-line, straight-on.eÍr�skw, eÍr skw, hÝron, eÕreka, eÕrhmai, eÍrèjhn : vb, find.âf�ptw : vb, bind to; mid, touch; � áfaptomènh, no, tangent;
see �ptw.âfarmìzw, âfarmìsw, âf rmosa, âf moka, âf mosmai, âf mìsjhn
: vb, coincide; pass, be applied.âfex¨
: adv, in order, adjacent.âf�sthmi : vb, set, stand, place upon; see Òsthmi.êqw, éxw, êsqon, êsqhka, -èsqhmai, — : vb, have.�gèomai, �g somai, �ghs�mhn, �ghmai, —, �g jhn : vb, lead.¢dh : ind, already, now.¡kw, ¡xw, —, —, —, — : vb, have come, be present.�mikÔklion, tì : no, semi-circle.�miìlio
-a -on : adj, containing one and a half, one and a half
times.¡misu
-eia -u : adj, half.¢per = ¢ + per : conj, than, than indeed.

541

STOIQEIWN GREEK–ENGLISH LEXICON¢toi . . . ¢ : par, surely, either . . . or; in fact, either . . . or.jèsi
-ew
, � : no, placing, setting, position.jewrhma -ato
, tì : no, theorem.Òdio
-a -on : adj, one’s own.Ês�ki
: adv, the same number of times; Ês�ki
pollapl�sia,
the same multiples, equal multiples.Êsog¸nio
-on : adj, equiangular.Êsìpleuro
-on : adj, equilateral.Êsoplhj
-è
: adj, equal in number.Òso
-h -on : adj, equal; âx Òsou, equally, evenly.Êsoskel
-è
: adj, isosceles.Òsthmi, st sw, êsthsa, —, —, âst�jhn : vb tr, stand (some-
thing).Òsthmi, st sw, êsthn, ésthka, éstamai, âstajhn : vb intr, stand
up (oneself); Note: perfect I have stood up can be taken
to mean present I am standing.Êso�y
-è
: adj, of equal height.kaj�per : ind, according as, just as.k�jeto
-on : adj, perpendicular.kajìlou : adv, on the whole, in general.kalèw : vb, call.k�keino
= kaÈ âkeØno
.k�n = kaÈ �n : ind, even if, and if.katagraf , � : no, diagram, figure.katagr�fw : vb, describe/draw, inscribe (a figure); see gr�fw.katakoloujèw : vb, follow after.katale�pw : vb, leave behind; see le�pw; t� kataleipìmena, no,
remainder.kat�llhlo
-on : adj, in succession, in corresponding order.katametrèw : vb, measure (exactly).katant�w : vb, come to, arrive at.kataskeu�zw : vb, furnish, construct.keØmai, keØsomai, —, —, —, — : vb, have been placed, lie, be
made; see t�jhmi.kèntron, tì : no, center.kl�w : vb, break off, inflect.kl�nw, kl�nw, êklina, kèklika, kèklimai, âkl�jhn : vb, lean, in-
cline.kl�si
-ew
, � : no, inclination, bending.koØlo
-h -on : adj, hollow, concave.koruf , � : no, top, summit, apex; kat� koruf n, vertically
opposite (of angles).kr�nw, krinÀ, êkrØna, kèkrika, kèkrimai, âkr�jhn : vb, judge.kÔbo
, å : no, cube.kÔklo
, å : no, circle.kÔlindro
, å : no, cylinder.kurtì
– -ìn : adj, convex.kÀno
, å : no, cone.

lamb�nw, l yomai, êlabon, eÒlhfa eÒlhmmai, âl fjhn : vb,
take.lègw : vb, say; pres pass part, legìmeno
-h -on, adj, so-called;
see êirw.le�pw, le�yw, êlipon, lèloipa, lèleimmai, âle�fjhn : vb, leave,
leave behind.lhmm�tion, tì : no, diminutive of l¨mma.l¨mma -ato
, tì : no, lemma.l¨yi
-ew
, � : no, taking, catching.lìgo
, å : no, ratio, proportion, argument.loipì
– -ìn : adj, remaining.manj�nw, maj somai, êmajon, mem�jhka, —, — : vb, learn.mègejo
-eo
, tì : no, magnitude, size.me�zwn -on : adj, greater.mènw, menÀ, êmeina, memènhka, —, — : vb, stay, remain.mèro
-ou
, tì : no, part, direction, side.mèso
-h -on : adj, middle, mean, medial; âk dÔo mèswn, bime-
dial.metalamb�nw : vb, take up.metaxÔ : adv, between.metèwro
-on : adj, raised off the ground.metrèw : vb, measure.mètron, tì : no, measure.mhde�
, mhdem�a, mhdèn : adj, not even one, (neut.) nothing.mhdèpote : adv, never.mhdètero
-a -on : pro, neither (of two).m¨ko
-eo
, tì : no, length.m n : par, truely, indeed.mon�
-�do
, � : no, unit, unity.monaqì
– -ìn : adj, unique.monaqÀ
: adv, uniquely.mìno
-h -on : adj, alone.noèw, —, nìhsa, nenìhka, nenìhmai, âno jhn : vb, apprehend,
conceive.oÙo
-a -on : pre, such as, of what sort.ækt�edro
-on : adj, eight-sided.ílo
-h -on : adj, whole.åmogen
-è
: adj, of the same kind.ímoio
-a -on : adj, similar.åmoioplhj
-è
: adj, similar in number.åmoiotag
-è
: adj, similarly arranged.åmoiìth
-hto
, � : no similarity.åmo�w
: adv, similarly.åmìlogo
-on : adj, corresponding, homologous.åmotag
-è
: adj, ranged in the same row or line.åm¸numo
-on : adj, having the same name.înoma -ato
, tì : no, name; âk dÔo ænom�twn, binomial.

542

STOIQEIWN GREEK–ENGLISH LEXICONæxug¸nio
-on : adj, acute-angled; tä æxug¸nion, no, acute an-
gle.æxÔ
-eØa -Ô : adj, acute.åpoiosoÜn = åpoØo
-a -on + oÞn : adj, of whatever kind, any
kind whatsoever.åpìso
-h -on : pro, as many, as many as.åpososdhpotoÜn = åpìso
-h -on + d + potè + oÞn : adj, of
whatever number, any number whatsoever.åpososoÜn = åpìso
-h -on + oÞn : adj, of whatever number,
any number whatsoever.åpìtero
-a -on : pro, either (of two), which (of two).ærjog¸nion, tì : no, rectangle, right-angle.ærjì
– -ìn : adj, straight, right-angled, perpendicular; prä
ærj�
gwn�a
, at right-angles.îro
, å : no, boundary, definition, term (of a ratio).åsadhpotoÜn = ísa + d + potè + oÞn : ind, any number
whatsoever.ås�ki
: ind, as many times as, as often as.åsapl�sio
-on : pro, as many times as.íso
-h -on : pro, as many as.ísper, ¡per, íper : pro, the very man who, the very thing
which.ísti
, ¡ti
, í ti : pro, anyone who, anything which.ítan : adv, when, whenever.åtioÜn : ind, whatsoever.oÎde�
, oÎdem�a, oÎdèn : pro, not one, nothing.oÍdètero
-a -on : pro, not either.oÍjètero
: see oÍdètero
.oÎjèn : ind, nothing.oÞn : adv, therefore, in fact.oÕtw
: adv, thusly, in this case.p�lin : adv, back, again.p�ntw
: adv, in all ways.par� : prep + acc, parallel to.parab�llw : vb, apply (a figure); see b�llw.parabol , � : no, application.par�keimai : vb, lie beside, apply (a figure); see keØmai.parall�ssw, parall�xw, —, par llaqa, —, — : vb, miss, fall
awry.parallhlep�pedo
, -on : adj, with parallel surfaces; tä paral-lhlep�pedon, no, parallelepiped.parallhlìgrammo
-on : adj, bounded by parallel lines; tä pa-rallhlìgrammon, no, parallelogram.par�llhlo
-on : adj, parallel; tä par�llhlon, no, parallel,
parallel-line.parapl rwma -ato
, tì : no, complement (of a parallelogram).paratèlueto
-on : adj, penultimate.parèk : prep + gen, except.paremp�ptw : vb, insert; see p�ptw.

p�sqw, pe�somai, êpajon, pèponja, —, — : vb, suffer.pent�gwno
-on : adj, pentagonal; tä pent�gwnon, no, pen-
tagon.pentapl�sio
-a -on : adj, five-fold, five-times.pentekaidek�gwnon, tì : no, fifteen-sided figure.peperasmèno
-h -on : adj, finite, limited; see pera�nw.pera�nw, peranÀ, âpèrana, —, peper�smai, âperan�njhn : vb, bring
to end, finish, complete; pass, be finite.pèra
-ato
, tì : no, end, extremity.peratìw, —, —, —, —, — : vb, bring to an end.perigr�fw : vb, circumscribe; see gr�fw.perièqw : vb, encompass, surround, contain, comprise; see êqw.perilamb�nw : vb, enclose; see lamb�nw.periss�ki
: adv, an odd number of times.perissì
– -ìn : adj, odd.perifèreia, � : no, circumference.perifèrw : vb, carry round; see fèrw.phlikìth
-hto
, � : no, magnitude, size.p�ptw, pesoÜmai, êpeson, pèptwka, —, — : vb, fall.pl�to
-eo
, tì : no, breadth, width.ple�wn -on : adj, more, several.pleur�, � : no, side.pl¨jo
-eo
, tì : no, great number, multitude, number.pl n : adv & prep + gen, more than.poiì
-� -ìn : adj, of a certain nature, kind, quality, type.pollaplasi�zw : vb, multiply.pollaplasiasmì
, å : no, multiplication.pollapl�sion, tì : no, multiple.polÔedro
-on : adj, polyhedral; tì polÔedron, no, polyhedron.polÔgwno
-on : adj, polygonal; tì polÔgwnon, no, polygon.polÔpleuro
-on : adj, multilateral.pìrisma -ato
, tì : no, corollary.potè : ind, at some time.prØsma -ato
, tì : no, prism.proba�nw : vb, step forward, advance.prode�knumi : vb, show previously; see de�knumi.proekt�jhmi : vb, set forth beforehand; see t�jhmi.proerèw : vb, say beforehand; perf pass part, proeirhmèno
-h-on, adj, aforementioned; see eÒrw.prosanaplhrìw : vb, fill up, complete.prosanagr�fw : vb, complete (tracing of); see gr�fw.prosarmìzw : vb, fit to, attach to.prosekb�llw : vb, produce (a line); see âkb�llw.proseur�skw : vb, find besides, find; see eÍr�skw.proslamb�mw : vb, add.prìkeimai : vb, set before, prescribe; see keØmai.prìskeimai : vb, be laid on, have been added to; see keØmai.

543

STOIQEIWN GREEK–ENGLISH LEXICONprosp�ptw : vb, fall on, fall toward, meet; see p�ptw.protasi
-ew
, � : no, proposition.prost�ssw : vb, prescribe, enjoin; tä trostaqjèn, no, the
thing prescribed; see t�ssw.prost�jhmi : vb, add; see t�jhmi.prìtero
-a -on : adj, first (comparative), before, former.prot�jhmi : vb, assign; see t�jhmi.proqwrèw : vb, go/come forward, advance.prÀto
-a -on : adj, first, prime.puram�
-�do
, � : no, pyramid.ûhtì
– -ìn : adj, expressible, rational.ûomboeid
-è
: adj, rhomboidal; tä ûomboeidè
, no, romboid.ûìmbo
, å no, rhombus.shmeØon, tì : no, point.skalhnì
– -ìn : adj, scalene.stereì
-� -ìn : adj, solid; tä stereìn, no, solid, solid body.stoiqeØon, tì : no, element.strèfw, -strèyw, êstreya, —, âstammai, âst�fhn : vb, turn.sÔgkeimai : vb, lie together, be the sum of, be composed;sugke�meno
-h -on, adj, composed (ratio), compounded;
see keØmai.sÔgkr�nw : vb, compare; see kr�nw.sumba�nw : vb, come to pass, happen, follow; see ba�nw.sumb�llw : vb, throw together, meet; see b�llw.sÔmmetro
-on : adj, commensurable.sÔmpa
-anto
, å : no, sum, whole.sump�ptw : vb, meet together (of lines); see p�ptw.sumplhrìw : vb, complete (a figure), fill in.sun�gw : vb, conclude, infer; see �gw.sunamfìteroi -ai -a : adj, both together; å sunamfìtero
, no,
sum (of two things).sunapode�knumi : no, demonstrate together; see de�knumi.sunaf , � : no, point of junction.sÔnduo, oÉ, aÉ, t� : no, two together, in pairs.suneq
-è
: adj, continuous; kat� tä suneqè
, continuously.sÔnjesi
-ew
, � : no, putting together, composition.sÔnjeto
-on : adj, composite.su[n℄�sthmi : vb, construct (a figure), set up together; perf im-
perat pass 3rd sg, sunest�tw; see Òsthmi.sunt�jhmi : vb, put together, add together, compound (ratio);
see t�jhmi.sqèsi
-ew
, � : no, state, condition.sq¨ma -ato
, tì : no, figure.sfaØra -a
, � : no, sphere.t�xi
-ew
, � : no, arrangement, order.tar�ssw, tar�xw, —, —, tet�ragmai, âtar�qjhn : vb, stir, trou-
ble, disturbe; tetaragmèno
-h -on, adj, disturbed, per-
turbed.

t�ssw, t�xw, êtaxa, tètaqa, tètagmai, ât�qjhn : vb, arrange,
draw up.tèleio
-a -on : adj, perfect.tèmnw, temnÀ, êtemon, -tètmhka, tètmhmai, âtm jhn : vb, cut;
pres/fut indic act 3rd sg, tèmei.tetarthmorion, tì : no, quadrant.tetr�gwno
-on : adj, square; tä tetr�gwnon, no, square.tetr�ki
: adv, four times.tetrapl�sio
-a -on : adj, quadruple.tetr�pleuro
-on : adj, quadrilateral.tetraplìo
-h -on : adj, fourfold.t�jhmi, j sw, êjhka, tèjhka, keØmai, âtèjhn : vb, place, put.tm¨ma -ato
, tì : no, part cut off, piece, segment.to�nun : par, accordingly.toioÜto
-aÔth -oÜto : pro, such as this.tomeÔ
-èw
, å : no, sector (of circle).tom , � : no, cutting, stump, piece.tìpo
, å : no, place, space.tosaut�ki
: adv, so many times.tosautapl�sio
-a -on : pro, so many times.tosoÜto
-aÔth -oÜto : pro, so many.toutèsti = toÜt> êsti : par, that is to say.trapèzion, tì : no, trapezium.tr�gwno
-on : adj, triangular; tä tr�gwnon, no, triangle.tripl�sio
-a -on : adj, triple, threefold.tr�pleuro
-on : adj, trilateral.tripl-ìo
-h -on : adj, triple.trìpo
, å : no, way.tugq�nw, teÔxomai, êtuqon, tetÔqhka, tèteugmai, âteÔqjhn :
vb, hit, happen to be at (a place).Íp�rqw : vb, begin, be, exist; see �rqw.Ípexa�resi
-ew
, � : no, removal.Íperb�llw : vb, overshoot, exceed; see b�llw.Íperoq , � : no, excess, difference.Íperèqw : vb, exceed; see êqw.Ípìjesi
-ew
, � : no, hypothesis.Ípìkeimai : vb, underlie, be assumed (as hypothesis); see keØmai.Ípole�pw : vb, leave remaining.Ípote�nw, ÍpotenÀ, Ípèteina, Ípotètaka, Ípotètamai, Ípet�jhn
: vb, subtend.Õyo
-eo
, tì : no, height.fanerì
-� -ìn : adj, visible, manifest.fhmÈ, f sw, êfhn, —, —, — : vb, say; êfamen, we said.fèrw, oÒsw, ¢negkon, ân noqa, ân negmai, �nèqjhn : vb, carry.q¸rion, tì : no, place, spot, area, figure.qwr�
: pre + gen, apart from.yaÔw : vb, touch.±
: par, as, like, for instance.±
étuqen : par, at random.±saÔtw
: adv, in the same manner, just so.¹ste : conj, so that (causal), hence.

544

Introduction
Book 1
Definitions
Proposition 1.1
Proposition 1.2
Proposition 1.3
Proposition 1.4
Proposition 1.5
Proposition 1.6
Proposition 1.7
Proposition 1.8
Proposition 1.9
Proposition 1.10
Proposition 1.11
Proposition 1.12
Proposition 1.13
Proposition 1.14
Proposition 1.15
Proposition 1.16
Proposition 1.17
Proposition 1.18
Proposition 1.19
Proposition 1.20
Proposition 1.21
Proposition 1.22
Proposition 1.23
Proposition 1.24
Proposition 1.25
Proposition 1.26
Proposition 1.27
Proposition 1.28
Proposition 1.29
Proposition 1.30
Proposition 1.31
Proposition 1.32
Proposition 1.33
Proposition 1.34
Proposition 1.35
Proposition 1.36
Proposition 1.37
Proposition 1.38
Proposition 1.39
Proposition 1.40
Proposition 1.41
Proposition 1.42
Proposition 1.43
Proposition 1.44
Proposition 1.45
Proposition 1.46
Proposition 1.47
Proposition 1.48

Book 2
Definitions
Proposition 2.1
Proposition 2.2
Proposition 2.3
Proposition 2.4
Proposition 2.5
Proposition 2.6
Proposition 2.7
Proposition 2.8
Proposition 2.9
Proposition 2.10
Proposition 2.11
Proposition 2.12
Proposition 2.13
Proposition 2.14

Book 3
Definitions
Proposition 3.1
Proposition 3.2
Proposition 3.3
Proposition 3.4
Proposition 3.5
Proposition 3.6
Proposition 3.7
Proposition 3.8
Proposition 3.9
Proposition 3.10
Proposition 3.11
Proposition 3.12
Proposition 3.13
Proposition 3.14
Proposition 3.15
Proposition 3.16
Proposition 3.17
Proposition 3.18
Proposition 3.19
Proposition 3.20
Proposition 3.21
Proposition 3.22
Proposition 3.23
Proposition 3.24
Proposition 3.25
Proposition 3.26
Proposition 3.27
Proposition 3.28
Proposition 3.29
Proposition 3.30
Proposition 3.31
Proposition 3.32
Proposition 3.33
Proposition 3.34
Proposition 3.35
Proposition 3.36
Proposition 3.37

Book 4
Definitions
Proposition 4.1
Proposition 4.2
Proposition 4.3
Proposition 4.4
Proposition 4.5
Proposition 4.6
Proposition 4.7
Proposition 4.8
Proposition 4.9
Proposition 4.10
Proposition 4.11
Proposition 4.12
Proposition 4.13
Proposition 4.14
Proposition 4.15
Proposition 4.16

Book 5
Definitions
Proposition 5.1
Proposition 5.2
Proposition 5.3
Proposition 5.4
Proposition 5.5
Proposition 5.6
Proposition 5.7
Proposition 5.8
Proposition 5.9
Proposition 5.10
Proposition 5.11
Proposition 5.12
Proposition 5.13
Proposition 5.14
Proposition 5.15
Proposition 5.16
Proposition 5.17
Proposition 5.18
Proposition 5.19
Proposition 5.20
Proposition 5.21
Proposition 5.22
Proposition 5.23
Proposition 5.24
Proposition 5.25

Book 6
Definitions
Proposition 6.1
Proposition 6.2
Proposition 6.3
Proposition 6.4
Proposition 6.5
Proposition 6.6
Proposition 6.7
Proposition 6.8
Proposition 6.9
Proposition 6.10
Proposition 6.11
Proposition 6.12
Proposition 6.13
Proposition 6.14
Proposition 6.15
Proposition 6.16
Proposition 6.17
Proposition 6.18
Proposition 6.19
Proposition 6.20
Proposition 6.21
Proposition 6.22
Proposition 6.23
Proposition 6.24
Proposition 6.25
Proposition 6.26
Proposition 6.27
Proposition 6.28
Proposition 6.29
Proposition 6.30
Proposition 6.31
Proposition 6.32
Proposition 6.33

Book 7
Definitions
Proposition 7.1
Proposition 7.2
Proposition 7.3
Proposition 7.4
Proposition 7.5
Proposition 7.6
Proposition 7.7
Proposition 7.8
Proposition 7.9
Proposition 7.10
Proposition 7.11
Proposition 7.12
Proposition 7.13
Proposition 7.14
Proposition 7.15
Proposition 7.16
Proposition 7.17
Proposition 7.18
Proposition 7.19
Proposition 7.20
Proposition 7.21
Proposition 7.22
Proposition 7.23
Proposition 7.24
Proposition 7.25
Proposition 7.26
Proposition 7.27
Proposition 7.28
Proposition 7.29
Proposition 7.30
Proposition 7.31
Proposition 7.32
Proposition 7.33
Proposition 7.34
Proposition 7.35
Proposition 7.36
Proposition 7.37
Proposition 7.38
Proposition 7.39

Book 8
Proposition 8.1
Proposition 8.2
Proposition 8.3
Proposition 8.4
Proposition 8.5
Proposition 8.6
Proposition 8.7
Proposition 8.8
Proposition 8.9
Proposition 8.10
Proposition 8.11
Proposition 8.12
Proposition 8.13
Proposition 8.14
Proposition 8.15
Proposition 8.16
Proposition 8.17
Proposition 8.18
Proposition 8.19
Proposition 8.20
Proposition 8.21
Proposition 8.22
Proposition 8.23
Proposition 8.24
Proposition 8.25
Proposition 8.26
Proposition 8.27

Book 9
Proposition 9.1
Proposition 9.2
Proposition 9.3
Proposition 9.4
Proposition 9.5
Proposition 9.6
Proposition 9.7
Proposition 9.8
Proposition 9.9
Proposition 9.10
Proposition 9.11
Proposition 9.12
Proposition 9.13
Proposition 9.14
Proposition 9.15
Proposition 9.16
Proposition 9.17
Proposition 9.18
Proposition 9.19
Proposition 9.20
Proposition 9.21
Proposition 9.22
Proposition 9.23
Proposition 9.24
Proposition 9.25
Proposition 9.26
Proposition 9.27
Proposition 9.28
Proposition 9.29
Proposition 9.30
Proposition 9.31
Proposition 9.32
Proposition 9.33
Proposition 9.34
Proposition 9.35
Proposition 9.36

Book 10
Definitions I
Proposition 10.1
Proposition 10.2
Proposition 10.3
Proposition 10.4
Proposition 10.5
Proposition 10.6
Proposition 10.7
Proposition 10.8
Proposition 10.9
Proposition 10.10
Proposition 10.11
Proposition 10.12
Proposition 10.13
Proposition 10.14
Proposition 10.15
Proposition 10.16
Proposition 10.17
Proposition 10.18
Proposition 10.19
Proposition 10.20
Proposition 10.21
Proposition 10.22
Proposition 10.23
Proposition 10.24
Proposition 10.25
Proposition 10.26
Proposition 10.27
Proposition 10.28
Proposition 10.29
Proposition 10.30
Proposition 10.31
Proposition 10.32
Proposition 10.33
Proposition 10.34
Proposition 10.35
Proposition 10.36
Proposition 10.37
Proposition 10.38
Proposition 10.39
Proposition 10.40
Proposition 10.41
Proposition 10.42
Proposition 10.43
Proposition 10.44
Proposition 10.45
Proposition 10.46
Proposition 10.47
Definitions II
Proposition 10.48
Proposition 10.49
Proposition 10.50
Proposition 10.51
Proposition 10.52
Proposition 10.53
Proposition 10.54
Proposition 10.55
Proposition 10.56
Proposition 10.57
Proposition 10.58
Proposition 10.59
Proposition 10.60
Proposition 10.61
Proposition 10.62
Proposition 10.63
Proposition 10.64
Proposition 10.65
Proposition 10.66
Proposition 10.67
Proposition 10.68
Proposition 10.69
Proposition 10.70
Proposition 10.71
Proposition 10.72
Proposition 10.73
Proposition 10.74
Proposition 10.75
Proposition 10.76
Proposition 10.77
Proposition 10.78
Proposition 10.79
Proposition 10.80
Proposition 10.81
Proposition 10.82
Proposition 10.83
Proposition 10.84
Definitions III
Proposition 10.85
Proposition 10.86
Proposition 10.87
Proposition 10.88
Proposition 10.89
Proposition 10.90
Proposition 10.91
Proposition 10.92
Proposition 10.93
Proposition 10.94
Proposition 10.95
Proposition 10.96
Proposition 10.97
Proposition 10.98
Proposition 10.99
Proposition 10.100
Proposition 10.101
Proposition 10.102
Proposition 10.103
Proposition 10.104
Proposition 10.105
Proposition 10.106
Proposition 10.107
Proposition 10.108
Proposition 10.109
Proposition 10.110
Proposition 10.111
Proposition 10.112
Proposition 10.113
Proposition 10.114
Proposition 10.115

Book 11
Definitions
Proposition 11.1
Proposition 11.2
Proposition 11.3
Proposition 11.4
Proposition 11.5
Proposition 11.6
Proposition 11.7
Proposition 11.8
Proposition 11.9
Proposition 11.10
Proposition 11.11
Proposition 11.12
Proposition 11.13
Proposition 11.14
Proposition 11.15
Proposition 11.16
Proposition 11.17
Proposition 11.18
Proposition 11.19
Proposition 11.20
Proposition 11.21
Proposition 11.22
Proposition 11.23
Proposition 11.24
Proposition 11.25
Proposition 11.26
Proposition 11.27
Proposition 11.28
Proposition 11.29
Proposition 11.30
Proposition 11.31
Proposition 11.32
Proposition 11.33
Proposition 11.34
Proposition 11.35
Proposition 11.36
Proposition 11.37
Proposition 11.38
Proposition 11.39

Book 12
Proposition 12.1
Proposition 12.2
Proposition 12.3
Proposition 12.4
Proposition 12.5
Proposition 12.6
Proposition 12.7
Proposition 12.8
Proposition 12.9
Proposition 12.10
Proposition 12.11
Proposition 12.12
Proposition 12.13
Proposition 12.14
Proposition 12.15
Proposition 12.16
Proposition 12.17
Proposition 12.18

Book 13
Proposition 13.1
Proposition 13.2
Proposition 13.3
Proposition 13.4
Proposition 13.5
Proposition 13.6
Proposition 13.7
Proposition 13.8
Proposition 13.9
Proposition 13.10
Proposition 13.11
Proposition 13.12
Proposition 13.13
Proposition 13.14
Proposition 13.15
Proposition 13.16
Proposition 13.17
Proposition 13.18

Greek-English Lexicon

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