EUCLID S ELEMENTS OF GEOMETRY

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1 ULI S LMNTS O GOMTRY The Greek text of J.L. Heiberg ( ) from uclidis lementa, edidit et Latine interpretatus est I.L. Heiberg, in aedibus.g. Teubneri, edited, and provided with a modern nglish translation, by Richard itzpatrick

2 irst edition Revised and corrected ISN

3 ontents Introduction 4 ook 1 5 ook 2 49 ook 3 69 ook ook ook ook ook ook ook ook ook ook Greek-nglish Lexicon 539

4 Introduction uclid s lements 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 lexandria around 300. The main subjects of the work are geometry, proportion, and number theory. Most of the theorems appearing in the lements were not discovered by uclid himself, but were the work of earlier Greek mathematicians such as Pythagoras (and his school), Hippocrates of hios, Theaetetus of thens, and udoxus of nidos. However, uclid 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. uclid is also credited with devising a number of particularly ingenious proofs of previously discovered theorems: e.g., Theorem 48 in ook 1. The geometrical constructions employed in the lements are restricted to those which can be achieved using a straight-rule and a compass. urthermore, 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 lements consists of thirteen books. ook 1 outlines the fundamental propositions of plane geometry, including 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. ook 2 is commonly said to deal with geometric algebra, since most of the theorems contained within it have simple algebraic interpretations. ook 3 investigates circles and their properties, and includes theorems on tangents and inscribed angles. ook 4 is concerned with regular polygons inscribed in, and circumscribed around, circles. ook 5 develops the arithmetic theory of proportion. ook 6 applies the theory of proportion to plane geometry, and contains theorems on similar figures. ook 7 deals with elementary number theory: e.g., prime numbers, greatest common denominators, etc. ook 8 is concerned with geometric series. ook 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. ook 10 attempts to classify incommensurable (i.e., irrational) magnitudes using the so-called method of exhaustion, an ancient precursor to integration. ook 11 deals with the fundamental propositions of three-dimensional geometry. ook 12 calculates the relative volumes of cones, pyramids, cylinders, and spheres using the method of exhaustion. inally, ook 13 investigates the five so-called Platonic solids. This edition of uclid s lements presents the definitive Greek text i.e., that edited by J.L. Heiberg ( ) accompanied by a modern nglish translation, as well as a Greek-nglish lexicon. Neither the spurious books 14 and 15, nor the extensive scholia which have been added to the lements 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 nglish) 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 nglish) indicates material which is implied, but not actually present, in the Greek text. My thanks to Mariusz Wodzicki (erkeley) for typesetting advice, and to Sam Watson & Jonathan enno (U. Mississippi), and Gregory Wong (US) for pointing out a number of errors in ook 1. 4

5 LMNTS OOK 1 undamentals of Plane Geometry Involving Straight-Lines 5

6 ÎÇÖÓ. efinitions αʹ.σημεῖόνἐστιν,οὗμέροςοὐθέν. 1. point is that of which there is no part. βʹ.ραμμὴδὲμῆκοςἀπλατές. 2. nd a line is a length without breadth. γʹ.ραμμῆςδὲπέρατασημεῖα. 3. nd the extremities of a line are points. δʹ.ὐθεῖαγραμμήἐστιν,ἥτιςἐξἴσουτοῖςἐφ ἑαυτῆς 4. straight-line is (any) one which lies evenly with σημείοιςκεῖται. points on itself. εʹ. πιφάνειαδέἐστιν,ὃμῆκοςκαὶπλάτοςμόνονἔχει. 5. nd a surface is that which has length and breadth ϛʹ. πιφανείας δὲ πέρατα γραμμαί. only. ζʹ. πίπεδος ἐπιφάνειά ἐστιν, ἥτις ἐξ ἴσου ταῖς ἐφ 6. nd the extremities of a surface are lines. ἑαυτῆςεὐθείαιςκεῖται. 7. plane surface is (any) one which lies evenly with ηʹ. πίπεδος δὲ γωνία ἐστὶν ἡ ἐν ἐπιπέδῳ δύο γραμμῶν the straight-lines on itself. ἁπτομένων ἀλλήλων καὶ μὴ ἐπ εὐθείας κειμένων πρὸς 8. nd 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. Οτανδὲαἱπεριέχουσαιτὴνγωνίανγραμμαὶεὐθεῖαι and are not lying in a straight-line. ὦσιν, εὐθύγραμμος καλεῖται ἡ γωνία. 9. nd 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. nd 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. Οξεῖαδὲἡἐλάσσωνὀρθῆς. upon which it stands. ιγʹ. Οροςἐστίν,ὅτινόςἐστιπέρας. 11. n obtuse angle is one greater than a right-angle. ιδʹ.σχῆμάἐστιτὸὑπότινοςἤτινωνὅρωνπεριεχόμενον. 12. nd an acute angle (is) one less than a right-angle. ιεʹ. Κύκλος ἐστὶ σχῆμα ἐπίπεδον ὑπὸ μιᾶς γραμμῆς 13. boundary is that which is the extremity of someπεριεχόμενον [ἣ καλεῖται περιφέρεια], πρὸς ἣν ἀφ ἑνὸς thing. σημείου τῶν ἐντὸς τοῦ σχήματος κειμένων πᾶσαι αἱ 14. figure is that which is contained by some boundπροσπίπτουσαι εὐθεῖαι[πρὸς τὴν τοῦ κύκλου περιφέρειαν] ary or boundaries. ἴσαι ἀλλήλαις εἰσίν. 15. 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. nd the point is called the center of the circle. ιηʹ. Ημικύκλιον δέ ἐστι τὸ περιεχόμενον σχῆμα ὑπό τε 17. nd 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. (nd) any κύκλουἐστίν. such (straight-line) also cuts the circle in half. ιθʹ. Σχήματα εὐθύγραμμά ἐστι τὰ ὑπὸ εὐθειῶν πε- 18. nd 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. nd the ὑπὸ τεσσάρων, πολύπλευρα δὲ τὰ ὑπὸ πλειόνων ἢ τεσσάρων center of the semi-circle is the same (point) as (the center εὐθειῶνπεριεχόμενα. 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 multiτὰςτρεῖςἀνίσουςἔχονπλευράς. lateral by more than four. καʹ τι δὲ τῶν τριπλεύρων σχημάτων ὀρθογώνιον μὲν 20. nd of the trilateral figures: an equilateral trianτρίγωνόν 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) ἔχονγωνίας. that having three unequal sides. 6

7 κβʹ. Τὼν δὲ τετραπλεύρων σχημάτων τετράγωνον μέν 21. nd 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. nd 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 τετράπλευρατραπέζιακαλείσθω. 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 anτὰ μέρη ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις. other which is neither right-angled nor equilateral. nd 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 direction, meet with one another in neither (of these directions). This should really be counted as a postulate, rather than as part of a definition. Ø Ñ Ø º Postulates αʹ. Ηιτήσθω ἀπὸ παντὸς σημείου ἐπὶ πᾶν σημεῖον 1. Let it have been postulated to draw a straight-line εὐθεῖανγραμμὴνἀγαγεῖν. from any point to any point. βʹ. Καὶ πεπερασμένην εὐθεῖαν κατὰ τὸ συνεχὲς ἐπ 2. nd to produce a finite straight-line continuously εὐθείαςἐκβαλεῖν. in a straight-line. γʹ.καὶπαντὶκέντρῳκαὶδιαστήματικύκλονγράφεσθαι. 3. nd to draw a circle with any center and radius. δʹ.καὶπάσαςτὰςὀρθὰςγωνίαςἴσαςἀλλήλαιςεἶναι. 4. nd that all right-angles are equal to one another. εʹ. Καὶ ἐὰν εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὰς ἐντὸς 5. nd 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 À Ø Û 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 nglish, 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. ÃÓ Ò ÒÒÓ º ommon Notions αʹ.τὰτῷαὐτῷἴσακαὶἀλλήλοιςἐστὶνἴσα. 1. Things equal to the same thing are also equal to βʹ. Καὶ ἐὰν ἴσοις ἴσα προστεθῇ, τὰ ὅλα ἐστὶν ἴσα. one another. γʹ.καὶἐὰνἀπὸἴσωνἴσαἀφαιρεθῇ,τὰκαταλειπόμενά 2. nd if equal things are added to equal things then ἐστινἴσα. the wholes are equal. δʹ. Καὶ τὰ ἐφαρμόζοντα ἐπ ἀλλήλα ἴσα ἀλλήλοις ἐστίν. 3. nd if equal things are subtracted from equal things εʹ.καὶτὸὅλοντοῦμέρουςμεῖζόν[ἐστιν]. then the remainders are equal. 4. nd things coinciding with one another are equal to one another. 5. nd the whole [is] greater than the part. s an obvious extension of.n.s 2 & 3 if equal things are added or subtracted from the two sides of an inequality then the inequality remains 7

8 an inequality of the same type.. Proposition 1 πὶ τῆς δοθείσης εὐθείας πεπερασμένης τρίγωνον To construct an equilateral triangle on a given finite ἰσόπλευρον συστήσασθαι. straight-line. στωἡδοθεῖσαεὐθεῖαπεπερασμένηἡ. Let be the given finite straight-line. Δεῖ δὴ ἐπὶ τῆς εὐθείας τρίγωνον ἰσόπλευρον So it is required to construct an equilateral triangle on συστήσασθαι. the straight-line. Κέντρῳ μὲν τῷ διαστήματι δὲ τῷ κύκλος Let the circle with center and radius have γεγράφθωὁδ,καὶπάλινκέντρῳμὲντῷδιαστήματιδὲ been drawn [Post. 3], and again let the circle with τῷ κύκλος γεγράφθω ὁ, καὶ ἀπὸ τοῦ σημείου, center and radius have been drawn [Post. 3]. nd καθ ὃτέμνουσινἀλλήλουςοἱκύκλοι,ἐπίτὰ,σημεῖα let the straight-lines and have been joined from ἐπεζεύχθωσανεὐθεῖαιαἱ,. the point, where the circles cut one another, to the Καὶ ἐπεὶ τὸ σημεῖον κέντρον ἐστὶ τοῦ Δ κύκλου, points and (respectively) [Post. 1]. ἴσηἐστὶνἡτῇ πάλιν,ἐπεὶτὸσημεῖονκέντρον nd since the point is the center of the circle, ἐστὶτοῦκύκλου,ἴσηἐστὶνἡτῇ.ἐδείχθηδὲ is equal to [ef. 1.15]. gain, since the point καὶἡτῇἴση ἑκατέραἄρατῶν,τῇἐστιν is the center of the circle, is equal to ἴση.τὰδὲτῷαὐτῷἴσακαὶἀλλήλοιςἐστὶνἴσα καὶἡἄρα [ef. 1.15]. ut was also shown (to be) equal to. τῇἐστινἴση αἱτρεῖςἄρααἱ,,ἴσαιἀλλήλαις Thus, and are each equal to. ut things equal εἰσίν. to the same thing are also equal to one another [.N. 1]. Ισόπλευρον ἄρα ἐστὶ τὸ τρίγωνον. καὶ συνέσταται Thus, is also equal to. Thus, the three (straight- ἐπὶτῆςδοθείσηςεὐθείαςπεπερασμένηςτῆς.ὅπερἔδει lines),, and are equal to one another. ποιῆσαι. Thus, the triangle is equilateral, and has been constructed on the given finite straight-line. (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.. 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). στωτὸμὲνδοθὲνσημεῖοντὸ,ἡδὲδοθεῖσαεὐθεῖα Let be the given point, and the given straight- ἡ δεῖδὴπρὸςτῷσημείῳτῇδοθείσῃεὐθείᾳτῇ line. So it is required to place a straight-line at point ἴσηνεὐθεῖανθέσθαι. equal to the given straight-line. πεζεύχθωγὰρἀπὸτοῦσημείουἐπίτὸσημεῖον or let the straight-line have been joined from εὐθεῖα ἡ, καὶ συνεστάτω ἐπ αὐτῆς τρίγωνον ἰσόπλευρον point to point [Post. 1], and let the equilateral trianτὸ Δ, καὶ ἐκβεβλήσθωσαν ἐπ εὐθείας ταῖς Δ, Δ gle have been been constructed upon it [Prop. 1.1]. 8

9 εὐθεῖαιαἱ,ζ,καὶκέντρῳμὲντῷδιαστήματιδὲτῷ nd let the straight-lines and have been pro- κύκλος γεγράφθω ὁ ΗΘ, καὶ πάλιν κέντρῳ τῷ Δ καὶ duced in a straight-line with and (respectively) διαστήματιτῷδηκύκλοςγεγράφθωὁηκλ. [Post. 2]. nd let the circle GH with center and radius have been drawn [Post. 3], and again let the circle GKL with center and radius G have been drawn [Post. 3]. Κ Θ K H Η G Ζ Λ L πεὶοὖντὸσημεῖονκέντρονἐστὶτοῦηθ,ἴσηἐστὶν Therefore, since the point is the center of (the cir- ἡτῇη.πάλιν,ἐπεὶτὸδσημεῖονκέντρονἐστὶτοῦ cle) GH, is equal to G [ef. 1.15]. gain, since ΗΚΛκύκλου,ἴσηἐστὶνἡΔΛτῇΔΗ,ὧνἡΔτῇΔἴση the point is the center of the circle GKL, L is equal ἐστίν. λοιπὴἄραἡλλοιπῇτῇηἐστινἴση. ἐδείχθηδὲ to G [ef. 1.15]. nd within these, is equal to. καὶἡτῇηἴση ἑκατέραἄρατῶνλ,τῇηἐστιν Thus, the remainder L is equal to the remainder G ἴση. τὰδὲτῷαὐτῷἴσακαὶἀλλήλοιςἐστὶνἴσα καὶἡλ [.N. 3]. ut was also shown (to be) equal to G. ἄρατῇἐστινἴση. Thus, L and are each equal to G. ut things equal Πρὸςἄρατῷδοθέντισημείῳτῷτῇδοθείσῃεὐθείᾳ to the same thing are also equal to one another [.N. 1]. τῇ ἴση εὐθεῖα κεῖται ἡ Λ ὅπερ ἔδει ποιῆσαι. Thus, L is also equal to. Thus, the straight-line L, equal to the given straightline, has been placed at the given point. (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 and the line. In such situations, uclid invariably only considers one particular case usually, the most difficult and leaves the remaining cases as exercises for the reader.. Proposition 3 Δύο δοθεισῶν εὐθειῶν ἀνίσων ἀπὸ τῆς μείζονος τῇ or two given unequal straight-lines, to cut off from ἐλάσσονιἴσηνεὐθεῖανἀφελεῖν. the greater a straight-line equal to the lesser. στωσαναἱδοθεῖσαιδύοεὐθεῖαιἄνισοιαἱ,,ὧν Let and be the two given unequal straight-lines, μείζωνἔστωἡ δεῖδὴἀπὸτῆςμείζονοςτῆςτῇ of which let the greater be. So it is required to cut off ἐλάσσονιτῇἴσηνεὐθεῖανἀφελεῖν. a straight-line equal to the lesser from the greater. ΚείσθωπρὸςτῷσημείῳτῇεὐθείᾳἴσηἡΔ καὶ Let the line, equal to the straight-line, have κέντρῳμὲντῷδιαστήματιδὲτῷδκύκλοςγεγράφθω been placed at point [Prop. 1.2]. nd let the circle ὁδζ. have been drawn with center and radius Καὶ ἐπεὶ τὸ σημεῖον κέντρον ἐστὶ τοῦ ΔΖ κύκλου, [Post. 3]. 9

10 ἴσηἐστὶνἡτῇδ ἀλλὰκαὶἡτῇδἐστινἴση. nd since point is the center of circle, ἑκατέραἄρατῶν,τῇδἐστινἴση ὥστεκαὶἡ is equal to [ef. 1.15]. ut, is also equal to. τῇἐστινἴση. Thus, and are each equal to. So is also equal to [.N. 1]. Ζ Δύοἄραδοθεισῶνεὐθειῶνἀνίσωντῶν,ἀπὸτῆς Thus, for two given unequal straight-lines, and, μείζονοςτῆςτῇἐλάσσονιτῇἴσηἀφῄρηταιἡ ὅπερ the (straight-line), equal to the lesser, has been cut ἔδειποιῆσαι. off from the greater. (Which is) the very thing it was required to do.. Proposition 4 ὰν 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 βάσιντῂβάσειἴσηνἕξει,καὶτὸτρίγωνοντῷτριγώνῳἴσον 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 angles. Ζ στω δύο τρίγωνα τὰ, ΔΖ τὰς δύο πλευρὰς Let and be two triangles having the two τὰς,ταῖςδυσὶπλευραῖςταῖςδ,δζἴσαςἔχοντα sides and equal to the two sides and, re- ἑκατέρανἑκατέρᾳτὴνμὲντῇδτὴνδὲτῇδζ spectively. (That is) to, and to. nd (let) καὶγωνίαντὴνὑπὸγωνίᾳτῇὑπὸδζἴσην. λέγω, the angle (be) equal to the angle. I say that ὅτικαὶβάσιςἡβάσειτῇζἴσηἐστίν, καὶτὸ the base is also equal to the base, and triangle τρίγωνοντῷδζτριγώνῳἴσονἔσται,καὶαἱλοιπαὶγωνίαι will be equal to triangle, 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 τῇ ὑπὸ ΔΖ, corresponding remaining angles. (That is) to, ἡδὲὑπὸτῇὑπὸδζ. and to. φαρμοζομένουγὰρτοῦτριγώνουἐπὶτὸδζ or if triangle is applied to triangle, the τρίγωνονκαὶτιθεμένουτοῦμὲνσημείουἐπὶτὸδσημεῖον point being placed on the point, and the straight-line 10

11 τῆςδὲεὐθείαςἐπὶτὴνδ,ἐφαρμόσεικαὶτὸσημεῖον on, then the point will also coincide with, ἐπὶτὸδιὰτὸἴσηνεἶναιτὴντῇδ ἐφαρμοσάσηςδὴ on account of being equal to. So (because of) τῆςἐπὶτὴνδἐφαρμόσεικαὶἡεὐθεῖαἐπὶτὴνδζ coinciding with, the straight-line will also διὰτὸἴσηνεἶναιτὴνὑπὸγωνίαντῇὑπὸδζ ὥστεκαὶ coincide with, on account of the angle being τὸ σημεῖον ἐπὶ τὸ Ζ σημεῖον ἐφαρμόσει διὰ τὸ ἴσην πάλιν equal to. So the point will also coincide with the εἶναιτὴντῇδζ.ἀλλὰμὴνκαὶτὸἐπὶτὸἐφηρμόκει point, again on account of being equal to. ut, ὥστε βάσις ἡ ἐπὶ βάσιν τὴν Ζ ἐφαρμόσει. εἰ γὰρ τοῦ point certainly also coincided with point, so that the μὲνἐπὶτὸἐφαρμόσαντοςτοῦδὲἐπὶτὸζἡβάσις base will coincide with the base. or if coin- ἐπὶτὴνζοὐκἐφαρμόσει,δύοεὐθεῖαιχωρίονπεριέξουσιν cides with, and with, and the base does not ὅπερ ἐστὶν ἀδύνατον. ἐφαρμόσει ἄρα ἡ βάσις ἐπὶ τὴν coincide with, then two straight-lines will encompass Ζκαὶἴσηαὐτῇἔσται ὥστεκαὶὅλοντὸτρίγωνον an area. The very thing is impossible [Post. 1]. Thus, ἐπὶὅλοντὸδζτρίγωνονἐφαρμόσεικαὶἴσοναὐτῷἔσται, the base will coincide with, and will be equal to καὶ αἱ λοιπαὶ γωνίαι ἐπὶ τὰς λοιπὰς γωνίας ἐφαρμόσουσι καὶ it [.N. 4]. So the whole triangle will coincide with ἴσαι αὐταῖς ἔσονται, ἡ μὲν ὑπὸ τῇ ὑπὸ ΔΖ ἡ δὲ ὑπὸ the whole triangle, and will be equal to it [.N. 4]. τῇ ὑπὸ ΔΖ. nd the remaining angles will coincide with the remain- ὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς [ταῖς] δύο ing angles, and will be equal to them [.N. 4]. (That is) πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν γωνίαν τῇ to, and to [.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 γωνίαιςἴσαιἔσονταιἑκατέραἑκατέρᾳ,ὑφ ἃςαἱἴσαιπλευραὶ 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 angles. (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.. Proposition 5 Τῶν ἰσοσκελῶν τριγώνων αἱ τρὸς τῇ βάσει γωνίαι ἴσαι or 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 ὑπὸτὴνβάσινγωνίαιἴσαιἀλλήλαιςἔσονται. the angles under the base will be equal to one another. Ζ Η G στω τρίγωνον ἰσοσκελὲς τὸ ἴσην ἔχον τὴν Let be an isosceles triangle having the side πλευρὰν τῇ πλευρᾷ, καὶ προσεκβεβλήσθωσαν ἐπ equal to the side, and let the straight-lines and εὐθείαςταῖς,εὐθεῖαιαἱδ, λέγω,ὅτιἡμὲν have been produced in a straight-line with and ὑπὸγωνίατῇὑπὸἴσηἐστίν,ἡδὲὑπὸδτῇ (respectively) [Post. 2]. I say that the angle is ὑπὸ. equal to, and (angle) to. ἰλήφθω γὰρ ἐπὶ τῆς Δ τυχὸν σημεῖον τὸ Ζ, καὶ or let the point have been taken at random on, ἀφῃρήσθωἀπὸτῆςμείζονοςτῆςτῇἐλάσσονιτῇζ and let G have been cut off from the greater, equal 11

12 ἴση ἡ Η, καὶ ἐπεζεύχθωσαν αἱ Ζ, Η εὐθεῖαι. to the lesser [Prop. 1.3]. lso, let the straight-lines πεὶοὖνἴσηἐστὶνἡμὲνζτῇηἡδὲτῇ, and G have been joined [Post. 1]. δύοδὴαἱζ,δυσὶταῖςη,ἴσαιεἰσὶνἑκατέρα In fact, since is equal to G, and to, ἑκατέρᾳ καὶ γωνίαν κοινὴν περιέχουσι τὴν ὑπὸ ΖΗ βάσις the two (straight-lines), are equal to the two ἄρα ἡ Ζ βάσει τῇ Η ἴση ἐστίν, καὶ τὸ Ζ τρίγωνον τῷ (straight-lines) G,, respectively. They also encom- Ητριγώνῳἴσονἔσται,καὶαἱλοιπαὶγωνίαιταῖςλοιπαῖς pass a common angle, G. Thus, the base is equal γωνίαιςἴσαιἔσονταιἑκατέραἑκατέρᾳ,ὑφ ἃςαἱἴσαιπλευραὶ to the base G, and the triangle will be equal to the ὑποτείνουσιν,ἡμὲνὑπὸζτῇὑπὸη,ἡδὲὑπὸζ triangle G, and the remaining angles subtendend by τῇὑπὸη.καὶἐπεὶὅληἡζὅλῃτῇηἐστινἴση,ὧν the equal sides will be equal to the corresponding remain- ἡτῇἐστινἴση,λοιπὴἄραἡζλοιπῇτῇηἐστιν ing angles [Prop. 1.4]. (That is) to G, and ἴση.ἐδείχθηδὲκαὶἡζτῇηἴση δύοδὴαἱζ,ζδυσὶ to G. nd since the whole of is equal to the whole ταῖςη,ηἴσαιεἰσὶνἑκατέραἑκατέρᾳ καὶγωνίαἡὑπὸ of G, within which is equal to, the remainder ΖγωνίᾳτῃὑπὸΗἴση,καὶβάσιςαὐτῶνκοινὴἡ is thus equal to the remainder G [.N. 3]. ut καὶτὸζἄρατρίγωνοντῷητριγώνῳἴσονἔσται,καὶ was also shown (to be) equal to G. So the two (straightαἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα lines), are equal to the two (straight-lines) G, ἑκατέρᾳ,ὑφ ἃςαἱἴσαιπλευραὶὑποτείνουσιν ἴσηἄραἐστὶν G, respectively, and the angle (is) equal to the ἡμὲνὑπὸζτῇὑπὸηἡδὲὑπὸζτῇὑπὸη. angle G, and the base is common to them. Thus, ἐπεὶ οὖν ὅλη ἡ ὑπὸ Η γωνία ὅλῃ τῇ ὑπὸ Ζ γωνίᾳ the triangle will be equal to the triangle G, and ἐδείχθηἴση,ὧνἡὑπὸητῇὑπὸζἴση,λοιπὴἄραἡ the remaining angles subtended by the equal sides will be ὑπὸ λοιπῇ τῇ ὑπὸ ἐστιν ἴση καί εἰσι πρὸς τῇ equal to the corresponding remaining angles [Prop. 1.4]. βάσειτοῦτριγώνου. ἐδείχθηδὲκαὶἡὑπὸζτῇ Thus, is equal to G, and to G. There- ὑπὸηἴση καίεἰσινὑπὸτὴνβάσιν. fore, since the whole angle G was shown (to be) equal Τῶνἄραἰσοσκελῶντριγώνωναἱτρὸςτῇβάσειγωνίαι to the whole angle, within which G is equal to ἴσαι ἀλλήλαις εἰσίν, καὶ προσεκβληθεισῶν τῶν ἴσων εὐθειῶν, the remainder is thus equal to the remainder αἱὑπὸτὴνβάσινγωνίαιἴσαιἀλλήλαιςἔσονται ὅπερἔδει [.N. 3]. nd they are at the base of triangle. δεῖξαι. nd was also shown (to be) equal to G. nd 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 another. (Which is) the very thing it was required to show.. Proposition 6 ὰν 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. στωτρίγωνοντὸἴσηνἔχοντὴνὑπὸγωνίαν Let be a triangle having the angle equal τῇὑπὸγωνίᾳ λέγω,ὅτικαὶπλευρὰἡπλευρᾷτῇ to the angle. I say that side is also equal to side ἐστιν ἴση.. 12

13 ἰγὰρἄνισόςἐστινἡτῇ,ἡἑτέρααὐτῶνμείζων or if is unequal to then one of them is ἐστίν. ἔστω μείζων ἡ, καὶ ἀφῃρήσθω ἀπὸ τῆς μείζονος greater. Let be greater. nd let, equal to τῆςτῇἐλάττονιτῇἴσηἡδ,καὶἐπεζεύχθωἡδ. the lesser, have been cut off from the greater πεὶ οὖν ἴση ἐστὶν ἡ Δ τῇ κοινὴ δὲ ἡ, δύο δὴ [Prop. 1.3]. nd let have been joined [Post. 1]. αἱδ,δύοταῖς,ἴσαιεἰσὶνἑκατέραἑκατέρᾳ,καὶ Therefore, since is equal to, and (is) comγωνίαἡὑπὸδγωνίᾳτῇὑπὸἐστινἴση βάσιςἄραἡ mon, the two sides, are equal to the two sides Δ βάσει τῇ ἴση ἐστίν, καὶ τὸ Δ τρίγωνον τῷ,, respectively, and the angle is equal to the τριγώνῳἴσονἔσται,τὸἔλασσοντῷμείζονι ὅπερἄτοπον angle. Thus, the base is equal to the base, οὐκἄραἄνισόςἐστινἡτῇ ἴσηἄρα. and the triangle will be equal to the triangle ὰν 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 [.N. 5]. Thus, is not unequal to. Thus, ἔσονται ὅπερἔδειδεῖξαι. (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. Þ. Proposition 7 πὶ 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. ἰγὰρδυνατόν,ἐπὶτῆςαὐτῆςεὐθείαςτῆςδύοταῖς or, if possible, let the two straight-lines,, αὐταῖς εὐθείαις ταῖς, ἄλλαι δύο εὐθεῖαι αἱ Δ, Δ equal to two other straight-lines,, respectively, ἴσαι ἑκατέρα ἑκατέρᾳ συνεστάτωσαν πρὸς ἄλλῳ καὶ ἄλλῳ have been constructed on the same straight-line, σημείῳτῷτεκαὶδἐπὶτὰαὐτὰμέρητὰαὐτὰπέρατα meeting at different points, and, on the same side ἔχουσαι,ὥστεἴσηνεἶναιτὴνμὲντῇδτὸαὐτὸπέρας (of ), and having the same ends (on ). So is ἔχουσαναὐτῇτὸ,τὴνδὲτῇδτὸαὐτὸπέραςἔχου- equal to, having the same end as it, and is σαναὐτῇτὸ,καὶἐπεζεύχθωἡδ. equal to, having the same end as it. nd let πεὶοὖνἴσηἐστὶνἡτῇδ,ἴσηἐστὶκαὶγωνίαἡ have been joined [Post. 1]. ὑπὸδτῇὑπὸδ μείζωνἄραἡὑπὸδτῆςὑπὸ Therefore, since is equal to, the angle Δ πολλῷἄραἡὑπὸδμείζωνἐστίτῆςὑπὸδ. is also equal to angle [Prop. 1.5]. Thus, (is) πάλιν ἐπεὶ ἴση ἐστὶν ἡ τῇ Δ, ἴση ἐστὶ καὶ γωνία ἡ greater than [.N. 5]. Thus, is much greater ὑπὸδγωνίᾳτῇὑπὸδ.ἐδείχθηδὲαὐτῆςκαὶπολλῷ than [.N. 5]. gain, since is equal to, the μείζων ὅπερ ἐστὶν ἀδύνατον. angle is also equal to angle [Prop. 1.5]. ut Οὐκἄραἐπὶτῆςαὐτῆςεὐθείαςδύοταῖςαὐταῖςεὐθείαις it was shown that the former (angle) is also much greater 13

14 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 different 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.. Proposition 8 ὰν 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,καὶτὴνγωνίαντῇγωνίᾳἴσηνἕξειτὴνὑπὸτῶνἴσων they will also have equal the angles encompassed by the εὐθειῶνπεριεχομένην. equal straight-lines. Η G Ζ στω δύο τρίγωνα τὰ, ΔΖ τὰς δύο πλευρὰς Let and be two triangles having the two τὰς,ταῖςδύοπλευραῖςταῖςδ,δζἴσαςἔχοντα sides and equal to the two sides and, ἑκατέρανἑκατέρᾳ,τὴνμὲντῇδτὴνδὲτῇδζ respectively. (That is) to, and to. Let ἐχέτωδὲκαὶβάσιντὴνβάσειτῇζἴσην λέγω,ὅτικαὶ them also have the base equal to the base. I say γωνία ἡ ὑπὸ γωνίᾳ τῇ ὑπὸ ΔΖ ἐστιν ἴση. that the angle is also equal to the angle. φαρμοζομένουγὰρτοῦτριγώνουἐπὶτὸδζ or if triangle is applied to triangle, the τρίγωνονκαὶτιθεμένουτοῦμὲνσημείουἐπὶτὸσημεῖον point being placed on point, and the straight-line τῆςδὲεὐθείαςἐπὶτὴνζἐφαρμόσεικαὶτὸσημεῖον on, then point will also coincide with, on ἐπὶτὸζδιὰτὸἴσηνεἶναιτὴντῇζ ἐφαρμοσάσηςδὴ account of being equal to. So (because of) τῆς ἐπὶ τὴν Ζ ἐφαρμόσουσι καὶ αἱ, ἐπὶ τὰς Δ, coinciding with, (the sides) and will also co- ΔΖ.εἰγὰρβάσιςμὲνἡἐπὶβάσιντὴνΖἐφαρμόσει,αἱ incide with and (respectively). or if base δὲ,πλευραὶἐπὶτὰςδ,δζοὐκἐφαρμόσουσιν coincides with base, but the sides and do not ἀλλὰ παραλλάξουσιν ὡς αἱ Η, ΗΖ, συσταθήσονται ἐπὶ τῆς coincide with and (respectively), but miss like αὐτῆςεὐθείαςδύοταῖςαὐταῖςεὐθείαιςἄλλαιδύοεὐθεῖαι G and G (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 straightαὐτὰ μέρη τὰ αὐτὰ πέρατα ἔχουσαι. οὐ συνίστανται δέ lines equal, respectively, to two (given) straight-lines, οὐκἄραἐφαρμοζομένηςτῆςβάσεωςἐπὶτὴνζβάσιν and (meeting) at a different point on the same side (of οὐκ ἐφαρμόσουσι καὶ αἱ, πλευραὶ ἐπὶ τὰς Δ, ΔΖ. the straight-line), but having the same ends. ut (such ἐφαρμόσουσιν ἄρα ὥστε καὶ γωνία ἡ ὑπὸ ἐπὶ γωνίαν straight-lines) cannot be constructed [Prop. 1.7]. Thus, τὴνὑπὸδζἐφαρμόσεικαὶἴσηαὐτῇἔσται. the base being applied to the base, the sides ὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς [ταῖς] δύο and cannot not coincide with and (respecπλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν βάσιν τῇ βάσει tively). Thus, they will coincide. So the angle will ἴσηνἔχῃ,καὶτὴνγωνίαντῇγωνίᾳἴσηνἕξειτὴνὑπὸτῶν also coincide with angle, and will be equal to it ἴσων εὐθειῶν περιεχομένην ὅπερ ἔδει δεῖξαι. [.N. 4]. Thus, if two triangles have two sides equal to two side, respectively, and have the base equal to the base, 14

15 then they will also have equal the angles encompassed by the equal straight-lines. (Which is) the very thing it was required to show.. Proposition 9 Τὴνδοθεῖσανγωνίανεὐθύγραμμονδίχατεμεῖν. To cut a given rectilinear angle in half. Ζ στωἡδοθεῖσαγωνίαεὐθύγραμμοςἡὑπὸ.δεῖ Let be the given rectilinear angle. So it is reδὴαὐτὴνδίχατεμεῖν. quired to cut it in half. ἰλήφθωἐπὶτῆςτυχὸνσημεῖοντὸδ,καὶἀφῃρήσθω Let the point have been taken at random on, ἀπὸτῆςτῇδἴσηἡ,καὶἐπεζεύχθωἡδ,καὶ and let, equal to, have been cut off from συνεστάτω ἐπὶ τῆς Δ τρίγωνον ἰσόπλευρον τὸ ΔΖ, καὶ [Prop. 1.3], and let have been joined. nd let the ἐπεζεύχθω ἡ Ζ λέγω, ὅτι ἡ ὑπὸ γωνία δίχα τέτμηται equilateral triangle have been constructed upon ὑπὸτῆςζεὐθείας. [Prop. 1.1], and let have been joined. I say that πεὶγὰρἴσηἐστὶνἡδτῇ,κοινὴδὲἡζ,δύοδὴ the angle has been cut in half by the straight-line αἱδ,ζδυσὶταῖς,ζἴσαιεἰσὶνἑκατέραἑκατέρᾳ.. καὶβάσιςἡδζβάσειτῇζἴσηἐστίν γωνίαἄραἡὑπὸ or since is equal to, and is common, ΔΖγωνίᾳτῇὑπὸΖἴσηἐστίν. the two (straight-lines), are equal to the two Η ἄρα δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ δίχα (straight-lines),, respectively. nd the base τέτμηταιὑπὸτῆςζεὐθείας ὅπερἔδειποιῆσαι. is equal to the base. Thus, angle is equal to angle [Prop. 1.8]. Thus, the given rectilinear angle has been cut in half by the straight-line. (Which is) the very thing it was required to do.. Proposition 10 Τὴνδοθεῖσανεὐθεῖανπεπερασμένηνδίχατεμεῖν. To cut a given finite straight-line in half. στωἡδοθεῖσαεὐθεῖαπεπερασμένηἡ δεῖδὴτὴν Let be the given finite straight-line. So it is re- εὐθεῖανπεπερασμένηνδίχατεμεῖν. quired to cut the finite straight-line in half. Συνεστάτω ἐπ αὐτῆς τρίγωνον ἰσόπλευρον τὸ, καὶ Let the equilateral triangle have been conτετμήσθωἡὑπὸγωνίαδίχατῇδεὐθείᾳ λέγω,ὅτι structed upon () [Prop. 1.1], and let the angle ἡ εὐθεῖα δίχα τέτμηται κατὰ τὸ Δ σημεῖον. have been cut in half by the straight-line [Prop. 1.9]. πεὶγὰρἴσηἐστὶνἡτῇ,κοινὴδὲἡδ,δύοδὴ I say that the straight-line has been cut in half at αἱ,δδύοταῖς,δἴσαιεἰσὶνἑκατέραἑκατέρᾳ καὶ point. γωνίαἡὑπὸδγωνίᾳτῇὑπὸδἴσηἐστίν βάσιςἄρα or since is equal to, and (is) common, 15

16 ἡδβάσειτῇδἴσηἐστίν. the two (straight-lines), are equal to the two (straight-lines),, respectively. nd the angle is equal to the angle. Thus, the base is equal to the base [Prop. 1.4]. Η ἄρα δοθεῖσα εὐθεῖα πεπερασμένη ἡ δίχα τέτμηται Thus, the given finite straight-line has been cut κατὰτὸδ ὅπερἔδειποιῆσαι. in half at (point). (Which is) the very thing it was required to do.. 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. Ζ στωἡμὲνδοθεῖσαεὐθεῖαἡτὸδὲδοθὲνσημεῖον Let be the given straight-line, and the given ἐπ αὐτῆςτὸ δεῖδὴἀπὸτοῦσημείουτῇεὐθείᾳ point on it. So it is required to draw a straight-line from πρὸς ὀρθὰς γωνίας εὐθεῖαν γραμμὴν ἀγαγεῖν. the point at right-angles to the straight-line. ἰλήφθωἐπὶτῆςτυχὸνσημεῖοντὸδ,καὶκείσθω Let the point be have been taken at random on, τῇ Δ ἴση ἡ, καὶ συνεστάτω ἐπὶ τῆς Δ τρίγωνον and let be made equal to [Prop. 1.3], and let the ἰσόπλευρον τὸ ΖΔ, καὶ ἐπεζεύχθω ἡ Ζ λέγω, ὅτι τῇ equilateral triangle have been constructed on δοθείσῃεὐθείᾳτῇἀπὸτοῦπρὸςαὐτῇδοθέντοςσημείου [Prop. 1.1], and let have been joined. I say that the τοῦπρὸςὀρθὰςγωνίαςεὐθεῖαγραμμὴἦκταιἡζ. straight-line has been drawn at right-angles to the πεὶ γὰρ ἴση ἐστὶν ἡ Δ τῇ, κοινὴ δὲ ἡ Ζ, δύο given straight-line from the given point on it. δὴαἱδ,ζδυσὶταῖς,ζἴσαιεἰσὶνἑκατέραἑκατέρᾳ or since is equal to, and is common, καὶβάσιςἡδζβάσειτῇζἴσηἐστίν γωνίαἄραἡὑπὸ the two (straight-lines), are equal to the two ΔΖ γωνίᾳ τῇ ὑπὸ Ζ ἴση ἐστίν καί εἰσιν ἐφεξῆς. ὅταν (straight-lines),,, respectively. nd the base δὲ εὐθεῖα ἐπ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας is equal to the base. Thus, the angle is equal ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν ὀρθὴ to the angle [Prop. 1.8], and they are adjacent. ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΔΖ, Ζ. ut when a straight-line stood on a(nother) straight-line 16

17 Τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ἀπὸ τοῦ πρὸς αὐτῇ makes the adjacent angles equal to one another, each of δοθέντος σημείου τοῦ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ the equal angles is a right-angle [ef. 1.10]. Thus, each ἦκταιἡζ ὅπερἔδειποιῆσαι. of the (angles) and is a right-angle. Thus, the straight-line has been drawn at rightangles to the given straight-line from the given point on it. (Which is) the very thing it was required to do.. Proposition 12 πὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον ἀπὸ τοῦ δοθέντος To draw a straight-line perpendicular to a given infiσημείου,ὃμήἐστινἐπ αὐτῆς,κάθετονεὐθεῖανγραμμὴν nite straight-line from a given point which is not on it. ἀγαγεῖν. Ζ Η Θ στωἡμὲνδοθεῖσαεὐθεῖαἄπειροςἡτὸδὲδοθὲν Let be the given infinite straight-line and the σημεῖον,ὃμήἐστινἐπ αὐτῆς,τὸ δεῖδὴἐπὶτὴνδοθεῖσαν given point, which is not on (). So it is required to εὐθεῖαν ἄπειρον τὴν ἀπὸ τοῦ δοθέντος σημείου τοῦ, draw a straight-line perpendicular to the given infinite ὃ μή ἐστιν ἐπ αὐτῆς, κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν. straight-line from the given point, which is not on ἰλήφθω γὰρ ἐπὶ τὰ ἕτερα μέρη τῆς εὐθείας τυχὸν (). σημεῖοντὸδ,καὶκέντρῳμὲντῷδιαστήματιδὲτῷδ or let point have been taken at random on the κύκλοςγεγράφθωὁζη,καὶτετμήσθωἡηεὐθεῖαδίχα other side (to ) of the straight-line, and let the κατὰ τὸ Θ, καὶ ἐπεζεύχθωσαν αἱ Η, Θ, εὐθεῖαι circle G have been drawn with center and radius λέγω, ὅτι ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον τὴν ἀπὸ [Post. 3], and let the straight-line G have been cut τοῦδοθέντοςσημείουτοῦ,ὃμήἐστινἐπ αὐτῆς,κάθετος in half at (point) H [Prop. 1.10], and let the straight- ἦκταιἡθ. lines G, H, and have been joined. I say that the πεὶγὰρἴσηἐστὶνἡηθτῇθ,κοινὴδὲἡθ,δύο (straight-line) H has been drawn perpendicular to the δὴαἱηθ,θδύοταῖςθ,θἴσαιεἱσὶνἑκατέραἑκατέρᾳ given infinite straight-line from the given point, καὶβάσιςἡηβάσειτῇἐστινἴση γωνίαἄραἡὑπὸ which is not on (). ΘΗγωνίᾳτῇὑπὸΘἐστινἴση.καίεἰσινἐφεξῆς.ὅταν or since GH is equal to H, and H (is) common, δὲ εὐθεῖα ἐπ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας the two (straight-lines) GH, H are equal to the two ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν, καὶ (straight-lines) H, H, respectively, and the base G ἡἐφεστηκυῖαεὐθεῖακάθετοςκαλεῖταιἐφ ἣνἐφέστηκεν. is equal to the base. Thus, the angle HG is equal πὶτὴνδοθεῖσανἄραεὐθεῖανἄπειροντὴνἀπὸτοῦ to the angle H [Prop. 1.8], and they are adjacent. δοθέντος σημείου τοῦ, ὃ μή ἐστιν ἐπ αὐτῆς, κάθετος ut when a straight-line stood on a(nother) straight-line ἦκται ἡ Θ ὅπερ ἔδει ποιῆσαι. makes the adjacent angles equal to one another, each of the equal angles is a right-angle, and the former straightline is called a perpendicular to that upon which it stands [ef. 1.10]. Thus, the (straight-line) H has been drawn perpendicular to the given infinite straight-line from the G H 17

18 given point, which is not on (). (Which is) the very thing it was required to do.. Proposition 13 ὰν 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 rightangles, or (angles whose sum is) equal to two rightangles. ὐθεῖα γάρ τις ἡ ἐπ εὐθεῖαν τὴν Δ σταθεῖσα or let some straight-line stood on the straightγωνίαςποιείτωτὰςὑπὸ,δ λὲγω,ὅτιαἱὑπὸ, line make the angles and. I say that Δ γωνίαι ἤτοι δύο ὀρθαί εἰσιν ἢ δυσὶν ὀρθαῖς ἴσαι. the angles and are certainly either two right- ἰ μὲν οὖν ἴση ἐστὶν ἡ ὑπὸ τῇ ὑπὸ Δ, δύο ὀρθαί angles, or (have a sum) equal to two right-angles. εἰσιν.εἰδὲοὔ,ἤχθωἀπὸτοῦσημείουτῇδ[εὐθείᾳ]πρὸς In fact, if is equal to then they are two ὀρθὰςἡ αἱἄραὑπὸ,δδύοὀρθαίεἰσιν καὶ right-angles [ef. 1.10]. ut, if not, let have been ἐπεὶἡὑπὸδυσὶταῖςὑπὸ,ἴσηἐστίν,κοινὴ drawn from the point at right-angles to [the straightπροσκείσθωἡὑπὸδ αἱἄραὑπὸ,δτρισὶταῖς line] [Prop. 1.11]. Thus, and are two ὑπὸ,,δἴσαιεἰσίν. πάλιν,ἐπεὶἡὑπὸδ right-angles. nd since is equal to the two (anδυσὶταῖςὑπὸδ,ἴσηἐστίν,κοινὴπροσκείσθωἡ gles) and, let have been added to both. ὑπὸ αἱ ἄρα ὑπὸ Δ, τρισὶ ταῖς ὑπὸ Δ,, Thus, the (sum of the angles) and is equal to ἴσαιεἰσίν.ἐδείχθησανδὲκαὶαἱὑπὸ,δτρισὶ the (sum of the) three (angles),, and ταῖςαὐταῖςἴσαι τὰδὲτῷαὐτῷἴσακαὶἀλλήλοιςἐστὶνἴσα [.N. 2]. gain, since is equal to the two (anκαὶαἱὑπὸ,δἄραταῖςὑπὸδ,ἴσαιεἰσίν gles) and, let have been added to both. ἀλλὰ αἱ ὑπὸ, Δ δύο ὀρθαί εἰσιν καὶ αἱ ὑπὸ Δ, Thus, the (sum of the angles) and is equal to ἄραδυσὶνὀρθαῖςἴσαιεἰσίν. the (sum of the) three (angles),, and ὰνἄραεὐθεῖαἐπ εὐθεῖανσταθεῖσαγωνίαςποιῇ,ἤτοι [.N. 2]. ut (the sum of) and was also δύοὀρθὰςἢδυσὶνὀρθαῖςἴσαςποιήσει ὅπερἔδειδεῖξαι. shown (to be) equal to the (sum of the) same three (angles). nd things equal to the same thing are also equal to one another [.N. 1]. Therefore, (the sum of) and is also equal to (the sum of) and. ut, (the sum of) and is two right-angles. Thus, (the sum of) and is also equal to two right-angles. Thus, if a straight-line stood on a(nother) straightline makes angles, it will certainly either make two rightangles, or (angles whose sum is) equal to two rightangles. (Which is) the very thing it was required to show. 18

19 . Proposition 14 ὰν πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ δύο If two straight-lines, not lying on the same side, make εὐθεῖαιμὴἐπὶτὰαὐτὰμέρηκείμεναιτὰςἐφεξῆςγωνίας 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 another. Πρὸςγάρτινιεὐθείᾳτῇκαὶτῷπρὸςαὐτῇσημείῳ or let two straight-lines and, not lying on the τῷ δύο εὐθεῖαι αἱ, Δ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι same side, make adjacent angles and (whose τὰς ἐφεξῆς γωνίας τὰς ὑπὸ, Δ δύο ὀρθαῖς ἴσας sum is) equal to two right-angles with some straight-line ποιείτωσαν λέγω,ὅτιἐπ εὐθείαςἐστὶτῇἡδ., at the point on it. I say that is straight-on with ἰγὰρμήἐστιτῇἐπ εὐθείαςἡδ,ἔστωτῇ respect to. ἐπ εὐθείαςἡ. or if is not straight-on to then let be πεὶ οὖν εὐθεῖα ἡ ἐπ εὐθεῖαν τὴν ἐφέστηκεν, straight-on to. αἱἄραὑπὸ,γωνίαιδύοὀρθαῖςἴσαιεἰσίν εἰσὶδὲ Therefore, since the straight-line stands on the καὶαἱὑπὸ,δδύοὀρθαῖςἴσαι αἱἄραὑπὸ, straight-line, the (sum of the) angles and ταῖςὑπὸ,δἴσαιεἰσίν. κοινὴἀφῃρήσθωἡ is thus equal to two right-angles [Prop. 1.13]. ut ὑπὸ λοιπὴἄραἡὑπὸλοιπῇτῇὑπὸδἐστιν (the sum of) and is also equal to two right- ἴση,ἡἐλάσσωντῇμείζονι ὅπερἐστὶνἀδύνατον. οὐκἄρα angles. Thus, (the sum of angles) and is equal ἐπ εὐθείας ἐστὶν ἡ τῇ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ to (the sum of angles) and [.N. 1]. Let (an- ἄλλη τις πλὴν τῆς Δ ἐπ εὐθείας ἄρα ἐστὶν ἡ τῇ Δ. gle) have been subtracted from both. Thus, the re- ὰν ἄρα πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇσημείῳ mainder is equal to the remainder [.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, is not straight-on with respect to. Simiεὐθεῖαι ὅπερ ἔδει δεῖξαι. larly, we can show that neither (is) any other (straightline) than. Thus, is straight-on with respect to. Thus, if two straight-lines, not lying on the same side, make adjacent angles (whose sum is) equal to two rightangles 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.. Proposition 15 ὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὰς κατὰ κορυφὴν If two straight-lines cut one another then they make γωνίαςἴσαςἀλλήλαιςποιοῦσιν. the vertically opposite angles equal to one another. 19

20 Δύο γὰρ εὐθεῖαι αἱ, Δ τεμνέτωσαν ἀλλήλας κατὰ or let the two straight-lines and cut one anτὸσημεῖον λέγω,ὅτιἴσηἐστὶνἡμὲνὑπὸγωνίατῇ other at the point. I say that angle is equal to ὑπὸ Δ, ἡ δὲ ὑπὸ τῇ ὑπὸ Δ. (angle), and (angle) to (angle). πεὶ γὰρ εὐθεῖα ἡ ἐπ εὐθεῖαν τὴν Δ ἐφέστηκε or since the straight-line stands on the straightγωνίαςποιοῦσατὰςὑπὸ,δ,αἱἄραὑπὸ,δ line, making the angles and, the (sum γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. πάλιν, ἐπεὶ εὐθεῖα ἡ Δ ἐπ of the) angles and is thus equal to two rightεὐθεῖαν τὴν ἐφέστηκε γωνίας ποιοῦσα τὰς ὑπὸ Δ, angles [Prop. 1.13]. gain, since the straight-line Δ, αἱ ἄρα ὑπὸ Δ, Δ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. stands on the straight-line, making the angles ἐδείχθησανδὲκαὶαἱὑπὸ,δδυσὶνὀρθαῖςἴσαι αἱ and, the (sum of the) angles and is ἄραὑπὸ,δταῖςὑπὸδ,δἴσαιεἰσίν.κοινὴ thus equal to two right-angles [Prop. 1.13]. ut (the sum ἀφῃρήσθωἡὑπὸδ λοιπὴἄραἡὑπὸλοιπῇτῇὑπὸ of) and was also shown (to be) equal to two Δ ἴση ἐστίν ὁμοίως δὴ δειχθήσεται, ὅτι καὶ αἱ ὑπὸ, right-angles. Thus, (the sum of) and is equal Δἴσαιεἰσίν. to (the sum of) and [.N. 1]. Let have ὰν ἄρα δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὰς κατὰ κο- been subtracted from both. Thus, the remainder is ρυφὴν γωνίας ἴσας ἀλλήλαις ποιοῦσιν ὅπερ ἔδει δεῖξαι. equal to the remainder [.N. 3]. Similarly, it can be shown that and 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.. Proposition 16 Παντὸς τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης or 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 μείζωνἐστίν. opposite angles. στωτρίγωνοντὸ,καὶπροσεκβεβλήσθωαὐτοῦ Let be a triangle, and let one of its sides μίαπλευρὰἡἐπὶτὸδ λὲγω,ὅτιἡἐκτὸςγωνίαἡὑπὸ have been produced to. I say that the external angle Δ μείζων ἐστὶν ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον τῶν is greater than each of the internal and opposite ὑπὸ, γωνιῶν. angles, and. Τετμήσθωἡδίχακατὰτὸ,καὶἐπιζευχθεῖσαἡ Let the (straight-line) have been cut in half at ἐκβεβλήσθω ἐπ εὐθείας ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ἴση ἡ (point) [Prop. 1.10]. nd being joined, let it have Ζ,καὶἐπεζεύχθωἡΖ,καὶδιήχθωἡἐπὶτὸΗ. been produced in a straight-line to (point). nd let πεὶοὖνἴσηἐστὶνἡμὲντῇ,ἡδὲτῇζ,δύο be made equal to [Prop. 1.3], and let have δὴαἱ,δυσὶταῖς,ζἴσαιεἰσὶνἑκατέραἑκατέρᾳ been joined, and let have been drawn through to καὶγωνίαἡὑπὸγωνίᾳτῇὑπὸζἴσηἐστίν κατὰ (point) G. κορυφὴνγάρ βάσιςἄραἡβάσειτῇζἴσηἐστίν,καὶτὸ Therefore, since is equal to, and to, τρίγωνοντῷζτριγώνῳἐστὶνἴσον,καὶαἱλοιπαὶ the two (straight-lines), are equal to the two 20

21 γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, ὑφ (straight-lines),, respectively. lso, angle ἃςαἱἴσαιπλευραὶὑποτείνουσιν ἴσηἄραἐστὶνἡὑπὸ is equal to angle, for (they are) vertically opposite τῇὑπὸζ.μείζωνδέἐστινἡὑπὸδτῆςὑπὸζ [Prop. 1.15]. Thus, the base is equal to the base, μείζωνἄραἡὑπὸδτῆςὑπὸ. Ομοίωςδὴτῆς and the triangle is equal to the triangle, and τετμημένης δίχα δειχθήσεται καὶ ἡ ὑπὸ Η, τουτέστιν ἡ the remaining angles subtended by the equal sides are ὑπὸ Δ, μείζων καὶ τῆς ὑπὸ. equal to the corresponding remaining angles [Prop. 1.4]. Thus, is equal to. ut is greater than. Thus, is greater than. Similarly, by having cut in half, it can be shown (that) G that is to say, (is) also greater than. Ζ Η G Παντὸς ἄρα τριγώνου μιᾶς τῶν πλευρῶν προσεκ- Thus, for any triangle, when one of the sides is proβληθείσης ἡ ἐκτὸς γωνία ἑκατέρας τῶν ἐντὸς καὶ ἀπε- duced, the external angle is greater than each of the inναντίον 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 lies in the interior of the angle should be counted as an additional postulate. Þ. Proposition 17 Παντὸ ςτριγώνουαἱδύογωνίαιδύοὀρθῶνἐλάσσονές or any triangle, (the sum of) two angles taken toεἰσι πάντῇ μεταλαμβανόμεναι. gether in any (possible way) is less than two right-angles. στωτρίγωνοντὸ λέγω,ὅτιτοῦτριγώνου Let be a triangle. I say that (the sum of) two αἱ δύο γωνίαι δύο ὀρθῶν ἐλάττονές εἰσι πάντῃ μεταλαμ- angles of triangle taken together in any (possible βανόμεναι. way) is less than two right-angles. 21

22 κβεβλήσθω γὰρ ἡ ἐπὶ τὸ Δ. or let have been produced to. Καὶἐπεὶτριγώνουτοῦἐκτόςἐστιγωνίαἡὑπὸ nd since the angle is external to triangle, Δ,μείζωνἐστὶτῆςἐντὸςκαὶἀπεναντίοντῆςὑπὸ. it is greater than the internal and opposite angle κοινὴπροσκείσθωἡὑπὸ αἱἄραὑπὸδ,τῶν [Prop. 1.16]. Let have been added to both. Thus, ὑπὸ,μείζονέςεἰσιν. ἀλλ αἱὑπὸδ, the (sum of the angles) and is greater than δύοὀρθαῖςἴσαιεἰσίν αἱἄραὑπὸ,δύοὀρθῶν the (sum of the angles) and. ut, (the sum of) ἐλάσσονές εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ὑπὸ, and is equal to two right-angles [Prop. 1.13]. δύοὀρθῶνἐλάσσονέςεἰσικαὶἔτιαἱὑπὸ,. Thus, (the sum of) and is less than two right- Παντὸ ςἄρατριγώνουαἱδύογωνίαιδύοὀρθῶνἐλάσς- angles. Similarly, we can show that (the sum of) ονές εἰσι πάντῇ μεταλαμβανόμεναι ὅπερ ἔδει δεῖξαι. and is also less than two right-angles, and further (that the sum of) and (is less than two rightangles). Thus, for any triangle, (the sum of) two angles taken together in any (possible way) is less than two rightangles. (Which is) the very thing it was required to show.. Proposition 18 Παντὸς τριγώνου ἡ μείζων πλευρὰ τὴν μείζονα γωνίαν In any triangle, the greater side subtends the greater ὑποτείνει. angle. στω γὰρ τρίγωνον τὸ μείζονα ἔχον τὴν or let be a triangle having side greater than πλευρὰν τῆς λέγω, ὅτι καὶ γωνία ἡ ὑπὸ μείζων. I say that angle is also greater than. ἐστὶτῆςὑπὸ or since is greater than, let be made πεὶ γὰρ μείζων ἐστὶν ἡ τῆς, κείσθω τῇ ἴση equal to [Prop. 1.3], and let have been joined. ἡδ,καὶἐπεζεύχθωἡδ. nd since angle is external to triangle, it ΚαὶἐπεὶτριγώνουτοῦΔἐκτόςἐστιγωνίαἡὑπὸ is greater than the internal and opposite (angle) Δ, μείζων ἐστὶ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ Δ [Prop. 1.16]. ut (is) equal to, since side ἴσηδὲἡὑπὸδτῇὑπὸδ,ἐπεὶκαὶπλευρὰἡτῇ is also equal to side [Prop. 1.5]. Thus, is Δἐστινἴση μείζωνἄρακαὶἡὑπὸδτῆςὑπὸ also greater than. Thus, is much greater than πολλῷἄραἡὑπὸμείζωνἐστὶτῆςὑπὸ.. 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.. Proposition 19 Παντὸς τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων In any triangle, the greater angle is subtended by the πλευρὰὑποτείνει. greater side. στω τρίγωνον τὸ μείζονα ἔχον τὴν ὑπὸ Let be a triangle having the angle greater γωνίαντῆςὑπὸ λέγω,ὅτικαὶπλευρὰἡπλευρᾶς than. I say that side is also greater than side τῆς μείζων ἐστίν.. 22