EUCLID S ELEMENTS OF GEOMETRY

Transcript

1 ULI S LMNTS OF 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 Fitzpatrick

2 c Richard Fitzpatrick, ll rights reserved. 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. 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 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. Finally, 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. 4

5 LMNTS OOK 1 Fundamentals of plane geometry involving straight-lines 5

6 Οροι. efinitions α. Σηµε όν στιν, ο µέρος ο θέν. 1. point is that of which there is no part. β. ραµµ δ µ κος πλατές. 2. nd a line is a length without breadth. γ. ραµµ ς δ πέρατα σηµε α. 3. nd the extremities of a line are points. δ. Ε θε α γραµµή στιν, τις ξ σου το ς φ αυτ ς 4. straight-line is whatever lies evenly with points σηµείοις κε ται. upon itself. ε. Επιφάνεια δέ στιν, µ κος κα πλάτος µόνον 5. nd a surface is that which has length and breadth χει. alone.. Επιφανείας δ πέρατα γραµµαί. 6. nd the extremities of a surface are lines. ζ. Επίπεδος πιφάνειά στιν, τις ξ σου τα ς φ 7. plane surface is whatever lies evenly with αυτ ς ε θείαις κε ται. straight-lines upon itself. η. Επίπεδος δ γωνία στ ν ν πιπέδ δύο 8. nd a plane angle is the inclination of the lines, γραµµ ν πτοµένων λλήλων κα µ π ε θείας when two lines in a plane meet one another, and are not κειµένων πρ ς λλήλας τ ν γραµµ ν κλίσις. laid down straight-on with respect to one another. θ. Οταν δ α περιέχουσαι τ ν γωνίαν γραµµα 9. nd when the lines containing the angle are ε θε αι σιν, ε θύγραµµος καλε ται γωνία. straight then the angle is called rectilinear. ι. Οταν δ ε θε α π ε θε αν σταθε σα τ ς φεξ ς 10. nd when a straight-line stood upon (another) γωνίας σας λλήλαις ποι, ρθ κατέρα τ ν σων straight-line makes adjacent angles (which are) equal to γωνι ν στι, κα φεστηκυ α ε θε α κάθετος καλε ται, one another, each of the equal angles is a right-angle, and φ ν φέστηκεν. the former straight-line is called perpendicular to that ια. µβλε α γωνία στ ν µείζων ρθ ς. upon which it stands. ιβ. Οξε α δ λάσσων ρθ ς. 11. n obtuse angle is greater than a right-angle. ιγ. Ορος στίν, τινός στι πέρας. 12. nd an acute angle is less than a right-angle. ιδ. Σχ µά στι τ πό τινος τινων ρων πε- 13. boundary is that which is the extremity of someριεχόµενον. thing. ιε. Κύκλος στ σχ µα πίπεδον π µι ς γραµµ ς 14. figure is that which is contained by some boundπεριεχόµενον [ καλε ται περιφέρεια], πρ ς ν φ ary or boundaries. ν ς σηµείου τ ν ντ ς το σχήµατος κειµένων π σαι 15. circle is a plane figure contained by a single α προσπίπτουσαι ε θε αι [πρ ς τ ν το κύκλου πε- line [which is called a circumference], (such that) all of ριφέρειαν] σαι λλήλαις ε σίν. the straight-lines radiating towards [the circumference] ι. Κέντρον δ το κύκλου τ σηµε ον καλε ται. from a single point lying inside the figure are equal to ιζ. ιάµετρος δ το κύκλου στ ν ε θε ά τις δι το one another. κέντρου γµένη κα περατουµένη φ κάτερα τ µέρη 16. nd the point is called the center of the circle. π τ ς το κύκλου περιφερείας, τις κα δίχα τέµνει 17. nd a diameter of the circle is any straight-line, τ ν κύκλον. being drawn through the center, which is brought to an ιη. Ηµικύκλιον δέ στι τ περιεχόµενον σχ µα πό end in each direction by the circumference of the circle. τε τ ς διαµέτρου κα τ ς πολαµβανοµένης π α τ ς nd any such (straight-line) cuts the circle in half. περιφερείας. κέντρον δ το µικυκλίου τ α τό, κα 18. nd a semi-circle is the figure contained by the το κύκλου στίν. diameter and the circumference it cuts off. nd the center ιθ. Σχήµατα ε θύγραµµά στι τ π ε θει ν πε- of the semi-circle is the same (point) as (the center of) the ριεχόµενα, τρίπλευρα µ ν τ π τρι ν, τετράπλευρα circle. δ τ π τεσσάρων, πολύπλευρα δ τ π πλειόνων 19. Rectilinear figures are those figures contained by τεσσάρων ε θει ν περιεχόµενα. straight-lines: trilateral figures being contained by three κ. Τ ν δ τριπλεύρων σχηµάτων σόπλευρον µ ν straight-lines, quadrilateral by four, and multilateral by τρίγωνόν στι τ τ ς τρε ς σας χον πλευράς, σοσκελ ς more than four. δ τ τ ς δύο µόνας σας χον πλευράς, σκαλην ν δ τ 20. nd of the trilateral figures: an equilateral trianτ ς τρε ς νίσους χον πλευράς. gle is that having three equal sides, an isosceles (triangle) κα Ετι δ τ ν τριπλεύρων σχηµάτων ρθογώνιον that having only two equal sides, and a scalene (triangle) µ ν τρίγωνόν στι τ χον ρθ ν γωνίαν, µβλυγώνιον that having three unequal sides. 6

7 δ τ χον µβλε αν γωνίαν, ξυγώνιον δ τ τ ς τρε ς 21. nd further of the trilateral figures: a right-angled ξείας χον γωνίας. triangle is that having a right-angle, an obtuse-angled κβ. Τ ν δ τετραπλεύρων σχηµάτων τετράγωνον µέν (triangle) that having an obtuse angle, and an acute- στιν, σόπλευρόν τέ στι κα ρθογώνιον, τερόµηκες angled (triangle) that having three acute angles. δέ, ρθογώνιον µέν, ο κ σόπλευρον δέ, όµβος δέ, 22. nd of the quadrilateral figures: a square is that σόπλευρον µέν, ο κ ρθογώνιον δέ, οµβοειδ ς δ which is right-angled and equilateral, a rectangle that τ τ ς πεναντίον πλευράς τε κα γωνίας σας λλήλαις which is right-angled but not equilateral, a rhombus that χον, ο τε σόπλευρόν στιν ο τε ρθογώνιον τ δ which is equilateral but not right-angled, and a rhomboid παρ τα τα τετράπλευρα τραπέζια καλείσθω. that having opposite sides and angles equal to one anκγ. Παράλληλοί ε σιν ε θε αι, α τινες ν τ α τ other which is neither right-angled nor equilateral. nd πιπέδ ο σαι κα κβαλλόµεναι ε ς πειρον φ κάτερα let quadrilateral figures besides these be called trapezia. τ µέρη π µηδέτερα συµπίπτουσιν λλήλαις. 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. τήµατα. Postulates α. Ηιτήσθω π παντ ς σηµείου π π ν σηµε ον 1. Let it have been postulated to draw a straight-line ε θε αν γραµµ ν γαγε ν. from any point to any point. β. Κα πεπερασµένην ε θε αν κατ τ συνεχ ς π 2. nd to produce a finite straight-line continuously ε θείας κβαλε ν. in a straight-line. γ. Κα παντ κέντρ κα διαστήµατι κύκλον γράφεσ- 3. nd to draw a circle with any center and radius. θαι. 4. nd that all right-angles are equal to one another. δ. Κα πάσας τ ς ρθ ς γωνίας σας λλήλαις ε ναι. 5. nd that if a straight-line falling across two (other) ε. Κα ν ε ς δύο ε θείας ε θε α µπίπτουσα straight-lines makes internal angles on the same side (of τ ς ντ ς κα π τ α τ µέρη γωνίας δύο ρθ ν itself whose sum is) less than two right-angles, then, be- λάσσονας ποι, κβαλλοµένας τ ς δύο ε θείας π πει- ing produced to infinity, the two (other) straight-lines ρον συµπίπτειν, φ µέρη ε σ ν α τ ν δύο ρθ ν meet on that side (of the original straight-line) that the λάσσονες. (sum of the internal angles) is less than two right-angles (and do not meet on the other side). This postulate effectively specifies that we are dealing with the geometry of flat, rather than curved, space. Κοινα ννοιαι. ommon Notions α. Τ τ α τ σα κα λλήλοις στ ν σα. 1. Things equal to the same thing are also equal to β. Κα ν σοις σα προστεθ, τ λα στ ν σα. one another. γ.κα ν π σων σα φαιρεθ, τ καταλειπόµενά 2. nd if equal things are added to equal things then στιν σα. the wholes are equal. δ. Κα τ φαρµόζοντα π λλήλα σα λλήλοις 3. nd if equal things are subtracted from equal things στίν. then the remainders are equal. ε. Κα τ λον το µέρους µε ζόν [ στιν]. 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 an inequality of the same type. 7

8 α. Proposition 1 Επ τ ς δοθείσης ε θείας πεπερασµένης τρίγωνον To construct an equilateral triangle on a given finite σόπλευρον συστήσασθαι. straight-line. Ε Εστω δοθε σα ε θε α πεπερασµένη. Let be the given finite straight-line. ε δ π τ ς ε θείας τρίγωνον σόπλευρον So it is required to construct an equilateral triangle on συστήσασθαι. the straight-line. Κέντρ µ ν τ διαστήµατι δ τ κύκλος Let the circle with center and radius have γεγράφθω, κα πάλιν κέντρ µ ν τ διαστήµατι been drawn [Post. 3], and again let the circle with δ τ κύκλος γεγράφθω Ε, κα π το center and radius have been drawn [Post. 3]. nd σηµείου, καθ τέµνουσιν λλήλους ο κύκλοι, πί τ, let the straight-lines and have been joined from σηµε α πεζεύχθωσαν ε θε αι α,. the point, where the circles cut one another, to the Κα πε τ σηµε ον κέντρον στ το κύκλου, points and (respectively) [Post. 1]. ση στ ν τ πάλιν, πε τ σηµε ον κέντρον nd since the point is the center of the circle, στ το Ε κύκλου, ση στ ν τ. δείχθη is equal to [ef. 1.15]. gain, since the point δ κα τ ση κατέρα ρα τ ν, τ is the center of the circle, is equal to στιν ση. τ δ τ α τ σα κα λλήλοις στ ν σα [ef. 1.15]. ut was also shown (to be) equal to. κα ρα τ στιν ση α τρε ς ρα α,, Thus, and are each equal to. ut things equal σαι λλήλαις ε σίν. to the same thing are also equal to one another [.N. 1]. Ισόπλευρον ρα στ τ τρίγωνον. κα Thus, is also equal to. Thus, the three (straightσυνέσταται π τ ς δοθείσης ε θείας πεπερασµένης τ ς lines),, and are equal to one another. περ δει ποι σαι. 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. β. Proposition 2 Πρ ς τ δοθέντι σηµεί τ δοθείσ ε θεί σην To place a straight-line equal to a given straight-line ε θε αν θέσθαι. at a given point. Εστω τ µ ν δοθ ν σηµε ον τ, δ δοθε σα Let be the given point, and the given straightε θε α δε δ πρ ς τ σηµεί τ δοθείσ line. So it is required to place a straight-line at point ε θεί τ σην ε θε αν θέσθαι. equal to the given straight-line. Επεζεύχθω γ ρ π το σηµείου πί τ For let the straight-line have been joined from σηµε ον ε θε α, κα συνεστάτω π α τ ς τρίγωνον point to point [Post. 1], and let the equilateral trian- σόπλευρον τ, κα κβεβλήσθωσαν π ε θείας gle have been been constructed upon it [Prop. 1.1]. τα ς, ε θε αι α Ε, Ζ, κα κέντρ µ ν τ nd let the straight-lines and F have been pro- διαστήµατι δ τ κύκλος γεγράφθω ΗΘ, duced in a straight-line with and (respectively) κα πάλιν κέντρ τ κα διαστήµατι τ Η κύκλος [Post. 2]. nd let the circle GH with center and ra- 8

9 γεγράφθω ΗΚΛ. dius 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 F Λ L Ε Επε ο ν τ σηµε ον κέντρον στ το ΗΘ, ση Therefore, since the point is the center of (the cir- στ ν τ Η. πάλιν, πε τ σηµε ον κέντρον cle) GH, is equal to G [ef. 1.15]. gain, since στ το ΗΚΛ κύκλου, ση στ ν Λ τ Η, ν the point is the center of the circle GKL, L is equal τ ση στίν. λοιπ ρα Λ λοιπ τ Η to G [ef. 1.15]. nd within these, is equal to. στιν ση. δείχθη δ κα τ Η ση κατέρα ρα Thus, the remainder L is equal to the remainder G τ ν Λ, τ Η στιν ση. τ δ τ α τ σα κα [.N. 3]. ut was also shown (to be) equal to G. λλήλοις στ ν σα κα Λ ρα τ στιν ση. Thus, L and are each equal to G. ut things equal Πρ ς ρα τ δοθέντι σηµεί τ τ δοθείσ to the same thing are also equal to one another [.N. 1]. ε θεί τ ση ε θε α κε ται Λ περ δει ποι σαι. 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. γ. Proposition 3 ύο δοθεισ ν ε θει ν νίσων π τ ς µείζονος τ For two given unequal straight-lines, to cut off from λάσσονι σην ε θε αν φελε ν. the greater a straight-line equal to the lesser. Εστωσαν α δοθε σαι δύο ε θε αι νισοι α,, Let and be the two given unequal straight-lines, ν µείζων στω δε δ π τ ς µείζονος τ ς of which let the greater be. So it is required to cut off τ λάσσονι τ σην ε θε αν φελε ν. a straight-line equal to the lesser from the greater. Κείσθω πρ ς τ σηµεί τ ε θεί ση Let the line, equal to the straight-line, have κα κέντρ µ ν τ διαστήµατι δ τ κύκλος been placed at point [Prop. 1.2]. nd let the circle γεγράφθω ΕΖ. F have been drawn with center and radius Κα πε τ σηµε ον κέντρον στ το ΕΖ [Post. 3]. κύκλου, ση στ ν Ε τ λλ κα τ nd since point is the center of circle F, στιν ση. κατέρα ρα τ ν Ε, τ στιν ση is equal to [ef. 1.15]. ut, is also equal to. στε κα Ε τ στιν ση. Thus, and are each equal to. So is also 9

10 equal to [.N. 1]. Ε Ζ ύο ρα δοθεισ ν ε θει ν νίσων τ ν, π Thus, for two given unequal straight-lines, and, τ ς µείζονος τ ς τ λάσσονι τ ση φ ρηται the (straight-line), equal to the lesser, has been cut Ε περ δει ποι σαι. off from the greater. (Which is) the very thing it was required to do. δ. Proposition 4 Ε ν δύο τρίγωνα τ ς δύο πλευρ ς [τα ς] δυσ If two triangles have two corresponding sides equal, πλευρα ς σας χ κατέραν κατέρ κα τ ν γωνίαν τ and have the angles enclosed by the equal sides equal, γωνί σην χ τ ν π τ ν σων ε θει ν περιεχοµένην, then they will also have equal bases, and the two trianκα τ ν βάσιν τ βάσει σην ξει, κα τ τρίγωνον τ gles will be equal, and the remaining angles subtended τριγών σον σται, κα α λοιπα γωνίαι τα ς λοιπα ς by the equal sides will be equal to the corresponding reγωνίαις σαι σονται κατέρα κατέρ, φ ς α σαι maining angles. πλευρα ποτείνουσιν. F Ε Ζ F Εστω δύο τρίγωνα τ, ΕΖ τ ς δύο πλευρ ς Let and F be two triangles having the two τ ς, τα ς δυσ πλευρα ς τα ς Ε, Ζ σας χοντα sides and equal to the two sides and F, re- κατέραν κατέρ τ ν µ ν τ Ε τ ν δ τ spectively. (That is) to, and to F. nd (let) Ζ κα γωνίαν τ ν π γωνί τ π Ε Ζ σην. the angle (be) equal to the angle F. I say that λέγω, τι κα βάσις βάσει τ ΕΖ ση στίν, κα the base is also equal to the base F, and triangle τ τρίγωνον τ ΕΖ τριγών σον σται, κα α will be equal to triangle F, and the remaining λοιπα γωνίαι τα ς λοιπα ς γωνίαις σαι σονται κατέρα angles subtended by the equal sides will be equal to the κατέρ, φ ς α σαι πλευρα ποτείνουσιν, µ ν π corresponding remaining angles. (That is) to F, τ π ΕΖ, δ π τ π ΖΕ. and to F. Εφαρµοζοµένου γ ρ το τριγώνου π τ ΕΖ Let the triangle be applied to the triangle τρίγωνον κα τιθεµένου το µ ν σηµείου π τ F, the point being placed on the point, and σηµε ον τ ς δ ε θείας π τ ν Ε, φαρµόσει κα the straight-line on. The point will also coinτ σηµε ον π τ Ε δι τ σην ε ναι τ ν τ cide with, on account of being equal to. So Ε φαρµοσάσης δ τ ς π τ ν Ε φαρµόσει (because of) coinciding with, the straight-line 10

11 κα ε θε α π τ ν Ζ δι τ σην ε ναι τ ν π will also coincide with F, on account of the angle γωνίαν τ π Ε Ζ στε κα τ σηµε ον π being equal to F. So the point will also coτ Ζ σηµε ον φαρµόσει δι τ σην πάλιν ε ναι τ ν incide with the point F, again on account of being τ Ζ. λλ µ ν κα τ π τ Ε φηρµόκει στε equal to F. ut, point certainly also coincided with βάσις π βάσιν τ ν ΕΖ φαρµόσει. ε γ ρ το point, so that the base will coincide with the base µ ν π τ Ε φαρµόσαντος το δ π τ Ζ F. For if coincides with, and with F, and the βάσις π τ ν ΕΖ ο κ φαρµόσει, δύο ε θε αι χωρίον base does not coincide with F, then two straightπεριέξουσιν περ στ ν δύνατον. φαρµόσει ρα lines will encompass an area. The very thing is impossible βάσις π τ ν ΕΖ κα ση α τ σται στε κα λον τ [Post. 1]. Thus, the base will coincide with F, and τρίγωνον π λον τ ΕΖ τρίγωνον φαρµόσει will be equal to it [.N. 4]. So the whole triangle κα σον α τ σται, κα α λοιπα γωνίαι π τ ς λοιπ ς will coincide with the whole triangle F, and will be γωνίας φαρµόσουσι κα σαι α τα ς σονται, µ ν π equal to it [.N. 4]. nd the remaining angles will co- τ π ΕΖ δ π τ π ΖΕ. incide with the remaining angles, and will be equal to Ε ν ρα δύο τρίγωνα τ ς δύο πλευρ ς [τα ς] δύο them [.N. 4]. (That is) to F, and to πλευρα ς σας χ κατέραν κατέρ κα τ ν γωνίαν τ F [.N. 4]. γωνί σην χ τ ν π τ ν σων ε θει ν περιεχοµένην, Thus, if two triangles have two corresponding sides κα τ ν βάσιν τ βάσει σην ξει, κα τ τρίγωνον τ equal, and have the angles enclosed by the equal sides τριγών σον σται, κα α λοιπα γωνίαι τα ς λοιπα ς equal, then they will also have equal bases, and the two γωνίαις σαι σονται κατέρα κατέρ, φ ς α σαι triangles will be equal, and the remaining angles subπλευρα ποτείνουσιν περ δει δε ξαι. tended 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. ε. Proposition 5 Τ ν σοσκελ ν τριγώνων α τρ ς τ βάσει γωνίαι σαι For isosceles triangles, the angles at the base are equal λλήλαις ε σίν, κα προσεκβληθεισ ν τ ν σων ε θει ν to one another, and if the equal sides are produced then α π τ ν βάσιν γωνίαι σαι λλήλαις σονται. the angles under the base will be equal to one another. Ζ Η F G Ε Εστω τρίγωνον σοσκελ ς τ σην χον τ ν Let be an isosceles triangle having the side πλευρ ν τ πλευρ, κα προσεκβεβλήσθωσαν equal to the side, and let the straight-lines and π ε θείας τα ς, ε θε αι α, Ε λέγω, τι have been produced in a straight-line with and µ ν π γωνία τ π ση στίν, δ π (respectively) [Post. 2]. I say that the angle is τ π Ε. equal to, and (angle) to. Ε λήφθω γ ρ π τ ς τυχ ν σηµε ον τ Ζ, κα For let the point F have been taken somewhere on φ ρήσθω π τ ς µείζονος τ ς Ε τ λάσσονι τ Ζ, and let G have been cut off from the greater, ση Η, κα πεζεύχθωσαν α Ζ, Η ε θε αι. equal to the lesser F [Prop. 1.3]. lso, let the straight- Επε ο ν ση στ ν µ ν Ζ τ Η δ τ lines F and G have been joined [Post. 1]. 11

12 , δύο δ α Ζ, δυσ τα ς Η, σαι ε σ ν In fact, since F is equal to G, and to, κατέρα κατέρ κα γωνίαν κοιν ν περιέχουσι τ ν π the two (straight-lines) F, are equal to the two ΖΗ βάσις ρα Ζ βάσει τ Η ση στίν, κα (straight-lines) G,, respectively. They also encomτ Ζ τρίγωνον τ Η τριγών σον σται, κα α pass a common angle FG. Thus, the base F is equal λοιπα γωνίαι τα ς λοιπα ς γωνίαις σαι σονται κατέρα to the base G, and the triangle F will be equal to the κατέρ, φ ς α σαι πλευρα ποτείνουσιν, µ ν π triangle G, and the remaining angles subtendend by Ζ τ π Η, δ π Ζ τ π Η. κα πε the equal sides will be equal to the corresponding remain- λη Ζ λ τ Η στιν ση, ν τ στιν ing angles [Prop. 1.4]. (That is) F to G, and F ση, λοιπ ρα Ζ λοιπ τ Η στιν ση. δείχθη δ to G. nd since the whole of F is equal to the whole κα Ζ τ Η ση δύο δ α Ζ, Ζ δυσ τα ς Η, of G, within which is equal to, the remainder Η σαι ε σ ν κατέρα κατέρ κα γωνία π Ζ F is thus equal to the remainder G [.N. 3]. ut F γωνί τ π Η ση, κα βάσις α τ ν κοιν was also shown (to be) equal to G. So the two (straightκα τ Ζ ρα τρίγωνον τ Η τριγών σον σται, lines) F, F are equal to the two (straight-lines) G, κα α λοιπα γωνίαι τα ς λοιπα ς γωνίαις σαι σονται G, respectively, and the angle F (is) equal to the κατέρα κατέρ, φ ς α σαι πλευρα ποτείνουσιν angle G, and the base is common to them. Thus, ση ρα στ ν µ ν π Ζ τ π Η δ π the triangle F will be equal to the triangle G, and Ζ τ π Η. πε ο ν λη π Η γωνία λ the remaining angles subtended by the equal sides will be τ π Ζ γωνί δείχθη ση, ν π Η τ π equal to the corresponding remaining angles [Prop. 1.4]. Ζ ση, λοιπ ρα π λοιπ τ π Thus, F is equal to G, and F to G. There- στιν ση καί ε σι πρ ς τ βάσει το τριγώνου. fore, since the whole angle G was shown (to be) equal δείχθη δ κα π Ζ τ π Η ση καί ε σιν to the whole angle F, within which G is equal to π τ ν βάσιν. F, the remainder is thus equal to the remainder Τ ν ρα σοσκελ ν τριγώνων α τρ ς τ βάσει [.N. 3]. nd they are at the base of triangle. γωνίαι σαι λλήλαις ε σίν, κα προσεκβληθεισ ν τ ν nd F was also shown (to be) equal to G. nd σων ε θει ν α π τ ν βάσιν γωνίαι σαι λλήλαις they are under the base. σονται περ δει δε ξαι. 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 Ε ν τριγώνου α δ ο γωνίαι σαι λλήλαις σιν, If a triangle has two angles equal to one another then κα α π τ ς σας γωνίας ποτείνουσαι πλευρα σαι the sides subtending the equal angles will also be equal λλήλαις σονται. to one another. Εστω τρίγωνον τ σην χον τ ν π Let be a triangle having the angle equal γωνίαν τ π γωνί λέγω, τι κα πλευρ to the angle. I say that side is also equal to side πλευρ τ στιν ση.. Ε γ ρ νισός στιν τ, τέρα α τ ν For if is unequal to then one of them is µείζων στίν. στω µείζων, κα φ ρήσθω π greater. Let be greater. nd let, equal to 12

13 τ ς µείζονος τ ς τ λάττονι τ ση, κα the lesser, have been cut off from the greater πεζεύχθω. [Prop. 1.3]. nd let have been joined [Post. 1]. Επε ο ν ση στ ν τ κοιν δ, Therefore, since is equal to, and (is) comδύο δ α, δύο τα ς, σαι ε σ ν κατέρα mon, the two sides, are equal to the two sides κατέρ, κα γωνία π γωνι τ π στιν,, respectively, and the angle is equal to the ση βάσις ρα βάσει τ ση στίν, κα τ angle. Thus, the base is equal to the base, τρίγωνον τ τριγών σον σται, τ λασσον τ and the triangle will be equal to the triangle µείζονι περ τοπον ο κ ρα νισός στιν τ [Prop. 1.4], the lesser to the greater. The very notion (is) ση ρα. absurd [.N. 5]. Thus, is not unequal to. Thus, Ε ν ρα τριγώνου α δ ο γωνίαι σαι λλήλαις σιν, (it is) equal. κα α π τ ς σας γωνίας ποτείνουσαι πλευρα σαι Thus, if a triangle has two angles equal to one another λλήλαις σονται περ δει δε ξαι. 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. ζ. Proposition 7 Επ τ ς α τ ς ε θείας δύο τα ς α τα ς ε θείαις λλαι On the same straight-line, two other straight-lines δύο ε θε αι σαι κατέρα κατέρ ο συσταθήσονται equal, respectively, to two (given) straight-lines (which πρ ς λλ κα λλ σηµεί π τ α τ µέρη τ α τ meet) cannot be constructed (meeting) at a different πέρατα χουσαι τα ς ξ ρχ ς ε θείαις. point on the same side (of the straight-line), but having the same ends as the given straight-lines. Ε γ ρ δυνατόν, π τ ς α τ ς ε θείας τ ς δύο For, if possible, let the two straight-lines,, τα ς α τα ς ε θείαις τα ς, λλαι δύο ε θε αι α equal to two (given) straight-lines,, respectively,, σαι κατέρα κατερ συνεστάτωσαν πρ ς λλ have been constructed on the same straight-line, κα λλ σηµεί τ τε κα π τ α τ µέρη τ meeting at different points, and, on the same side α τ πέρατα χουσαι, στε σην ε ναι τ ν µ ν τ (of ), and having the same ends (on ). So and τ α τ πέρας χουσαν α τ τ, τ ν δ τ are equal, having the same ends at, and and τ α τ πέρας χουσαν α τ τ, κα πεζεύχθω are equal, having the same ends at. nd let. have been joined [Post. 1]. Επε ο ν ση στ ν τ, ση στ κα γωνία Therefore, since is equal to, the angle π τ π µείζων ρα π τ ς is also equal to angle [Prop. 1.5]. Thus, (is) π πολλ ρα π µείζων στί τ ς π greater than [.N. 5]. Thus, is much greater. πάλιν πε ση στ ν τ, ση στ κα than [.N. 5]. gain, since is equal to, the γωνία π γωνί τ π. δείχθη δ α τ ς angle is also equal to angle [Prop. 1.5]. ut κα πολλ µείζων περ στ ν δύατον. it was shown that the former (angle) is also much greater Ο κ ρα π τ ς α τ ς ε θείας δύο τα ς α τα ς (than the latter). The very thing is impossible. ε θείαις λλαι δύο ε θε αι σαι κατέρα κατέρ συ- Thus, on the same straight-line, two other straight- 13

14 σταθήσονται πρ ς λλ κα λλ σηµεί π τ α τ lines equal, respectively, to two (given) straight-lines µέρη τ α τ πέρατα χουσαι τα ς ξ ρχ ς ε θείαις (which meet) cannot be constructed (meeting) at a dif- περ δει δε ξαι. 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. η. Proposition 8 Ε ν δύο τρίγωνα τ ς δύο πλευρ ς [τα ς] δύο If two triangles have two corresponding sides equal, πλευρα ς σας χ κατέραν κατέρ, χ δ κα τ ν and also have equal bases, then the angles encompassed βάσιν τ βάσει σην, κα τ ν γωνίαν τ γωνί σην ξει by the equal straight-lines will also be equal. τ ν π τ ν σων ε θει ν περιεχοµένην. Η G Ζ F Ε Εστω δύο τρίγωνα τ, ΕΖ τ ς δύο πλευρ ς Let and F be two triangles having the two τ ς, τα ς δύο πλευρα ς τα ς Ε, Ζ σας χοντα sides and equal to the two sides and F, κατέραν κατέρ, τ ν µ ν τ Ε τ ν δ τ Ζ respectively. (That is) to, and to F. Let χέτω δ κα βάσιν τ ν βάσει τ ΕΖ σην λέγω, τι them also have the base equal to the base F. I say κα γωνία π γωνί τ π Ε Ζ στιν ση. that the angle is also equal to the angle F. Εφαρµοζοµένου γ ρ το τριγώνου π τ ΕΖ For if triangle is applied to triangle F, the τρίγωνον κα τιθεµένου το µ ν σηµείου π τ Ε point being placed on point, and the straight-line σηµε ον τ ς δ ε θείας π τ ν ΕΖ φαρµόσει κα on F, point will also coincide with F, on account τ σηµε ον π τ Ζ δι τ σην ε ναι τ ν τ ΕΖ of being equal to F. So (because of) coinciding φαρµοσάσης δ τ ς π τ ν ΕΖ φαρµόσουσι κα with F, (the sides) and will also coincide with α, π τ ς Ε, Ζ. ε γ ρ βάσις µ ν and F (respectively). For if base coincides with π βάσιν τ ν ΕΖ φαρµόσει, α δ, πλευρα base F, but the sides and do not coincide with π τ ς Ε, Ζ ο κ φαρµόσουσιν λλ παραλλάξουσιν and F (respectively), but miss like G and GF (in ς α ΕΗ, ΗΖ, συσταθήσονται π τ ς α τ ς ε θείας the above figure), then we will have constructed upon δύο τα ς α τα ς ε θείαις λλαι δύο ε θε αι σαι κατέρα the same straight-line, two other straight-lines equal, re- κατέρ πρ ς λλ κα λλ σηµεί π τ α τ µέρη spectively, to two (given) straight-lines, and (meeting) at τ α τ πέρατα χουσαι. ο συνίστανται δέ ο κ ρα a different point on the same side (of the straight-line), φαρµοζοµένης τ ς βάσεως π τ ν ΕΖ βάσιν ο κ but having the same ends. ut (such straight-lines) can- φαρµόσουσι κα α, πλευρα π τ ς Ε, Ζ. not be constructed [Prop. 1.7]. Thus, the base being φαρµόσουσιν ρα στε κα γωνία π π applied to the base F, the sides and cannot not γωνίαν τ ν π Ε Ζ φαρµόσει κα ση α τ σται. coincide with and F (respectively). Thus, they will Ε ν ρα δύο τρίγωνα τ ς δύο πλευρ ς [τα ς] δύο coincide. So the angle will also coincide with angle πλευρα ς σας χ κατέραν κατέρ κα τ ν βάσιν τ F, and they will be equal [.N. 4]. βάσει σην χ, κα τ ν γωνίαν τ γωνί σην ξει τ ν Thus, if two triangles have two corresponding sides π τ ν σων ε θει ν περιεχοµένην περ δει δε ξαι. equal, and have equal bases, then the angles encompassed by the equal straight-lines will also be equal. (Which is) the very thing it was required to show. 14

15 θ. Proposition 9 Τ ν δοθε σαν γωνίαν ε θύγραµµον δίχα τεµε ν. To cut a given rectilinear angle in half. Ε Ζ Εστω δοθε σα γωνία ε θύγραµµος π. Let be the given rectilinear angle. So it is reδε δ α τ ν δίχα τεµε ν. quired to cut it in half. Ε λήφθω π τ ς τυχ ν σηµε ον τ, κα Let the point have been taken somewhere on, φ ρήσθω π τ ς τ ση Ε, κα πεζεύχθω and let, equal to, have been cut off from Ε, κα συνεστάτω π τ ς Ε τρίγωνον σόπλευρον [Prop. 1.3], and let have been joined. nd let the τ ΕΖ, κα πεζεύχθω Ζ λέγω, τι π equilateral triangle F have been constructed upon γωνία δίχα τέτµηται π τ ς Ζ ε θείας. [Prop. 1.1], and let F have been joined. I say that Επε γ ρ ση στ ν τ Ε, κοιν δ Ζ, the angle has been cut in half by the straight-line δύο δ α, Ζ δυσ τα ς Ε, Ζ σαι ε σ ν κατέρα F. κατέρ. κα βάσις Ζ βάσει τ ΕΖ ση στίν γωνία For since is equal to, and F is common, ρα π Ζ γωνί τ π ΕΖ ση στίν. the two (straight-lines), F are equal to the two Η ρα δοθε σα γωνία ε θύγραµµος π δίχα (straight-lines), F, respectively. nd the base F τέτµηται π τ ς Ζ ε θείας περ δει ποι σαι. is equal to the base F. Thus, angle F is equal to angle F [Prop. 1.8]. Thus, the given rectilinear angle has been cut in half by the straight-line F. (Which is) the very thing it was required to do. ι. Proposition 10 Τ ν δοθε σαν ε θε αν πεπερασµένην δίχα τεµε ν. To cut a given finite straight-line in half. Εστω δοθε σα ε θε α πεπερασµένη δε δ Let be the given finite straight-line. So it is reτ ν ε θε αν πεπερασµένην δίχα τεµε ν. quired to cut the finite straight-line in half. Συνεστάτω π α τ ς τρίγωνον σόπλευρον τ, Let the equilateral triangle have been conκα τετµήσθω π γωνία δίχα τ ε θεί structed upon () [Prop. 1.1], and let the angle λέγω, τι ε θε α δίχα τέτµηται κατ τ σηµε ον. have been cut in half by the straight-line [Prop. 1.9]. Επε γ ρ ση στ ν τ, κοιν δ, I say that the straight-line has been cut in half at δύο δ α, δύο τα ς, σαι ε σ ν κατέρα point. κατέρ κα γωνία π γωνί τ π ση For since is equal to, and (is) common, στίν βάσις ρα βάσει τ ση στίν. 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]. F 15

16 Η ρα δοθε σα ε θε α πεπερασµένη δίχα τέτµηται κατ τ περ δει ποι σαι. Thus, the given finite straight-line has been cut in half at (point). (Which is) the very thing it was required to do. ια. Proposition 11 ΠΤ δοθείσ ε θεί π το πρ ς α τ δοθέντος σηµείου πρ ς ρθ ς γωνίας ε θε αν γραµµ ν γαγε ν. Ζ To draw a straight-line at right-angles to a given straight-line from a given point on it. F Ε Εστω µ ν δοθε σα ε θε α τ δ δοθ ν Let be the given straight-line, and the given σηµε ον π α τ ς τ δε δ π το σηµείου τ point on it. So it is required to draw a straight-line from ε θεί πρ ς ρθ ς γωνίας ε θε αν γραµµ ν γαγε ν. the point at right-angles to the straight-line. Ε λήφθω π τ ς τυχ ν σηµε ον τ, κα κείσθω Let the point be have been taken somewhere on τ ση Ε, κα συνεστάτω π τ ς Ε τρίγωνον, and let be made equal to [Prop. 1.3], and σόπλευρον τ Ζ Ε, κα πεζεύχθω Ζ λέγω, τι let the equilateral triangle F have been constructed τ δοθείσ ε θεί τ π το πρ ς α τ δοθέντος on [Prop. 1.1], and let F have been joined. I say σηµείου το πρ ς ρθ ς γωνίας ε θε α γραµµ κται that the straight-line F has been drawn at right-angles Ζ. to the given straight-line from the given point on Επε γ ρ ση στ ν τ Ε, κοιν δ Ζ, it. δύο δ α, Ζ δυσ τα ς Ε, Ζ σαι ε σ ν κατέρα For since is equal to, and F is common, κατέρ κα βάσις Ζ βάσει τ ΖΕ ση στίν γωνία the two (straight-lines), F are equal to the two ρα π Ζ γωνί τ π ΕΖ ση στίν καί (straight-lines),, F, respectively. nd the base F ε σιν φεξ ς. ταν δ ε θε α π ε θε αν σταθε σα τ ς is equal to the base F. Thus, the angle F is equal φεξ ς γωνίας σας λλήλαις ποι, ρθ κατέρα τ ν to the angle F [Prop. 1.8], and they are adjacent. σων γωνι ν στιν ρθ ρα στ ν κατέρα τ ν π ut when a straight-line stood on a(nother) straight-line Ζ, ΖΕ. makes the adjacent angles equal to one another, each of Τ ρα δοθείσ ε θεί τ π το πρ ς α τ the equal angles is a right-angle [ef. 1.10]. Thus, each δοθέντος σηµείου το πρ ς ρθ ς γωνίας ε θε α of the (angles) F and F is a right-angle. γραµµ κται Ζ περ δει ποι σαι. Thus, the straight-line F has been drawn at right- 16

17 angles to the given straight-line from the given point on it. (Which is) the very thing it was required to do. ιβ. Proposition 12 Επ τ ν δοθε σαν ε θε αν πειρον π το δοθέντος To draw a straight-line perpendicular to a given infiσηµείου, µή στιν π α τ ς, κάθετον ε θε αν γραµµ ν nite straight-line from a given point which is not on it. γαγε ν. Ζ F Η Θ Ε Εστω µ ν δοθε σα ε θε α πειρος τ δ Let be the given infinite straight-line and the δοθ ν σηµε ον, µή στιν π α τ ς, τ δε δ π given point, which is not on (). So it is required to τ ν δοθε σαν ε θε αν πειρον τ ν π το δοθέντος draw a straight-line perpendicular to the given infinite σηµείου το, µή στιν π α τ ς, κάθετον ε θε αν straight-line from the given point, which is not on γραµµ ν γαγε ν. (). Ε λήφθω γ ρ π τ τερα µέρη τ ς ε θείας For let point have been taken somewhere on the τυχ ν σηµε ον τ, κα κέντρ µ ν τ διαστήµατι δ other side (to ) of the straight-line, and let the cirτ κύκλος γεγράφθω ΕΖΗ, κα τετµήσθω ΕΗ cle FG have been drawn with center and radius ε θε α δίχα κατ τ Θ, κα πεζεύχθωσαν α Η, Θ, [Post. 3], and let the straight-line G have been cut in Ε ε θε αι λέγω, τι π τ ν δοθε σαν ε θε αν πειρον half at (point) H [Prop. 1.10], and let the straight-lines τ ν π το δοθέντος σηµείου το, µή στιν π G, H, and have been joined. I say that a (straightα τ ς, κάθετος κται Θ. line) H has been drawn perpendicular to the given in- Επε γ ρ ση στ ν ΗΘ τ ΘΕ, κοιν δ Θ, finite straight-line from the given point, which is δύο δ α ΗΘ, Θ δύο τα ς ΕΘ, Θ σαι ε σ ν κατέρα not on (). κατέρ κα βάσις Η βάσει τ Ε στιν ση γωνία For since GH is equal to H, and H (is) common, ρα π ΘΗ γωνί τ π ΕΘ στιν ση. καί the two (straight-lines) GH, H are equal to the two ε σιν φεξ ς. ταν δ ε θε α π ε θε αν σταθε σα τ ς straight-lines H, H, respectively, and the base G is φεξ ς γωνίας σας λλήλαις ποι, ρθ κατέρα τ ν equal to the base. Thus, the angle HG is equal σων γωνι ν στιν, κα φεστηκυ α ε θε α κάθετος to the angle H [Prop. 1.8], and they are adjacent. καλε ται φ ν φέστηκεν. ut when a straight-line stood on a(nother) straight-line Επ τ ν δοθε σαν ρα ε θε αν πειρον τ ν π makes the adjacent angles equal to one another, each of το δοθέντος σηµείου το, µή στιν π α τ ς, the equal angles is a right-angle, and the former straightκάθετος κται Θ περ δει ποι σαι. line is called 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 given point, which is not on (). (Which is) the very thing it was required to do. G H 17

18 ιγ. Proposition 13 Ε ν ε θε α π ε θε αν σταθε σα γωνίας ποι, τοι δύο ρθ ς δυσ ν ρθα ς σας ποιήσει. Ε If a straight-line stood on a(nother) straight-line makes angles, it will certainly either make two rightangles, or (angles whose sum is) equal to two rightangles. Ε θε α γάρ τις π ε θε αν τ ν σταθε σα For let some straight-line stood on the straightγωνίας ποιείτω τ ς π, λ γω, τι α π line make the angles and. I say that, γωνίαι τοι δύο ρθαί ε σιν δυσ ν ρθα ς the angles and are certainly either two right- σαι. angles, or (have a sum) equal to two right-angles. Ε µ ν ο ν ση στ ν π τ π, δύο In fact, if is equal to then they are two ρθαί ε σιν. ε δ ο, χθω π το σηµείου τ right-angles [ef. 1.10]. ut, if not, let have been [ε θεί ] πρ ς ρθ ς Ε α ρα π Ε, Ε δύο drawn from the point at right-angles to [the straight- ρθαί ε σιν κα πε π Ε δυσ τα ς π, line] [Prop. 1.11]. Thus, and are two Ε ση στίν, κοιν προσκείσθω π Ε α ρα right-angles. nd since is equal to the two (an- π Ε, Ε τρισ τα ς π, Ε, Ε σαι gles) and, let have been added to both. ε σίν. πάλιν, πε π δυσ τα ς π Ε, Ε Thus, the (sum of the angles) and is equal to ση στίν, κοιν προσκείσθω π α ρα πό the (sum of the) three (angles),, and, τρισ τα ς π Ε, Ε, σαι ε σίν. [.N. 2]. gain, since is equal to the two (an- δείχθησαν δ κα α π Ε, Ε τρισ τα ς α τα ς gles) and, let have been added to both. σαι τ δ τ α τ σα κα λλήλοις στ ν σα κα α π Thus, the (sum of the angles) and is equal to Ε, Ε ρα τα ς π, σαι ε σίν λλ α the (sum of the) three (angles),, and π Ε, Ε δύο ρθαί ε σιν κα α π, [.N. 2]. ut (the sum of) and was also ρα δυσ ν ρθα ς σαι ε σίν. shown (to be) equal to the (sum of the) same three (an- Ε ν ρα ε θε α π ε θε αν σταθε σα γωνίας ποι, gles). nd things equal to the same thing are also equal τοι δύο ρθ ς δυσ ν ρθα ς σας ποιήσει περ δει to one another [.N. 1]. Therefore, (the sum of) δε ξαι. 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. ιδ. Proposition 14 Ε ν πρός τινι ε θεί κα τ πρ ς α τ σηµεί δύο If two straight-lines, not lying on the same side, make ε θε αι µ π τ α τ µέρη κείµεναι τ ς φεξ ς γωνίας adjacent angles (whose sum is) equal to two right-angles 18

19 δυσ ν ρθα ς σας ποι σιν, π ε θείας σονται λλήλαις at the same point on some straight-line, then the two α ε θε αι. straight-lines will be straight-on (with respect) to one another. Ε Πρ ς γάρ τινι ε θεί τ κα τ πρ ς α τ For let two straight-lines and, not lying on the σηµεί τ δύο ε θε αι α, µ π τ α τ same side, make adjacent angles and (whose µέρη κείµεναι τ ς φεξ ς γωνίας τ ς π, sum is) equal to two right-angles at the same point on δύο ρθα ς σας ποιείτωσαν λέγω, τι π ε θείας στ some straight-line. I say that is straight-on with τ. respect to. Ε γ ρ µή στι τ π ε θείας, στω τ For if is not straight-on to then let be π ε θείας Ε. straight-on to. Επε ο ν ε θε α π ε θε αν τ ν Ε Therefore, since the straight-line stands on the φέστηκεν, α ρα π, Ε γωνίαι δύο ρθα ς straight-line, the (sum of the) angles and σαι ε σίν ε σ δ κα α π, δύο ρθα ς σαι is thus equal to two right-angles [Prop. 1.13]. ut α ρα π, Ε τα ς π, σαι ε σίν. (the sum of) and is also equal to two rightκοιν φ ρήσθω π λοιπ ρα π Ε angles. Thus, (the sum of angles) and is equal λοιπ τ π στιν ση, λάσσων τ µείζονι to (the sum of angles) and [.N. 1]. Let (an- περ στ ν δύνατον. ο κ ρα π ε θείας στ ν Ε gle) have been subtracted from both. Thus, the reτ. µοίως δ δείξοµεν, τι ο δ λλη τις πλ ν τ ς mainder is equal to the remainder [.N. 3], π ε θείας ρα στ ν τ. the lesser to the greater. The very thing is impossible. Ε ν ρα πρός τινι ε θεί κα τ πρ ς α τ σηµεί Thus, is not straight-on with respect to. Simiδύο ε θε αι µ π α τ µέρη κείµεναι τ ς φεξ ς γωνίας larly, we can show that neither (is) any other (straightδυσ ν ρθα ς σας ποι σιν, π ε θείας σονται λλήλαις line) than. Thus, is straight-on with respect to α ε θε αι περ δει δε ξαι.. Thus, if two straight-lines, not lying on the same side, make adjacent angles (whose sum is) equal to two rightangles at the same point on some straight-line, then the two straight-lines will be straight-on (with respect) to one another. (Which is) the very thing it was required to show. ιε. Proposition 15 Ε ν δύο ε θε αι τέµνωσιν λλήλας, τ ς κατ κο- If two straight-lines cut one another then they make ρυφ ν γωνίας σας λλήλαις ποιο σιν. the vertically opposite angles equal to one another. ύο γ ρ ε θε αι α, τεµνέτωσαν λλήλας For let the two straight-lines and cut one anκατ τ Ε σηµε ον λέγω, τι ση στ ν µ ν π Ε other at the point. I say that angle is equal to γωνία τ π Ε, δ π Ε τ π Ε. (angle), and (angle) to (angle). Επε γ ρ ε θε α Ε π ε θε αν τ ν φέστηκε For since the straight-line stands on the straightγωνίας ποιο σα τ ς π Ε, Ε, α ρα π Ε, line, making the angles and, the (sum Ε γωνίαι δυσ ν ρθα ς σαι ε σίν. πάλιν, πε ε θε α of the) angles and is thus equal to two right- 19

20 Ε π ε θε αν τ ν φέστηκε γωνίας ποιο σα τ ς angles [Prop. 1.13]. gain, since the straight-line π Ε, Ε, α ρα π Ε, Ε γωνίαι δυσ ν stands on the straight-line, making the angles ρθα ς σαι ε σίν. δείχθησαν δ κα α π Ε, Ε and, the (sum of the) angles and is δυσ ν ρθα ς σαι ι ρα π Ε, Ε τα ς π thus equal to two right-angles [Prop. 1.13]. ut (the sum Ε, Ε σαι ε σίν. κοιν φ ρήσθω π Ε of) and was also shown (to be) equal to two λοιπ ρα π Ε λοιπ τ π Ε ση στίν right-angles. Thus, (the sum of) and is equal µοίως δ δειχθήσεται, τι κα α π Ε, Ε σαι to (the sum of) and [.N. 1]. Let have ε σίν. been subtracted from both. Thus, the remainder is equal to the remainder [.N. 3]. Similarly, it can be shown that and are also equal. Ε Ε ν ρα δύο ε θε αι τέµνωσιν λλήλας, τ ς κατ Thus, if two straight-lines cut one another then they κορυφ ν γωνίας σας λλήλαις ποιο σιν περ δει make the vertically opposite angles equal to one another. δε ξαι. (Which is) the very thing it was required to show. ι. Proposition 16 Παντ ς τριγώνου µι ς τ ν πλευρ ν προσεκβληθείσης For any triangle, when one of the sides is produced, κτ ς γωνία κατέρας τ ν ντ ς κα πεναντίον γωνι ν the external angle is greater than each of the internal and µείζων στίν. opposite angles. Εστω τρίγωνον τ, κα προσεκβεβλήσθω Let be a triangle, and let one of its sides α το µία πλευρ π τ λ γω, τι κτ ς have been produced to. I say that the external angle γωνία π µείζων στ ν κατέρας τ ν ντ ς κα is greater than each of the internal and opposite πεναντίον τ ν π, γωνι ν. angles, and. Τετµήσθω δίχα κατ τ Ε, κα πιζευχθε σα Let the (straight-line) have been cut in half at Ε κβεβλήσθω π ε θείας π τ Ζ, κα κείσθω τ (point) [Prop. 1.10]. nd being joined, let it have Ε ση ΕΖ, κα πεζεύχθω Ζ, κα διήχθω been produced in a straight-line to (point) F. nd let π τ Η. F be made equal to [Prop. 1.3], and let F have Επε ο ν ση στ ν µ ν Ε τ Ε, δ Ε τ ΕΖ, been joined, and let have been drawn through to δύο δ α Ε, Ε δυσ τα ς Ε, ΕΖ σαι ε σ ν κατέρα (point) G. κατέρ κα γωνία π Ε γωνί τ π ΖΕ ση Therefore, since is equal to, and to F, στίν κατ κορυφ ν γάρ βάσις ρα βάσει τ Ζ the two (straight-lines), are equal to the two ση στίν, κα τ Ε τρίγωνον τ ΖΕ τριγών στ ν (straight-lines), F, respectively. lso, angle σον, κα α λοιπα γωνίαι τα ς λοιπα ς γωνίαις σαι ε σ ν is equal to angle F, for (they are) vertically opposite κατέρα κατέρ, φ ς α σας πλευρα ποτείνουσιν [Prop. 1.15]. Thus, the base is equal to the base F, ση ρα στ ν π Ε τ π ΕΖ. µείζων δέ στιν and the triangle is equal to the triangle F, and π Ε τ ς π ΕΖ µείζων ρα π τ ς π the remaining angles subtended by the equal sides are Ε. Οµοίως δ τ ς τετµηµένης δίχα δειχθήσεται equal to the corresponding remaining angles [Prop. 1.4]. 20