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 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 anumberofparticularlyingeniousproofsofpreviously discovered theorems: e.g., Theorem48inook1. 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, varioustheoremsinvolvingparallellines,thetheoremregarding 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 themethodofexhaustion.finally,ook13investigatesthe five so-called Platonic solids. This edition of uclid s lements presents the definitive Greek text i.e., thateditedbyj.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

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

4 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. 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. πιφάνεια δέ στιν, µ κος κα πλάτος µόνον 5. nd a surface is that which has length and breadth χει. alone.. πιφανείας δ πέρατα γραµµαί. 6. nd the extremities of a surface are lines. ζ. πίπεδος 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. πίπεδος 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 τι 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

5 δ τ χον µβλε αν γωνίαν, ξυγώνιον δ τ τ ς τρε ς ξείας χον γωνίας. κβ. Τ ν δ τετραπλεύρων σχηµάτων τετράγωνον µέν στιν, σόπλευρόν τέ στι κα ρθογώνιον, τερόµηκες δέ, ρθογώνιον µέν, ο κ σόπλευρον δέ, όµβος δέ, σόπλευρον µέν, ο κ ρθογώνιον δέ, οµβοειδ ς δ τ τ ς πεναντίον πλευράς τε κα γωνίας σας λλήλαις χον, ο τε σόπλευρόν στιν ο τε ρθογώνιον τ δ παρ τα τα τετράπλευρα τραπέζια καλείσθω. κγ. Παράλληλοί ε σιν ε θε αι, α τινες ν τ α τ πιπέδ ο σαι κα κβαλλόµεναι ε ς πειρον φ κάτερα τ µέρη π µηδέτερα συµπίπτουσιν λλήλαις. 21. nd further of the trilateral figures: a right-angled triangle is that having a right-angle, an obtuse-angled (triangle) that having an obtuse angle, and an acuteangled (triangle) that having three acute angles. 22. nd of the quadrilateral figures: a square is that which is right-angled and equilateral, a rectangle that which is right-angled but not equilateral, a rhombus that which is equilateral but not right-angled, and a rhomboid that having opposite sides and angles equal to one an- 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. 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 thegeometryofflat, ratherthancurved,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 addedorsubtractedfromthetwosidesofaninequalitythentheinequality remains an inequality of the same type. 7

6 α. Proposition 1 π τ ς δοθείσης ε θείας πεπερασµένης τρίγωνον σόπλευρον συστήσασθαι. To construct an equilateral triangle on a given finite straight-line. στω 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, 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, wherethecirclescutoneanother, 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 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. utthingsequal 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,thethree(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. στω 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. πεζεύχθω 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, 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

7 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 πε 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. utthingsequal 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,equaltothegivenstraightline, hasbeenplacedatthegivenpoint. (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. Insuchsituations, 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. στωσαν 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. Soitisrequiredtocutoff τ λάσσονι τ σην ε θε αν φελε ν. astraight-lineequaltothelesser from the greater. Κείσθω πρ ς τ σηµεί τ ε θεί ση Let the line, equaltothestraight-line, have κα κέντρ µ ν τ διαστήµατι δ τ κύκλος been placed at point [Prop. 1.2]. nd let the circle γεγράφθω Ζ. F have been drawn with center and radius Κα πε τ σηµε ον κέντρον στ το Ζ [Post. 3]. κύκλου, ση στ ν τ λλ κα τ nd since point is the center of circle F, στιν ση. κατέρα ρα τ ν, τ στιν ση is equal to [ef. 1.15]. ut, is also equal to. στε κα τ στιν ση. Thus, and are each equal to. So is also 9

8 equal to [.N. 1]. 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), equaltothelesser, has been cut περ δει ποι σαι. off from the greater. (Whichis)theverythingitwas required to do. δ. Proposition 4 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 Ζ F στω δύο τρίγωνα τ, Ζ τ ς δύο πλευρ ς Let and F be two triangles having the two τ ς, τα ς δυσ πλευρα ς τα ς, Ζ σας χοντα sides and equal to the two sides and F, re- κατέραν κατέρ τ ν µ ν τ τ ν δ τ spectively. (That is) to, and to F. nd(let) Ζ κα γωνίαν τ ν π γωνί τ π Ζ σην. the angle (be) equal to the angle F. Isaythat λέγω, τι κα βάσις βάσει τ Ζ ση στίν, κα the base is also equal to the base F, andtriangle τ τρίγωνον τ Ζ τριγών σον σται, κα α will be equal to triangle F, andtheremaining 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 Ζ, δ π τ π Ζ. and to F. φαρµοζοµένου γ ρ το τριγώνου π τ Ζ 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, φαρµόσει κα the straight-line on. Thepoint will also coinτ σηµε ον π τ δι τ σην ε ναι τ ν τ cide with, onaccountof being equal to. So φαρµοσάσης δ τ ς π τ ν φαρµόσει (because of) coinciding with, thestraight-line 10

9 κα ε θε α π τ ν Ζ δι τ σην ε ναι τ ν π will also coincide with F, onaccountoftheangle γωνίαν τ π 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,againonaccountof being τ Ζ. λλ µ ν κα τ π τ φηρµόκει στε equal to F. ut,point certainly also coincided with βάσις π βάσιν τ ν Ζ φαρµόσει. ε γ ρ το point, sothatthebase will coincide with the base µ ν π τ φαρµόσαντος το δ π τ Ζ F. For if coincides with, and with F,andthe βάσις π τ ν Ζ ο κ φαρµόσει, δύο ε θε αι χωρίον base does not coincide with F, thentwostraightπεριέξουσιν περ στ ν δύνατον. φαρµόσει ρα lines will encompass an area. The very thing is impossible βάσις π τ ν Ζ κα ση α τ σται στε κα λον τ [Post. 1]. Thus, the base will coincide with F,and τρίγωνον π λον τ Ζ 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, andwillbe γωνίας φαρµόσουσι κα σαι α τα ς σονται, µ ν π equal to it [.N. 4]. nd the remaining angles will co- τ π Ζ δ π τ π Ζ. incide with the remaining angles, and will be equal to 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 additionalpostulate. 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 λλήλαις ε σίν, κα προσεκβληθεισ ν τ ν σων ε θει ν α π τ ν βάσιν γωνίαι σαι λλήλαις σονται. 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. Ζ Η F G στω 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, andletthestraight-lines and π ε θείας τα ς, ε θε αι α, 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. equal to, and(angle) to. 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 τ λάσσονι τ Ζ, andletg 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- πε ο ν ση στ ν µ ν Ζ τ Η δ τ lines F and G have been joined [Post. 1]. 11

10 , 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. Theyalsoencomτ Ζ τρίγωνον τ Η τριγών σον σται, κα α pass a common angle FG.Thus,thebaseF is equal λοιπα γωνίαι τα ς λοιπα ς γωνίαις σαι σονται κατέρα to the base G,andthetriangleF will be equal to the κατέρ, φ ς α σαι πλευρα ποτείνουσιν, µ ν π triangle G, andtheremaininganglessubtendendby Ζ τ π Η, δ π Ζτ 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, andf ση, λοιπ ρα Ζ λοιπ τ Η στιν ση. δείχθη δ to G. ndsincethewholeoff is equal to the whole κα Ζ τ Η ση δύο δ α Ζ, Ζ δυσ τα ς Η, of G, withinwhich is equal to, theremainder 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. Sothetwo(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,andtheanglef (is) equal to the κατέρα κατέρ, φ ς α σαι πλευρα ποτείνουσιν angle G,andthebase 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, andf 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,withinwhichG 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 ν τριγώνου α δ ο γωνίαι σαι λλήλαις σιν, κα α π τ ς σας γωνίας ποτείνουσαι πλευρα σαι λλήλαις σονται. If a triangle has two angles equal to one another then the sides subtending the equal angles will also be equal to one another. στω 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. Isaythatside is also equal to side πλευρ τ στιν ση.. 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

11 τ ς µείζονος τ ς τ λάττονι τ ση, κα the lesser, havebeencutofffromthegreater πεζεύχθω. [Prop. 1.3]. nd let have been joined [Post. 1]. πε 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,thebase 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, 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 thatiftwoquantitiesarenotunequalthentheymustbeequal. 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 π τ ς α τ ς ε θείας δύο τα ς α τα ς ε θείαις λλαι δύο ε θε αι σαι κατέρα κατέρ ο συσταθήσονται πρ ς λλ κα λλ σηµεί π τ α τ µέρη τ α τ πέρατα χουσαι τα ς ξ ρχ ς ε θείαις. On the same straight-line, two other straight-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. 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, onthesameside 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]. πε ο ν ση στ ν τ, ση στ κα γωνία Therefore, since is equal to, theangle 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

12 σταθήσονται πρ ς λλ κα λλ σηµεί π τ α τ µέρη τ α τ πέρατα χουσαι τα ς ξ ρχ ς ε θείαις περ δει δε ξαι. 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. η. Proposition 8 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 στω δύο τρίγωνα τ, Ζ τ ς δύο πλευρ ς Let and F be two triangles having the two τ ς, τα ς δύο πλευρα ς τα ς, Ζ σας χοντα sides and equal to the two sides and F, κατέραν κατέρ, τ ν µ ν τ τ ν δ τ Ζ respectively. (That is) to, and to F. Let χέτω δ κα βάσιν τ ν βάσει τ Ζ σην λέγω, τι them also have the base equal to the base F. Isay κα γωνία π γωνί τ π Ζ στιν ση. that the angle is also equal to the angle F. φαρµοζοµένου γ ρ το τριγώνου π τ Ζ For if triangle is applied to triangle F, the τρίγωνον κα τιθεµένου το µ ν σηµείου π τ point being placed on point, andthestraight-line σηµε ον τ ς δ ε θείας π τ ν Ζ 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 Ζ of being equal to F. So(becauseof) coinciding φαρµοσάσης δ τ ς π τ ν Ζ φαρµόσουσι κα with F, (thesides) and will also coincide with α, π τ ς, Ζ. ε γ ρ βάσις µ ν and F (respectively). For if base coincides with π βάσιν τ ν Ζ φαρµόσει, α δ, πλευρα base F, butthesides and do not coincide with π τ ς, Ζ ο κ φαρµόσουσιν λλ παραλλάξουσιν and F (respectively), but miss like G and GF (in ς α Η, 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 adifferentpointonthesameside(ofthestraight-line), φαρµοζοµένης τ ς βάσεως π τ ν Ζ βάσιν ο κ but having the same ends. ut (such straight-lines) can- φαρµόσουσι κα α, πλευρα π τ ς, Ζ. not be constructed [Prop. 1.7]. Thus, the base being φαρµόσουσιν ρα στε κα γωνία π π applied to the base F,thesides and cannot not γωνίαν τ ν π Ζ φαρµόσει κα ση α τ σται. coincide with and F (respectively). Thus, they will 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,andtheywillbeequal[.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

13 θ. Proposition 9 Τ ν δοθε σαν γωνίαν ε θύγραµµον δίχα τεµε ν. To cut a given rectilinear angle in half. Ζ στω 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. 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, κα πεζεύχθω and let, equalto, havebeencutofffrom, κα συνεστάτω π τ ς τρίγωνον σόπλευρον [Prop. 1.3], and let have been joined. nd let the τ Ζ, 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 πε γ ρ ση στ ν τ, κοιν δ Ζ, the angle has been cut in half by the straight-line δύο δ α, Ζ δυσ τα ς, Ζ σαι ε σ ν κατέρα F. κατέρ. κα βάσις Ζ βάσει τ Ζ ση στίν γωνία For since is equal to, andf is common, ρα π Ζ γωνί τ π Ζ 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. ndthebasef 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.(Whichis)theverythingit 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. στω 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],andlettheangle λέγω, τι ε θε α δίχα τέτµηται κατ τ σηµε ον. have been cut in half by the straight-line [Prop. 1.9]. πε γ ρ ση στ ν τ, κοιν δ, Isaythatthestraight-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

14 Η ρα δοθε σα ε θε α πεπερασµένη δίχα Thus, the given finite straight-line has been cut τέτµηται κατ τ περ δει ποι σαι. 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 στω 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. λήφθω π τ ς τυχ ν σηµε ον τ, κα κείσθω Let the point be have been taken somewhere on τ ση, κα συνεστάτω π τ ς τρίγωνον, andlet be made equal to [Prop. 1.3], and σόπλευρον τ Ζ, 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 πε γ ρ ση στ ν τ, κοιν δ Ζ, it. δύο δ α, Ζ δυσ τα ς, Ζ σαι ε σ ν κατέρα For since is equal to, andf is common, κατέρ κα βάσις Ζ βάσει τ Ζ ση στίν γωνία the two (straight-lines), F are equal to the two ρα π Ζ γωνί τ π Ζ ση στίν καί (straight-lines),, F,respectively.ndthebaseF 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, Ζ. 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

15 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 To draw a straight-line perpendicular to a given infi- nite straight-line from a given point which is not on it. π τ ν δοθε σαν ε θε αν πειρον π το δοθέντος σηµείου, µή στιν π α τ ς, κάθετον ε θε αν γραµµ ν γαγε ν. Ζ F Η Θ στω 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, whichisnoton γραµµ ν γαγε ν. (). 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 ) ofthestraight-line, andletthecirτ κύκλος γεγράφθω ΖΗ, κα τετµήσθω Η 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 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- πε γ ρ ση στ ν ΗΘ τ Θ, κοιν δ Θ, finite straight-line from the given point, whichis δύο δ α ΗΘ, Θ δύο τα ς Θ, Θ σαι ε σ ν κατέρα not on (). κατέρ κα βάσις Η βάσει τ στιν ση γωνία For since GH is equal to H,andH (is) common, ρα π ΘΗ γωνί τ π Θ 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,andthebaseG 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 π 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, whichisnoton(). (Which is) the very thing it was required to do. G H 17

16 ιγ. Proposition 13 ν ε θε α π ε θε αν σταθε σα γωνίας ποι, τοι δύο ρθ ς δυσ ν ρθα ς σας ποιήσει. 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. 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. 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 α ρα π, δύο drawn from the point at right-angles to [the straight- ρθαί ε σιν κα πε π δυσ τα ς π, line] [Prop. 1.11]. Thus, and are two ση στίν, κοιν προσκείσθω π α ρα right-angles. nd since is equal to the two (an- π, τρισ τα ς π,, σαι gles) and, let have been added to both. ε σίν. πάλιν, πε π δυσ τα ς π, 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,, σαι ε σίν. [.N. 2]. gain, since is equal to the two (an- δείχθησαν δ κα α π, 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, ρα τα ς π, σαι ε σίν λλ α the (sum of the) three (angles),, and π, 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- 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 ν πρός τινι ε θεί κα τ πρ ς α τ σηµεί δύο ε θε αι µ π τ α τ µέρη κείµεναι τ ς φεξ ς γωνίας If two straight-lines, not lying on the same side, make adjacent angles (whose sum is) equal to two right-angles 18

17 δυσ ν ρθα ς σας ποι σιν, π ε θείας σονται λλήλαις α ε θε αι. at the same point on some straight-line, then the two straight-lines will be straight-on (with respect) to one another. Πρ ς γάρ τινι ε θεί τ κα τ πρ ς α τ For let two straight-lines and,notlyingonthe 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. Isaythat is straight-on with τ. respect to. γ ρ µή στι τ π ε θείας, στω τ For if is not straight-on to then let be π ε θείας. straight-on to. πε ο ν ε θε α π ε θε αν τ ν Therefore, since the straight-line stands on the φέστηκεν, α ρα π, 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, τα ς π, σαι ε σίν. (the sum of) and is also equal to two rightκοιν φ ρήσθω π λοιπ ρα π 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 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. 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 If two straight-lines cut one another then they make the vertically opposite angles equal to one another. For let the two straight-lines and cut one an- other at the point. I say that angle is equal to (angle), and(angle) to (angle). For since the straight-line stands on the straight- line, makingtheangles and, the(sum of the) angles and is thus equal to two right- ν δύο ε θε αι τέµνωσιν λλήλας, τ ς κατ κορυφ ν γωνίας σας λλήλαις ποιο σιν. ύο γ ρ ε θε αι α, τεµνέτωσαν λλήλας κατ τ σηµε ον λέγω, τι ση στ ν µ ν π γωνία τ π, δ π τ π. πε γ ρ ε θε α π ε θε αν τ ν φέστηκε γωνίας ποιο σα τ ς π,, α ρα π, γωνίαι δυσ ν ρθα ς σαι ε σίν. πάλιν, πε ε θε α 19

18 π ε θε αν τ ν φέστηκε γωνίας ποιο σα τ ς angles [Prop. 1.13]. gain, since the straight-line π,, α ρα π, γωνίαι δυσ ν stands on the straight-line, makingtheangles ρθα ς σαι ε σίν. δείχθησαν δ κα α π, and, the(sumofthe)angles and is δυσ ν ρθα ς σαι ι ρα π, τα ς π thus equal to two right-angles [Prop. 1.13]. ut (the sum, σαι ε σίν. κοιν φ ρήσθω π of) and was also shown (to be) equal to two λοιπ ρα π λοιπ τ π ση στίν right-angles. Thus, (the sum of) and is equal µοίως δ δειχθήσεται, τι κα α π, 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. 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. στω 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, κα πιζευχθε σα Let the (straight-line) have been cut in half at κβεβλήσθω π ε θείας π τ Ζ, κα κείσθω τ (point) [Prop. 1.10]. nd being joined, let it have ση Ζ, 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 πε ο ν ση στ ν µ ν τ, δ τ Ζ, been joined, and let have been drawn through to δύο δ α, δυσ τα ς, Ζ σαι ε σ ν κατέρα (point) G. κατέρ κα γωνία π γωνί τ π Ζ 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 τρίγωνον τ Ζ 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(theyare)verticallyopposite 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 τ π Ζ. µείζων δέ στιν and the triangle is equal to the triangle F,and π τ ς π Ζ µείζων ρα π τ ς π the remaining angles subtended by the equal sides are. Οµοίως δ τ ς τετµηµένης δίχα δειχθήσεται equal to the corresponding remaining angles [Prop. 1.4]. 20

19 κα π Η, τουτέστιν π, µείζων κα τ ς π. Ζ Thus, is equal to F. ut is greater than F. Thus, is greater than. Similarly, by having cut in half, it can be shown (that) G that is to say, (is) also greater than. F G Thus, for any triangle, when one of the sides is pro- duced, the external angle is greater than each of the in- ternal and opposite angles. (Which is) the very thing it was required to show. Η Παντ ς ρα τριγώνου µι ς τ ν πλευρ ν προσεκβληθείσης κτ ς γωνία κατέρας τ ν ντ ς κα πεναντίον γωνι ν µείζων στίν περ δει δε ξαι. The implicit assumption that the point F lies in the interior of the angle should be counted as an additional postulate. ιζ. Proposition 17 Παντ ς τριγώνου α δύο γωνίαι δύο ρθ ν λάσσονές ε σι πάντ µεταλαµβανόµεναι. For any triangle, (the sum of any) two angles is less than two right-angles, (the angles) being taken up in any (possible way). στω τρίγωνον τ λέγω, τι το Let be a triangle. I say that (the sum of any) τριγώνου α δύο γωνίαι δύο ρθ ν λάττονές ε σι πάντ two angles of triangle is less than two right-angles, µεταλαµβανόµεναι. (the angles) being taken up in any (possible way). κβεβλήσθω γ ρ π τ. For let have been produced to. Κα πε τριγώνου το κτός στι γωνία π nd since the angle is external to triangle,, µείζων στ τ ς ντ ς κα πεναντίον τ ς π. it is greater than the internal and opposite angle κοιν προσκείσθω π α ρα π, [Prop. 1.16]. Let have been added to both. Thus, τ ν π, µείζονές ε σιν. λλ α π, the (sum of the angles) and is greater than 21

20 δύο ρθα ς σαι ε σίν α ρα π, δύο the (sum of the angles) and. ut,(thesumof) ρθ ν λάσσονές ε σιν. µοίως δ δείξοµεν, τι κα α and is equal to two right-angles [Prop. 1.13]. π, δύο ρθ ν λάσσονές ε σι κα τι α π Thus, (the sum of) and is less than two right-,. angles. Similarly, we can show that (the sum of) Παντ ς ρα τριγώνου α δύο γωνίαι δύο ρθ ν and is also less than two right-angles, and again λάσσονές ε σι πάντ µεταλαµβανόµεναι περ δει (that the sum of) and (is less than two rightδε ξαι. angles). Thus, for any triangle, (the sum of any) two angles is less than two right-angles, (the angles) being taken up in any (possible way). (Which is) the very thing it was required to show. ιη. Proposition 18 Παντ ς τριγώνου µείζων πλευρ τ ν µείζονα For any triangle, the greater side subtends the greater γωνίαν ποτείνει. angle. στω γ ρ τρίγωνον τ µείζονα χον τ ν For let be a triangle having side greater than πλευρ ν τ ς λέγω, τι κα γωνία π µείζων. Isaythatangle is also greater than. στ τ ς π For since is greater than, let be made πε γ ρ µείζων στ ν τ ς, κείσθω τ equal to [Prop. 1.3], and let have been joined. ση, κα πεζεύχθω. nd since angle is external to triangle, it Κα πε τριγώνου το κτός στι γωνία π is greater than the internal and opposite (angle), µείζων στ τ ς ντ ς κα πεναντίον τ ς π [Prop. 1.16]. ut (is) equal to, sinceside ση δ π τ π, πε κα πλευρ is also equal to side [Prop. 1.5]. Thus, is τ στιν ση µείζων ρα κα π τ ς π also greater than. Thus, is much greater than πολλ ρα π µείζων στ τ ς π.. Παντ ς ρα τριγώνου µείζων πλευρ τ ν µείζονα Thus, for any triangle, the greater side subtends the γωνίαν ποτείνει περ δει δε ξαι. greater angle. (Which is) the very thing it was required to show. ιθ. Proposition 19 Παντ ς τριγώνου π τ ν µείζονα γωνίαν µείζων For any triangle, the greater angle is subtended by the πλευρ ποτείνει. greater side. στω τρίγωνον τ µείζονα χον τ ν π Let be a triangle having the angle greater γωνίαν τ ς π λέγω, τι κα πλευρ than. I say that side is also greater than side πλευρ ς τ ς µείζων στίν.. γ ρ µή, τοι ση στ ν τ λάσσων For if not, is certainly either equal to, or less than, ση µ ν ο ν ο κ στιν τ ση γ ρ ν ν κα. Infact, is not equal to. Forthenangle γωνία π τ π ο κ στι δέ ο κ ρα would also have been equal to [Prop. 1.5]. ut it is ση στ ν τ. ο δ µ ν λάσσων στ ν not. Thus, is not equal to. Neither,indeed,is 22

21 τ ς λάσσων γ ρ ν ν κα γωνία π τ ς π ο κ στι δέ ο κ ρα λάσσων στ ν τ ς. δείχθη δέ, τι ο δ ση στίν. µείζων ρα στ ν τ ς. less than. Forthenangle would also have been less than [Prop. 1.18]. ut it is not. Thus, is not less than. utitwasshownthat() isalsonot equal (to ). Thus, is greater than. Παντ ς ρα τριγώνου π τ ν µείζονα γωνίαν µείζων πλευρ ποτείνει περ δει δε ξαι. Thus, for any triangle, the greater angle is subtended by the greater side. (Which is) the very thing it was required to show. κ. Proposition 20 Παντ ς τριγώνου α δύο πλευρα τ ς λοιπ ς µείζονές ε σι πάντ µεταλαµβανόµεναι. For any triangle, (the sum of any) two sides is greater than the remaining (side), (the sides) being taken up in any (possible way). στω γ ρ τρίγωνον τ λέγω, τι το τριγώνου α δύο πλευρα τ ς λοιπ ς µείζονές ε σι παντ µεταλαµβανόµεναι, α µ ν, τ ς, α δ, τ ς, α δ, τ ς. ιήχθω γ ρ π τ σηµε ον, κα κείσθω τ ση, κα πεζεύχθω. πε ο ν ση στ ν τ, ση στ κα γωνία π τ π µείζων ρα π τ ς For let be a triangle. I say that for triangle (the sum of any) two sides is greater than the remaining (side), (the sides) being taken up in any (possible way). (So), (the sum of) and (is greater) than, (the sum of) and than, and(thesumof) and than. For let have been drawn through to point, and let be made equal to [Prop. 1.3], and let 23

22 π κα πε τρίγωνόν στι τ µείζονα χον τ ν π γωνίαν τ ς π, π δ τ ν µείζονα γωνίαν µείζων πλευρ ποτείνει, ρα τ ς στι µείζων. ση δ τ µείζονες ρα α, τ ς µοίως δ δείξοµεν, τι κα α µ ν, τ ς µείζονές ε σιν, α δ, τ ς. Παντ ς ρα τριγώνου α δύο πλευρα τ ς λοιπ ς µείζονές ε σι πάντ µεταλαµβανόµεναι περ δει δε ξαι. have been joined. Therefore, since is equal to, theangle is also equal to [Prop. 1.5]. Thus, is greater than. ndsincetriangle has the angle greater than, andthegreateranglesubtendsthe greater side [Prop. 1.19], is thus greater than. ut is equal to. Thus,(thesumof) and is greater than. Similarly,wecanshowthat(thesum of) and is also greater than,and(thesumof) and than. Thus, for any triangle, (the sum of any) two sides is greater than the remaining (side), (the sides) being taken up in any (possible way). (Which is) the very thing it was required to show. κα. Proposition 21 ν τριγώνου π µι ς τ ν πλευρ ν π τ ν If two internal straight-lines are constructed on one περάτων δύο ε θε αι ντ ς συσταθ σιν, α συσταθε σαι of the sides of a triangle, from its ends, the constructed τ ν λοιπ ν το τριγώνου δύο πλευρ ν λάττονες µ ν (straight-lines) will be less than the two remaining sides σονται, µείζονα δ γωνίαν περιέξουσιν. of the triangle, but will encompass a greater angle. Τριγώνου γ ρ το π µι ς τ ν πλευρ ν τ ς For let the two internal straight-lines and π τ ν περάτων τ ν, δύο ε θε αι ντ ς συ- have been constructed on one of the sides of the triνεστάτωσαν α, λέγω, τι α, τ ν λοιπ ν angle, fromitsends and (respectively). I say το τριγώνου δύο πλευρ ν τ ν, λάσσονες µέν that and are less than the (sum of the) two reε σιν, µείζονα δ γωνίαν περιέχουσι τ ν π τ ς maining sides of the triangle and, but encompass π. an angle greater than. ιήχθω γ ρ π τ. κα πε παντ ς τριγώνου For let have been drawn through to. ndsince α δύο πλευρα τ ς λοιπ ς µείζονές ε σιν, το ρα for every triangle (the sum of any) two sides is greater τριγώνου α δύο πλευρα α, τ ς µείζονές than the remaining (side) [Prop. 1.20], for triangle ε σιν κοιν προσκείσθω α ρα, τ ν, the (sum of the) two sides and is thus greater µείζονές ε σιν. πάλιν, πε το τριγώνου α than. Let have been added to both. Thus, (the δύο πλευρα α, τ ς µείζονές ε σιν, κοιν sum of) and is greater than (the sum of) and προσκείσθω α, ρα τ ν, µείζονές. gain,sinceintriangle the (sum of the) two ε σιν. λλ τ ν, µείζονες δείχθησαν α, sides and is greater than,let have been πολλ ρα α, τ ν, µείζονές ε σιν. added to both. Thus, (the sum of) and is greater Πάλιν, πε παντ ς τριγώνου κτ ς γωνία τ ς ντ ς than (the sum of) and. ut, (the sum of) κα πεναντίον µείζων στίν, το ρα τριγώνου and was shown (to be) greater than (the sum of) κτ ς γωνία π µείζων στ τ ς π. δι and. Thus,(thesumof) and is much greater τα τ τοίνυν κα το τριγώνου κτ ς γωνία than (the sum of) and. π µείζων στ τ ς π. λλ τ ς π gain, since for every triangle the external angle is µείζων δείχθη π πολλ ρα π greater than the internal and opposite (angles) [Prop. 24