EUCLID S ELEMENTS IN GREEK

EUCLID S ELEMENTS IN GREEK The Greek text of J.L. Heiberg (1883 1884) from Euclidis Elementa, edidit et Latine interpretatus est I.L. Heiberg, Lipsiae, in aedibus B.G. Teubneri, 1883 1884 with an accompanying English translation by Richard Fitzpatrick

For Faith

Preface Euclid s Elements is by far the most famous mathematical work of classical antiquity, and also has the distinction of being the world s oldest continuously used mathematical textbook. Little is known about the author, beyond the fact that he lived in Alexandria around 300 BCE. The main subject of this work is Geometry, which was something of an obsession for the Ancient Greeks. Most of the theorems appearing in Euclid s Elements were not discovered by Euclid himself, but were the work of earlier Greek mathematicians such as Pythagoras (and his school), Hippocrates of Chios, Theaetetus, and Eudoxus of Cnidos. However, Euclid is generally credited with arranging these theorems in a logical manner, so as to demonstrate (admittedly, not always with the rigour demanded by modern mathematics) that they necessarily follow from five simple axioms. Euclid is also credited with devising a number of particularly ingenious proofs of previously discovered theorems: e.g., Theorem 48 in Book 1. It is natural that anyone with a knowledge of Ancient Greek, combined with a general interest in Mathematics, would wish to read the Elements in its original form. It is therefore extremely surprizing that, whilst translations of this work into modern languages are easily available, the Greek text has been completely unobtainable (as a book) for many years. This purpose of this publication is to make the definitive Greek text of Euclid s Elements i.e., that edited by J.L. Heiberg (1883-1888) again available to the general public in book form. The Greek text is accompanied by my own English translation. The aim of my translation is to be as literal as possible, whilst still (approximately) remaining within the bounds of idiomatic English. Text within square parenthesis (in both Greek and English) indicates material identified by Heiberg as being later interpolations to the original text (some particularly obvious or unhelpful interpolations are omitted altogether). Text within round parenthesis (in English) indicates material which is implied, but but not actually present, in the Greek text. My thanks goes to Mariusz Wodzicki for advice regarding the typesetting of this work. Richard Fitzpatrick; Austin, Texas; December, 2005. References Euclidus Opera Ominia, J.L. Heiberg & H. Menge (editors), Teubner (1883-1916). Euclid in Greek, Book 1, T.L. Heath (translator), Cambridge (1920). Euclid s Elements, T.L. Heath (translator), Dover (1956). History of Greek Mathematics, T.L. Heath, Dover (1981).

ΣΤΟΙΧΕΙΩΝ α

ELEMENTS BOOK 1 Fundamentals of plane geometry involving straight-lines

ΣΤΟΙΧΕΙΩΝ α Οροι α Σηµε όν στιν, ο µέρος ο θέν. β Γραµµ δ µ κος πλατές. γ Γραµµ ς δ πέρατα σηµε α. δ Ε θε α γραµµή στιν, τις ξ σου το ς φ αυτ ς σηµείοις κε ται. ε Επιφάνεια δέ στιν, µ κος κα πλάτος µόνον χει. Επιφανείας δ πέρατα γραµµαί. ζ Επίπεδος πιφάνειά στιν, τις ξ σου τα ς φ αυτ ς ε θείαις κε ται. η Επίπεδος δ γωνία στ ν ν πιπέδ δύο γραµµ ν πτοµένων λλήλων κα µ π ε θείας κειµένων πρ ς λλήλας τ ν γραµµ ν κλίσις. θ Οταν δ α περιέχουσαι τ ν γωνίαν γραµµα ε θε αι σιν, ε θύγραµµος καλε ται γωνία. ι Οταν δ ε θε α π ε θε αν σταθε σα τ ς φεξ ς γωνίας σας λλήλαις ποι, ρθ κατέρα τ ν σων γωνι ν στι, κα φεστηκυ α ε θε α κάθετος καλε ται, φ ν φέστηκεν. ια Αµβλε α γωνία στ ν µείζων ρθ ς. ιβ Οξε α δ λάσσων ρθ ς. ιγ Ορος στίν, τινός στι πέρας. ιδ Σχ µά στι τ πό τινος τινων ρων περιεχόµενον. ιε Κύκλος στ σχ µα πίπεδον π µι ς γραµµ ς περιεχόµενον [ καλε ται περιφέρεια], πρ ς ν φ ν ς σηµείου τ ν ντ ς το σχήµατος κειµένων π σαι α προσπίπτουσαι ε θε αι [πρ ς τ ν το κύκλου περιφέρειαν] σαι λλήλαις ε σίν. ι Κέντρον δ το κύκλου τ σηµε ον καλε ται. ιζ ιάµετρος δ το κύκλου στ ν ε θε ά τις δι το κέντρου γµένη κα περατουµένη φ κάτερα τ µέρη π τ ς το κύκλου περιφερείας, τις κα δίχα τέµνει τ ν κύκλον. ιη Ηµικύκλιον δέ στι τ περιεχόµενον σχ µα πό τε τ ς διαµέτρου κα τ ς πολαµβανοµένης π α τ ς περιφερείας. κέντρον δ το µικυκλίου τ α τό, κα το κύκλου στίν. ιθ Σχήµατα ε θύγραµµά στι τ π ε θει ν περιεχόµενα, τρίπλευρα µ ν τ π τρι ν, τετράπλευρα δ τ π τεσσάρων, πολύπλευρα δ τ π πλειόνων τεσσάρων ε θει ν περιεχόµενα. 6

1 A point is that of which there is no part. 2 And a line is a length without breadth. 3 And the extremities of a line are points. ELEMENTS BOOK 1 Definitions 4 A straight-line is whatever lies evenly with points upon itself. 5 And a surface is that which has length and breadth alone. 6 And the extremities of a surface are lines. 7 A plane surface is whatever lies evenly with straight-lines upon itself. 8 And a plane angle is the inclination of the lines, when two lines in a plane meet one another, and are not laid down straight-on with respect to one another. 9 And when the lines containing the angle are straight then the angle is called rectilinear. 10 And when a straight-line stood upon (another) straight-line makes adjacent angles (which are) equal to one another, each of the equal angles is a right-angle, and the former straightline is called perpendicular to that upon which it stands. 11 An obtuse angle is greater than a right-angle. 12 And an acute angle is less than a right-angle. 13 A boundary is that which is the extremity of something. 14 A figure is that which is contained by some boundary or boundaries. 15 A circle is a plane figure contained by a single line [which is called a circumference], (such that) all of the straight-lines radiating towards [the circumference] from a single point lying inside the figure are equal to one another. 16 And the point is called the center of the circle. 17 And a diameter of the circle is any straight-line, being drawn through the center, which is brought to an end in each direction by the circumference of the circle. And any such (straight-line) cuts the circle in half. 1 18 And a semi-circle is the figure contained by the diameter and the circumference it cuts off. And the center of the semi-circle is the same (point) as (the center of) the circle. 19 Rectilinear figures are those figures contained by straight-lines: trilateral figures being contained by three straight-lines, quadrilateral by four, and multilateral by more than four. 1 This should really be counted as a postulate, rather than as part of a definition. 7

ΣΤΟΙΧΕΙΩΝ α κ Τ ν δ τριπλεύρων σχηµάτων σόπλευρον µ ν τρίγωνόν στι τ τ ς τρε ς σας χον πλευράς, σοσκελ ς δ τ τ ς δύο µόνας σας χον πλευράς, σκαλην ν δ τ τ ς τρε ς νίσους χον πλευράς. κα Ετι δ τ ν τριπλεύρων σχηµάτων ρθογώνιον µ ν τρίγωνόν στι τ χον ρθ ν γωνίαν, µβλυγώνιον 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. Α τήµατα α Ηιτήσθω π παντ ς σηµείου π π ν σηµε ον ε θε αν γραµµ ν γαγε ν. β Κα πεπερασµένην ε θε αν κατ τ συνεχ ς π ε θείας κβαλε ν. γ Κα παντ κέντρ κα διαστήµατι κύκλον γράφεσθαι. δ Κα πάσας τ ς ρθ ς γωνίας σας λλήλαις ε ναι. ε Κα ν ε ς δύο ε θείας ε θε α µπίπτουσα τ ς ντ ς κα π τ α τ µέρη γωνίας δύο ρθ ν λάσσονας ποι, κβαλλοµένας τ ς δύο ε θείας π πειρον συµπίπτειν, φ µέρη ε σ ν α τ ν δύο ρθ ν λάσσονες. Κοινα ννοιαι α Τ τ α τ σα κα λλήλοις στ ν σα. β Κα ν σοις σα προστεθ, τ λα στ ν σα. γ Κα ν π σων σα φαιρεθ, τ καταλειπόµενά στιν σα. δ Κα τ φαρµόζοντα π λλήλα σα λλήλοις στίν. ε Κα τ λον το µέρους µε ζόν [ στιν]. 8

ELEMENTS BOOK 1 20 And of the trilateral figures: an equilateral triangle is that having three equal sides, an isosceles (triangle) that having only two equal sides, and a scalene (triangle) that having three unequal sides. 21 And 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 acute-angled (triangle) that having three acute angles. 22 And 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 another which is neither right-angled nor equilateral. And let quadrilateral figures besides these be called trapezia. 23 Parallel lines are straight-lines which, being in the same plane, and being produced to infinity in each direction, meet with one another in neither (of these directions). Postulates 1 Let it have been postulated to draw a straight-line from any point to any point. 2 And to produce a finite straight-line continuously in a straight-line. 3 And to draw a circle with any center and radius. 4 And that all right-angles are equal to one another. 5 And that if a straight-line falling across two (other) straight-lines makes internal angles on the same side (of itself) less than two right-angles, being produced to infinity, the two (other) straight-lines meet on that side (of the original straight-line) that the (internal angles) are less than two right-angles (and do not meet on the other side). 2 Common Notions 1 Things equal to the same thing are also equal to one another. 2 And if equal things are added to equal things then the wholes are equal. 3 And if equal things are subtracted from equal things then the remainders are equal. 3 4 And things coinciding with one another are equal to one another. 5 And the whole [is] greater than the part. 2 This postulate effectively specifies that we are dealing with the geometry of flat, rather than curved, space. 3 As an obvious extension of C.N.s 2 & 3 if equal things are added or subtracted from the two sides of an inequality then the inequality remains an inequality of the same type. 9

ΣΤΟΙΧΕΙΩΝ α α Γ Α Β Ε Επ τ ς δοθείσης ε θείας πεπερασµένης τρίγωνον σόπλευρον συστήσασθαι. Εστω δοθε σα ε θε α πεπερασµένη ΑΒ. ε δ π τ ς ΑΒ ε θείας τρίγωνον σόπλευρον συστήσασθαι. Κέντρ µ ν τ Α διαστήµατι δ τ ΑΒ κύκλος γεγράφθω ΒΓ, κα πάλιν κέντρ µ ν τ Β διαστήµατι δ τ ΒΑ κύκλος γεγράφθω ΑΓΕ, κα π το Γ σηµείου, καθ τέµνουσιν λλήλους ο κύκλοι, πί τ Α, Β σηµε α πεζεύχθωσαν ε θε αι α ΓΑ, ΓΒ. Κα πε τ Α σηµε ον κέντρον στ το Γ Β κύκλου, ση στ ν ΑΓ τ ΑΒ πάλιν, πε τ Β σηµε ον κέντρον στ το ΓΑΕ κύκλου, ση στ ν ΒΓ τ ΒΑ. δείχθη δ κα ΓΑ τ ΑΒ ση κατέρα ρα τ ν ΓΑ, ΓΒ τ ΑΒ στιν ση. τ δ τ α τ σα κα λλήλοις στ ν σα κα ΓΑ ρα τ ΓΒ στιν ση α τρε ς ρα α ΓΑ, ΑΒ, ΒΓ σαι λλήλαις ε σίν. Ισόπλευρον ρα στ τ ΑΒΓ τρίγωνον. κα συνέσταται π τ ς δοθείσης ε θείας πεπερασµένης τ ς ΑΒ περ δει ποι σαι. 10

ELEMENTS BOOK 1 Proposition 1 C D A B E To construct an equilateral triangle on a given finite straight-line. Let AB be the given finite straight-line. So it is required to construct an equilateral triangle on the straight-line AB. Let the circle BCD with center A and radius AB have been drawn [Post. 3], and again let the circle ACE with center B and radius BA have been drawn [Post. 3]. And let the straight-lines CA and CB have been joined from the point C, where the circles cut one another, 4 to the points A and B (respectively) [Post. 1]. And since the point A is the center of the circle CDB, AC is equal to AB [Def. 1.15]. Again, since the point B is the center of the circle CAE, BC is equal to BA [Def. 1.15]. But CA was also shown (to be) equal to AB. Thus, CA and CB are each equal to AB. But things equal to the same thing are also equal to one another [C.N. 1]. Thus, CA is also equal to CB. Thus, the three (straight-lines) CA, AB, and BC are equal to one another. Thus, the triangle ABC is equilateral, and has been constructed on the given finite straight-line AB. (Which is) the very thing it was required to do. 4 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. 11

ΣΤΟΙΧΕΙΩΝ α β Γ Κ Θ Β Α Η Ζ Λ Πρ ς τ δοθέντι σηµεί τ δοθείσ ε θεί σην ε θε αν θέσθαι. Εστω τ µ ν δοθ ν σηµε ον τ Α, δ δοθε σα ε θε α ΒΓ δε δ πρ ς τ Α σηµεί τ δοθείσ ε θεί τ ΒΓ σην ε θε αν θέσθαι. Επεζεύχθω γ ρ π το Α σηµείου πί τ Β σηµε ον ε θε α ΑΒ, κα συνεστάτω π α τ ς τρίγωνον σόπλευρον τ ΑΒ, κα κβεβλήσθωσαν π ε θείας τα ς Α, Β ε θε αι α ΑΕ, ΒΖ, κα κέντρ µ ν τ Β διαστήµατι δ τ ΒΓ κύκλος γεγράφθω ΓΗΘ, κα πάλιν κέντρ τ κα διαστήµατι τ Η κύκλος γεγράφθω ΗΚΛ. Επε ο ν τ Β σηµε ον κέντρον στ το ΓΗΘ, ση στ ν ΒΓ τ ΒΗ. πάλιν, πε τ σηµε ον κέντρον στ το ΗΚΛ κύκλου, ση στ ν Λ τ Η, ν Α τ Β ση στίν. λοιπ ρα ΑΛ λοιπ τ ΒΗ στιν ση. δείχθη δ κα ΒΓ τ ΒΗ ση. κατέρα ρα τ ν ΑΛ, ΒΓ τ ΒΗ στιν ση. τ δ τ α τ σα κα λλήλοις στ ν σα κα ΑΛ ρα τ ΒΓ στιν ση. Πρ ς ρα τ δοθέντι σηµεί τ Α τ δοθείσ ε θεί τ ΒΓ ση ε θε α κε ται ΑΛ περ δει ποι σαι. Ε 12

ELEMENTS BOOK 1 Proposition 2 5 C K H D B A G F L To place a straight-line equal to a given straight-line at a given point. Let A be the given point, and BC the given straight-line. So it is required to place a straight-line at point A equal to the given straight-line BC. For let the line AB have been joined from point A to point B [Post. 1], and let the equilateral triangle DAB have been been constructed upon it [Prop. 1.1]. And let the straight-lines AE and BF have been produced in a straight-line with DA and DB (respectively) [Post. 2]. And let the circle CGH with center B and radius BC have been drawn [Post. 3], and again let the circle GKL with center D and radius DG have been drawn [Post. 3]. Therefore, since the point B is the center of (the circle) CGH, BC is equal to BG [Def. 1.15]. Again, since the point D is the center of the circle GKL, DL is equal to DG [Def. 1.15]. And within these, DA is equal to DB. Thus, the remainder AL is equal to the remainder BG [C.N. 3]. But BC was also shown (to be) equal to BG. Thus, AL and BC are each equal to BG. But things equal to the same thing are also equal to one another [C.N. 1]. Thus, AL is also equal to BC. Thus, the straight-line AL, equal to the given straight-line BC, has been placed at the given point A. (Which is) the very thing it was required to do. 5 This proposition admits of a number of different cases, depending on the relative positions of the point A and the line BC. In such situations, Euclid invariably only considers one particular case usually, the most difficult and leaves the remaining cases as exercises for the reader. 13 E

ΣΤΟΙΧΕΙΩΝ α γ Γ Α Ε Β ύο δοθεισ ν ε θει ν νίσων π τ ς µείζονος τ λάσσονι σην ε θε αν φελε ν. Ζ Εστωσαν α δοθε σαι δύο ε θε αι νισοι α ΑΒ, Γ, ν µείζων στω ΑΒ δε δ π τ ς µείζονος τ ς ΑΒ τ λάσσονι τ Γ σην ε θε αν φελε ν. Κείσθω πρ ς τ Α σηµεί τ Γ ε θεί ση Α κα κέντρ µ ν τ Α διαστήµατι δ τ Α κύκλος γεγράφθω ΕΖ. Κα πε τ Α σηµε ον κέντρον στ το ΕΖ κύκλου, ση στ ν ΑΕ τ Α λλ κα Γ τ Α στιν ση. κατέρα ρα τ ν ΑΕ, Γ τ Α στιν ση στε κα ΑΕ τ Γ στιν ση. ύο ρα δοθεισ ν ε θει ν νίσων τ ν ΑΒ, Γ π τ ς µείζονος τ ς ΑΒ τ λάσσονι τ Γ ση φ ρηται ΑΕ περ δει ποι σαι. 14

ELEMENTS BOOK 1 Proposition 3 C D A E B F For two given unequal straight-lines, to cut off from the greater a straight-line equal to the lesser. Let AB and C be the two given unequal straight-lines, of which let the greater be AB. So it is required to cut off a straight-line equal to the lesser C from the greater AB. Let the line AD, equal to the straight-line C, have been placed at point A [Prop. 1.2]. And let the circle DEF have been drawn with center A and radius AD [Post. 3]. And since point A is the center of circle DEF, AE is equal to AD [Def. 1.15]. But, C is also equal to AD. Thus, AE and C are each equal to AD. So AE is also equal to C [C.N. 1]. Thus, for two given unequal straight-lines, AB and C, the (straight-line) AE, equal to the lesser C, has been cut off from the greater AB. (Which is) the very thing it was required to do. 15

ΣΤΟΙΧΕΙΩΝ α δ Α Β Γ Ε Ζ Ε ν δύο τρίγωνα τ ς δύο πλευρ ς [τα ς] δυσ πλευρα ς σας χ κατέραν κατέρ κα τ ν γωνίαν τ γωνί σην χ τ ν π τ ν σων ε θει ν περιεχοµένην, κα τ ν βάσιν τ βάσει σην ξει, κα τ τρίγωνον τ τριγών σον σται, κα α λοιπα γωνίαι τα ς λοιπα ς γωνίαις σαι σονται κατέρα κατέρ, φ ς α σαι πλευρα ποτείνουσιν. Εστω δύο τρίγωνα τ ΑΒΓ, ΕΖ τ ς δύο πλευρ ς τ ς ΑΒ, ΑΓ τα ς δυσ πλευρα ς τα ς Ε, Ζ σας χοντα κατέραν κατέρ τ ν µ ν ΑΒ τ Ε τ ν δ ΑΓ τ Ζ κα γωνίαν τ ν π ΒΑΓ γωνί τ π Ε Ζ σην. λέγω, τι κα βάσις ΒΓ βάσει τ ΕΖ ση στίν, κα τ ΑΒΓ τρίγωνον τ ΕΖ τριγών σον σται, κα α λοιπα γωνίαι τα ς λοιπα ς γωνίαις σαι σονται κατέρα κατέρ, φ ς α σαι πλευρα ποτείνουσιν, µ ν π ΑΒΓ τ π ΕΖ, δ π ΑΓΒ τ π ΖΕ. Εφαρµοζοµένου γ ρ το ΑΒΓ τριγώνου π τ ΕΖ τρίγωνον κα τιθεµένου το µ ν Α σηµείου π τ σηµε ον τ ς δ ΑΒ ε θείας π τ ν Ε, φαρµόσει κα τ Β σηµε ον π τ Ε δι τ σην ε ναι τ ν ΑΒ τ Ε φαρµοσάσης δ τ ς ΑΒ π τ ν Ε φαρµόσει κα ΑΓ ε θε α π τ ν Ζ δι τ σην ε ναι τ ν π ΒΑΓ γωνίαν τ π Ε Ζ στε κα τ Γ σηµε ον π τ Ζ σηµε ον φαρµόσει δι τ σην πάλιν ε ναι τ ν ΑΓ τ Ζ. λλ µ ν κα τ Β π τ Ε φηρµόκει στε βάσις ΒΓ π βάσιν τ ν ΕΖ φαρµόσει. ε γ ρ το µ ν Β π τ Ε φαρµόσαντος το δ Γ π τ Ζ ΒΓ βάσις π τ ν ΕΖ ο κ φαρµόσει, δύο ε θε αι χωρίον περιέξουσιν περ στ ν δύνατον. φαρµόσει ρα ΒΓ βάσις π τ ν ΕΖ κα ση α τ σται στε κα λον τ ΑΒΓ τρίγωνον π λον τ ΕΖ τρίγωνον φαρµόσει κα σον α τ σται, κα α λοιπα γωνίαι π τ ς λοιπ ς γωνίας φαρµόσουσι κα σαι α τα ς σονται, µ ν π ΑΒΓ τ π ΕΖ δ π ΑΓΒ τ π ΖΕ. Ε ν ρα δύο τρίγωνα τ ς δύο πλευρ ς [τα ς] δύο πλευρα ς σας χ κατέραν κατέρ κα τ ν γωνίαν τ γωνί σην χ τ ν π τ ν σων ε θει ν περιεχοµένην, κα τ ν βάσιν τ βάσει σην ξει, κα τ τρίγωνον τ τριγών σον σται, κα α λοιπα γωνίαι τα ς λοιπα ς γωνίαις σαι σονται κατέρα κατέρ, φ ς α σαι πλευρα ποτείνουσιν περ δει δε ξαι. 16

ELEMENTS BOOK 1 Proposition 4 A D B C E F If two triangles have two corresponding sides equal, and have the angles enclosed by the equal sides equal, then they will also have equal bases, and the two triangles will be equal, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles. Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF, respectively. (That is) AB to DE, and AC to DF. And (let) the angle BAC (be) equal to the angle EDF. I say that the base BC is also equal to the base EF, and triangle ABC will be equal to triangle DEF, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles. (That is) ABC to DEF, and ACB to DFE. Let the triangle ABC be applied to the triangle DEF, 6 the point A being placed on the point D, and the straight-line AB on DE. The point B will also coincide with E, on account of AB being equal to DE. So (because of) AB coinciding with DE, the straight-line AC will also coincide with DF, on account of the angle BAC being equal to EDF. So the point C will also coincide with the point F, again on account of AC being equal to DF. But, point B certainly also coincided with point E, so that the base BC will coincide with the base EF. For if B coincides with E, and C with F, and the base BC does not coincide with EF, then two straight-lines will encompass a space. The very thing is impossible [Post. 1]. 7 Thus, the base BC will coincide with EF, and will be equal to it [C.N. 4]. So the whole triangle ABC will coincide with the whole triangle DEF, and will be equal to it [C.N. 4]. And the remaining angles will coincide with the remaining angles, and will be equal to them [C.N. 4]. (That is) ABC to DEF, and ACB to DFE [C.N. 4]. Thus, if two triangles have two corresponding sides equal, and have the angles enclosed by the equal sides equal, then they will also have equal bases, and the two triangles will be equal, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles. (Which is) the very thing it was required to show. 6 The application of one figure to another should be counted as an additional postulate. 7 Since Post. 1 implicitly assumes that the straight-line joining two given points is unique. 17

ΣΤΟΙΧΕΙΩΝ α ε Α Β Γ Ζ Η Ε 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. Εστω τρίγωνον σοσκελ ς τ ΑΒΓ σην χον τ ν ΑΒ πλευρ ν τ ΑΓ πλευρ, κα προσεκβεβλήσθωσαν π ε θείας τα ς ΑΒ, ΑΓ ε θε αι α Β, ΓΕ λέγω, τι µ ν π ΑΒΓ γωνία τ π ΑΓΒ ση στίν, δ π ΓΒ τ π ΒΓΕ. Ε λήφθω γ ρ π τ ς Β τυχ ν σηµε ον τ Ζ, κα φ ρήσθω π τ ς µείζονος τ ς ΑΕ τ λάσσονι τ ΑΖ ση ΑΗ, κα πεζεύχθωσαν α ΖΓ, ΗΒ ε θε αι. Επε ο ν ση στ ν µ ν ΑΖ τ ΑΗ δ ΑΒ τ ΑΓ, δύο δ α ΖΑ, ΑΓ δυσ τα ς ΗΑ, ΑΒ σαι ε σ ν κατέρα κατέρ κα γωνίαν κοιν ν περιέχουσι τ ν π ΖΑΗ βάσις ρα ΖΓ βάσει τ ΗΒ ση στίν, κα τ ΑΖΓ τρίγωνον τ ΑΗΒ τριγών σον σται, κα α λοιπα γωνίαι τα ς λοιπα ς γωνίαις σαι σονται κατέρα κατέρ, φ ς α σαι πλευρα ποτείνουσιν, µ ν π ΑΓΖ τ π ΑΒΗ, δ π ΑΖΓ τ π ΑΗΒ. κα πε λη ΑΖ λ τ ΑΗ στιν ση, ν ΑΒ τ ΑΓ στιν ση, λοιπ ρα ΒΖ λοιπ τ ΓΗ στιν ση. δείχθη δ κα ΖΓ τ ΗΒ ση δύο δ α ΒΖ, ΖΓ δυσ τα ς ΓΗ, ΗΒ σαι ε σ ν κατέρα κατέρ κα γωνία π ΒΖΓ γωνί τ π ΓΗΒ ση, κα βάσις α τ ν κοιν ΒΓ κα τ ΒΖΓ ρα τρίγωνον τ ΓΗΒ τριγών σον σται, κα α λοιπα γωνίαι τα ς λοιπα ς γωνίαις σαι σονται κατέρα κατέρ, φ ς α σαι πλευρα ποτείνουσιν ση ρα στ ν µ ν π ΖΒΓ τ π ΗΓΒ δ π ΒΓΖ τ π ΓΒΗ. πε ο ν λη π ΑΒΗ γωνία λ τ π ΑΓΖ γωνί δείχθη ση, ν π ΓΒΗ τ π ΒΓΖ ση, λοιπ ρα π ΑΒΓ λοιπ τ π ΑΓΒ στιν ση καί ε σι πρ ς τ βάσει το ΑΒΓ τριγώνου. δείχθη δ κα π ΖΒΓ τ π ΗΓΒ ση καί ε σιν π τ ν βάσιν. Τ ν ρα 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 περ δει δε ξαι. 18

ELEMENTS BOOK 1 Proposition 5 A B C F G D E 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. Let ABC be an isosceles triangle having the side AB equal to the side AC, and let the straightlines BD and CE have been produced in a straight-line with AB and AC (respectively) [Post. 2]. I say that the angle ABC is equal to ACB, and (angle) CBD to BCE. For let the point F have been taken somewhere on BD, and let AG have been cut off from the greater AE, equal to the lesser AF [Prop. 1.3]. Also, let the straight-lines FC and GB have been joined [Post. 1]. In fact, since AF is equal to AG and AB to AC, the two (straight-lines) FA, AC are equal to the two (straight-lines) GA, AB, respectively. They also encompass a common angle F AG. Thus, the base FC is equal to the base GB, and the triangle AFC will be equal to the triangle AGB, and the remaining angles subtendend by the equal sides will be equal to the corresponding remaining angles [Prop. 1.4]. (That is) ACF to ABG, and AFC to AGB. And since the whole of AF is equal to the whole of AG, within which AB is equal to AC, the remainder BF is thus equal to the remainder CG [C.N. 3]. But FC was also shown (to be) equal to GB. So the two (straightlines) BF, FC are equal to the two (straight-lines) CG, GB, respectively, and the angle BFC (is) equal to the angle CGB, and the base BC is common to them. Thus, the triangle BFC will be equal to the triangle CGB, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles [Prop. 1.4]. Thus, F BC is equal to GCB, and BCF to CBG. Therefore, since the whole angle ABG was shown (to be) equal to the whole angle ACF, within which CBG is equal to BCF, the remainder ABC is thus equal to the remainder ACB [C.N. 3]. And they are at the base of triangle ABC. And FBC was also shown (to be) equal to GCB. And they are under the base. Thus, for isosceles triangles, the angles at the base are equal to one another, and if the equal sides are produced then the angles under the base will be equal to one another. (Which is) the very thing it was required to show. 19

ΣΤΟΙΧΕΙΩΝ α Α Β Γ Ε ν τριγώνου α δ ο γωνίαι σαι λλήλαις σιν, κα α π τ ς σας γωνίας ποτείνουσαι πλευρα σαι λλήλαις σονται. Εστω τρίγωνον τ ΑΒΓ σην χον τ ν π ΑΒΓ γωνίαν τ π ΑΓΒ γωνί λέγω, τι κα πλευρ ΑΒ πλευρ τ ΑΓ στιν ση. Ε γ ρ νισός στιν ΑΒ τ ΑΓ, τέρα α τ ν µείζων στίν. στω µείζων ΑΒ, κα φ ρήσθω π τ ς µείζονος τ ς ΑΒ τ λάττονι τ ΑΓ ση Β, κα πεζεύχθω Γ. Επε ο ν ση στ ν Β τ ΑΓ κοιν δ ΒΓ, δύο δ α Β, ΒΓ δύο τα ς ΑΓ, ΓΒ σαι ε σ ν κατέρα κατέρ, κα γωνία π ΒΓ γωνι τ π ΑΓΒ στιν ση βάσις ρα Γ βάσει τ ΑΒ ση στίν, κα τ ΒΓ τρίγωνον τ ΑΓΒ τριγών σον σται, τ λασσον τ µείζονι περ τοπον ο κ ρα νισός στιν ΑΒ τ ΑΓ ση ρα. Ε ν ρα τριγώνου α δ ο γωνίαι σαι λλήλαις σιν, κα α π τ ς σας γωνίας ποτείνουσαι πλευρα σαι λλήλαις σονται περ δει δε ξαι. 20