euclid's elements in greek_ vol. i - richard fitzpatrick

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 twoothe

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

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), Hip-pocrates of Chios, Theaetetus, and Eudoxus of Cnidos However, Euclid is generally creditedwith arranging these theorems in a logical manner, so as to demonstrate (admittedly, not alwayswith the rigour demanded by modern mathematics) that they necessarily follow from five sim-ple axioms Euclid is also credited with devising a number of particularly ingenious proofs ofpreviously 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 extremelysurprizing that, whilst translations of this work into modern languages are easily available, theGreek 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 TheGreek text is accompanied by my own English translation

The aim of my translation is to be as literal as possible, whilst still (approximately) ing within the bounds of idiomatic English Text within square parenthesis (in both Greek andEnglish) indicates material identified by Heiberg as being later interpolations to the original text(some particularly obvious or unhelpful interpolations are omitted altogether) Text within roundparenthesis (in English) indicates material which is implied, but but not actually present, in theGreek text

remain-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)

Fundamentals of plane geometry involving

straight-lines

θ΄ “Οταν δ αƒ περιέχουσαι τ¾ν γωνίαν γρᵵሠεÙθε‹αι ðσιν, εÙθύγραµµος καλε‹ται ¹ γωνία.ι΄ “Οταν δ εÙθε‹α ™π' εÙθε‹αν σταθε‹σα τ¦ς ™φεξÁς γωνίας ‡σας ¢λλήλαις ποιÍ, Ñρθ¾ ˜κατέρατîν ‡σων γωνιîν ™στι, κሠ¹ ™φεστηκυ‹α εÙθε‹α κάθετος καλε‹ται, ™φ' ¿ν ™φέστηκεν.

ι$΄ Κέντρον δ τοà κύκλου τÕ σηµε‹ον καλε‹ται

ιζ΄ ∆ιάµετρος δ τοà κύκλου ™στˆν εÙθε‹ά τις δι¦ τοà κέντρου ºγµένη κሠπερατουµένη ™φ'

˜κάτερα τ¦ µέρη ØπÕ τÁς τοà κύκλου περιφερείας, ¼τις κሠδίχα τέµνει τÕν κύκλον

ιη΄ `Ηµικύκλιον δέ ™στι τÕ περιεχόµενον σχÁµα Øπό τε τÁς διαµέτρου κሠτÁς νοµένης Øπ' αÙτÁς περιφερείας κέντρον δ τοà ¹µικυκλίου τÕ αÙτό, Ö κሠτοà κύκλου

¢πολαµβα-™στίν

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

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 (whichare) equal to one another, each of the equal angles is a right-angle, and the former straight-line 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], (suchthat) all of the straight-lines radiating towards [the circumference] from a single point lyinginside 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 tained by three straight-lines, quadrilateral by four, and multilateral by more than four

con-1 This should really be counted as a postulate, rather than as part of a definition.

κ΄ Τîν δ τριπλεύρων σχηµάτων „σόπλευρον µν τρίγωνόν ™στι τÕ τ¦ς τρε‹ς ‡σας œχονπλευράς, „σοσκελς δ τÕ τ¦ς δύο µόνας ‡σας œχον πλευράς, σκαληνÕν δ τÕ τ¦ς τρε‹ς

¢νίσους œχον πλευράς

κα΄ ”Ετι δ τîν τριπλεύρων σχηµάτων Ñρθογώνιον µν τρίγωνόν ™στι τÕ œχον Ñρθ¾ν γωνίαν,

¢µβλυγώνιον δ τÕ œχον ¢µβλε‹αν γωνίαν, Ñξυγώνιον δ τÕ τ¦ς τρε‹ς Ñξείας œχον γωνίας.κβ΄ Τëν δ τετραπλεύρων σχηµάτων τετράγωνον µέν ™στιν, Ö „σόπλευρόν τέ ™στι κሠÑρθο-γώνιον, ˜τερόµηκες δέ, Ö Ñρθογώνιον µέν, οÙκ „σόπλευρον δέ, ·όµβος δέ, Ö „σόπλευρονµέν, οÙκ Ñρθογώνιον δέ, ·οµβοειδς δ τÕ τ¦ς ¢πεναντίον πλευράς τε κሠγωνίας ‡σας

¢λλήλαις œχον, Ö οÜτε „σόπλευρόν ™στιν οÜτε Ñρθογώνιον· τ¦ δ παρ¦ ταàτα τετράπλευρατραπέζια καλείσθω

three unequal sides.

21 And further of the trilateral figures: a right-angled triangle is that having a right-angle, anobtuse-angled (triangle) that having an obtuse angle, and an acute-angled (triangle) thathaving three acute angles

22 And of the quadrilateral figures: a square is that which is right-angled and equilateral, arectangle that which is right-angled but not equilateral, a rhombus that which is equilateralbut not right-angled, and a rhomboid that having opposite sides and angles equal to oneanother which is neither right-angled nor equilateral And let quadrilateral figures besidesthese be called trapezia

23 Parallel lines are straight-lines which, being in the same plane, and being produced to ity in each direction, meet with one another in neither (of these directions)

infin-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 an-gles) 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.

‡ση· ˜κατέρα ¥ρα τîν ΓΑ, ΓΒ τÍ ΑΒ ™στιν ‡ση τ¦ δ τù αÙτù ‡σα κሠ¢λλήλοις ™στˆν ‡σα· καˆ

¹ ΓΑ ¥ρα τÍ ΓΒ ™στιν ‡ση· αƒ τρε‹ς ¥ρα αƒ ΓΑ, ΑΒ, ΒΓ ‡σαι ¢λλήλαις ε„σίν

'Ισόπλευρον ¤ρα ™στˆ τÕ ΑΒΓ τρίγωνον κሠσυνέσταται ™πˆ τÁς δοθείσης εÙθείας πεπερασµένηςτÁς ΑΒ· Óπερ œδει ποιÁσαι

D

C

To construct an equilateral triangle on a given finite straight-line

LetAB be the given finite straight-line

So it is required to construct an equilateral triangle on the straight-lineAB

Let the circle BCD with center A and radius AB have been drawn [Post 3], and again let thecircleACE 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 wasalso shown (to be) equal toAB Thus, CA and CB are each equal to AB But things equal to thesame 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 triangleABC 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.

ΘΚ

Ε

ΠρÕς τù δοθέντι σηµείJ τÍ δοθείσV εÙθείv ‡σην εÙθε‹αν θέσθαι

”Εστω τÕ µν δοθν σηµε‹ον τÕ Α, ¹ δ δοθε‹σα εÙθε‹α ¹ ΒΓ· δε‹ δ¾ πρÕς τù Α σηµείJ τÍδοθείσV εÙθείv τÍ ΒΓ ‡σην εÙθε‹αν θέσθαι

'Επεζεύχθω γ¦ρ ¢πÕ τοà Α σηµείου ™πί τÕ Β σηµε‹ον εÙθε‹α ¹ ΑΒ, κሠσυνεστάτω ™π' αÙτÁςτρίγωνον „σόπλευρον τÕ ∆ΑΒ, κሠ™κβεβλήσθωσαν ™π' εÙθείας τα‹ς ∆Α, ∆Β εÙθε‹αι αƒ ΑΕ, ΒΖ,κሠκέντρJ µν τù Β διαστήµατι δ τù ΒΓ κύκλος γεγράφθω Ð ΓΗΘ, κሠπάλιν κέντρJ τù ∆κሠδιαστήµατι τù ∆Η κύκλος γεγράφθω Ð ΗΚΛ

'Επεˆ οâν τÕ Β σηµε‹ον κέντρον ™στˆ τοà ΓΗΘ, ‡ση ™στˆν ¹ ΒΓ τÍ ΒΗ πάλιν, ™πεˆ τÕ ∆ σηµε‹ονκέντρον ™στˆ τοà ΗΚΛ κύκλου, ‡ση ™στˆν ¹ ∆Λ τÍ ∆Η, ïν ¹ ∆Α τÍ ∆Β ‡ση ™στίν λοιπ¾ ¥ρα

¹ ΑΛ λοιπÍ τÍ ΒΗ ™στιν ‡ση ™δείχθη δ κሠ¹ ΒΓ τÍ ΒΗ ‡ση ˜κατέρα ¥ρα τîν ΑΛ, ΒΓ τÍ

ΒΗ ™στιν ‡ση τ¦ δ τù αÙτù ‡σα κሠ¢λλήλοις ™στˆν ‡σα· κሠ¹ ΑΛ ¥ρα τÍ ΒΓ ™στιν ‡ση

G F

E

To place a straight-line equal to a given straight-line at a given point

LetA be the given point, and BC the given straight-line So it is required to place a straight-line

at pointA 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 equilateraltriangleDAB have been been constructed upon it[Prop 1.1] And let the straight-linesAE and

BF have been produced in a straight-line with DA and DB (respectively)[Post 2] And let thecircle CGH with center B and radius BC have been drawn [Post 3], and again let the circleGKL 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] Andwithin these,DA is equal to DB Thus, the remainder AL is equal to the remainder BG[C.N 3].ButBC was also shown (to be) equal to BG Thus, AL and BC are each equal to BG But thingsequal to the same thing are also equal to one another[C.N 1] Thus,AL is also equal to BC

Thus, the straight-lineAL, 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.

Κείσθω πρÕς τù Α σηµείJ τÍ Γ εÙθείv ‡ση ¹ Α∆· κሠκέντρJ µν τù Α διαστήµατι δ τù Α∆κύκλος γεγράφθω Ð ∆ΕΖ.

Κሠ™πεˆ τÕ Α σηµε‹ον κέντρον ™στˆ τοà ∆ΕΖ κύκλου, ‡ση ™στˆν ¹ ΑΕ τÍ Α∆· ¢λλ¦ κሠ¹ Γ τÍΑ∆ ™στιν ‡ση ˜κατέρα ¥ρα τîν ΑΕ, Γ τÍ Α∆ ™στιν ‡ση· éστε κሠ¹ ΑΕ τÍ Γ ™στιν ‡ση

∆ύο ¥ρα δοθεισîν εÙθειîν ¢νίσων τîν ΑΒ, Γ ¢πÕ τÁς µείζονος τÁς ΑΒ τÍ ™λάσσονι τÍ Γ ‡ση

¢φÇρηται ¹ ΑΕ· Óπερ œδει ποιÁσαι

E D

C

A

F

B

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 isrequired to cut off a straight-line equal to the lesserC from the greater AB

Let the lineAD, equal to the straight-line C, have been placed at point A[Prop 1.2] And let thecircleDEF have been drawn with center A and radius AD[Post 3]

And since pointA is the center of circle DEF , AE is equal to AD[Def 1.15] But,C is also equal

toAD 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

›ξει, κሠτÕ τρίγωνον τù τριγώνJ ‡σον œσται, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι

œσονται ˜κατέρα ˜κατέρv, Øφ' §ς αƒ ‡σαι πλευρሠØποτείνουσιν

”Εστω δύο τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ τ¦ς δύο πλευρ¦ς τ¦ς ΑΒ, ΑΓ τα‹ς δυσˆ πλευρα‹ς τα‹ς ∆Ε,

∆Ζ ‡σας œχοντα ˜κατέραν ˜κατέρv τ¾ν µν ΑΒ τÍ ∆Ε τ¾ν δ ΑΓ τÍ ∆Ζ κሠγωνίαν τ¾ν ØπÕΒΑΓ γωνίv τÍ ØπÕ Ε∆Ζ ‡σην λέγω, Óτι κሠβάσις ¹ ΒΓ βάσει τÍ ΕΖ ‡ση ™στίν, κሠτÕ ΑΒΓτρίγωνον τù ∆ΕΖ τριγώνJ ‡σον œσται, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται

˜κατέρα ˜κατέρv, Øφ' §ς αƒ Šσαι πλευρሠØποτείνουσιν, ¹ µν ØπÕ ΑΒΓ τÍ ØπÕ ∆ΕΖ, ¹ δ ØπÕΑΓΒ τÍ ØπÕ ∆ΖΕ

'Εφαρµοζοµένου γ¦ρ τοà ΑΒΓ τριγώνου ™πˆ τÕ ∆ΕΖ τρίγωνον κሠτιθεµένου τοà µν Ασηµείου ™πˆ τÕ ∆ σηµε‹ον τÁς δ ΑΒ εÙθείας ™πˆ τ¾ν ∆Ε, ™φαρµόσει κሠτÕ Β σηµε‹ον ™πˆ τÕ

Ε δι¦ τÕ ‡σην εναι τ¾ν ΑΒ τÍ ∆Ε· ™φαρµοσάσης δ¾ τÁς ΑΒ ™πˆ τ¾ν ∆Ε ™φαρµόσει κሠ¹ ΑΓεÙθε‹α ™πˆ τ¾ν ∆Ζ δι¦ τÕ ‡σην εναι τ¾ν ØπÕ ΒΑΓ γωνίαν τÍ ØπÕ Ε∆Ζ· éστε κሠτÕ Γ σηµε‹ον

™πˆ τÕ Ζ σηµε‹ον ™φαρµόσει δι¦ τÕ ‡σην πάλιν εναι τ¾ν ΑΓ τÍ ∆Ζ ¢λλ¦ µ¾ν κሠτÕ Β ™πˆ

τÕ Ε ™φηρµόκει· éστε βάσις ¹ ΒΓ ™πˆ βάσιν τ¾ν ΕΖ ™φαρµόσει ε„ γ¦ρ τοà µν Β ™πˆ τÕ Ε

™φαρµόσαντος τοà δ Γ ™πˆ τÕ Ζ ¹ ΒΓ βάσις ™πˆ τ¾ν ΕΖ οÙκ ™φαρµόσει, δύο εÙθε‹αι χωρίονπεριέξουσιν· Óπερ ™στˆν ¢δύνατον ™φαρµόσει ¥ρα ¹ ΒΓ βάσις ™πˆ τ¾ν ΕΖ κሠ‡ση αÙτÍ œσται·éστε κሠÓλον τÕ ΑΒΓ τρίγωνον ™πˆ Óλον τÕ ∆ΕΖ τρίγωνον ™φαρµόσει κሠ‡σον αÙτù œσται,καˆ αƒ λοιπሠγωνίαι ™πˆ τ¦ς λοιπ¦ς γωνίας ™φαρµόσουσι κሠ‡σαι αÙτα‹ς œσονται, ¹ µν ØπÕ ΑΒΓ

τÍ ØπÕ ∆ΕΖ ¹ δ ØπÕ ΑΓΒ τÍ ØπÕ ∆ΖΕ

F B

LetABC and DEF be two triangles having the two sides AB and AC equal to the two sides DEandDF , respectively (That is) AB to DE, and AC to DF And (let) the angle BAC (be) equal

to the angleEDF I say that the base BC is also equal to the base EF , and triangle ABC will beequal to triangle DEF , and the remaining angles subtended by the equal sides will be equal tothe corresponding remaining angles (That is)ABC to DEF , and ACB to DF E

Let the triangleABC be applied to the triangle DEF ,6the pointA being placed on the point D,and the straight-lineAB on DE The point B will also coincide with E, on account of AB beingequal toDE 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 withthe point F , again on account of AC being equal to DF But, point B certainly also coincidedwith pointE, 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 baseBC will coincide with EF , andwill be equal to it [C.N 4] So the whole triangle ABC will coincide with the whole triangleDEF , and will be equal to it[C.N 4] And the remaining angles will coincide with the remainingangles, and will be equal to them[C.N 4] (That is)ABC to DEF , and ACB to DF E [C.N 4].Thus, if two triangles have two corresponding sides equal, and have the angles enclosed by theequal sides equal, then they will also have equal bases, and the two triangles will be equal, andthe remaining angles subtended by the equal sides will be equal to the corresponding remainingangles (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.

Ε

Α

Β Ζ

∆

Γ Η

Τîν „σοσκελîν τριγώνων αƒ τρÕς τÍ βάσει γωνίαι ‡σαι ¢λλήλαις ε„σίν, κሠπροσεκβληθεισîν τîν

‡σων εÙθειîν αƒ ØπÕ τ¾ν βάσιν γωνίαι ‡σαι ¢λλήλαις œσονται

”Εστω τρίγωνον „σοσκελς τÕ ΑΒΓ ‡σην œχον τ¾ν ΑΒ πλευρ¦ν τÍ ΑΓ πλευρ´, κሠβεβλήσθωσαν ™π' εÙθείας τα‹ς ΑΒ, ΑΓ εÙθε‹αι αƒ Β∆, ΓΕ· λέγω, Óτι ¹ µν ØπÕ ΑΒΓ γωνία τÍØπÕ ΑΓΒ ‡ση ™στίν, ¹ δ ØπÕ ΓΒ∆ τÍ ØπÕ ΒΓΕ

ïν ¹ ΑΒ τÍ ΑΓ ™στιν ‡ση, λοιπ¾ ¥ρα ¹ ΒΖ λοιπÍ τÍ ΓΗ ™στιν ‡ση ™δείχθη δ κሠ¹ ΖΓ τÍ

ΗΒ ‡ση· δύο δ¾ αƒ ΒΖ, ΖΓ δυσˆ τα‹ς ΓΗ, ΗΒ „σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕΒΖΓ γωνίv τV ØπÕ ΓΗΒ ‡ση, κሠβάσις αÙτîν κοιν¾ ¹ ΒΓ· κሠτÕ ΒΖΓ ¥ρα τρίγωνον τù ΓΗΒτριγώνJ ‡σον œσται, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι œσονται ˜κατέρα ˜κατέρv, Øφ'

§ς αƒ ‡σαι πλευρሠØποτείνουσιν· ‡ση ¥ρα ™στˆν ¹ µν ØπÕ ΖΒΓ τÍ ØπÕ ΗΓΒ ¹ δ ØπÕ ΒΓΖ τÍØπÕ ΓΒΗ ™πεˆ οâν Óλη ¹ ØπÕ ΑΒΗ γωνία ÓλV τÍ ØπÕ ΑΓΖ γωνίv ™δείχθη ‡ση, ïν ¹ ØπÕ ΓΒΗ

τÍ ØπÕ ΒΓΖ ‡ση, λοιπ¾ ¥ρα ¹ ØπÕ ΑΒΓ λοιπÍ τÍ ØπÕ ΑΓΒ ™στιν ‡ση· καί ε„σι πρÕς τÍ βάσει

D

F

C G A

EFor isosceles triangles, the angles at the base are equal to one another, and if the equal sides areproduced then the angles under the base will be equal to one another

LetABC be an isosceles triangle having the side AB equal to the side AC, and let the linesBD and CE have been produced in a straight-line with AB and AC (respectively)[Post 2]

straight-I say that the angleABC 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 thegreaterAE, equal to the lesser AF [Prop 1.3] Also, let the straight-linesF C and GB have beenjoined[Post 1]

In fact, sinceAF is equal to AG and AB to AC, the two (straight-lines) F A, AC are equal to thetwo (straight-lines)GA, AB, respectively They also encompass a common angle F AG Thus, thebaseF C is equal to the base GB, and the triangle AF C will be equal to the triangle AGB, andthe remaining angles subtendend by the equal sides will be equal to the corresponding remainingangles [Prop 1.4] (That is) ACF to ABG, and AF C to AGB And since the whole of AF isequal to the whole ofAG, within which AB is equal to AC, the remainder BF is thus equal tothe remainderCG [C.N 3] ButF C was also shown (to be) equal to GB So the two (straight-lines)BF , F C are equal to the two (straight-lines) CG, GB, respectively, and the angle BF C (is)equal to the angle CGB, and the base BC is common to them Thus, the triangle BF C will beequal to the triangleCGB, 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 toCBG Therefore, since the whole angle ABG was shown (to be) equal to the whole angle ACF ,within whichCBG 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 F BC was also shown (to be) equal toGCB And they are under the base

Thus, for isosceles triangles, the angles at the base are equal to one another, and if the equal sidesare produced then the angles under the base will be equal to one another (Which is) the verything it was required to show

”Εστω τρίγωνον τÕ ΑΒΓ ‡σην œχον τ¾ν ØπÕ ΑΒΓ γωνίαν τÍ ØπÕ ΑΓΒ γωνίv· λέγω, Óτι καˆπλευρ¦ ¹ ΑΒ πλευρ´ τÍ ΑΓ ™στιν ‡ση.

D A

C B

If a triangle has two angles equal to one another then the sides subtending the equal angles willalso be equal to one another

LetABC be a triangle having the angle ABC equal to the angle ACB I say that side AB is alsoequal to sideAC

For ifAB is unequal to AC then one of them is greater Let AB be greater And let DB, equal tothe lesserAC, have been cut off from the greater AB [Prop 1.3] And letDC have been joined[Post 1]

Therefore, sinceDB is equal to AC, and BC (is) common, the two sides DB, BC are equal to thetwo sidesAC, CB, respectively, and the angle DBC is equal to the angle ACB Thus, the base

DC is equal to the base AB, and the triangle DBC will be equal to the triangle ACB[Prop 1.4],the lesser to the greater The very notion (is) absurd[C.N 5] Thus, AB is not unequal to AC.Thus, (it is) equal.8

Thus, if a triangle has two angles equal to one another then the sides subtending the equal angleswill also be equal to one another (Which is) the very thing it was required to show

8 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.

Β Α

Γ

∆

'Επˆ τÁς αÙτÁς εÙθείας δύο τα‹ς αÙτα‹ς εÙθείαις ¥λλαι δύο εÙθε‹αι ‡σαι ˜κατέρα ˜κατέρv οÙσυσταθήσονται πρÕς ¥λλJ κሠ¥λλJ σηµείJ ™πˆ τ¦ αÙτ¦ µέρη τ¦ αÙτ¦ πέρατα œχουσαι τα‹ς ™ξ

¢ρχÁς εÙθείαις

Ε„ γ¦ρ δυνατόν, ™πˆ τÁς αÙτÁς εÙθείας τÁς ΑΒ δύο τα‹ς αÙτα‹ς εÙθείαις τα‹ς ΑΓ, ΓΒ ¥λλαι δύοεÙθε‹αι αƒ Α∆, ∆Β ‡σαι ˜κατέρα ˜κατερv συνεστάτωσαν πρÕς ¥λλJ κሠ¥λλJ σηµείJ τù τε Γκሠ∆ ™πˆ τ¦ αÙτ¦ µέρη τ¦ αÙτ¦ πέρατα œχουσαι, éστε ‡σην εναι τÁν µν ΓΑ τÍ ∆Α τÕ αÙτÕπέρας œχουσαν αÙτÍ τÕ Α, τ¾ν δ ΓΒ τÍ ∆Β τÕ αÙτÕ πέρας œχουσαν αÙτÍ τÕ Β, κሠ™πεζεύχθω

¹ Γ∆

'Επεˆ οâν ‡ση ™στˆν ¹ ΑΓ τÍ Α∆, ‡ση ™στˆ κሠγωνία ¹ ØπÕ ΑΓ∆ τÍ ØπÕ Α∆Γ· µείζων ¥ρα ¹ØπÕ Α∆Γ τÁς ØπÕ ∆ΓΒ· πολλù ¥ρα ¹ ØπÕ Γ∆Β µείζων ™στί τÁς ØπÕ ∆ΓΒ πάλιν ™πεˆ ‡ση ™στˆν

¹ ΓΒ τÍ ∆Β, ‡ση ™στˆ κሠγωνία ¹ ØπÕ Γ∆Β γωνίv τÍ ØπÕ ∆ΓΒ ™δείχθη δ αÙτÁς κሠπολλùµείζων· Óπερ ™στˆν ¢δύατον

ΟÙκ ¥ρα ™πˆ τÁς αÙτÁς εÙθείας δύο τα‹ς αÙτα‹ς εÙθείαις ¥λλαι δύο εÙθε‹αι ‡σαι ˜κατέρα ˜κατέρvσυσταθήσονται πρÕς ¥λλJ κሠ¥λλJ σηµείJ ™πˆ τ¦ αÙτ¦ µέρη τ¦ αÙτ¦ πέρατα œχουσαι τα‹ς ™ξ

¢ρχÁς εÙθείαις· Óπερ œδει δε‹ξαι

B A

C

D

On the same line, two other lines equal, respectively, to two (given) lines (which meet) cannot be constructed (meeting) at different points on the same side (of thestraight-line), but having the same ends as the given straight-lines

straight-For, if possible, let the two straight-lines AD, DB, equal to two (given) straight-lines AC, CB,respectively, have been constructed on the same straight-lineAB, meeting at different points, CandD, on the same side (of AB), and having the same ends (on AB) So CA and DA are equal,having the same ends atA, and CB and DB are equal, having the same ends at B And let CDhave been joined[Post 1]

Therefore, sinceAC is equal to AD, the angle ACD is also equal to angle ADC[Prop 1.5] Thus,ADC (is) greater than DCB [C.N 5] Thus, CDB is much greater than DCB [C.N 5] Again,since CB is equal to DB, the angle CDB is also equal to angle DCB [Prop 1.5] But it wasshown that the former (angle) is also much greater (than the latter) The very thing is impossi-ble

Thus, on the same line, two other lines equal, respectively, to two (given) lines (which meet) cannot be constructed (meeting) at different points on the same side (of thestraight-line), but having the same ends as the given straight-lines (Which is) the very thing itwas required to show

”Εστω δύο τρίγωνα τ¦ ΑΒΓ, ∆ΕΖ τ¦ς δύο πλευρ¦ς τ¦ς ΑΒ, ΑΓ τα‹ς δύο πλευρα‹ς τα‹ς ∆Ε,

∆Ζ ‡σας œχοντα ˜κατέραν ˜κατέρv, τ¾ν µν ΑΒ τÍ ∆Ε τ¾ν δ ΑΓ τÍ ∆Ζ· ™χέτω δ κሠβάσιντ¾ν ΒΓ βάσει τÍ ΕΖ ‡σην· λέγω, Óτι κሠγωνία ¹ ØπÕ ΒΑΓ γωνίv τÍ ØπÕ Ε∆Ζ ™στιν ‡ση

'Εφαρµοζοµένου γ¦ρ τοà ΑΒΓ τριγώνου ™πˆ τÕ ∆ΕΖ τρίγωνον κሠτιθεµένου τοà µν Β σηµείου

™πˆ τÕ Ε σηµε‹ον τÁς δ ΒΓ εÙθείας ™πˆ τ¾ν ΕΖ ™φαρµόσει κሠτÕ Γ σηµε‹ον ™πˆ τÕ Ζ δι¦ τÕ

‡σην εναι τ¾ν ΒΓ τÍ ΕΖ· ™φαρµοσάσης δ¾ τÁς ΒΓ ™πˆ τ¾ν ΕΖ ™φαρµόσουσι καˆ αƒ ΒΑ, ΓΑ ™πˆτ¦ς Ε∆, ∆Ζ ε„ γ¦ρ βάσις µν ¹ ΒΓ ™πˆ βάσιν τ¾ν ΕΖ ™φαρµόσει, αƒ δ ΒΑ, ΑΓ πλευρሠ™πˆ τ¦ςΕ∆, ∆Ζ οÙκ ™φαρµόσουσιν ¢λλ¦ παραλλάξουσιν æς αƒ ΕΗ, ΗΖ, συσταθήσονται ™πˆ τÁς αÙτÁςεÙθείας δύο τα‹ς αÙτα‹ς εÙθείαις ¥λλαι δύο εÙθε‹αι ‡σαι ˜κατέρα ˜κατέρv πρÕς ¥λλJ κሠ¥λλJσηµείJ ™πˆ τ¦ αÙτ¦ µέρη τ¦ αÙτ¦ πέρατα œχουσαι οÙ συνίστανται δέ· οÙκ ¥ρα ™φαρµοζοµένηςτÁς ΒΓ βάσεως ™πˆ τ¾ν ΕΖ βάσιν οÙκ ™φαρµόσουσι καˆ αƒ ΒΑ, ΑΓ πλευρሠ™πˆ τ¦ς Ε∆, ∆Ζ

™φαρµόσουσιν ¥ρα· éστε κሠγωνία ¹ ØπÕ ΒΑΓ ™πˆ γωνίαν τ¾ν ØπÕ Ε∆Ζ ™φαρµόσει κሠ‡σηαÙτÍ œσται

'Ε¦ν ¥ρα δύο τρίγωνα τ¦ς δύο πλευρ¦ς [τα‹ς] δύο πλευρα‹ς ‡σας œχV ˜κατέραν ˜κατέρv καˆτ¾ν βάσιν τÍ βάσει ‡σην œχV, κሠτ¾ν γωνίαν τÍ γωνίv ‡σην ›ξει τ¾ν ØπÕ τîν ‡σων εØθειîνπεριεχοµένην· Óπερ œδει δε‹ξαι

G

F C

to the baseEF I say that the angle BAC is also equal to the angle EDF

For if triangle ABC is applied to triangle DEF , the point B being placed on point E, and thestraight-lineBC on EF , point C will also coincide with F on account of BC being equal to EF

So (because of)BC coinciding with EF , (the sides) BA and CA will also coincide with ED and

DF (respectively) For if base BC coincides with base EF , but the sides AB and AC do notcoincide withED and DF (respectively), but miss like EG and GF (in the above figure), then

we will have constructed upon the same straight-line, two other straight-lines equal, respectively,

to two (given) lines, and (meeting) at different points on the same side (of the line), but having the same ends But (such straight-lines) cannot be constructed [Prop 1.7].Thus, the baseBC being applied to the base EF , the sides BA and AC cannot not coincide with

straight-ED and DF (respectively) Thus, they will coincide So the angle BAC will also coincide withangleEDF , and they will be equal[C.N 4]

Thus, if two triangles have two corresponding sides equal, and have equal bases, then the anglesencompassed by the equal straight-lines will also be equal (Which is) the very thing it wasrequired to show

Ε Α

∆

ΖΤ¾ν δοθε‹σαν γωνίαν εÙθύγραµµον δίχα τεµε‹ν

”Εστω ¹ δοθε‹σα γωνία εÙθύγραµµος ¹ ØπÕ ΒΑΓ δε‹ δ¾ αÙτ¾ν δίχα τεµε‹ν

Ε„λήφθω ™πˆ τÁς ΑΒ τυχÕν σηµε‹ον τÕ ∆, κሠ¢φVρήσθω ¢πÕ τÁς ΑΓ τÍ Α∆ ‡ση ¹ ΑΕ, καˆ

™πεζεύχθω ¹ ∆Ε, κሠσυνεστάτω ™πˆ τÁς ∆Ε τρίγωνον „σόπλευρον τÕ ∆ΕΖ, κሠ™πεζεύχθω ¹ΑΖ· λέγω, Óτι ¹ ØπÕ ΒΑΓ γωνία δίχα τέτµηται ØπÕ τÁς ΑΖ εØθείας

'Επεˆ γ¦ρ ‡ση ™στˆν ¹ Α∆ τÍ ΑΕ, κοιν¾ δ ¹ ΑΖ, δύο δ¾ αƒ ∆Α, ΑΖ δυσˆ τα‹ς ΕΑ, ΑΖ ‡σαιε„σˆν ˜κατέρα ˜κατέρv κሠβάσις ¹ ∆Ζ βάσει τÍ ΕΖ ‡ση ™στίν· γωνία ¥ρα ¹ ØπÕ ∆ΑΖ γωνίv

τÍ ØπÕ ΕΑΖ ‡ση ™στίν

`Η ¥ρα δοθε‹σα γωνία εÙθύγραµµος ¹ ØπÕ ΒΑΓ δίχα τέτµηται ØπÕ τÁς ΑΖ εÙθείας· Óπερ œδειποιÁσαι

F D

E A

To cut a given rectilinear angle in half

LetBAC be the given rectilinear angle So it is required to cut it in half

Let the pointD have been taken somewhere on AB, and let AE, equal to AD, have been cut offfrom AC [Prop 1.3], and let DE have been joined And let the equilateral triangle DEF havebeen constructed uponDE [Prop 1.1], and letAF have been joined I say that the angle BAChas been cut in half by the straight-lineAF

For sinceAD is equal to AE, and AF is common, the two (straight-lines) DA, AF are equal tothe two (straight-lines)EA, AF , respectively And the base DF is equal to the base EF Thus,angleDAF is equal to angle EAF [Prop 1.8]

Thus, the given rectilinear angle BAC has been cut in half by the straight-line AF (Which is)the very thing it was required to do

'Επεˆ γ¦ρ ‡ση ™στˆν ¹ ΑΓ τÍ ΓΒ, κοιν¾ δ ¹ Γ∆, δύο δ¾ αƒ ΑΓ, Γ∆ δύο τα‹ς ΒΓ, Γ∆ ‡σαι ε„σˆν

˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ ΑΓ∆ γωνίv τÍ ØπÕ ΒΓ∆ ‡ση ™στίν· βάσις ¥ρα ¹ Α∆ βάσει

τÍ Β∆ ‡ση ™στίν

`Η ¥ρα δοθε‹σα εÙθε‹α πεπερασµένη ¹ ΑΒ δίχα τέτµηται κατ¦ τÕ ∆· Óπερ œδει ποιÁσαι

B A

D C

To cut a given finite straight-line in half

LetAB be the given finite straight-line So it is required to cut the finite straight-line AB in half

Let the equilateral triangleABC have been constructed upon (AB) [Prop 1.1], and let the angleACB have been cut in half by the straight-line CD[Prop 1.9] I say that the straight-lineAB hasbeen cut in half at pointD

For since AC is equal to CB, and CD (is) common, the two (straight-lines) AC, CD are equal

to the two (straight-lines)BC, CD, respectively And the angle ACD is equal to the angle BCD.Thus, the baseAD is equal to the base BD [Prop 1.4]

Thus, the given finite straight-lineAB has been cut in half at (point) D (Which is) the very thing

it was required to do

¢λλήλαις ποιÍ, Ñρθ¾ ˜κατέρα τîν ‡σων γωνιîν ™στιν· Ñρθ¾ ¥ρα ™στˆν ˜κατέρα τîν ØπÕ ∆ΓΖ,ΖΓΕ.

ΤÍ ¥ρα δοθείσV εÙθείv τÍ ΑΒ ¢πÕ τοà πρÕς αÙτÍ δοθέντος σηµείου τοà Γ πρÕς Ñρθ¦ς γωνίαςεÙθε‹α γραµµ¾ Ãκται ¹ ΓΖ· Óπερ œδει ποιÁσαι

D A

F

B

To draw a straight-line at right-angles to a given straight-line from a given point on it

LetAB be the given line, and C the given point on it So it is required to draw a line from the pointC at right-angles to the straight-line AB

straight-Let the point D be have been taken somewhere on AC, and let CE be made equal to CD[Prop 1.3], and let the equilateral triangle F DE have been constructed on DE [Prop 1.1], andlet F C have been joined I say that the straight-line F C has been drawn at right-angles to thegiven straight-lineAB from the given point C on it

For since DC is equal to CE, and CF is common, the two (straight-lines) DC, CF are equal

to the two (straight-lines), EC, CF , respectively And the base DF is equal to the base F E.Thus, the angle DCF is equal to the angle ECF [Prop 1.8], and they are adjacent But when

a straight-line stood on a(nother) straight-line makes the adjacent angles equal to one another,each of the equal angles is a right-angle [Def 1.10] Thus, each of the (angles) DCF and F CE

is a right-angle

Thus, the straight-lineCF has been drawn at right-angles to the given straight-line AB from thegiven pointC on it (Which is) the very thing it was required to do

”Εστω ¹ µν δοθε‹σα εÙθε‹α ¥πειρος ¹ ΑΒ τÕ δ δοθν σηµε‹ον, Ö µή ™στιν ™π' αÙτÁς, τÕ Γ·δε‹ δ¾ ™πˆ τ¾ν δοθε‹σαν εÙθε‹αν ¥πειρον τ¾ν ΑΒ ¢πÕ τοà δοθέντος σηµείου τοà Γ, Ö µή ™στιν

™π' αÙτÁς, κάθετον εÙθε‹αν γραµµ¾ν ¢γαγε‹ν

Ε„λήφθω γ¦ρ ™πˆ τ¦ ›τερα µέρη τÁς ΑΒ εÙθείας τυχÕν σηµε‹ον τÕ ∆, κሠκέντρJ µν τù Γδιαστήµατι δ τù Γ∆ κύκλος γεγράφθω Ð ΕΖΗ, κሠτετµήσθω ¹ ΕΗ εÙθε‹α δίχα κατ¦ τÕ Θ,κሠ™πεζεύχθωσαν αƒ ΓΗ, ΓΘ, ΓΕ εØθε‹αι· λέγω, Óτι ™πˆ τ¾ν δοθε‹σαν εÙθε‹αν ¥πειρον τ¾ν ΑΒ

¢πÕ τοà δοθέντος σηµείου τοà Γ, Ö µή ™στιν ™π' αÙτÁς, κάθετος Ãκται ¹ ΓΘ

'Επεˆ γ¦ρ ‡ση ™στˆν ¹ ΗΘ τÍ ΘΕ, κοιν¾ δ ¹ ΘΓ, δύο δ¾ αƒ ΗΘ, ΘΓ δύο τα‹ς ΕΘ, ΘΓ ‡σαιεƒσˆν ˜κατέρα ˜κατέρv· κሠβάσις ¹ ΓΗ βάσει τÍ ΓΕ ™στιν ‡ση· γωνία ¥ρα ¹ ØπÕ ΓΘΗ γωνίv τÍØπÕ ΕΘΓ ™στιν ‡ση καί ε„σιν ™φεξÁς Óταν δ εÙθε‹α ™π' εÙθε‹αν σταθε‹σα τ¦ς ™φεξÁς γωνίας

‡σας ¢λλήλαις ποιÍ, Ñρθ¾ ˜κατέρα τîν ‡σων γωνιîν ™στιν, κሠ¹ ™φεστηκυ‹α εÙθε‹α κάθετοςκαλε‹ται ™φ' ¿ν ™φέστηκεν

'Επˆ τ¾ν δοθε‹σαν ¥ρα εÙθε‹αν ¥πειρον τ¾ν ΑΒ ¢πÕ τοà δοθέντος σηµείου τοà Γ, Ö µή ™στιν

™π' αÙτÁς, κάθετος Ãκται ¹ ΓΘ· Óπερ œδει ποιÁσαι

To draw a straight-line perpendicular to a given infinite straight-line from a given point which isnot on it.

Let AB be the given infinite straight-line and C the given point, which is not on (AB) So it

is required to draw a straight-line perpendicular to the given infinite straight-lineAB from thegiven pointC, which is not on (AB)

For let pointD have been taken somewhere on the other side (to C) of the straight-line AB, andlet the circleEF G have been drawn with center C and radius CD[Post 3], and let the straight-lineEG have been cut in half at (point) H [Prop 1.10], and let the straight-linesCG, CH, and

CE have been joined I say that a (straight-line) CH has been drawn perpendicular to the giveninfinite straight-lineAB from the given point C, which is not on (AB)

For sinceGH is equal to HE, and HC (is) common, the two (straight-lines) GH, HC are equal tothe two straight-linesEH, HC, respectively, and the base CG is equal to the base CE Thus, theangleCHG is equal to the angle EHC [Prop 1.8], and they are adjacent But when a straight-line stood on a(nother) straight-line makes the adjacent angles equal to one another, each of theequal angles is a right-angle, and the former straight-line is called perpendicular to that uponwhich it stands[Def 1.10]

Thus, the (straight-line)CH has been drawn perpendicular to the given infinite straight-line ABfrom the given pointC, which is not on (AB) (Which is) the very thing it was required to do

Ε„ µν οâν ‡ση ™στˆν ¹ ØπÕ ΓΒΑ τÍ ØπÕ ΑΒ∆, δύο Ñρθαί ε„σιν ε„ δ οÜ, ½χθω ¢πÕ τοà Βσηµείου τÍ Γ∆ [εÙθείv] πρÕς Ñρθ¦ς ¹ ΒΕ· αƒ ¥ρα ØπÕ ΓΒΕ, ΕΒ∆ δύο Ñρθαί ε„σιν· κሠ™πεˆ ¹ØπÕ ΓΒΕ δυσˆ τα‹ς ØπÕ ΓΒΑ, ΑΒΕ ‡ση ™στίν, κοιν¾ προσκείσθω ¹ ØπÕ ΕΒ∆· αƒ ¥ρα ØπÕ ΓΒΕ,ΕΒ∆ τρισˆ τα‹ς ØπÕ ΓΒΑ, ΑΒΕ, ΕΒ∆ ‡σαι ε„σίν πάλιν, ™πεˆ ¹ ØπÕ ∆ΒΑ δυσˆ τα‹ς ØπÕ ∆ΒΕ,ΕΒΑ ‡ση ™στίν, κοιν¾ προσκείσθω ¹ ØπÕ ΑΒΓ· αƒ ¥ρα Øπό ∆ΒΑ, ΑΒΓ τρισˆ τα‹ς ØπÕ ∆ΒΕ,ΕΒΑ, ΑΒΓ ‡σαι ε„σίν ™δείχθησαν δ καˆ αƒ ØπÕ ΓΒΕ, ΕΒ∆ τρισˆ τα‹ς αÙτα‹ς ‡σαι· τ¦ δ τùαÙτù ‡σα κሠ¢λλήλοις ™στˆν ‡σα· καˆ αƒ ØπÕ ΓΒΕ, ΕΒ∆ ¥ρα τα‹ς ØπÕ ∆ΒΑ, ΑΒΓ ‡σαι ε„σίν·

¢λλ¦ αƒ ØπÕ ΓΒΕ, ΕΒ∆ δύο Ñρθαί ε„σιν· καˆ αƒ ØπÕ ∆ΒΑ, ΑΒΓ ¥ρα δυσˆν Ñρθα‹ς ‡σαι ε„σίν.'Ε¦ν ¥ρα εÙθε‹α ™π' εÙθε‹αν σταθε‹σα γωνίας ποιÍ, ½τοι δύο Ñρθ¦ς À δυσˆν Ñρθα‹ς ‡σας ποιήσει·Óπερ œδει δε‹ξαι

A E

to the three (angles)DBE, EBA, and ABC [C.N 2] ButCBE and EBD were also shown (tobe) equal to the same three (angles) And things equal to the same thing are also equal to oneanother[C.N 1] Therefore, CBE and EBD are also equal to DBA and ABC But, CBE andEBD are two right-angles Thus, ABD and ABC are also equal to two right-angles

Thus, if a straight-line stood on a(nother) straight-line makes angles, it will certainly either maketwo right-angles, or (angles whose sum is) equal to two right-angles (Which is) the very thing itwas required to show

Ε„ γ¦ρ µή ™στι τÍ ΒΓ ™π' εÙθείας ¹ Β∆, œστω τÍ ΓΒ ™π' εÙθείας ¹ ΒΕ.

'Επεˆ οâν εÙθε‹α ¹ ΑΒ ™π' εÙθε‹αν τ¾ν ΓΒΕ ™φέστηκεν, αƒ ¥ρα ØπÕ ΑΒΓ, ΑΒΕ γωνίαι δύοÑρθα‹ς ‡σαι ε„σίν· ε„σˆ δ καˆ αƒ ØπÕ ΑΒΓ, ΑΒ∆ δύο Ñρθα‹ς ‡σαι· αƒ ¥ρα ØπÕ ΓΒΑ, ΑΒΕ τα‹ςØπÕ ΓΒΑ, ΑΒ∆ ‡σαι ε„σίν κοιν¾ ¢φVρήσθω ¹ ØπÕ ΓΒΑ· λοιπ¾ ¥ρα ¹ ØπÕ ΑΒΕ λοιπÍ τÍ ØπÕΑΒ∆ ™στιν ‡ση, ¹ ™λάσσων τÍ µείζονι· Óπερ ™στˆν ¢δύνατον οÙκ ¥ρα ™π' εÙθείας ™στˆν ¹ ΒΕ τÍ

ΓΒ Ðµοίως δ¾ δείξοµεν, Óτι οÙδ ¥λλη τις πλ¾ν τÁς Β∆· ™π' εÙθείας ¥ρα ™στˆν ¹ ΓΒ τÍ Β∆

'Ε¦ν ¥ρα πρός τινι εÙθείv κሠτù πρÕς αÙτÍ σηµείJ δύο εÙθε‹αι µ¾ ™πˆ αÙτ¦ µέρη κείµεναιτ¦ς ™φεξÁς γωνίας δυσˆν Ñρθα‹ς ‡σας ποιîσιν, ™π' εÙθείας œσονται ¢λλήλαις αƒ εÙθε‹αι· Óπερ œδειδε‹ξαι

E A

If two straight-lines, not lying on the same side, make adjacent angles equal to two right-angles atthe same point on some straight-line, then the two straight-lines will be straight-on (with respect)

to one another

For let two straight-lines BC and BD, not lying on the same side, make adjacent angles ABCandABD equal to two right-angles at the same point B on some straight-line AB I say that BD

is straight-on with respect toCB

For ifBD is not straight-on to BC then let BE be straight-on to CB

Therefore, since the straight-lineAB stands on the straight-line CBE, the angles ABC and ABEare thus equal to two right-angles[Prop 1.13] ButABC and ABD are also equal to two right-angles Thus, (angles)CBA and ABE are equal to (angles) CBA and ABD[C.N 1] Let (angle)CBA have been subtracted from both Thus, the remainder ABE is equal to the remainder ABD[C.N 3], the lesser to the greater The very thing is impossible Thus,BE is not straight-on withrespect toCB Similarly, we can show that neither (is) any other (straight-line) than BD Thus,

CB is straight-on with respect to BD

Thus, if two straight-lines, not lying on the same side, make adjacent angles equal to two angles 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

Ε

∆ Α

Β Γ

'Ε¦ν δύο εÙθε‹αι τέµνωσιν ¢λλήλας, τ¦ς κατ¦ κορυφ¾ν γωνίας ‡σας ¢λλήλαις ποιοàσιν

∆ύο γ¦ρ εÙθε‹αι αƒ ΑΒ, Γ∆ τεµνέτωσαν ¢λλήλας κατ¦ τÕ Ε σηµε‹ον· λέγω, Óτι ‡ση ™στˆν ¹ µνØπÕ ΑΕΓ γωνία τÍ ØπÕ ∆ΕΒ, ¹ δ ØπÕ ΓΕΒ τÍ ØπÕ ΑΕ∆

'Επεˆ γ¦ρ εÙθε‹α ¹ ΑΕ ™π' εÙθε‹αν τ¾ν Γ∆ ™φέστηκε γωνίας ποιοàσα τ¦ς ØπÕ ΓΕΑ, ΑΕ∆, αƒ

¥ρα ØπÕ ΓΕΑ, ΑΕ∆ γωνίαι δυσˆν Ñρθα‹ς ‡σαι ε„σίν πάλιν, ™πεˆ εÙθε‹α ¹ ∆Ε ™π' εÙθε‹αν τ¾ν

ΑΒ ™φέστηκε γωνίας ποιοàσα τ¦ς ØπÕ ΑΕ∆, ∆ΕΒ, αƒ ¥ρα ØπÕ ΑΕ∆, ∆ΕΒ γωνίαι δυσˆν Ñρθα‹ς

‡σαι ε„σίν ™δείχθησαν δ καˆ αƒ ØπÕ ΓΕΑ, ΑΕ∆ δυσˆν Ñρθα‹ς ‡σαι· ¡ι ¥ρα ØπÕ ΓΕΑ, ΑΕ∆ τα‹ςØπÕ ΑΕ∆, ∆ΕΒ ‡σαι ε„σίν κοιν¾ ¢φVρήσθω ¹ ØπÕ ΑΕ∆· λοιπ¾ ¥ρα ¹ ØπÕ ΓΕΑ λοιπÍ τÍ ØπÕΒΕ∆ ‡ση ™στίν· еοίως δ¾ δειχθήσεται, Óτι καˆ αƒ ØπÕ ΓΕΒ, ∆ΕΑ ‡σαι ε„σίν

'Ε¦ν ¥ρα δύο εÙθε‹αι τέµνωσιν ¢λλήλας, τ¦ς κατ¦ κορυφ¾ν γωνίας ‡σας ¢λλήλαις ποιοàσιν· Óπερ

œδει δε‹ξαι

A

E

B C

If two straight-lines cut one another then they make the vertically opposite angles equal to oneanother

For let the two straight-linesAB and CD cut one another at the point E I say that angle AEC isequal to (angle)DEB, and (angle) CEB to (angle) AED

For since the straight-lineAE stands on the straight-line CD, making the angles CEA and AED,the angles CEA and AED are thus equal to two right-angles [Prop 1.13] Again, since thestraight-line DE stands on the straight-line AB, making the angles AED and DEB, the anglesAED and DEB are thus equal to two right-angles [Prop 1.13] But CEA and AED were alsoshown (to be) equal to two right-angles Thus, CEA and AED are equal to AED and DEB[C.N 1] Let AED have been subtracted from both Thus, the remainder CEA is equal to theremainderBED [C.N 3] Similarly, it can be shown thatCEB and DEA are also equal

Thus, if two straight-lines cut one another then they make the vertically opposite angles equal toone another (Which is) the very thing it was required to show

Τετµήσθω ¹ ΑΓ δίχα κατ¦ τÕ Ε, κሠ™πιζευχθε‹σα ¹ ΒΕ ™κβεβλήσθω ™π' εÙθείας ™πˆ τÕ Ζ, καˆκείσθω τÍ ΒΕ ‡ση ¹ ΕΖ, κሠ™πεζεύχθω ¹ ΖΓ, κሠδιήχθω ¹ ΑΓ ™πˆ τÕ Η.

'Επεˆ οâν ‡ση ™στˆν ¹ µν ΑΕ τÍ ΕΓ, ¹ δ ΒΕ τÍ ΕΖ, δύο δ¾ αƒ ΑΕ, ΕΒ δυσˆ τα‹ς ΓΕ, ΕΖ

‡σαι ε„σˆν ˜κατέρα ˜κατέρv· κሠγωνία ¹ ØπÕ ΑΕΒ γωνίv τÍ ØπÕ ΖΕΓ ‡ση ™στίν· κατ¦ κορυφ¾νγάρ· βάσις ¥ρα ¹ ΑΒ βάσει τÍ ΖΓ ‡ση ™στίν, κሠτÕ ΑΒΕ τρίγωνον τù ΖΕΓ τριγώνJ ™στˆν

‡σον, καˆ αƒ λοιπሠγωνίαι τα‹ς λοιπα‹ς γωνίαις ‡σαι ε„σˆν ˜κατέρα ˜κατέρv, Øφ' §ς αƒ ‡σας πλευραˆØποτείνουσιν· ‡ση ¥ρα ™στˆν ¹ ØπÕ ΒΑΕ τÍ ØπÕ ΕΓΖ µείζων δέ ™στιν ¹ ØπÕ ΕΓ∆ τÁς ØπÕ ΕΓΖ·µείζων ¥ρα ¹ ØπÕ ΑΓ∆ τÁς ØπÕ ΒΑΕ `Οµοίως δ¾ τÁς ΒΓ τετµηµένης δίχα δειχθήσεται κሠ¹ØπÕ ΒΓΗ, τουτέστιν ¹ ØπÕ ΑΓ∆, µείζων κሠτÁς ØπÕ ΑΒΓ

ΠαντÕς ¥ρα τριγώνου µι©ς τîν πλευρîν προσεκβληθείσης ¹ ™κτÕς γωνία ˜κατέρας τîν ™ντÕς