I say that they will also have the remaining sides equal to the remaining sides respectively, namely AB to DE and AC to DF , and the remaining angle to the remaining angle, namely the angle BAC to the angle EDF .

For, if AB is unequal to DE , one of them is greater.

Let AB be greater, and let BG be made equal to DE ; and let GC be joined.

Then, since BG is equal to DE , and BC to EF , the two sides GB , BC are equal to the two sides DE , EF respectively; and the angle GBC is equal to the angle DEF ; therefore the base GC is equal to the base DF , and the triangle GBC is equal to the triangle DEF , and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [[I. 4](elem.1.4)] therefore the angle GCB is equal to the angle DFE .

But the angle DFE is by hypothesis equal to the angle BCA ; therefore the angle BCG is equal to the angle BCA , the less to the greater: which is impossible. Therefore AB is not unequal to DE , and is therefore equal to it.

But BC is also equal to EF ; therefore the two sides AB , BC are equal to the two sides DE , EF respectively, and the angle ABC is equal to the angle DEF ; therefore the base AC is equal to the base DF , and the remaining angle BAC is equal to the remaining angle EDF . [[I. 4](elem.1.4)]

Again, let sides subtending equal angles be equal, as AB to DE ;

I say again that the remaining sides will be equal to the remaining sides, namely AC to DF and BC to EF , and further the remaining angle BAC is equal to the remaining angle EDF .

For, if BC is unequal to EF , one of them is greater.

Let BC be greater, if possible, and let BH be made equal to EF ; let AH be joined.

Then, since BH is equal to EF , and AB to DE , the two sides AB , BH are equal to the two sides DE , EF respectively, and they contain equal angles; therefore the base AH is equal to the base DF , and the triangle ABH is equal to the triangle DEF , and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [[I. 4](elem.1.4)] therefore the angle BHA is equal to the angle EFD .

But the angle EFD is equal to the angle BCA ; therefore, in the triangle AHC , the exterior angle BHA is equal to the interior and opposite angle BCA : which is impossible. [[I. 16](elem.1.16)]

Therefore BC is not unequal to EF , and is therefore equal to it.

But AB is also equal to DE ; therefore the two sides AB , BC are equal to the two sides DE , EF respectively, and they contain equal angles; therefore the base AC is equal to the base DF , the triangle ABC equal to the triangle DEF , and the remaining angle BAC equal to the remaining angle EDF . [[I. 4](elem.1.4)]

Therefore etc.

Q. E. D.

πλευρὰν τὴν πρὸς ταῖς ἴσαις γωνίαις .

ὑπόκειται ἴση , according to the elegant Greek idiom. ὑπόκειμαι is used for the passive of ὑποτίθημι , as κεῖμαι is used for the passive of τίθημι, and so with the other compounds. Cf. προσκεῖσθαι , to be added.

Proposition 27.
If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.

For let the straight line EF falling on the two straight lines AB , CD make the alternate angles AEF , EFD equal to one another;

I say that AB is parallel to CD .

For, if not, AB , CD when produced will meet either in the direction of B , D or towards A , C .

Let them be produced and meet, in the direction of B , D , at G .

Then, in the triangle GEF , the exterior angle AEF is equal to the interior and opposite angle EFG : which is impossible. [[I. 16](elem.1.16)]

Therefore AB , CD when produced will not meet in the direction of B , D .

Similarly it can be proved that neither will they meet towards A , C .

But straight lines which do not meet in either direction are parallel; [[Def. 23](elem.1.def.23)] therefore AB is parallel to CD .

Therefore etc.

Q. E. D.

εὶς δύο εὐθείας ἐμπίπτουσα , the phrase being the same as that used in [Post. 5](elem.1.post.5), meaning a transversal .

αἱ ἐναλλὰξ γωνίαι . Proclus (p. 357, 9 ) explains that Euclid uses the word alternate (or, more exactly, alternately , ἐναλλάξ ) in two connexions, (1) of a certain transformation of a proportion, as in Book V. and the arithmetical Books, (2) as here, of certain of the angles formed by parallels with a straight line crossing them.

Alternate angles are, according to Euclid as interpreted by Proclus, those which are not on the same side of the transversal, and are not adjacent, but are separated by the transversal, both being within the parallels but one above

and the other below.

The meaning is natural enough if we imagine the four internal angles to be taken in cyclic order and alternate angles to be any two of them not successive but separated by one angle of the four.
literally towards the parts B , D or towards A , C ,

ἐπὶ τὰ Β, Δ μέρη ἢ ἐπὶ τὰ Α Γ .

Proposition 28.
If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.

For let the straight line EF falling on the two straight lines AB , CD make the exterior angle EGB equal to the interior and opposite angle GHD , or the interior angles on the same side, namely BGH , GHD , equal to two right angles;

I say that AB is parallel to CD .

For, since the angle EGB is equal to the angle GHD , while the angle EGB is equal to the angle AGH , [[I. 15](elem.1.15)]

the angle AGH is also equal to the angle GHD ; and they are alternate; therefore AB is parallel to CD . [[I. 27](elem.1.27)]

Again, since the angles BGH , GHD are equal to two right angles, and the angles AGH , BGH are also equal to two right angles, [[I. 13](elem.1.13)] the angles AGH , BGH are equal to the angles BGH , GHD .

Let the angle BGH be subtracted from each; therefore the remaining angle AGH is equal to the remaining angle GHD ; and they are alternate; therefore AB is parallel to CD . [[I. 27](elem.1.27)]

Therefore etc.

Q. E. D.

Proposition 29.
A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the interior angles on the same side equal to two right angles.

For let the straight line EF fall on the parallel straight lines AB , CD ;

I say that it makes the alternate angles AGH , GHD equal, the exterior angle EGB equal to the interior and opposite angle GHD , and the interior angles on the same side, namely BGH , GHD , equal to two right angles.

For, if the angle AGH is unequal to the angle GHD , one of them is greater.

Let the angle AGH be greater.

Let the angle BGH be added to each; therefore the angles AGH , BGH are greater than the angles BGH , GHD .

But the angles AGH , BGH are equal to two right angles; [[I. 13](elem.1.13)] therefore the angles BGH , GHD are less than two right angles.

But straight lines produced indefinitely from angles less than two right angles meet; [[Post. 5](elem.1.post.5)] therefore AB , CD , if produced indefinitely, will meet; but they do not meet, because they are by hypothesis parallel.

Therefore the angle AGH is not unequal to the angle GHD , and is therefore equal to it.

Again, the angle AGH is equal to the angle EGB ; [[I. 15](elem.1.15)] therefore the angle EGB is also equal to the angle GHD . [[C.N. 1](elem.1.c.n.1)]

Let the angle BGH be added to each; therefore the angles EGB , BGH are equal to the angles BGH , GHD . [[C.N. 2](elem.1.c.n.2)]

But the angles EGB , BGH are equal to two right angles; [[I. 13](elem.1.13)] therefore the angles BGH , GHD are also equal to two right angles.

Therefore etc.

Q. E. D.

αἰ δὲ ἀπ: ὲλασσόνων ἢ δύο ὀρθῶν ἐκβαλλόμεναι εἰς ἄπειπον συμπίπτουσιν , a variation from the more explicit language of [Postulate 5](elem.1.post.5). A good deal is left to be understood, namely that the straight lines begin from points at which they meet a transversal, and make with it internal angles on the same side the sum of which is less than two right angles.

literally because they are supposed parallel,

διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι .

Proposition 30.
Straight lines parallel to the same straight line are also parallel to one another.

Let each of the straight lines AB , CD be parallel to EF ; I say that AB is also parallel to CD .

For let the straight line GK fall upon them;

Then, since the straight line GK has fallen on the parallel straight lines AB , EF , the angle AGK is equal to the angle GHF . [[I. 29](elem.1.29)]

Again, since the straight line GK has fallen on the parallel straight lines EF , CD , the angle GHF is equal to the angle GKD . [[I. 29](elem.1.29)]

But the angle AGK was also proved equal to the angle GHF ; therefore the angle AGK is also equal to the angle GKD ; [[C.N. 1](elem.1.c.n.1)] and they are alternate.

Therefore AB is parallel to CD .

Q. E. D.

The usual conclusion in general terms (Therefore etc.

) repeating the enunciation is, curiously enough, wanting at the end of this proposition.

Proposition 31.
Through a given point to draw a straight line parallel to a given straight line.

Let A be the given point, and BC the given straight line; thus it is required to draw through the point A a straight line parallel to the straight line BC .

Let a point D be taken at random on BC , and let AD be joined; on the straight line DA , and at the point A on it, let the angle DAE be constructed equal to the angle ADC [[I. 23](elem.1.23)]; and let the straight line AF be produced in a straight line with EA .

Then, since the straight line AD falling on the two straight lines BC , EF has made the alternate angles EAD , ADC equal to one another, therefore EAF is parallel to BC . [[I. 27](elem.1.27)]

Therefore through the given point A the straight line EAF has been drawn parallel to the given straight line BC .

Q. E. F.

Proposition 32.
In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.

Let ABC be a triangle, and let one side of it BC be produced to D ;

I say that the exterior angle ACD is equal to the two interior and opposite angles CAB , ABC , and the three interior angles of the triangle ABC , BCA , CAB are equal to two right angles.

For let CE be drawn through the point C parallel to the straight line AB . [[I. 31](elem.1.31)]

Then, since AB is parallel to CE , and AC has fallen upon them, the alternate angles BAC , ACE are equal to one another. [[I. 29](elem.1.29)]

Again, since AB is parallel to CE , and the straight line BD has fallen upon them, the exterior angle ECD is equal to the interior and opposite angle ABC . [[I. 29](elem.1.29)]

But the angle ACE was also proved equal to the angle BAC ; therefore the whole angle ACD is equal to the two interior and opposite angles BAC , ABC .

Let the angle ACB be added to each; therefore the angles ACD , ACB are equal to the three angles ABC , BCA , CAB .

But the angles ACD , ACB are equal to two right angles; [[I. 13](elem.1.13)] therefore the angles ABC , BCA , CAB are also equal to two right angles.

Therefore etc.

Q. E. D.

Proposition 33.
The straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are themselves also equal and parallel.

Let AB , CD be equal and parallel, and let the straight lines AC , BD join them (at the extremities which are) in the same directions (respectively); I say that AC , BD are also equal and parallel.

Let BC be joined.

Then, since AB is parallel to CD , and BC has fallen upon them, the alternate angles ABC , BCD are equal to one another. [[I. 29](elem.1.29)]

And, since AB is equal to CD , and BC is common, the two sides AB , BC are equal to the two sides DC , CB ; and the angle ABC is equal to the angle BCD ; therefore the base AC is equal to the base BD , and the griangle ABC is equal to the triangle DCB , and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; [[I. 4](elem.1.4)] therefore the angle ACB is equal to the angle CBD .

And, since the straight line BC falling on the two straight lines AC , BD has made the alternate angles equal to one another, AC is parallel to BD . [[I. 27](elem.1.27)]

And it was also proved equal to it.

Therefore etc.

Q. E. D.

I have for clearness' sake inserted the words in brackets though they are not in the original Greek, which has joining...in the same directions

or on the same sides,

ἐπὶ τὰ αυτὰ μέρη ἐπιζευγνύουσαι . The expression tiwards the same parts,

though usage has sanctioned it, is perhaps not quite satisfactory.

and 18. DCB. The Greek has BC , CD

and BCD

in these places respectively. Euclid is not always careful to write in corresponding order the letters denoting corresponding points in congruent figures. On the contrary, he evidently prefers the alphabetical order, and seems to disdain to alter it for the sake of beginners or others who might be confused by it. In the case of angles alteration is perhaps unnecessary; but in the case of triangles and pairs of corresponding sides I have ventured to alter the order to that which the mathematician of to-day expects.

Proposition 34.
In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.

Let ACDB be a parallelogrammic area, and BC its diameter; I say that the opposite sides and angles of the parallelogram ACDB are equal to one another, and the diameter BC bisects it.

For, since AB is parallel to CD , and the straight line BC has fallen upon them, the alternate angles ABC , BCD are equal to one another. [[I. 29](elem.1.29)]

Again, since AC is parallel to BD , and BC has fallen upon them, the alternate angles ACB , CBD are equal to one another. [[I. 29](elem.1.29)]

Therefore ABC , DCB are two triangles having the two angles ABC , BCA equal to the two angles DCB , CBD respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, BC ; therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [[I. 26](elem.1.26)] therefore the side AB is equal to CD , and AC to BD ,

and further the angle BAC is equal to the angle CDB .

And, since the angle ABC is equal to the angle BCD , and the angle CBD to the angle ACB , the whole angle ABD is equal to the whole angle ACD . [[C.N. 2](elem.1.c.n.2)]
And the angle BAC was also proved equal to the angle CDB .

Therefore in parallelogrammic areas the opposite sides and angles are equal to one another.

I say, next, that the diameter also bisects the areas.

For, since AB is equal to CD , and BC is common, the two sides AB , BC are equal to the two sides DC , CB respectively; and the angle ABC is equal to the angle BCD ; therefore the base AC is also equal to DB , and the triangle ABC is equal to the triangle DCB . [[I. 4](elem.1.4)]

Therefore the diameter BC bisects the parallelogram ACDB .

Q. E. D.

It is to be observed that, when parallelograms have to be mentioned for the first time, Euclid calls them parallelogrammic areas

or, more exactly, parallelogram

areas (παραλληλόγραμμα χωρία ). The meaning is simply areas bounded by parallel straight lines with the further limitation placed upon the term by Euclid that only four-sided figures are so called, although of course there are certain regular polygons which have opposite sides parallel, and which therefore might be said to be areas bounded by parallel straight lines. We gather from Proclus (p. 393 ) that the word parallelogram

was first introduced by Euclid, that its use was suggested by [I. 33](elem.1.33), and that the formation of the word παραλληλόγραμμος (parallel-lined) was analogous to that of εὐθύγραμμος (straight-lined or rectilineal).

and 36. DC, CB. The Greek has in these places BCD

and CD , BC

respectively. Cf. note on [I. 33](elem.1.33), lines 15, 18.

Proposition 35.
Parallelograms which are on the same base and in the same parallels are equal to one another.

Let ABCD , EBCF be parallelograms on the same base BC and in the same parallels AF , BC ; I say that ABCD is equal to the parallelogram EBCF .

For, since ABCD is a parallelogram, AD is equal to BC . [[I. 34](elem.1.34)]

For the same reason also EF is equal to BC , so that AD is also equal to EF ; [[C.N. 1](elem.1.c.n.1)]

and DE is common; therefore the whole AE is equal to the whole DF . [[C.N. 2](elem.1.c.n.2)]

But AB is also equal to DC ; [[I. 34](elem.1.34)] therefore the two sides EA , AB are equal to the two sides FD , DC respectively,

and the angle FDC is equal to the angle EAB , the exterior to the interior; [[I. 29](elem.1.29)] therefore the base EB is equal to the base FC , and the triangle EAB will be equal to the triangle FDC . [[I. 4](elem.1.4)]

Let DGE be subtracted from each; therefore the trapezium ABGD which remains is equal to the trapezium EGCF which remains. [[C.N. 3](elem.1.c.n.3)]

Let the triangle GBC be added to each; therefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF . [[C.N. 2](elem.1.c.n.2)]

Therefore etc.

Q. E. D.

The text has DFC .

Euclid speaks of the triangle DGE without any explanation that, in the case which he takes (where AD , EF have no point in common), BE , CD must meet at a point G between the two parallels. He allows this to appear from the figure simply.

Proposition 36.
Parallelograms which are on equal bases and in the same parallels are equal to one another.

Let ABCD , EFGH be parallelograms which are on equal bases BC , FG and in the same parallels AH , BG ;

I say that the parallelogram ABCD is equal to EFGH .

For let BE , CH be joined.

Then, since BC is equal to FG while FG is equal to EH , BC is also equal to EH . [[C.N. 1](elem.1.c.n.1)]

But they are also parallel.

And EB , HC join them; but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [[I. 33](elem.1.33)]

Therefore EBCH is a parallelogram. [[I. 34](elem.1.34)]

And it is equal to ABCD ; for it has the same base BC with it, and is in the same parallels BC , AH with it. [[I. 35](elem.1.35)]

For the same reason also EFGH is equal to the same EBCH ; [[I. 35](elem.1.35)] so that the parallelogram ABCD is also equal to EFGH . [[C.N. 1](elem.1.c.n.1)]

Therefore etc. Q. E. D.

Proposition 37.
Triangles which are on the same base and in the same parallels are equal to one another.

Let ABC , DBC be triangles on the same base BC and in the same parallels AD , BC ; I say that the triangle ABC is equal to the triangle DBC .

Let AD be produced in both directions to E , F ; through B let BE be drawn parallel to CA , [[I. 31](elem.1.31)] and through C let CF be drawn parallel to BD . [[I. 31](elem.1.31)]

Then each of the figures EBCA , DBCF is a parallelogram; and they are equal,

for they are on the same base BC and in the same parallels BC , EF . [[I. 35](elem.1.35)]

Moreover the triangle ABC is half of the parallelogram EBCA ; for the diameter AB bisects it. [[I. 34](elem.1.34)]

And the triangle DBC is half of the parallelogram DBCF ; for the diameter DC bisects it. [[I. 34](elem.1.34)]

[But the halves of equal things are equal to one another.]

Therefore the triangle ABC is equal to the triangle DBC .

Therefore etc.

Q. E. D.

Here and in the next proposition Heiberg brackets the words But the halves of equal things are equal to one another

on the ground that, since the

Common Notion which asserted this fact was interpolated at a very early date (before the time of Theon), it is probable that the words here were interpolated at the same time. Cf. note above (p. 224) on the interpolated Common Notion .
Proposition 38.
Triangles which are on equal bases and in the same parallels are equal to one another.

Let ABC , DEF be triangles on equal bases BC , EF and in the same parallels BF , AD ; I say that the triangle ABC is equal to the triangle DEF .

For let AD be produced in both directions to G , H ; through B let BG be drawn parallel to CA , [[I. 31](elem.1.31)] and through F let FH be drawn parallel to DE .

Then each of the figures GBCA , DEFH is a parallelogram; and GBCA is equal to DEFH ;

for they are on equal bases BC , EF and in the same parallels BF , GH . [[I. 36](elem.1.36)]

Moreover the triangle ABC is half of the parallelogram GBCA ; for the diameter AB bisects it. [[I. 34](elem.1.34)]

And the triangle FED is half of the parallelogram DEFH ; for the diameter DF bisects it. [[I. 34](elem.1.34)]

[But the halves of equal things are equal to one another.]

Therefore the triangle ABC is equal to the triangle DEF .

Therefore etc.

Q. E. D.

Proposition 39.
Equal triangles which are on the same base and on the same side are also in the same parallels.

Let ABC , DBC be equal triangles which are on the same base BC and on the same side of it; [I say that they are also in the same parallels.]

And [For] let AD be joined; I say that AD is parallel to BC .

For, if not, let AE be drawn through the point A parallel to the straight line BC , [[I. 31](elem.1.31)] and let EC be joined.

Therefore the triangle ABC is equal to the triangle EBC ; for it is on the same base BC with it and in the same parallels. [[I. 37](elem.1.37)]

But ABC is equal to DBC ; therefore DBC is also equal to EBC , [[C.N. 1](elem.1.c.n.1)] the greater to the less: which is impossible.

Therefore AE is not parallel to BC .

Similarly we can prove that neither is any other straight line except AD ; therefore AD is parallel to BC .

Therefore etc.

Q. E. D.

[I say that they are also in the same parallels.] Heiberg has proved (

Hermes , XXXVIII., 1903, p. 50) from a recently discovered papyrus-fragment (Fayūm towns and their papyri , p. 96, No. IX.) that these words are an interpolation by some one who did not observe that the words And let AD be joined

are part of the setting-out (ἔκθεσις ), but took them as belonging to the construction (κατασκευή ) and consequently thought that a διορισμός or definition

(of the thing to be proved) should precede. The interpolator then altered And

into For

in the next sentence.
[Proposition 40.
Equal triangles which are on equal bases and on the same side are also in the same parallels .

Let ABC , CDE be equal triangles on equal bases BC , CE and on the same side.

I say that they are also in the same parallels.

For let AD be joined; I say that AD is parallel to BE .

For, if not, let AF be drawn through A parallel to BE [[I. 31](elem.1.31)], and let FE be joined.

Therefore the triangle ABC is equal to the triangle FCE ; for they are on equal bases BC , CE and in the same parallels BE , AF . [[I. 38](elem.1.38)]

But the triangle ABC is equal to the triangle DCE ; therefore the triangle DCE is also equal to the triangle FCE , [[C.N. 1](elem.1.c.n.1)] the greater to the less: which is impossible. Therefore AF is not parallel to BE .

Similarly we can prove that neither is any other straight line except AD ; therefore AD is parallel to BE .

Therefore etc. Q. E. D.]

Proposition 41.
If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.

For let the parallelogram ABCD have the same base BC with the triangle EBC , and let it be in the same parallels BC , AE ;

I say that the parallelogram ABCD is double of the triangle BEC .

For let AC be joined.

Then the triangle ABC is equal to the triangle EBC ; for it is on the same base BC with it and in the same parallels BC , AE . [[I. 37](elem.1.37)]

But the parallelogram ABCD is double of the triangle ABC ; for the diameter AC bisects it; [[I. 34](elem.1.34)] so that the parallelogram ABCD is also double of the triangle EBC .

Therefore etc.

Q. E. D.

Proposition 42.
To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.

Let ABC be the given triangle, and D the given rectilineal angle; thus it is required to construct in the rectilineal angle D a parallelogram equal to the triangle ABC .

Let BC be bisected at E , and let AE be joined; on the straight line EC , and at the point E on it, let the angle CEF be constructed equal to the angle D ; [[I. 23](elem.1.23)] through A let AG be drawn parallel to EC , and [[I. 31](elem.1.31)] through C let CG be drawn parallel to EF .

Then FECG is a parallelogram.

And, since BE is equal to EC , the triangle ABE is also equal to the triangle AEC , for they are on equal bases BE , EC and in the same parallels BC , AG ; [[I. 38](elem.1.38)] therefore the triangle ABC is double of the triangle AEC .

But the parallelogram FECG is also double of the triangle AEC , for it has the same base with it and is in the same parallels with it; [[I. 41](elem.1.41)] therefore the parallelogram FECG is equal to the triangle ABC .

And it has the angle CEF equal to the given angle D .

Therefore the parallelogram FECG has been constructed equal to the given triangle ABC , in the angle CEF which is equal to D . Q. E. F.

Proposition 43.
In any parallelogram the complements of the parallelograms about the diameter are equal to one another.

Let ABCD be a parallelogram, and AC its diameter; and about AC let EH , FG be parallelograms, and BK , KD
the so-called complements;

I say that the complement BK is equal to the complement KD .

For, since ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ACD . [[I. 34](elem.1.34)]

Again, since EH is a parallelogram, and AK is its diameter, the triangle AEK is equal to the triangle AHK .

For the same reason the triangle KFC is also equal to KGC .

Now, since the triangle AEK is equal to the triangle AHK , and KFC to KGC ,
the triangle AEK together with KGC is equal to the triangle AHK together with KFC . [[C.N. 2](elem.1.c.n.2)]

And the whole triangle ABC is also equal to the whole ADC ; therefore the complement BK which remains is equal to the complement KD which remains. [[C.N. 3](elem.1.c.n.3)]

Therefore etc.

Q. E. D.

complements, παραπληρώματα , the figures put in to fill up (interstices).

Euclid's phraseology here and in the next proposition implies that the complements as well as the other parallelograms are about

the diagonal. The words are here περὶ δὲ τὴν ΑΓ παραλληλόγραμμα μὲν ἔστω τὰ ΕΘ, ΖΗ, τὰ δὲ λεγόμενα παραπληρώματα τὰ ΒΚ, ΚΔ . The expression the so-called complements

indicates that this technical use of παραπληρώματα was not new, though it might not be universally known.

Proposition 44.
To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.

Let AB be the given straight line, C the given triangle and D the given rectilineal angle; thus it is required to apply to the given straight line AB , in an angle equal to the angle D , a parallelogram equal to the given triangle C .

Let the parallelogram BEFG be constructed equal to the triangle C , in the angle EBG which is equal to D [[I. 42](elem.1.42)]; let it be placed so that BE is in a straight line with AB ; letFG be drawn through to H , and let AH be drawn through A parallel to either BG or EF . [[I. 31](elem.1.31)]

Let HB be joined.

Then, since the straight line HF falls upon the parallels AH , EF , the angles AHF , HFE are equal to two right angles. [[I. 29](elem.1.29)] Therefore the angles BHG , GFE are less than two right angles; and straight lines produced indefinitely from angles less than two right angles meet; [[Post. 5](elem.1.post.5)] therefore HB , FE , when produced, will meet.

Let them be produced and meet at K ; through the point K let KL be drawn parallel to either EA or FH , [[I. 31](elem.1.31)] and let HA , GB be produced to the points L , M .

Then HLKF is a parallelogram, HK is its diameter, and AG , ME are parallelograms. and LB , BF the so-called complements, about HK ; therefore LB is equal to BF . [[I. 43](elem.1.43)]

But BF is equal to the triangle C ; therefore LB is also equal to C . [[C.N. 1](elem.1.c.n.1)]

And, since the angle GBE is equal to the angle ABM , [[I. 15](elem.1.15)] while the angle GBE is equal to D , the angle ABM is also equal to the angle D .

Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB , in the angle ABM which is equal to D .

Q. E. F.

The verb is in the aorist (ὲνέπεσεν ) here and in similar expressions in the following propositions.

Proposition 45.
To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.

Let ABCD be the given rectilineal figure and E the given rectilineal angle; thus it is required to construct, in the given angle E , a parallelogram equal to the rectilineal figure ABCD .

Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD , in the angle HKF which is equal to E ; [[I. 42](elem.1.42)] let the parallelogram GM equal to the triangle DBC be applied to the straight line GH , in the angle GHM which is equal to E . [[I. 44](elem.1.44)]

Then, since the angle E is equal to each of the angles HKF , GHM , the angle HKF is also equal to the angle GHM . [[C.N. 1](elem.1.c.n.1)]

Let the angle KHG be added to each; therefore the angles FKH , KHG are equal to the angles KHG , GHM .

But the angles FKH , KHG are equal to two right angles; [[I. 29](elem.1.29)] therefore the angles KHG , GHM are also equal to two right angles.

Thus, with a straight line GH , and at the point H on it, two straight lines KH , HM not lying on the same side make the adjacent angles equal to two right angles; therefore KH is in a straight line with HM . [[I. 14](elem.1.14)]

And, since the straight line HG falls upon the parallels KM , FG , the alternate angles MHG , HGF are equal to one another. [[I. 29](elem.1.29)]

Let the angle HGL be added to each; therefore the angles MHG , HGL are equal to the angles HGF , HGL . [[C.N. 2](elem.1.c.n.2)]

But the angles MHG , HGL are equal to two right angles; [[I. 29](elem.1.29)] therefore the angles HGF , HGL are also equal to two right angles. [[C.N. 1](elem.1.c.n.1)] Therefore FG is in a straight line with GL . [[I. 14](elem.1.14)]

And, since FK is equal and parallel to HG , [[I. 34](elem.1.34)] and HG to ML also,
KF is also equal and parallel to ML ; [[C.N. 1](elem.1.c.n.1);[ I. 30](elem.1.30)] and the straight lines KM , FL join them (at their extremities); therefore KM , FL are also equal and parallel. [[I. 33](elem.1.33)] Therefore KFLM is a parallelogram.

And, since the triangle ABD is equal to the parallelogram FH , and DBC to GM ,
the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM .

Therefore the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD , in the angle FKM which is equal to the given angle E .

Q. E. F.

rectilineal

simply, without figure,

εὐθύγραμμον being here used as a substantive, like the similarly formed παραλληλόγραμμον .

Proposition 46.
On a given straight line to describe a square.

Let AB be the given straight line; thus it is required to describe a square on the straight line AB .

Let AC be drawn at right angles to the straight line AB from the point A on it [[I. 11](elem.1.11)], and let AD be made equal to AB ; through the point D let DE be drawn parallel to AB , and through the point B let BE be drawn parallel to AD . [[I. 31](elem.1.31)]

Therefore ADEB is a parallelogram; therefore AB is equal to DE , and AD to BE . [[I. 34](elem.1.34)]

But AB is equal to AD ; therefore the four straight lines BA , AD , DE , EB are equal to one another; therefore the parallelogram ADEB is equilateral.

I say next that it is also right-angled.

For, since the straight line AD falls upon the parallels AB , DE , the angles BAD , ADE are equal to two right angles. [[I. 29](elem.1.29)]

But the angle BAD is right; therefore the angle ADE is also right.

And in parallelogrammic areas the opposite sides and angles are equal to one another; [[I. 34](elem.1.34)] therefore each of the opposite angles ABE , BED is also right. Therefore ADEB is right-angled.

And it was also proved equilateral.

Therefore it is a square; and it is described on the straight line AB .

Q. E. F.

Proclus (p. 423, 18 sqq. ) notes the difference between the word construct (συστἡσασθαι ) applied by Euclid to the construction of a triangle (and, he might have added, of an angle ) and the words describe on (ἀναγράφειν ἀπό ) used of drawing a square on a given straight line as one side. The triangle (or angle ) is, so to say, pieced together, while the describing of a square on a given straight line is the making of a figure from

one side, and corresponds to the multiplication of the number representing the side by itself.

Proposition 47.
In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

Let ABC be a right-angled triangle having the angle BAC right;

I say that the square on BC is equal to the squares on BA , AC .

For let there be described on BC the square BDEC , and on BA , AC the squares GB , HC ; [[I. 46](elem.1.46)] through A let AL be drawn parallel to either BD or CE , and let AD , FC be joined.

Then, since each of the angles BAC , BAG is right, it follows that with a straight line BA , and at the point A on it, the two straight lines AC , AG not lying on the same side make the adjacent angles equal to two right angles; therefore CA is in a straight line with AG . [[I. 14](elem.1.14)]

For the same reason BA is also in a straight line with AH .

And, since the angle DBC is equal to the angle FBA : for each is right: let the angle ABC be added to each; therefore the whole angle DBA is equal to the whole angle FBC . [[C.N. 2](elem.1.c.n.2)]

And, since DB is equal to BC , and FB to BA , the two sides AB , BD are equal to the two sides FB , BC respectively, and the angle ABD is equal to the angle FBC ; therefore the base AD is equal to the base FC , and the triangle ABD is equal to the triangle FBC . [[I. 4](elem.1.4)]

Now the parallelogram BL is double of the triangle ABD , for they have the same base BD and are in the same parallels BD , AL . [[I. 41](elem.1.41)]

And the square GB is double of the triangle FBC , for they again have the same base FB and are in the same parallels FB , GC . [[I. 41](elem.1.41)]

[But the doubles of equals are equal to one another.] Therefore the parallelogram BL is also equal to the square GB .

Similarly, if AE , BK be joined, the parallelogram CL can also be proved equal to the square HC ; therefore the whole square BDEC is equal to the two squares GB , HC . [[C.N. 2](elem.1.c.n.2)]

And the square BDEC is described on BC , and the squares GB , HC on BA , AC .

Therefore the square on the side BC is equal to the squares on the sides BA , AC .

Therefore etc.

Q. E. D.

, τὸ ἀπὸ...τετρἁγωνον , the word ἀναγραφέν or ἀναγεγραμμένον being understood.

subtending the right angle. Here ὑποτεινούσης , subtending,

is used with the simple accusative (τὴν ὀρθὴν γωνίαν ) instead of being followed by ὑπό and the accusative, which seems to be the original and more orthodox construction. Cf. [I. 18](elem.1.18), note.

Euclid actually writes DB , BA ,

and therefore the equal sides in the two triangles are not mentioned in corresponding order, though he adheres to the words ἑκατέρα ἑκατέρα respectively.

Here DB is equal to BC and BA to FB .

[But the doubles of equals are equal to one another.] Heiberg brackets these words as an interpolation, since it quotes a

Common Notion which is itself interpolated. Cf. notes on [I. 37](elem.1.37), p. 332, and on interpolated Common Notions , pp. 223-4.
Proposition 48.
If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.

For in the triangle ABC let the square on one side BC be equal to the squares on the sides BA , AC ;

I say that the angle BAC is right.

For let AD be drawn from the point A at right angles to the straight line AC , let AD be made equal to BA , and let DC be joined.

Since DA is equal to AB , the square on DA is also equal to the square on AB .

Let the square on AC be added to each; therefore the squares on DA , AC are equal to the squares on BA , AC .

But the square on DC is equal to the squares on DA , AC , for the angle DAC is right; [[I. 47](elem.1.47)] and the square on BC is equal to the squares on BA , AC , for this is the hypothesis; therefore the square on DC is equal to the square on BC , so that the side DC is also equal to BC .

And, since DA is equal to AB , and AC is common, the two sides DA , AC are equal to the two sides BA , AC ; and the base DC is equal to the base BC ; therefore the angle DAC is equal to the angle BAC . [[I. 8](elem.1.8)] But the angle DAC is right; therefore the angle BAC is also right.

Therefore etc.

Q. E. D.