Then because the two right lines DC, CE are equal to the two right lines F A, A G, each to each; and the base DE is equal to the base FG; the angle D C E [by prop. 8.] is equal to the angle F A G.

Therefore at the given right line AB, and at the given point A therein, is made the right-lined angle F A G, equal to the given right-lined angle D CE. Which was to be done.

• There is an easier folution of this problem, by means of But as Ecluid wanted this problem to help prop. 27. lib. iii. to demonftrate the 24th, he could not give that folution in this place, because it could not be here demonftrated; and fo he gave the only one that could be demonftrated upon the axioms and propofitions already established. He has done the fame all along. And whoever believes that Euclid was ignorant of shorter conftructions of several of his problems that might be given, because he did not give them, I really think, are much miftaken; he rather omitted them, because they could not be demonftrated but by propofitions he was not yet come to; his conftant purpose being to admit nothing without first demonftrating it.

PROP. XXIV. THEOR.

If two triangles have two fides of the one equal to two fides of the other, each to each; and the angle of the one contained under the equal right lines, greater than the angle of the other; then will the bafe of the one be greater than the base of the

other.

Let there be two triangles A B C, DEF, having the two fides A B, AC equal to the two fides DE, DF, each to each; that is, the fide A B equal to the fide D E, and the

B

D

F

fide A c equal to the fide DF; and let the angle B A C be greater than the angle EDF: I fay, the base B C is greater than the base E F. For because the angle BAC is greater than the Gangle E D F, at the right line DE, and at the point

D, therein make [by prop. 23.] the angle DG equal to

the

the angle BAC; and put D G [by prop. 3.] equal to either of the right lines A C, DF: alfo draw G E, F G.

Now because A B is equal to the right line DE, and the right line AC equal to the right line D G, therefore there are two right lines BA, AC equal to two right lines E D, DG, each to each; and the angle B A C [by conftr.] is equal to the angle EDG; therefore the bafe BC [by prop. 4.] will be equal to the bafe E G. Again, becaufe DG is equal to the right line D F, and the angle D F G is [by prop. 5.] equal to the angle DG F, the angle DFG will be greater than the angle E G F ; and therefore the angle E F G will be much greater than the angle E G F. And because the triangle EFG has the angle EFG greater than the angle EGF; and [by prop. 19.] the greater fide is oppofite to the greater angle; the fide EG is greater than the fide E F. But [by conftr.] the fide E G is equal to the fide BC; and fo the fide B C is greater than the fide EF.

If therefore two triangles have two fides of the one equal to two fides of the other, each to each; and the angle of the one contained by the equal right lines greater than the angle of the other; then will the base of the one be greater than the base of the other. Which was to be demonftrated.

PROP. XXV. THEOR.

If two triangles have two fides of the one equal to two fides of the other, each to each; and the bafe of the one be greater than the base of the other; they fhall also have the angles contained by the equal fides, the one greater than the other h.

D

Let there be two triangles A B C, DE F, having two fides A B, AC, equal to two fides DE, DF, each to each, viz. the fide A B equal to the fide D E, and the fide AC equal to the fide DF; but the bafe BC greater than the base EF: I fay, the angle B A C is greater than the angle E D F.

B

CE F

For if it be not greater, it will be either equal to, or lefs than it. But the angle B A C is not equal to the angle

EDF;

Book I. EDF; for if it were [by prop. 4.] the bafe B C would be equal to the base E F. But it is not fo [by fuppofition]. Therefore the angle B A C is not equal to the angle E D F, neither is it lefs; for if it were, the bafe B C [by prop. 24.] would be less than the base EF: but it is not; therefore the angle B A C is not lefs than the angle E D F. But it has been proved not be equal to it neither. Wherefore the angle B A C is greater than the angle E D F.

If therefore two triangles have two fides of the one equal to two fides of the other, each to each; and the base of the one be greater than the base of the other; they fhall also have the angles contained by the equal fides, the one greater than the other. Which was to be demonftrated.

This propofition is demonftrated directly by Menelaus of Alexandria and by Hero. See Proclus's Commentary upon the first book of Euclid; wherein are extant their demonftrations. Clavius has also tranflated them into Latin.

PROP. XXVI. THEOR.

If two triangles have two angles of the one equal to two angles of the other, each to each, and one fide of the one equal to one fide of the other, either that fide which is between the equal angles, or that which is oppofite to one of them; then will the remaining fides of the one triangle be equal to the remaining fides of the other, each to each; and the remaining angle of the one will be equal to the remaining angle of the other.

Let there be two triangles A B C, DE F, having the two angles A B C, B C A of the one equal to the two angles D E F, EFD of the other, each to each, viz. the angle ABC equal to the angle DE F, and the angle B C A equal to the angle E FD; alfo let one fide of the one be equal to one fide of the other. And first, Let these fides be в C, E F, lying between the equal angles: I fay also, the remaining fides of the one will be equal to the remaining fides of the other, each to each, viz. the fide A B equal to the fide D E, and the fide A C to the fide D F ; and the remaining angle B A C equal to the remaining angle E D F.

For

For if the fide A B be unequal to the fide DE, one of

B

A

G

E

H C

D

F

them will be the greateft. Let A B be the greater; and [by prop. 3.] put the right line B G equal to E D, and join G c. Then because the fide B G is equal to the fide D E, and the fide B C to the fide E F, viz. the two fides B G, B C equal to the two fides DE, EF, each to each; and the angle G B C is equal to the angle DEF; therefore will the bafe GC be equal to the bafe D F [by prop. 4.]; the triangle G C B will be equal to the triangle DEF, and the remaining angles will be equal to the remaining angles which are oppofite to the equal fides. Therefore the angle GCB is equal to the angle DFE; but the angle DFE is put equal to the angle B C A wherefore the angle B C G is equal to the angle B CA; the lefs equal to the greater, which is impoffible: therefore the fide A B is not unequal to the fide DE; and fo it is equal to it. But the Ade BC is equal to the fide E F, and the two fides A B, B C are equal to the two fides D E, EF, each to each; and the angle A B C is equal to the angle DEF. Wherefore the bafe A c [by prop. 4.] will be equal to the base D F, and the remaining angle B A C is equal to the remaining angle E D F

Secondly, Let the fides A B, D E oppofite to the equal angles be equal; that is, the fide A B to the fide D E : I say, the remaining fides of the one triangle will be equal to the remaining fides of the other; that is, the fide A c equal to the fide D F, and B C equal to the fide EF; and also the remaining angles BAC and E D F equal.

For if B C be unequal to E F, one of them is the greater, which let be B C (if poffible); and put [by prop. 3.] BH equal to E F, and join A H.

Then because the fide в H is equal to the fide E F, and the fide A B equal to the fide D E; viz. the two fides A B, BH equal to the two fides D E, EF, each to each; and the angles contained by them equal; therefore the bafes AH, DF [by prop. 4.] are equal; and the triangle ABH is equal to the triangle D, and the remaining angles equal

equal to the remaining angles, each to each; which are oppofite to the equal fides; therefore the angle BHA is equal to the angle E F D. But [by fuppofition] the angle EFD is equal to the angle BCA; therefore the angle BHA is equal to the angle B C A; viz. the outward angle BH A of the triangle AC H equal to the inward oppofite angle B CA; which [by prop. 16.] cannot be. Therefore the fide BC is not unequal to the fide E F, and fo it is equal to it; but AB is equal to the fide DE: wherefore two fides A B, BC are equal to two fides D E, E F, each to each; and they contain equal angles: therefore the base AC [by prop. 4.] is equal to the base D F, and the triangle ABC is equal to the triangle DEF, and the remaining angle BAC equal to the remaining angle E DF.

If therefore two triangles have two angles of the one equal to two angles of the other, each to each; and one fide of the one equal to one fide of the other, either that which is between the equal angles, or that which is oppofite to one of them; the remaining fides of the one triangle will be equal to the remaining fides of the other, each to each; and the remaining angle of the one will be equal to the remaining angle of the other. Which was to be demonftrated.

PROP. XXVII. THEOR.

If a right line falling upon two right lines makes the alternate angles equal to one another, thefe right lines will be parallel to one another.

For let the right line E F, falling upon the right lines A B, C D, make the alternate angles A E F, E F D equal to one another: I say, the right line A B is parallel to the right line CD.

For if it be not parallel, the lines A B, DC produced will

meet either towards B D or A C; let them be produced towards B D, and meet in the point G.

Now the angle AEF being the external angle of the triangle EGF, is greater [by prop. 16.] than the inward oppofite angle

EFC