is equal to the outward angle EAD, and the angle ACF to the Book VI. alternate angle CAD; therefore alfo EAD is equal to the angle CAD. Wherefore if the outward, &c. Q. E. D.

TH

PROP. IV. THEOR.

HE fides about the equal angles of equiangular triangles are proportionals; and those which are oppofite to the equal angles are homologous fides, that is, are the antecedents or confequents of the ratios.

Let ABC, DCE be equiangular triangles, having the angle ABC equal to the angle DCE, and the angle ACB to the angle DEC, and confequently a the angle BAC equal to the angle CDE. The a. 32. I. fides about the equal angles of the triangles ABC, DCE are proportionals; and those are the homologous fides which are oppofite to the equal angles.

F

Let the triangle DCE be placed fo that its fide CE may be contiguous to BC, and in the fame ftraight line with it. and because the angles ABC, ACB are together lefs than two right angles "; ABC and DEC, which is equal to ACB, are alfo less than two right angles. wherefore BA, ED produced fhall meet; let them be produced and meet in the point F. and because the angle ABC is equal to the angle DCE, BF is parallel to CD. Again, because the angle ACB is equal to

the angle DEC, AC is parallel to

B

E

FEd. therefore FACD is a parallelogram; and confequently AF is equal to CD, and AC to FD. and because AC is parallel to FE e. 34. L. one of the fides of the triangle FBE, BA is to AF, as BC to CE f. f. 2. 6. but AF is equal to CD, therefore & as BA to CD, fo is BC to CE; g⋅ 7⋅ 5′ and alternately, as AB to BC, so DC to CE. Again, because CD is parallel to BF, as BC to CE, fo is FD to DE f; but FD is equal to AC; therefore as BC to CE, fo is AC to DE. and alternately, as BC to CA, fo CE to ED. therefore because it has been proved that AB is to BC, as DC to CE; and as BC to CA, fo CE to ED,

ex aequali 1, BA is to AC, as CD to DE. Therefore the fides, &c, h. 22, 5, Q. E. D.

154

Book VI.

2. 23. I.

IF

PROP. V. THEOR.

F the fides of two triangles, about each of their angles, be proportionals, the triangles fhall be equiangular, and have their equal angles oppofite to the homologous fides.

Let the triangles ABC, DEF have their fides proportionals, so that AB is to BC, as DE to EF; and BC to CA, as EF to FD; and confequently, ex aequali, BA to AC, as ED to DF. the triangle ABC is equiangular to the triangle DEF, and their equal angles are oppofite to the homologous fides, viz. the angle ABC equal to the angle DEF, and BCA to EFD, and also BAC to EDF.

At the points E, F in the straight line EF make the angle FEG equal to the angle ABC, and the angle EFG equal to BCA; wherefore the remaining angle BAC

is equal to the remaining angle b. 32. 1. EGF, and the triangle ABC is therefore equiangular to the triangle GEF; and confequently they have their fides opposite to the equal angles proportionals. B wherefore as AB to BC, fo is GE to EF; but as AB to BC,

c. 4. 6.

e. 9. 5.

f. 8. I.

A

d. 11. 5. fo is DE to EF; therefore as DE to EF, fod GE to EF. therefore DE and GE have the fame ratio to EF, and confequently are equal. for the fame reafon DF is equal to FG. and because, in the triangles DEF, GEF, DE is equal to EG, and EF common, the two fides DE, EF are equal to the two GE, EF, and the bafe DF is equal to the base GF; therefore the angle DEF is equal to the angle GEF, and the other angles to the other angles which are fubtended by the equal fides. Wherefore the angle DFE is equal to the angle GFE, and EDF to EGF. and because the angle DEF is equal to the angle GEF, and GEF to the angle ABC; therefore the angle ABC is equal to the angle DEF. for the fame reason, the angle ACB is equal to the angle DFE, and the angle at A to the angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore if the fides, &c. Q. E. D.

g. 4. I.

F

I'

PROP. VI. THEOR.

two triangles have one angle of the one equal to one angle of the other, and the fides about the equal angles proportionals, the triangles fhall be equiangular, and fhall have those angles equal which are oppofite to the homologous fides.

Let the triangles ABC, DEF have the angle BAC in the one equal to the angle EDF in the other, and the fides about those angles proportionals; that is, BA to AC, as ED to DF. The triangles ABC, DEF are equiangular, and have the angle ABC equal to the angle DEF, and ACB to DFE.

Book VI.

At the points D, F, in the straight line DF, make a the angle a. 23. I. FDG equal to either of the angles BAC, EDF; and the angle DFG

equal to the angle ACB. where

fore the remaining angle at B is A

equal to the remaining one at

f. 4. L.

equal to DG; and DF is common to the two triangles EDF, GDF. e. 9 5. therefore the two fides ED, DF are equal to the two fides GD, DF; and the angle EDF is equal to the angle GDF, wherefore the base EF is equal to the bafe FG f, and the triangle EDF to the triangle GDF, and the remaining angles to the remaining angles, each to each, which are fubtended by the equal fides. therefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E. but the angle DFG is equal to the angle ACB; therefore the angle ACB is equal to the angle DFE. and the angle BAC is equal to the angle EDF ; wherefore also the remaining g. Hype angle at B is equal to the remaining angle at E. Therefore the triangle ABC is equiangular to the triangle DEF. two triangles, &c. Q. E. D.

Wherefore if

Book VI.

See N.

IF

two triangles have one angle of the one equal to one angle of the other, and the fides about two other angles proportionals; then if each of the remaining angles be either lefs, or not less than a right angle ; or if one of them be a right angle: the triangles fhall be equiangular, and have those angles equal about which the fides are proportionals.

Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viz. the angle BAC to the angle EDF, and the fides about two other angles ABC, DEF proportionals, so that AB is to BC, as DE to EF; and, in the first case, let each of the remaining angles at C, F be lefs than a right angle. The triangle ABC is equiangular to the triangle DEF, viz. the angle ABC is equal to the angle DEF, and the remaining angle at C, to the remaining angle at F.

For if the angles ABC, DEF be not equal, one of them is greater than the other; let ABC be the greater, and at the point B in the ftraight line AB make the an

a. 23. 1. gle ABG equal to the angle a DEF. and because the angle at A is equal to the angle at D, and the angle ABG to the angle DEF; the remaining angle

b. 32. 1. AGB is equal to the remain- B ing angle DFE. therefore the

c. 4. 6.

e. 9. 5.

f. 5. I.

CE F

triangle ABG is equiangular to the triangle DEF; wherefore as AB is to BG, fo is DE to EF; but as DE to EF, fo, by Hypothed. 11. 5. fis, is AB to BC; therefore as AB to BC, fo is AB to BG"; and because AB has the fame ratio to each of the lines BC, BG; BC is equal to BG, and therefore the angle EGC is equal to . the angle BCG . but the angle BCG is, by Hypothefis, lefs than a right angle; therefore also the angle BGC is less than a right angle, and the adjacent angle AGB must be greater than a right . 13. 1. angle 8. But it was proved that the angle AGB is equal to the angle at F; therefore the angle at F is greater than a right angle. but, by the Hypothefis, it is less than a right angle, which is ab

furd. Therefore the angles ABC, DEF are not unequal, that is, Book VI. they are equal. and the angle at A is equal to the angle at D; wherefore the remaining angle at C is equal to the remaining angle at F. therefore the triangle ABC is equiangular to the triangle DEF.

Next, Let each of the angles at C, F be not lefs than a right angle. the triangle ABC is alfo in this cafe equiangular to the triangle DEF.

BGC is not less than a right angle. wherefore two angles of the triangle BGC are together not less than two right angles; which

is impoffible 1; and therefore the triangle ABC may be proved to h. 17. 1. be equiangular to the triangle DEF, as in the first case.

Laftly, Let one of the angles at C, F, viz. the angle at C be a right angle; in this cafe likewise the triangle ABC is equiangular to the triangle DEF.

For if they be not equiangular, make at the point B of the straight line AB the angle ABG aqual to the angle DEF; then it may be proved, as in the firstcafe, that BG is equal to BC. but the angle BCG is a right angle, therefore f the angle BGC is alfo a right angle; whence two of the angles of the triangle BGC are together not less than twoj WOB right angles; which is impoflible h, therefore the triangle ABC

A

A

D

f. 5. x.

E

C

is equiangular to the triangle DEF. Wherefore if two triangles,

&c. Q. E. D.