Thro' the point C draw CE parallel to DA, and let BA pro- Book VI. duced meet CE in E. Because the straight line AC meets the pa

rallels AD, EC, the angle ACE is equal to the alternate angle a. 31. 1. CAD. but CAD, by the Hypothefis, is equal to the angle BAD; b. 29. x. wherefore BAD is equal to the angle ACE. Again, because the

ftraight line BAE meets the pa-
rallels AD, EC, the outward an-
gle BAD is equal to the inward
and oppofite angle AEC. but the
angle ACE has been proved e-
qual to the angle BAD; there-
fore alfo ACE is equal to the
angle AEC, and confequently the
fide AE is equal to the fide AC. B

and because AD is drawn parallel to one of the fides of the triangle BCE, viz. to EC, BD is to DC, as BA to AE; but AE is d. 2. 6. equal to AC; therefore as BD to DC, fo is BA to AC.

Let now BD be to DC, as BA to AC, and join AD; the angle BAC is divided into two equal angles by the straight line AD.

c. 7. Sp

The fame construction being made, because as BD to DC, fo is BA to AC; and as BD to DC, fo is BA to AE, because AD is parallel to EC; therefore BA is to AC, as BA to AE f. con- f. 11. 5. fequently AC is equal to AE, and the angle AEC is therefore g. 9. 5. equal to the angle ACE b. but the angle AEC is equal to the out- h. 5. 1, ward and oppofite angle BAD; and the angle ACE is equal to the alternate angle CAD b. wherefore alfo the angle BAD is equal to the angle CAD. therefore the angle BAC is cut into two equal angles by the straight line AD.

&c. Q. E. D.

Therefore if the angle,

K 4

Book VI.

IFt

F the outward angle of a triangle made by producing one of its fides, be divided into two equal angles, by a ftraight line which alfo cuts the bafe produced; the fegments between the dividing line and the extremities of the bafe have the fame ratio which the other fides of the triangle have to one another, and if the fegments of the bafe produced have the fame ratio which the other fides of the triangle have, the ftraight line drawn from the vertex to the point of fection divides the outward angle of the triangle into two equal angles.

Let the outward angle CAE of any triangle ABC be divided into two equal angles by the ftraight line AD which meets the bale produced in D. BD is to DC, as BA to AC.

E

Through C draw CF parallel to AD2; and because the straight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CAD. but CAD is equal to the angle DAES; therefore alfo DAE is equal to the angle ACF. Again, because the ftraight line FAE meets the parallels AD, FC, the outward angle DAE is equal to the inward and oppofite angle CFA. but the angle ACF has been proved equal to the angle DAE; therefore alfo the angle ACF is equal to the angle CFA, and confequently the. ide AF is equal to the fide B

T

C

D

AC, and becaufe AD is parallel to FC a fide of the triangle BCF, O is to DC, as BA to AF; but AF is equal to AC; as therefore BD is to DC, fo is BA to AC.

Let now BD be to DC, as BA to AC, and join AD; the angle CAD is equal to the angle DAE.

The fame conftruction being made, because BD is to DC, as BA to AC; and that BD is alfo to DC, as BA to AF; therefore BA is to AC, as BA to AF f. wherefore AC is equal to AF 3, and the angle AFC equal to the angle ACF. but the angle AFC

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.

THE 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. 1. fides about the equal angles of the triangles ABC, DCE are proportionals; and thote are the homologous fides which are oppofite to the equal angles.

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 becaufe

the angles ABC, ACB are together less than two right angles b; b. 17. 1. ABC and DEC, which is equal to ACB, are alfo lefs 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 4 to CD. Again, because the angle ACB is equal to B

the angle DEC, AC is parallel to

A

C. 12. Ax. I.

D

d. 28. I.

FE. therefore FACD is a parallelogram; and confequently AF is equal to CD, and AC to FD. and becaufe AC is parallel to FE e. 34. 1. one of the fides of the triangle FBE, BA is to AF, as BC to CEf. 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, fo 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, so CE to ED, ex aequali 1, BA is to AC, as CD to DE. Therefore the fides, &c. h. 22. 5. Q. E. D.

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, fo 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 appofite to the homologous fides, viz. the angle ABC equal to the angle DEF, and BCA to EFD, and alfo BAC to EDF.

A

a

At the points E, F in the ftraight 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 EGF, and the triangle ABC is therefore equiangular to the triangle GEF; and confequently they have their fides oppofite to the equal angles proportionals. B

wherefore as AB to BC, fo is
GE to EF; but as AB to BC,

D

E

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 bafe GF; therefore the angle DEF is equal f to the angle GEF, and the other angles to the other angles which are fubtended by the equal fides 8. 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 reafon, the angle ACB is equal to the angle DFE, and the angle at Ato the angle at D. Therefore the triangle ABC is equiangular Wherefore if the fides, &c. Q. E: D,

to the triangle DEF.

PROP. VI.

THEOR.

Book VI.

IF

F 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 thofe 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.

a

At the points D, F, in the ftraight line DF, make the angle a. 23. 1, FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the angle ACB. where

equal to DG; and DF is common to the two triangles EDF,GDF. c. 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 bafe EF is equal to the bafe FG f, and the triangle EDF to f. 4. 1. 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 alfo the remaining g. Hyp. angle at B is equal to the remaining angle at E. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore if two triangles, &c. Q. E. D.

1