PROP. III. THEOR.

Book VI.

IF the angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base will have the same ratio which the other sides of the triangle have to each other; and if the segments of the base have the same ratio which the other sides of the triangle have to each other, the straight line drawn from the vertex to the point of section bisects the vertical angle.

Let the angle BAC, of any triangle ABC, be divided into two equal angles by the straight line AD; BD is to DC as BA to AC.

Through the point C draw CE parallela to DA, and let a 31. 1. BA produced meet CE in E. Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CADb. But CAD, by the hypothesis, is b 29. 1.. equal to the angle BAD; wherefore BAD is equal to the angle ACE. Again, be

cause the straight line BAE meets the parallels AD, EC, the exterior angle BAD is equal to the interior and opposite angle AEC. But the angle ACE has been proved equal to the angle BAD; therefore also ACE is equal to the angle AEC, and conse

E

quently the side AE is

equal to the side AC.

c 6. 1..

Because AD is drawn parallel to EC, one of the sides of the triangle BCE, BD is to DC, as BA to AEd or AC.

d 2.6.

Book VI.

d 2. 6.

e 11.5. f 9. 5. g 5. 1.

b 29. 1.

Next, let 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.

The same construction being made, as BD to DC, so is BA to AC; and as BD to DC, so is BA to AEd, because AD is parallel to EC; therefore AB is to AC, as AB to AE; consequently AC is equal to AEf, therefore the angle. AEC is equal to the angle ACEs. But the angle AEC is equal to the exterior and opposite angle BAD, and the angle ACE is equal to the alternate angle CAD; wherefore also the angle BAD is equal to the angle CAD; therefore the angle BAC is cut into two equal angles by the straight line AD. Therefore, if the angle, &c. Q. E. D.

PROP. A. THEOR.

3 31. 1.

b 29. 1.

c Hyp.

IF the exterior angle of a triangle be bisected by a straight line which also cuts the base produced, the segments between the bisecting line and the extremities of the base have the same ratio which the other sides of the triangle have to each other; and if the segments of the base produced have the same ratio which the other sides of the triangle have, the straight line, drawn from the vertex to the point of section, bisects the exterior angle of the triangle.

Let the exterior angle CAE, of any triangle ABC, be bisected by the straight line AD, which meets the base produced in D; BD is to DC, as BA to AC.

Through C draw CF parallel to ADa. 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 DAE; therefore also DAE is equal to the angle ACF. Again, because the straight line FAE meets the parallels AD, FC, the exterior angle DAE is equal to the interior and opposite angle CFA. But the angle ACF has been

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

d 6. 1.

e 2. 6.

The same construction being made, BD is to DC, as BA to AC, and also BD is to DC, as BA to AFe; therefore BA is to AC, as BA to AFf; wherefore AC is equal to AF, f 11. 5. therefore the angle AFC is equal to the angle ACF. But g 9. 5. the angle AFC is equal to the exterior angle EAD, and the h 5.1. angle ACF to the alternate angle CAD; therefore also EAD is equal to the angle CAD. Wherefore, if the exterior, &c. Q. E. D.

PROP. IV. THEOR.

THE sides about the equal angles of equiangular triangles are proportional; and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or consequents of the ratios.

Let ABC, DCE be equiangular triangles, having the angle ABC equal to the angle DČE, and the angle ACB to the angle DEC, and consequentlya the angle BAC equal to the a 32. 1. angle CDE. The sides about the equal angles of the triangles ABC, DCE are proportional; and those are the homologous sides which are opposite to the equal angles.

Book VI.

b 17.1.

Let the triangle DCE be placed so that its side CE may be contiguous to BC, and in the same straight line with it. Because the angles ABC, ACB are together less than two right angles, ABC and DEC are also less than two right angles; wherefore BA, ED produced will c Cor. 29.1. meet. Let them be produced, and A meet in the point F.

d 28. 1.

e 34. 1.

f 2.6.

g 16. 5.

h 22. 5.

F

C

E

Because the angle ABC is equal to the angle DČE, BF is paralleld to CD; and because the angle ACB is equal to the angle DEC, AC is parallel to FEd. Therefore FACD is a paral- B lelogram; consequently AF is equal to CD, and AC to FDe. Because AC is parallel to FE, one of the sides of the triangle FBE, BA: AF BC CEf, or, because AF is equal to CD, BA: CD: BC: CE; therefore, alternately, BA: BC:: DC: CE. Again, because CD is parallel to BF, BC: CE :: FD: DE, or, because FD is equal to AC, BC : CE :: AC: DE; therefore BC: CA : : CE : ED.

But AB: BC :: DC: CE; therefore, ex æquali, BA: AC:: CD: DE1. Therefore, the sides, &c. Q. E. D.

a22. 5.

IF the sides of two triangles, about each of their angles, be proportional, the triangles will be equiangular, and have their equal angles opposite to the homologous sides.

Let the triangles ABC, DEF have their sides proportional, so that AB is to BC, as DE to EF, and BC to CA, as EF to FD; and consequently, ex æqualia, BA to AC, as ED to DF; the triangle ABC is equiangular to the triangle DEF, and their equal angles are opposite to the homologous

sides, viz. the angle ABC being equal to the angle DEF, and Book VI. BCA to EFD, and also BAC to EDF.

DE: EF:: GE: EF. Therefore DE and e 11. 5. GE are equal. For the same reason DF is equal to FG. In f9. 5. the triangles DEF, GEF, DE is equal to EG, and EF common, and DF equal to GF; therefore the angle DEF is equals to the angle GEF; wherefore the angle DFE is equalh g 8. 1. to the angle GFE, and EDF to EGF. Now the angle DEF h 4.1, is equal to the angle GEF, and GEF to the angle ABC; therefore the angle ABC is equal to the angle DEF. For the same 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 sides, &c. Q. E. D.

PROP. VI. THEOR.

IF two triangles have one angle of one equal to one angle of the other, and the sides about the equal angles proportional, the triangles will be equiangular, and will have those angles equal which are opposite to the homologous sides.