Book VI. ADE, so is CE to EA. Therefore, as BD to DA, so is CE to EA d.

d 11. 5.

Next, Let the sides AB, AC, of the triangle ABC, or these

e l. 6.

f 9. 5.

g 39. 1.

See N.

sides produced, be cut proportionally in the points D, E, that is, so that BD be to DA, as CE to EA, and join DE; then DE is parallel to BC.

The same construction being made, because as BD to DA, so is CE to EA; and as BD to DA, so is the triangle BDE to the triangle ADE; and as CE to EA, so is the triangle CDE to the triangle ADE; therefore, the triangle BDE is to the triangle ADE, as the triangle CDE to the triangle ADE; that is, the triangles BDE, CDE have the same ratio to the triangle ADE; and therefore, the triangle BDE is equal to the triangle CDE: and they are on the same base DE; but equal triangles on the same base are between the same parallels; therefore, DE is parallel to BC. Wherefore, " if a "straight line," &c. Q. E. D.

દ

If the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another: And if the segments of the base have the same ratio which the other sides of the triangle have to one another, the straight line drawn from the vertex to the point of section, divides the vertical angle into two equal angles.

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

Through the point C draw CE parallel to DA, and let BA Book VI. produced meet CE in E. Because the straight line AC meets

the parallels AD, EC, the angle ACE is equal to the alternate a 31. 1. angle CAD: But CAD, by the hypothesis, is equal to the b 29. 1. angle BAD: wherefore, BAD is equal to the angle ACE.

Again, because the straight line BAE meets the parallels AD, EC, the outward angle BAD is equal to the inward and opposite angle AEC. But the angle ACE has been proved equal to the angle BAD; therefore also, ACE is equal to the angle A EC, and consequently the side AE is equal to the side AC: and because

AD is drawn parallel to one of the sides of the triangle BCE, viz. to EC, BD is to DC, as BA to AEd: but AE is equal d 2. 6. to AC; therefore, as BD to DC, so is BA to AC.

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

The same construction being made; because, as BD to DC, so is BA to AC; and as BD to DC, so is BA to AE, because AD is parallel to EC; therefore BA is to AC, as BA to AE: Consequently AC is equal to AE, and the angle AEC is therefore equal to the angle ACE: But the angle AEC is equal to the outward 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.

e 7. 5.

f 11. 5.

9.5.

h 5. 1.

Book VI.

a 31. 1.

b 29. 1. с Нур.

d 6. 1.

e 2. 6.

f 11. 5.

g 9. 5. b 5. 1.

PROP. A. THEOR.

If the outward angle of a triangle made by producing one of its sides, be divided into two equal angles, by a straight line which also cuts the base produced; the segments between the dividing line and the extremities of the base have the same ratio which the other sides of the triangle have to one another: 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 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 straight line AD, which meets the base produced in D; then BD is to DC, as BA to AC.

a

Through C draw CF parallel to AD; 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 DAE; therefore also DAE is equal to the angle ACF. Again, because the straight line FAE meets the parallels AD, FC, the outward angle DAE is equal to the inward and opposite angle CFA: But the angle ACF has been proved to be equal to the angle DAE; therefore also the angle ACF is equal to the angle CFA, and conse- B

quently the side AF is

a

E

A

C

D

equal to the side AC: And because AD is parallel to FC, side of the triangle BCF, BD is to DC, as BA to AF; but AF is equal to AC; as therefore BD is to DC, so is BA to AC. Now, let BD be to DC, as BA to AC, and join AD; then the angle CAD is equal to the angle DAE.

The same construction being made, because BD is to DC, as BA to AC; and also BD to DC, as BA to AF; therefore BA is to AC, as BA to AF; wherefore AC is equal to AF, and the angle AFC equal to the angle ACF: But

h

OF EUCLID.

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

PROP. IV. THEOR.

The sides about the equal angles of equiangular triangles are proportionals; 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 DCE, and the angle ACB to the angle DEC, and consequently a the angle BAC equal to the a 32. 1. angle CDE. The sides about the equal angles of the triangles ABC, DCE are proportionals; and those are the homologous sides which are opposite to the equal angles.

F

Let the triangle DCE be placed, so that its side CE may be contiguous to BC, and in the same straight line with it: And because the angles ABC, ACB are together less than two right angles ABC, and DEC which is equal to ACB, are also less than two right angles; wherefore BA, ED produced shall meet; let them be produced and meet in the point F: And because the angle ABC is equal to the angle DČE, BF is parallel to CD. Again, because the angle ACB is equal to the angle DEC, AC is rallel to FEd: Therefore FACD

pa

A

B

b 17. 1.

c 12. Ax. 1.

D

d 28. 1.

C

E

is a parallelogram; and consequently AF is equal to CD, and AC to FD: And because AC is parallel to FE, one of the e 34. 1. sides of the triangle FBE, BA is to AF, as BC to CE: But f 2. 6. AF is equal to CD; therefore 8, as BA to CD, so is BC to g 7. 5. CE; and alternately, as AB to BC, so is DC to CE: Again, h 16. 5. because CD is parallel to BF, as BC to CE, so is FD to DE': but FD is equal to AC; therefore, as BC to CE, so is AC to DE: And alternately, as BC to CA, so is CE to ED: Therefore, because it has been proved that AB is to BC, as DC to CE, and as BC to CA, so is CE to ED, ex æquali i BA is to i 22. 5. AC as CD to DE. Therefore," the sides," &c. Q. E. D.

Book VI.

a 23. 1.

b 32. 1.

c 4. 6.

d 11. 5. e 9. 5.

f 8. 1.

g 4. 1.

PROP. V. THEOR.

If the sides of two triangles, about each of their angles be proportionals, the triangles shall be equiangular, and have their equal angles opposite to the ho mologous sides.

Let the triangles ABC, DEF have their sides proportionals, so that AB is to BC, as DE to EF: and BC to CA, as EF to FD; and consequently, ex æquali, 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 equal to the angle DEF, and BCA to EFD, and also BAC to EDF.

A

D

E

F

C

Ꮐ

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 EGF, and the triangle ABC is therefore equiangular to the triangle GEF: and consequently they have their sides opposite to the B equal angles proportionals. Wherefore, as AB to BC, so is GE to EF; but by supposition, as AB to BC, so is DE to EF; therefore as DE to EF, so is a GE to EF: Therefore, DE and GE have the same ratio to EF, and consequently are equal: For the same reason, DF is equal to FG: And because in the triangles DEF, GEF, DE is equal to EG, and EF common, the two sides DE, EF are equal to the two GE, EF, and also the base DF 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 subtended by the equal sides ". 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 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.