that is, the triangles BDE, CDE have the same ratio to the triangle ADE:

therefore the triangle BDE is equal to the triangle CDE: (v. 9.) and they are on the same base DE:

but equal triangles on the same base and on the same side of it, are between the same parallels; (1. 39.)

therefore DE is parallel to BC.

Wherefore, if a straight line, &c. Q.E.D.

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 conversely, 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 ABC be a triangle, and let the angle BAC be divided into two equal angles by the straight line AD.

Then BD shall be to DC, as BA to AC.

Through the point C draw CE parallel to DA, (I. 31.)
and let 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 CAD: (1. 29.) but CAD, by the hypothesis, is equal to the angle BAD; wherefore BAD is equal to the angle ACE. (ax. 1.) Again, because the straight line BAE meets the parallels AD, EC, the outward angle BÅD is equal to the inward and opposite angle AEC: (1. 29.)

but the angle ACE has been proved equal to the angle BAD; therefore also ACE is equal to the angle AEC, (ax. 1.) and consequently, the side AE is equal to the side AC: (1. 6.) and because AD is drawn parallel to EC, one of the sides of the triangle BCE,

therefore BD is to DC, as BA to AE: (vI. 2.)

but AE is equal to AC;

therefore, as BD to DC, so is BA to AC. (v. 7.) Next, let BD be to DC, as BA to AC, and join AD. Then the angle BAC shall be 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; (VI. 2.)

therefore BA is to AC, as BA to AE: (v. 11.)

consequently AC is equal to AE, (v. 9.)

and therefore the angle AEC is equal to the angle ACE: (1. 5.) but the angle AEC is equal to the outward and opposite angle BÁD; and the angle ACE is equal to the alternate angle CAD: (1. 29.) wherefore also the angle BAD is equal to the angle CAD; (ax. 1.) that is, the angle BAC is cut into two equal angles by the straight line AD.

Therefore, if the angle, &c. Q. E.D.

PROPOSITION A. THEOREM.

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 conversely, 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 ABC be a triangle, and let one of its sides BA be produced to E; and let the outward angle CAE be divided into two equal angles by the straight line AD which meets the base produced in D. Then BD shall be to DC, as BA to AC.

Through C draw CF parallel to AD: (1. 31.)

and because the straight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CAD: (1. 29.) but CAD is equal to the angle DAE; (hyp.)

therefore also DAE is equal to the angle ACF. (ax. 1.) Again, because the straight line FAE meets the parallels AD, FC, the outward angle DAE is equal to the inward and opposite angle CFA: (1. 29.)

but the angle ACF has been proved equal to the angle DAE; therefore also the angle ACF is equal to the angle CFA; (ax. 1.) and consequently the side AF is equal to the side AC: (1. 6.) and because AD is parallel to FC, a side of the triangle BCF, therefore BD is to DC, as BA to AF: (VI. 2.) but AF is equal to AC;

therefore, as BD is to DC, so is BA to AC. (v. 7.)
Next, let BD be to DC, as BA to AC, and join AD.
The angle CAD, shall be equal to the angle DAE.
The same construction being made,

because BD is to DC, as BA to AC;

and that BD is also to DC, as BA to AF; (v1. 2.)
therefore BA is to AC, as BA to AF: (v. 11.)
wherefore AC is equal to AF, (v. 9.)

and the angle AFC equal to the angle ACF: (1. 5.) but the angle AFC is equal to the outward angle EAD, (1. 29.) and the angle ACF to the alternate angle CAD; therefore also EAD is equal to the angle CAD. (ax. 1.) Wherefore, if the outward, &c. Q. E.D.

PROPOSITION IV. THEOREM.

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 the angle BAC equal to the angle CDE. (1. 32.)

The sides about the equal angles of the triangles ABC, DCE shall be proportionals;

and those shall be the homologous sides which are opposite to the equal angles.

F

B

Let the triangle DCE be placed, so that its side CE may be contiguous to BC, and in the same straight line with it. (1. 22.)

Then, because the angle BCA is equal to the angle CED, (hyp.) add to each the angle ABC;

therefore the two angles ABC, BCA are equal to the two angles ABC, CED: (ax. 2.)

but the angles ABC, BCA are together less than two right angles; (1. 17.)

therefore the angles ABC, CED are also less than two right angles : wherefore BA, ED if produced will meet: (I. ax. 12.)

let them be produced and meet in the point F:

then because the angle ABC is equal to the angle DCE, (hyp.) BF is parallel to CD; (1. 28.)

and because the angle ACB is equal to the angle DEC,
AC is parallel to FE: (1. 28.)
therefore FACD is a parallelogram;

and consequently AF is equal to CD, and AC to FD: (1. 34.) and because A Cis parallel to FE, one of the sides of the triangle FBE, BA is to AF, as BC to CE: (VI. 2.)

but AF is equal to CD;

therefore, as BA to CD, so is BC to CE: (v. 7.) and alternately, as AB to BC, so is DC to CE; (v. 16.)

again, because CD is parallel to BF,
as BC to CE, so is FD to DE: (vI. 2.)
but FD is equal to AC;

therefore, as BC to CE, so is AC to DE; (v. 7.) and alternately, as BC to CA, so CE to ED: (v. 16.) 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 æquali, BA is to AC, as CD to DE. (v. 22.)

Therefore the sides, &c.

PROPOSITION V.

Q.E.D.

THEOREM.

If the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular; and the equal angles shall be those which are opposite to the homologous 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.

Then the triangle ABC shall be equiangular to the triangle DEF, and the angles which are opposite to the homologous sides shall be equal, 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: (1. 23.) wherefore the remaining angle EGF, is equal to the remaining angle BAC, (1. 32.)

and the triangle GEF is therefore equiangular to the triangle ABC: consequently they have their sides opposite to the equal angles proportional: (vi. 4.)

wherefore, as AB to BC, so is GE to EF; but as AB to BC, so is DE to EF: (hyp.) therefore as DE to EF, so GE to EF; (v. 11.) that is, DE and GE have the same ratio to EF, and consequently are equal. (v. 9.)

For the same reason, DF is equal to FG:

and because, in the triangles DEF, GEF, DE is equal to EG, and EF is common,

the two sides DE, EF are equal to the two GE, EF, each to each; and the base DF is equal to the base GF;

therefore the angle DEF is equal to the angle GEF, (1. 8.) and the other angles to the other angles which are subtended by the equal sides; (1. 4.)

therefore the angle DFE is equal to the angle GFE, and EDF to EGF.

N

and because the angle DEF is equal to the angle GEF,
and GEF equal to the angle ABC; (constr.)

therefore the angle ABC is equal to the angle DEF: (ax. 1.) for the same reason, the angle ACB is equal to the anglè DFÉ, and the angle at A equal to the angle at D:

therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the sides, &c.

Q.E. D.

PROPOSITION VI. THEOREM.

If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.

Let the triangles ABC, DEF have the angle BA Cin the one equal to the angle EDF in the other, and the sides about those angles proportionals; that is, BA to AC, as ED to DF.

Then the triangles ABC, DEF shall be equiangular, and shall have the angle ABC equal to the angle DEF, and AČB to DFE.

At the points D, F, in the straight line DF, make the angle FDG equal to either of the angles BAC, EDF; (1. 23.)

and the angle DFG equal to the angle ACB:

wherefore the remaining angle at B is equal to the remaining angle at G: (1. 32.)

and consequently the triangle DGF is equiangular to the triangle ABC; therefore as BA to AC, so is GD to DF: (VI. 4.)

but, by the hypothesis, as BA to AC, so is ED to DF;
therefore as ED to DF, so is GD to DF; (v. 11.)
wherefore ED is equal to DG; (v. 9.)

and DF is common to the two triangles EDF, GDF: therefore the two sides ED, DF are equal to the two sides GD, DF, each to each;

and the angle EDF is equal to the angle GDF; (constr.)
wherefore the base EF is equal to the base FG, (I. 4.)
and the triangle EDF to the triangle GDF,

and the remaining angles to the remaining angles, each to each, which are subtended by the equal sides:

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; (constr.) therefore the angle ACB is equal to the angle DFE; (ax. 1.) and the angle BAC is equal to the angle EDF: (hyp.) wherefore also the remaining angle at B is equal to the remaining angle at E; (1. 32.)

therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q. E. D.