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, DCEshall be proportionals;

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

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; (I. 17.)

therefore the angles ABC, CED are also less than two right angles : wherefore BA, ED if produced will meet: (1. 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. Q. E.D.

PROPOSITION V. 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.

G

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 angle DFE, 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, DEFhave the angle BACin 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 ACB 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.

PROPOSITION VII. THEOREM.

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

The triangle ABC shall be equiangular to the triangle DEF, viz. the angle ABC shall be equal to the angle DEF, and the remaining angle at C equal to the remaining angle at F.

For if the angles ABC, DEF be not equal,

one of them must be greater than the other:
let ABC be the greater,

and at the point B, in the straight line AB,
make the angle ABG equal to the angle DEF; (1. 23.)
and because the angle at A is equal to the angle at D, (hyp.)
and the angle ABG to the angle DEF;

the remaining angle AGB is equal to the remaining angle DFE: (1. 32.)

therefore the triangle ABG is equiangular to the triangle DEF: wherefore as AB is to BG, so is DE to EF: (VI. 4.)

but as DE to EF, so, by hypothesis, is AB to BC; therefore as AB to BC, so is AB to BG: (v. 11.) and because AB has the same ratio to each of the lines BC, BG, BC is equal to BG; (v. 9.)

and therefore the angle BGC is equal to the angle BCG: (1. 5.) but the angle BCG is, by hypothesis, less than a right angle;

therefore also the angle BGC is less than a right angle ; and therefore the adjacent angle AGB must be greater than a right angle; (1. 13.)

but it was proved that the angle AGB is equal to the angle at F; therefore the angle at Fis greater than a right angle;

but, by the hypothesis, it is less than a right angle; which is absurd. Therefore the angles ABC, DEF are not unequal,

that is, they are equal:

and the angle at A is equal to the angle at D: (hyp.) wherefore the remaining angle at C is equal to the remaining angle at F: (1. 32.)

therefore the triangle ABC is equiangular to the triangle DEF.

Next, let each of the angles at C, Fbe not less than a right angle. Then the triangle ABC shall also in this case be equiangular to the triangle DEF.

G

E F

D

it

B

The same construction being made,

may be proved in like manner that BC is equal to BG, and therefore the angle at Cequal to the angle BGC: but the angle at Cis not less than a right angle; (hyp.) therefore the angle 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 impossible; (1. 17.)

and therefore the triangle ABC may be proved to be equiangular to the triangle DEF, as in the first case.

Lastly, let one of the angles at C, F, viz. the angle at C, be a right angle in this case likewise the triangle ABC shall be equiangular to the triangle DEF.

For, if they be not equiangular,

at the point B in the straight line AB make the angle ABG equal to the angle DEF;

then it may be proved, as in the first case, that BG is equal to BC: and therefore the angle BCG equal to the angle BGC: (1. 5.) but the angle BCG is a right angle, (hyp.)

therefore the angle BGC is also a right angle; (ax. 1.)

whence two of the angles of the triangle BGC are together not less than two right angles;

which is impossible: (I. 17.)

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

Q. E.D.

PROPOSITION VIII. THEOREM.

In a right-angled triangle, if a perpendicular be drawn from the rightangle to the base; the triangles on each side of it are similar to the whole triangle, and to one another.

Let ABC be a right angled-triangle, having the right angle BAC; and from the point A let AD be drawn perpendicular to the base BC. Then the triangles ABD, ADC shall be similar to the whole triangle ABC, and to one another.