In the two triangles ABC, DEF, let the sides AC, AB, and the included angle at A, be respectively equal to the sides DF, DE, and the included angle at D; then we have to prove that the two triangles are equal. Fig. 5.

A

B D

E

Suppose the triangle DEF to be placed upon the triangle ABC, so that DE lies upon AB, the point D at A, the point E at B; now, since the angle FDE is equal to the angle CAB, the line DF will take the direction AC, and, being equal to AC in length, its extremity F will fall on C, and consequently the line FE will coincide with CB, the angle at F with the angle at C, and the angle at E with the angle at B.

PROP. V. THEOREM.

If two angles and an included side in one triangle be equal respectively to two angles and an included side in another, the triangles will be equal.

Let, (as in the figure above,) the angle A be equal to the angle D, the angle B equal to the angle E, and the side AB equal to DE; then we have to prove that the triangles ABC, DEF, are equal.

Suppose the triangle DEF to be placed upon ABC, so that DE shall coincide with AB; now, since the angle FDE is equal to the angle CAB, DF will take the direction AC; and, since the angle DEF is equal to the angle ABC, EF will take the direction BC; hence

DF, EF, will intersect at C, and DF be equal to AC, EF equal to BC, and the angle at F equal to the angle at C.

PROP. VI. THEOREM.

In an isosceles triangle, the angles opposite the equal sides are equal.

Let AB, BC, be the equal

sides; then we have to prove

that the angle A is equal to

C.

Fig. 6.

B

C D A

Draw from B the line BD, so as to divide the angle B into two equal parts; then in the two triangles ABD, CBD, we have the two sides AB, BD, and the included angle ABD, in the one, respectively equal to the two sides BC, BD, and the included angle CBD, in the other; hence the two triangles are equal (Prop. 4), and the angle A is equal to the angle C.

Cor. 1. From the equality of the two triangles ABD, CBD, we see further that AD is equal to DC, and the angles BDA, BDC, are equal, and hence (Def. 6) right angles.

Cor. 2. Every equilateral triangle is therefore equiangular.

Scholium. In a triangle any side may be assumed as the base, and then the vertex is the vertex of the opposite angle.

In an isosceles triangle, however, we generally assume as the base the side which is not equal to either of the other two.

PROP. VII. THEOREM.

(Converse of Prop. VI.)

If two angles of a triangle are equal, the sides opposite them are also equal, and the triangle is isosceles.

In the triangle ABC, let the angles ABC, BAC, be equal, then we have to prove that the sides AC, BC, will be also equal.

Fig. 7.

C

B

If they are not equal, one of them must be the longer; let us suppose it to be AC, and make AD-CB, and draw the line BD; then we have the side AD of the triangle ABD equal to the side BC of the triangle ACB, the side AB common to both triangles, and the angle DAB of one equal to the angle CBA of the other (by hypothesis), hence the two triangles ABD, ABC, will be equal (Prop. 4); but this is manifestly impossible, since the triangle ABD is only a part of ABC; therefore the line AC cannot be longer than CB.

In a similar manner it may be proved that BC cannot be longer than AC; therefore they must be equal. Cor. Every equiangular triangle is equilateral.

PROP. VIII. THEOREM.

Two angles are equal when they have their sides parallel and directed the same way.

Let AB be parallel to DE, and CB parallel to FE; then we have to prove that the angle ABC is equal to the angle DEF.

The lines BA, ED, have the same direction, as have also BC, EF (Def. 9); therefore, if we suppose DEF applied to ABC, so that the point E shall fall on B, and EF lie on BC, the line ED will pursue the same direction as BA; and hence the angle DEF is equal to ABC.

PROP. IX. THEOREM.

If two parallel lines are cut by a third straight line, the alternate angles will be equal.

Let the parallels

AB, CD, be cut by A

the line EF; then we

Fig. 9.

E

The angles EIB, EOD, are equal by the last proposition; but EIB is equal to AIF (Prop. 3); hence EOD must be equal to AIF (Ax. 1).

Cor. 1. We see by the demonstration that EIB, EOD, are equal; these are called opposite exterior and interior angles.

Cor. 2. A line which is perpendicular to one of two

parallels will be perpendicu

lar to the other. Let EF

be drawn perpendicular to ▲AB; now, because it cuts

the two parallels, it makes the alternate angles AIF,

Fig. 10.

E

I

B

-D

EOD, equal, and since AIO is a right angle, EOD must be one also.

PROP. X. THEOREM.

(Converse of Prop. IX.)

If two straight lines are cut by a third, making the alternate angles equal, the two first lines are parallel.

Let the angles AIF, EOD, be equal; then we have to prove that AB, CD, are parallel. If they are not

HIO must be equal to AIF, which is manifestly impossible, since HIF is only a part of AIF; therefore HK cannot be parallel to CD ; and, as the same can be proved with regard to any other line than AB, it follows that AB is parallel to CD.

Cor. 1. When the exterior and interior angles EIB, EOD, are equal, AB, CD, must be parallel; because when those angles are equal the alternate angles are equal also. Fig. 12.

being right angles, are equal to each other.

Cor. 3. Hence, if to each of two parallels, perpendic