Corollary.-If the triangle is to be equilateral (Def. 16), it is only necessary to know one side, AB.

From A and B, with radii equal to AB, describe (Post. 3) arcs of circles cutting in C. ABC is an equilateral triangle.

Theorem. If two triangles have two sides and the included angle of one, equal to two sides and the included angle of the other, each to each, the triangles will be equal in all their parts.

Let ABC, DEF be two triangles which have BA, AC, and the angle A, equal to ED, DF and the angle D, each to each; then will the triangles be equal in all their parts.

Let the triangle ABC be applied* to the triangle DEF, so that A shall be on D, and AB on DE. Then, because AB is equal to DE, B will coincide with E; also because the angle BAC is equal to the angle EDF, AC will coincide with DF, B and because AC is equal to DF, C

will coincide with F. Hence, B coinciding with E, and C with F, BC will coincide with EF, and the triangles will coincide, and are therefore equal (Ax. 8) in all their parts.

Scholium.―This proposition shows that if we have two triangles of any shape, in which we can find two sides of one equal to two sides of the other, and also the angles contained by these equal sides equal to each other, it is safe to conclude that the triangles are every way equal. It is important to remember that those angles will be equal which are opposite to the equal sides.

Thus, suppose we know that in the two triangles ABC and

ABD, AB is the same in both, AC is equal to AD, and the angle CAB, included by the sides CA and AB, is equal to the angle DAB, included by the sides DA and AB. We thence infer, from the preceding proof,

that the triangles are equal in the following respects:

1. Their areas are equal.

2. Their other sides, CB and DB, are equal to each other.

A

B

D

3. Their other angles are equal to each other, viz., the angles ACB and ADB, which are opposite the common side AB, are equal to each other; and the angles ABC and ABD, which are opposite the equal sides AC and AD, are equal to each other.

Proposition 5.

Theorem.-If two triangles have two angles and the included side of one, equal to two angles and the included side of the other, each to each, the triangles will be equal in all their parts.

Let ABC, DEF be two triangles which have the two angles ABC, ACB, and the side BC, respectively equal to DEF, DFE, and the side EF. The triangles will be equal in all their parts.

A

D

Apply the triangle ABC to the triangle DEF, so that B will be on E, and BC on EF; and because BC is equal to EF, C will coincide with F. Because the angle ABC is equal to the angle DEF, BA will coincide with ED; and because the angle BCA is equal to the angle EFD, CA will coincide with FD; hence the point A, which is the intersection of the two lines BA and AC, will agree with the point D, which is

B

CE

the intersection of the two lines ED and DF.

Therefore the

triangles will coincide altogether, and will be equal in all their parts; and the equal sides will be opposite to the equal angles.

Proposition 6.

Theorem.-If two sides of a triangle are equal to each other, the angles which are opposite to them will also be equal to each other.

Let ABC be a triangle having the side AB equal to the side AC; then will the angle ACB be equal to the angle ABC.

For if ACB be not equal to ABC, one of them must be the greater. Let ABC be the greater, and suppose BD to be drawn making the angle ABD equal to the angle ACB.

B

Α

D

C

In the two triangles ACB and ABD, AC is equal (Hyp.) to AB, the angle A is common to the two triangles, and, by the supposition, the angle ACB is equal to the angle ABD. Hence the two triangles ABD and ACB have two angles and the included side of one, equal to the two angles and the included side of the other. They are therefore equal (I. 5) in all their parts. Hence the area of ADB is equal to the area of ABC, a part to the whole, which is impossible (Ax. 9). The supposition that one of the angles ABC and ACB is greater than the other thus leads to an impossibility; it is therefore untrue, and ABC is equal to ACB.

NOTE. This method of proof is called an indirect demonstration— sometimes reductio ad absurdum. The theorem is proven to be true by showing that any contrary supposition leads to an absurdity or impossibility.

Corollary.-Every equilateral triangle is also equiangular.

For, all the sides being equal, the three opposite angles must all be equal.

Proposition 7.

Problem. To bisect a given rectilineal angle.

Let BAC be a rectilineal angle. It is required to bisect it.

Take any point D, in AB, and from AC cut off (I. 1) AE equal to AD; join DE, and on it describe (I. 3, Cor.) the equilateral triangle. DFE; join AF. AF will bisect BAC.

Because the two sides AD and AE of the triangle ADE are equal to each other, the opposite angles AED B and ADE are equal (I. 6) to each other; also because FD and FE of

A

E

C

the triangle FDE are equal (Def. 16), the opposite angles FED and FDE are equal (I. 6) to each other; therefore the whole angles AEF and ADF are equal (Ax. 2) to each other. Now the triangles ADF and AEF have (Constr.) AD equal to AE, and DF to EF, also the included angle ADF equal to the included angle AEF; hence (I. 4) the triangles are equal in all their parts, and the equal angles are opposite to the equal sides, viz., the angle DAF, opposite DF, is equal to the angle EAF opposite to EF. Hence BAC is bisected by. the line AF.

The proof is summarized in the following

NOTE. The inclined lines show that the preceding statements unite to prove the following one, with the aid of the reference at the apex. All the different parts of the proof are shown to be necessary to the conclusion.

Scholium.-By the same construction, each of the halves DAF, EAF may be again bisected, and thus by successive bisections a given angle may be divided into 4, 8, 16, etc. equal parts.

Proposition 8.

Theorem.—If two triangles have two sides of one equal to two sides of the other, each to each, but the angles contained by these sides unequal, the third side of that which has the greater included angle will be greater than the third side of the other.

Let ABC, DEF be two triangles, having the sides BA, AC equal to ED, DF, each to each, and the angle BAC greater than EDF; then will BC be greater than EF.

Construct (I. 3) the triangle DEG, having its sides equal to the sides of ABC. Bisect (I. 7) the angle FDG by DH, meeting EG in H; join FH.

Because DG and DF are both equal to AC, they are equal to each other. Then in the tri

Α

D

HG

angles DHF, DHG, DF is equal to DG, DH common, and the angle FDH equal to GDH, therefore (I. 4) FH is equal to GH.

Now (I. 2), EH+HF> EF, or EH+HG>EF;

EG or BC is greater than EF.