Corollary.-If the triangle is to be equilateral (Def. 15), 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 the three sides of the one equal to three sides of the other, each to each, the angles of one will be equal to the angles of the other, each to each; the equal angles being opposite to the equal sides; and the areas of the triangles will be equal.

Restatement.

Let the triangles ABC, DEF, have their sides equal, each to each, AB to DE, BC to EF, and AC to DF; then will the angle ABC be equal to the angle DEF, ACB to DFE, and BAC to

EDF; and the triangles ABC, DEF will be equal in area.

A

D

BC on EF; and because BC is equal B to EF, C will be on F. Also, if A

CE

will not fall on D, it will take a position either within the triangle DEF, on a side of DEF other than D, or outside of DEF. We will consider the three cases separately, and show they each lead to an impossibility.

CASE 1.

1. Suppose the triangle BAC to take the position EGF, A falling at G inside the triangle DEF.

Produce EG to meet DF in H.

Still more, then, is ED+DF>EG+ GF.

But because (Hyp.) the sides of BAC, that is, EGF, are equal to the sides of EDF,

ED+DF-EG+ GF.

But it has been proved greater, which is absurd. Hence the sup

Да

CE

position that A would fall inside of EDF is impossible.

CASE 2.

But

CASE 3.

2. Suppose G to fall at H in last case. We have shown that

ED+DF> EH + HF.

ED+DF = EH+HF, which is impossible.

3. Suppose G to fall outside of DEF. Then (I. 2),

AA

CE

Therefore the triangle BAC would take no other position than EDF.

Hence the two triangles would coincide entirely, and be equal in all their parts (Ax. 8).

Scholium.-Triangles whose sides are equal, each to each, are said to be mutually equilateral, and triangles whose angles are equal, each to each, are mutually equiangular.

NOTE. This is an example of an indirect demonstration. We show that all suppositions, contrary to the theorem, lead to contradictions. Hence, the theorem must be true. This method is sometimes called reductio ad absurdum.

Because we have proved that mutually equilateral triangles are mutually equiangular, it must not be supposed that the converse is trueviz., that mutually equiangular triangles are mutually equilateral. Two triangles may have their angles equal, each to each, and their sides unequal. In this case they will, however, be similar.

When a theorem is true, the converse is generally true, though not always; it is never safe to assume it without proof.

Proposition 5.

Problem.-At a given point in a straight line to make an angle equal to a given angle.

Restatement.

Let A be a given point in the line AB, and DEF a given angle. It is required to make an

angle at A equal to DEF.

the angles DEF, GAH, opposite the equal sides, are equal

(I. 4). Therefore at A the angle BAH has been constructed equal to DEF.

Problem.-To construct a triangle

1. Having two sides and the included angle given ;

2. Having two angles and the included side given.

1. Let A, B, and CDE be two sides and the included angle of a triangle. It is required to construct it.

2. Let A, B, C be a side and the two adjacent angles of a triangle. It is required to construct it.

Proposition 7.

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, and because AC is equal to DF, C

AA

CE

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.

F

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