24 II. ON TRIANGLES

1st

case: ASA. Two triangles are congruent if they have an equal side con-

tained between corresponding equal angles.

Suppose the triangles are ABC, A B C (ﬁg. 22) in which BC = B C , B = B ,

and C = C . Let us move angle B onto angle B so that side B A lies along the

line of BA and B C along the line of BC. Since BC = B C , the point C falls

on C. Now, since C = C, side C A assumes the direction of CA, and therefore

point A falls on the intersection of BA and CA; that is, on A. This establishes the

congruence of the two ﬁgures.

2nd

case: SAS. Two triangles are congruent if they have an equal angle con-

tained between corresponding equal sides.

Suppose the two triangles are ABC, A B C (Fig. 22) in which A = A , AB =

AC, and AC = A C .

Let us move angle A over angle A so that A B assumes the direction of AB and

A C assumes the direction of AC. Since A B = AB, point B will fall on B, and

likewise C on C. Therefore side B C coincides with BC, and the superposition of

the two ﬁgures is complete.

3rd case: SSS. Two triangles are congruent if they have three equal corre-

sponding sides.

Figure 23

Consider triangles ABC and A B C such that AB = A B , AC = A C , and

BC = B C , and let us move the second triangle so that side B C coincides with

BC, and the two triangles are on the same side of line BC. Denote the new position

of A by A1. We claim that point A1 coincides with A. This is obvious if BA1 has

the same direction as BA or if CA1 has the same direction as CA. If this were

not the case, we would have formed isosceles triangles BAA1 and CAA1 (Fig. 23)

and the perpendicular bisector of AA1 would have to pass through B and C (23,

Corollary); in other words, this bisector would have to be line BC. This however

is not possible, because points A and A1 are on the same side of BC, so that BC

cannot pass through the midpoint of AA1. Thus the only possibility is that points

A and A1 must be the same point; in other words, the given triangles coincide.

Remarks. I. In order to establish that A = A1, we have investigated what

would happen if these two points were distinct. Arriving, in this case, at a conclu-

sion which is clearly false, we concluded that this possibility does not arise. This

is a very useful method of reasoning, called proof by contradiction.

II. In a triangle there are six main elements to consider; namely, the three

angles and the three sides. We have seen that if we establish the equality of three

of these elements (properly chosen) in two triangles, we can conclude that the two