24 II. ON TRIANGLES
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.
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-
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