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

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