CHAPTER VII Congruent Lines in a Triangle 52. Theorem. In any triangle, the perpendicular bisectors of the three sides are concurrent. Figure 49 Suppose the triangle is ABC (Fig. 49). The perpendicular bisectors of sides AB, AC are not parallel (otherwise lines AB, AC would coincide), so they intersect at some point O. We must show that point O is also on the perpendicular bisector of BC. Because point O, is on the perpendicular bisector of AB, it is equidistant from A and B likewise, because it is on the perpendicular bisector of AC, it is equidistant from A and C. It is therefore equidistant from B and C and thus it is on the perpendicular bisector of BC. 53. Theorem. In any triangle, the three altitudes are concurrent. Suppose the triangle is ABC (Fig. 50). We draw a parallel to BC through A, a parallel to AC through B, and a parallel to AB through C. This forms a new triangle A B C . We will show that the altitudes of ABC are the perpendic- ular bisectors of the sides of the new triangle, from which it follows that they are concurrent. Parallelogram ABCB gives us BC = AB , and parallelogram ACBC gives BC = AC , so that A is indeed the midpoint of B C . Altitude AD of ABC thus passes through the midpoint of B C , and is perpendicular to it because it is perpendicular to the parallel line BC. Since this reasoning can be repeated for the other altitudes, the proof is com- plete. 54. Theorem. In any triangle: 1◦. the three angle bisectors are concurrent 2◦. the bisector of an angle, and the bisectors of the two non-adjacent exterior angles, are concurrent. 53
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