Congruent Lines in a Triangle
52. Theorem. In any triangle, the perpendicular bisectors of the three sides
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
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
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-
54. Theorem. In any triangle:
the three angle bisectors are concurrent;
the bisector of an angle, and the bisectors of the two non-adjacent exterior
angles, are concurrent.