Proof of Concurrency
of
Perpendicular Bisectors of a Triangle

by
Holly Anthony

Given: D ABC; the ^ of AB, BC, and AC

Prove: The ^ bisectors intersect in a point: that point equidistant from A, B, C.

Plan for Proof: Show that C, the point of intersection of the perpendicular bisectors of BC and AC, also lies on the perpendicular bisector of AB. Then show that C is equidistant from the vertices of the triangle, A, B, C.

 Statements: 1.  The perpendicular bisectors of BC and AC intersect at some point P.           2. Draw PA, PB, and PC.   3.  PA = PC, PC = PB       4.  PA = PB 5.  P is on the perpendicular bisector of AB   6.  PA = PB = PC, so P is equidistant from the vertices of the triangle Reasons: 1.  ABC is a triangle, so its sides BC and AC cannot be parallel; therefore, segments perpendicular to those sides cannot be parallel.  So, the perpendicular bisectors must intersect in some point.  Call it P. 2.  Throught any two points there is exactly one line. 3.  If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. (Theorem) 4.  Substitution property of Equality 5.  If a point is equidistant from the endpoints fo a segment, then it is on the perpendicular bisector of the segment. 6.  Steps 3 and 4 and the definition of equidistant.