Perpendicular Bisector Theorem

By Nikhat Parveen

Theorem

The Perpendicular bisectors of the sides of a triangle are concurrent at a point that is equidistant from the vertices of the triangle.

Given: Triangle ABC with l1, l2, and l3 the perpendicular bisectors of AB, AC and CB

Prove: a: L1,L2, and L3 are concurrent at point P.

b: P is equidistant from A, B, and C.

Proof:     Let P be the point where L1, and L2 intersect. [ Perpendiculars to intersecting lines also intersect]

Because P is on line L1, the perpendicular bisector of AB, P is equidistant from points A and B.

Hence,

PA = PB,  PA = PC [ Any point on the perpendicular bisector of a line segment is equidistant from the end points of that segment. P is on L1 and on L2]

Similarly, P is equidistant from points A and C because it lies on L2, the perpendicular bisector of AC.

By transitivity, P is equidistant from B and C.

PB = PC [By substitution]

P is on the perpendicular bisector of BC. ( P is on L3) [ A point equidistant from the endpoints of a segment  is on the perpendicular bisectors, PB = PC]

L1, L1, and L3 are concurrent at P. [ P is on L1, L1, and L3]

P is equidistant from A, B, and C. [PA = PB = PC]

***** The point of concurrency P of the perpendicular bisectors is called the circumcenter of the triangle.*****