**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.*******

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