Perpendicular Bisector Theorem
By Nikhat Parveen
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.
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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