ABC is an isosceles triangle in which AB=AC, circumscribed about a circle.Prove that the base is bisected by the point of contact.

Asked by snehamapink | 18th Feb, 2012, 11:42: AM

Expert Answer:

Let the circle touch the sides AB, BC and CA of triangle at points R, Q, and P.

We know that the tangents drawn from a point outside a circle are equal.
Therefore, AR = AQ (1)
BP = BR (2)
CP = CQ (3)
AB = AC (given) (4)
Subtracting (1) from (4), we get
BR = CQ
Using (2) and (3), we get,
BP = CP
Hence, proved.

Answered by  | 18th Feb, 2012, 04:48: PM

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