Question
Sat February 18, 2012 By:

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

Expert Reply
Sat February 18, 2012

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.
Sat March 25, 2017

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