Question

In the given figure, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.

Answer 1

Her is the answer to your question
points name are slightly changed

The tangents drawn from an externel point to a circle are equal in length.

∴ AP = AQ ............ (1)

BP = BR ................ (2)

CQ = CR .............. (3)

Given that ABC is an isosceles triangle with AB = AC.

Subtract AP on both sides, we obtain

AB – AP = AC – AP

⇒ AB – AP = AC – AQ (from (1))

∴ BP = CQ.

⇒ BR = CQ (from (2))

⇒ BR = CR (from (3))

∴BR = CR, it shows that BC is bisected at the point of contact R