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

Given: An isosceles ΔABC with AB=AC, circumscribing a circle.
To prove: P bisects BC.
Proof: AR and AQ are the tangents drawn from an external point A to the circle. Therefore, AR=AQ (Tangents drawn from an external point to the circle are equal).
Similarly, BR=BP and CP=CQ.
It is given that in ΔABC, AB=AC.
⇒ AR+RB = AQ+QC
⇒ BR=QC (As AR=AQ)
⇒ BP=CP (As BR=BP and CP=CQ)
⇒ P bisects BC.
Hence, the result is proved.