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 $\Delta ABC$ 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 $\Delta ABC$, $AB=AC.$ 
$\Rightarrow AR+RB=AQ+QC$
\Rightarrow BR = QC (As\space\  AR = AQ)
\Rightarrow BP = CP (As \space\ BR = BP \space\ and\space\  CP = CQ)
$\Rightarrow$ $P$ bisects $BC.$
Hence, the result is proved.