Let $ABC$ be a triangle with $AB = AC$. With center in a point of the side $BC$, the circle $S$ is constructed that is tangent to the sides $AB$ and $AC$. Let $P$ and $Q$ be any points on the sides $AB$ and $AC$ respectively, such that $PQ$ is tangent to $S$. Show that $PB \cdot CQ = \left(\frac{BC}{2}\right)^2$