Let $ABC$ be an isosceles triangle with base $AB$. A semicircle $\Gamma$ is constructed with its center on the segment AB and which is tangent to the two legs, $AC$ and $BC$. Consider a line tangent to $\Gamma$ which cuts the segments $AC$ and $BC$ at $D$ and $E$, respectively. The line perpendicular to $AC$ at $D$ and the line perpendicular to $BC$ at $E$ intersect each other at $P$. Let $Q$ be the foot of the perpendicular from $P$ to $AB$. Show that $\frac{PQ}{CP}=\frac{1}{2}\frac{AB}{AC}$.