Point $D$ is the midpoint of $BC$, where $ABC$ is an isosceles triangle ($AB=AC$). On circle $(ABD)$ a point $P \neq A$ is chosen. $O$ is the circumcenter of $ACP$, $Q$ is the foot of the perpendicular from $C$ onto $AO$. Prove that the circumcenter of triangle $ABQ$ lies on the line $AP$