As shown in figure , $\odot O$ is the inscribed circle of trapezoid $ABCD$, and the tangent points are $E, F, G, H$, $AB \parallel CD$. The line passing through$ B$, parallel to $AD$ intersects extension of $DC$ at point $P$. The extension of $AO$ intersects $CP$ at point $Q$. If $AE=BE$ , prove that $\angle CBQ = \angle PBQ$.
