Question 82
Let $ABC$ be a right triangle with $\angle ACB=90^{\circ}$ and let $M$ be the midpoint of $AB$. Let $G$ be an arbitrary point on the segment $MC$ and let $P$ be a point on the line $AG$ such that $\angle CPA=\angle BAC$. Furthermore, let $Q$ be a point on the line $BG$ such that $\angle BQC=\angle CBA$. Show that the circumcircles of the triangles $AQG$ and $BPG$ intersect on the segment $AB$.
