Let $ABC$ be a right triangle with $\angle ACB = 90^o$ and M the center of $AB$. Let $G$ br any point on the line $MC$ and $P$ a point on the line $AG$, such that $\angle CPA = \angle BAC$ . Further let $Q$ be a point on the straight line $BG$, such that $\angle BQC = \angle CBA$ . Show that the circles of the triangles $AQG$ and $BPG$ intersect on the segment $AB$.