2013 Canadian MO Problems/Problem 3

Problem

Let $G$ be the centroid of a right-angled triangle $ABC$ with $\angle BCA = 90^\circ$ . Let $P$ be the point on ray $AG$ such that $\angle CPA = \angle CAB$ , and let $Q$ be the point on ray $BG$ such that $\angle CQB = \angle ABC$ . Prove that the circumcircles of triangles $AQG$ and $BPG$ meet at a point on side $AB$ .

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2013 Canadian MO Problems/Problem 3

Problem

Solution