BMO Round 1, 2017

Let $ABC$ be a triangle with $\angle A < \angle B < 90^\circ$ and let $\Gamma$ be the circle through $A$, $B$ and $C$. The tangents to $\Gamma$ at $A$ and $C$ meet at $P$. The line segments $AB$ and $PC$ produced meet at $Q$. It is given that $$[ACP] = [ABC] = [BQC].$$ Prove that $\angle BCA = 90^\circ$. Here $[XYZ]$ denotes the area of triangle $XYZ$.