Question 178
A convex quadrilateral $ABCD$ with $AC \neq BD$ is inscribed in a circle with centre $O$. Let $E$ be the intersection of diagonals $AC$ and $BD$. If $P$ is a point inside $ABCD$ such that $\angle PAB + \angle PCB = \angle PBC + \angle PDC = 90^\circ$, prove that $O, P$ and $E$ are collinear.