Problem

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Tags: geometry, circumcircle, symmetry, power of a point, radical axis, geometry unsolved



A convex quadrilateral $ABCD$ with $AC \neq BD$ is inscribed in a circle with center $O$. 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.