MathNet 2023-part-05 — 191 problems (EN–VI) Question 107

Let $ABCD$ be a convex quadrilateral and let $P$ be a point inside such that $\angle APB + \angle CPD = \angle APD + \angle BPC$, $\angle PAD + \angle PCD = \angle PAB + \angle PCB$ and $\angle PDC + \angle PBC = \angle PDA + \angle PBA$. Prove that the quadrilateral $ABCD$ is circumscriptible.

Flavian Georgescu