Problem

Source: Stars of Mathematics 2022, J4/S2

Tags: geometry



Let $ABCD$ be a convex quadrilateral and $P$ be a point in its interior, such that $\angle APB+\angle CPD=\angle BPC+\angle DPA$, $\angle PAD+\angle PCD=\angle PAB+\angle PCB$ and $\angle PDC+ \angle PBC= \angle PDA+\angle PBA$. Prove that the quadrilateral is circumscribed.