Let the interior point $P$ of the convex quadrilateral $ABCD$ be such that $$|\angle PAD| = |\angle ADP| = |\angle CBP| = |\angle PCB| = |\angle CPD|.$$Let $O$ be the center of the circumcircle of the triangle $CPD$. Prove that $|OA| = |OB|$.
Source: 2024 Czech and Slovak Olympiad III A p2
Tags: equal angles, geometry
Let the interior point $P$ of the convex quadrilateral $ABCD$ be such that $$|\angle PAD| = |\angle ADP| = |\angle CBP| = |\angle PCB| = |\angle CPD|.$$Let $O$ be the center of the circumcircle of the triangle $CPD$. Prove that $|OA| = |OB|$.