Question 157
Suppose that a point $P$ lying in the interior of a convex quadrilateral $ABCD$ satisfies
$$
\angle PAD = \angle ADP = \angle CBP = \angle PCB = \angle CPD.
$$
Let $O$ be the circumcentre of the triangle $CPD$. Prove that $OA = OB$.
$$
\angle PAD = \angle ADP = \angle CBP = \angle PCB = \angle CPD.
$$
Let $O$ be the circumcentre of the triangle $CPD$. Prove that $OA = OB$.