The point P lies on the side $CD$ of the parallelogram $ABCD$ with $\angle DBA = \angle CBP$. Point $O$ is the center of the circle passing through the points $D$ and $P$ and tangent to the straight line $AD$ at point $D$. Prove that $AO = OC$.
Problem
Source: Polish Math Olympiad 2021 2nd round p2 day 1
Tags: equal segments, parallelogram, equal angles, geometry