Let $ABCD$ be a convex quadrilateral inscibed in circle $(O)$ such that $DB = DA + DC$. The point $P$ lies on the ray $AC$ such that $AP = BC$. The point $E$ is on $(O)$ such that $BE \perp AD$. Prove that $DP$ is parallel to the angle bisector of $\angle BEC$.
Problem
Source: 2018 Saudi Arabia IMO TST I p3
Tags: geometry, angle bisector, cyclic quadrilateral, parallel, equal segments, perpendicular