Let $ABCDE$ be a convex pentagon such that $\angle BAC = \angle CAD = \angle DAE$ and $\angle ABC = \angle ACD = \angle ADE$. Diagonals $BD$ and $CE$ meet at $P$. Prove that $AP$ bisects side $CD$.
Problem
Source: 2011 Saudi Arabia BMO TST 3.3 - Balkan MO
Tags: geometry, bisects segment, equal angles, pentagon