The vertex $F$ of the parallelogram $ACEF$ lies on the side $BC$ of parallelogram $ABCD$. It is known that $AC = AD$ and $AE = 2CD$. Prove that $\angle CDE = \angle BEF$.
Problem
Source: Champions Tournament (Ukraine) - Турнір чемпіонів - 2018 Seniors p3
Tags: geometry, parallelogram, equal angles, equal segments, Champions Tournament