Let $ABCDE$ be a convex pentagon such that $AE\parallel BC$ and $\angle ADE = \angle BDC$. The diagonals $AC$ and $BE$ intersect at point $F$. Prove that $\angle CBD= \angle ADF$.
Problem
Source: 2009 All-Ukrainian Correspondence MO of magazine ''In the World of Mathematics'', grades 5-12 p7
Tags: geometry, pentagon, equal angles, Ukraine Correspondence