RU-ALLRUSSIAN 2015 Problem 102

Given is a parallelogram $ABCD$ , with $AB <AC <BC$ . Points $E$ and $F$ are selected on the circumcircle $\omega$ of $ABC$ so that the tangenst to $\omega$ at these points pass through point $D$ and the segments $AD$ and $CE$ intersect. It turned out that $\angle ABF = \angle DCE$ . Find the angle $\angle{ABC}$ . A. Yakubov, S. Berlov