OIM 2018 Problem 2

Let $ABC$ be a triangle such that $\angle BAC = 90^{\circ}$ and $AB = AC$ . Let $M$ be the midpoint of $BC$ . A point $D \neq A$ is chosen on the semicircle with diameter $BC$ that contains $A$ . The circumcircle of triangle $DAM$ cuts lines $DB$ and $DC$ at $E$ and $F$ respectively. Show that $BE = CF$ .