66th Belarusian Mathematical Olympiad

Let $△ABC$ be an acute triangle with the orthocenter $H$. Let $D$ be the point such that $HABD$ is a parallelogram ($AB\parallel HD$, $AH\parallel BD$). Let $E$ be the point on the line $DH$ such that $AC$ bisects $HE$. The line $AC$ meets the circumcircle of the triangle $DCE$ at $C$ and $F$. Prove that $EF=AH$.