Question 141
Let $\triangle ABC$ 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 $EF = AH$.
Prove that $EF = AH$.