OMCC 2011 Problem 6

Let $ABC$ be an acute triangle and $D$ , $E$ , $F$ be the feet of the altitudes through $A$ , $B$ , $C$ respectively. Call $Y$ and $Z$ the feet of the perpendicular lines from $B$ and $C$ to $FD$ and $DE$ , respectively. Let $F_1$ be the symmetric of $F$ with respect to $E$ and $E_1$ be the symmetric of $E$ with respect to $F$ . If $3EF=FD+DE$ , prove that $\angle BZF_1=\angle CYE_1$ .