Harvard-MIT Math Tournament

Solution:

$H, I$ be the respective midpoints of sides $BC, AB$, and also extend $CB$ and $EF$ to intersect at $J$. By equal angles, $\triangle EIF \sim \triangle JBF$. However, $BF=12-9=3=9-6=IF$, so in fact $\triangle EIF \cong \triangle JBF$, and then $JB=6$. Now let $HI$ intersect $EF$ at $K$, and notice that $\triangle EIK \sim \triangle JHK \Rightarrow IK/HK=EI/JH=6/12=1/2 \Rightarrow HK=4$, since $IK+HK=HI=6$. Now consider the $60^{\circ}$ rotation about $E$ carrying triangle $CHE$ to triangle $HIE$; we see that it also takes $D$ to $K$, and thus $\angle DEF=\angle DEK=60^{\circ}$.