10. Let $\triangle A B C$ be equilateral, and let $D, E, F$ be points on sides $B C, C A, A B$ respectively, with $F A=9, A E=E C=6, C D=4$. Determine the measure (in degrees) of $\angle D E F$.
Solution: 60 . Let $H, I$ be the respective midpoints of sides $B C, A B$, and also extend $C B$ and $E F$ to intersect at $J$. By equal angles, $\triangle E I F \sim \triangle J B F$. However, $B F=12-9=$ $3=9-6=I F$, so in fact $\triangle E I F \cong \triangle J B F$, and then $J B=6$. Now let $H I$ intersect $E F$ at $K$, and notice that $\triangle E I K \sim \triangle J H K \Rightarrow I K / H K=E I / J H=6 / 12=1 / 2 \Rightarrow H K=4$, since $I K+H K=H I=6$. Now consider the $60^{\circ}$ rotation about $E$ carrying triangle $C H E$ to triangle $H I E$; we see that it also takes $D$ to $K$, and thus $\angle D E F=\angle D E K=60^{\circ}$.