8. Hexagon $A B C D E F$ has a circumscribed circle and an inscribed circle. If $A B=9, B C=6, C D=2$, and $E F=4$. Find $\{D E, F A\}$.
Answer: $\left\{\frac{9+\sqrt{33}}{2}, \frac{9-\sqrt{33}}{2}\right\}$ By Brianchon's Theorem, $A D, B E, C F$ concur at some point $P$. Also, it follows from the fact that tangents from a point to a circle have equal lengths that $A B+C D+E F=$ $B C+D E+F A$. Let $D E=x$, so that $F A=9-x$.
Note that $A P B \sim E P D, B P C \sim F P E$, and $C P D \sim A P F$. The second similarty gives $B P / F P=3 / 2$, so that $B P=3 y, F P=2 y$ for some $y$. From here, the first similarity gives $D P=x y / 3$. Now, the third similarity gives $4 y=\frac{(9-x) x y}{3}$, so that $x^{2}-9 x+12=0$. It follows that $x=\frac{9 \pm \sqrt{33}}{2}$, giving our answer.