9. Given triangle $A B C$, let $D$ be a point on side $A B$ and $E$ be a point on side $A C$. Let $F$ be the intersection of $B E$ and $C D$. If $\triangle D B F$ has an area of $4, \triangle B F C$ has an area of 6 , and $\triangle F C E$ has an area of 5 , find the area of quadrilateral $A D F E$.
Answer: $\frac{105}{4}$ or 26.25
Solution: Let the area of quadrilateral $A D F E$ be $x$. By Menelaus' Theorem, $\frac{A D}{D B} \cdot \frac{B F}{F E} \cdot \frac{E C}{C A}=1$. Since $\frac{A D}{D B}=\frac{x+5}{10}, \frac{B F}{F E}=\frac{6}{5}$, and $\frac{E C}{C A}=\frac{11}{x+15}$, we have $\frac{66(x+5)}{50(x+15)}=1$, or $x=\frac{105}{4}$ or 26.25 .