9. Given that the area of square $A B C D$ is 35 square centimeters, $E, F$ are points on sides $A B$, $B C$ respectively, $A F$ and $C E$ intersect at $G$, and the area of $\triangle A B F$ is 5 square centimeters, the area of $\triangle B C E$ is 14 square centimeters. Then, the area of quadrilateral $B E G F$ is $\qquad$ square centimeters.
9. $\frac{B F}{B C}=\frac{S_{\triangle A B F}}{S_{\triangle A B C}}=\frac{2}{7}$, similarly $\frac{B E}{B A}=\frac{4}{5}$. Connect $B G$,

Let $S_{\triangle A G E}=a, S_{\triangle E C B}=$ $b, S_{\triangle B G F}=c, S_{\triangle F C C}=d$.

Given $a+b+c=5$, $b+c+d=14$,

Solving, we get $b=\frac{28}{27}, c=\frac{100}{27}$. $\therefore S_{B H C i F}=b+c=\frac{128}{27}=4 \frac{20}{27}$ (square centimeters).