Example 5 Given that the area of square $A B C D$ is 35 square centimeters, $E$ and $F$ are points on sides $A B$ and $B C$ 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.
Solution: As shown in Figure 5, connect $A C, B G$, then
$$
\begin{array}{l}
\frac{B F}{B C}=\frac{S_{\triangle A B F}}{S_{\triangle A B C}} \\
=\frac{2}{7} .
\end{array}
$$

Similarly, $\frac{B E}{B A}=\frac{4}{5}$.
Let $S_{\triangle A G E}=a, S_{\triangle E B G}=b, S_{\triangle B G F}=c$, $S_{\triangle A G C}=d$, then
$$
\begin{array}{l}
a+b+c=5, \\
b+c+d=14, \\
\frac{c}{c+d}=\frac{S_{\triangle B G F}}{S_{\triangle A B C}}=\frac{B F}{B C}=\frac{2}{7}, \\
\frac{b}{a+b}=\frac{S_{\triangle G E B}}{S_{\triangle G A B}}=\frac{B E}{B A}=\frac{4}{5} .
\end{array}
$$
Solving (1), (2), (3), and (4) simultaneously, we get
$$
a=\frac{7}{27}, b=\frac{28}{27}, c=\frac{100}{27}, d=\frac{250}{27} \text {. }
$$
$\therefore S_{\text {quadrilateral } B E C F}=b+c=4 \frac{20}{27}$ (square centimeters).