Example 8 Let the area of $\triangle ABC$ be $1, B_{1}, B_{2}$ and $C_{1}, C_{2}$ be the trisection points of sides $AB, AC$, respectively. Connect $B_{1}C, B_{2}C, BC_{1}, BC_{2}$. Find the area of the quadrilateral they form.
As shown in Figure 1.3.7, let $B C_{2}$ intersect $C B_{2}$ at $R$, and intersect $B_{1} C$ at $S$; $B C_{1}$ intersects $B_{1} C$ at $P$, and intersects $B_{2} C$ at $Q$. By Menelaus' theorem, we have
$$
\frac{B S}{S C_{2}} \cdot \frac{C_{2} C}{C A} \cdot \frac{A B_{1}}{B_{1} B}=1
$$

Also, because
$$
\frac{C_{2} C}{C A}=\frac{1}{3}, \frac{A B_{1}}{B_{1} B}=\frac{1}{2}
$$

We have
$$
\frac{B S}{S C_{2}}=6, \frac{B S}{B C_{2}}=\frac{6}{7}
$$

By Menelaus' theorem, we have

Also, because
$$
\begin{array}{l}
\frac{B R}{R C_{2}} \cdot \frac{C_{2} C}{C A} \cdot \frac{A B_{2}}{B_{2} B}=1 \Rightarrow \frac{B R}{B C_{2}}=\frac{3}{5} \\
\frac{B Q}{Q C} \cdot \frac{C_{1} C}{C A} \cdot \frac{A B_{2}}{B_{2} B}=1 \Rightarrow \frac{B Q}{B C_{1}}=\frac{3}{7} \\
\frac{B P}{P C_{1}} \cdot \frac{C_{1} C}{C A} \cdot \frac{A B_{1}}{B_{1} B}=1 \Rightarrow \frac{B P}{B C_{1}}=\frac{3}{4}
\end{array}
$$

Therefore,
$$
\begin{array}{c}
S_{\triangle B C_{1} C_{2}}=\frac{C_{1} C_{2}}{A C} \cdot S_{\triangle A B C}=\frac{1}{3} \\
S_{P Q R S}=S_{\triangle B P S}-S_{\triangle B Q R}=\left(\frac{B P}{B C_{1}} \cdot \frac{B S}{B C_{2}}-\frac{B Q}{B C_{1}} \cdot \frac{B R}{B C_{2}}\right) \cdot S_{\triangle B C_{1} C_{2}}=\frac{9}{70}
\end{array}
$$