Example 6 Divide the sides $AB$ and $CD$ of the convex quadrilateral $ABCD$ into $m$ equal parts, and connect the corresponding division points. Then divide the sides $AD$ and $BC$ into $n$ equal parts, and connect the corresponding division points, where $m$ and $n$ are both odd numbers greater than 1. Find the ratio of the area of the middle quadrilateral to the area of quadrilateral $ABCD$.
Solution: As shown in Figure 7, from Example 1, it is easy to know that each intersection point equally divides the horizontal line $m$ and the vertical line $n$.

Let $\overrightarrow{A B}=\vec{a}$, $\overrightarrow{A C}=\vec{b}$, $\overrightarrow{A D}=\vec{c}$. Then the horizontal line where $\overrightarrow{A_{0} B_{0}}$ lies is
$$
\begin{array}{l}
\overrightarrow{D_{\frac{n-1}{2 n}} B_{\frac{n-1}{2 n}}}=\overrightarrow{D_{\frac{n-1}{2 n}}}+\overrightarrow{A B}+\overrightarrow{B B_{\frac{n-1}{2 n}}^{2 n}} \\
=\frac{n+1}{2 n} \vec{a}+\frac{n-1}{2 n} \vec{b}-\frac{n-1}{2 n} \vec{c} \text {. } \\
\end{array}
$$

Therefore, $\overrightarrow{A_{0} B_{0}}=\frac{1}{2 m n}[(n+1) \vec{a}+(n-1) \vec{b}$
$$
-(n-1) \vec{c}] \text {. }
$$
$$
\begin{aligned}
\text { Similarly, } \overrightarrow{D_{0}} \vec{C}_{0}= & \frac{1}{2 m}[(n-1) \vec{a}+(n+1) \\
& \cdot \vec{b}-(n+1) \vec{c}], \\
\overrightarrow{A_{0} D_{0}}= & \frac{1}{2 m n}[-(m-1) \vec{a}+(m-1) \vec{b} \\
& +(m+1) \vec{c}], \\
\overrightarrow{B_{0} C_{0}}= & \frac{1}{2 m n}[-(m+1) \vec{a}+(m+1) \vec{b} \\
& +(m-1) \vec{c}] .
\end{aligned}
$$

Therefore, $2 S_{\text {quadrilateral } A_{0} B_{0} C_{0} D_{0}}$
$$
\begin{aligned}
= & \left|\overrightarrow{A_{0} B_{0}} \times \overrightarrow{A_{0} D_{0}}+\overrightarrow{C_{0} D_{0}} \times \overrightarrow{C_{0} B_{0}}\right| \\
= & \left|\overrightarrow{A_{0} B_{0}} \times \overrightarrow{A_{0} D_{0}}+\overrightarrow{D_{0} C_{0}} \times \overrightarrow{B_{0} C_{0}}\right| \\
= & \left.\frac{1}{4 m^{2} n^{2}} \right\rvert\,(n+1)(m-1) \vec{a} \times \vec{b} \\
& +(n+1)(m+1) \vec{a} \times \vec{c}+\cdots \mid \\
= & \frac{1}{m n}|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}| \\
= & \frac{1}{m n} \times 2 S_{\text {quadrilateral } A B C D} . \\
\therefore & \frac{S_{\text {quadrilateral } A_{0} B_{0} C_{0} D_{0}}}{S_{\text {quadrilateral } A B C D}}=\frac{1}{m n} .
\end{aligned}
$$