Question 96, In quadrilateral $\mathrm{ABCD}$ on a plane, it is known that $\mathrm{AB}=1, \mathrm{BC}=4, \mathrm{CD}=2, \mathrm{DA}=3$. Try to find the value of $\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BD}}$.
Question 96, Solution: Noting that $A B+C B=A D+C D=5$, it follows that points $B$ and $D$ lie on an ellipse with foci at $A$ and $C$, and a major axis of 5.

Let $A(-c, 0), G(c, 0)$, then $\mathrm{a}=\frac{5}{2}, \mathrm{e}=\frac{2 c}{5}$. Let $\mathrm{B}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right), \mathrm{D}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$, then according to the focal radius formula, we have
$$
\left\{\begin{array}{l}
1=|B A|=\operatorname{ex}_{1}+a \\
3=|D A|=\operatorname{ex}_{2}+a \Rightarrow x_{2}-x_{1}=\frac{2}{e} \\
\text { Since } \overline{A C}=(2 c, 0), \overrightarrow{B D}=\left(x_{2}-x_{1}, y_{2}-y_{1}\right), \text { thus } \\
\overrightarrow{A C} \cdot B D=2 c\left(x_{2}-x_{1}\right)=2 c \cdot \frac{2}{2 c}=10
\end{array}\right.
$$

In summary, the value of $\overrightarrow{A C} \cdot \overrightarrow{B D}$ is 10.