Example 9 As shown in Figure
1, in $\square A B C D$, $\angle B A D$ is an obtuse angle, and $A C^{2} \cdot B D^{2}$ $=A B^{4}+A D^{4}$. Find the degree measure of $\angle B A D$.
(1986, Chengdu Junior High School Mathematics Competition)
Solution: By the properties of a parallelogram, we have
$$
\begin{array}{l}
A C^{2}+B D^{2}=2\left(A B^{2}+A D^{2}\right) \\
\text { Let } A C^{2}=A B^{2}+A D^{2}+t, \\
B D^{2}=A B^{2}+A D^{2}-t,
\end{array}
$$

Then, from the given equation, we get
$$
\begin{array}{l}
\left(A B^{2}+A D^{2}+t\right)\left(A B^{2}+A D^{2}-t\right) \\
=A B^{4}+A D^{4} .
\end{array}
$$

Simplifying, we get $t^{2}=2 A B^{2} \cdot A D^{2}$.
Since $\angle B A D$ is an obtuse angle, then $B D>A C$,
$$
\therefore B D^{2}>A C^{2} \text {, }
$$

i.e., $A B^{2}+A D^{2}-t>A B^{2}+A D^{2}+t$.
$$
\begin{array}{l}
\therefore t<0, \\
\therefore t=-\sqrt{2} A B \cdot A D .
\end{array}
$$

In $\triangle A B D$, by the cosine rule, we get
$$
\begin{array}{l}
\cos \angle B A D=\frac{A B^{2}+A D^{2}-B D^{2}}{2 A B \cdot A D} \\
=\frac{t}{2 A B \cdot A D}=\frac{-\sqrt{2} A B \cdot A D}{2 A B \cdot A D}=-\frac{\sqrt{2}}{2} .
\end{array}
$$

Thus, $\angle B A D=135^{\circ}$.