II. (25 points) In $\square A B C D$, connect the diagonals $A C$ and $B D$. If $\angle B A D$ is an acute angle, and
$$
(A C \cdot B D)^{2}=A B^{4}+B C^{4} \text {, }
$$

find the degree measure of $\angle B A D$.
$$
\begin{array}{l}
\angle B C E=\angle A C D, \\
\angle C B E=\angle A D C .
\end{array}
$$

Then we have $\angle A B E$
$$
\begin{array}{l}
=360^{\circ}-\angle A D C-\angle A B C \\
=2 \angle B A D \\
\angle A C E=\angle A C B+\angle B C E \\
=\angle A C B+\angle A C D=\angle D C B .
\end{array}
$$

Also, $\triangle E C B \backsim \triangle A C D$, thus, $\frac{B E}{D A}=\frac{B C}{C D}$, which means
$$
\begin{array}{l}
B E=\frac{B C \cdot D A}{C D}=\frac{B C^{2}}{C D}, \\
\frac{E C}{A C}=\frac{B C}{C D} .
\end{array}
$$

Furthermore, from equations (2) and (4), we can get $\triangle A C E \backsim \triangle D C B$, if
$$
\frac{A E}{B D}=\frac{A C}{C D} \text {, }
$$

which means $A E=\frac{A C \cdot B D}{C D}$.
Given that $(A C \cdot B D)^{2}=A B^{4}+B C^{4}$, dividing both sides by $C D^{2}\left(=A B^{2}\right)$, we get
$$
\left(\frac{A C \cdot B D}{C D}\right)^{2}=A B^{2}+\left(\frac{B C^{2}}{C D}\right)^{2} \text {. }
$$

Substituting equations (3) and (5) into the above equation, we get
$$
A E^{2}=A B^{2}+B E^{2} \text {. }
$$

In $\triangle A B E$, by the converse of the Pythagorean theorem, we know
$$
\angle A B E=90^{\circ} \text {. }
$$

Substituting equation (1) into the above, we get
$$
2 \angle B A D=90^{\circ}, \angle B A D=45^{\circ} \text {. }
$$