Let $D$ be a point inside an acute triangle $\triangle A B C$ such that
$$
\begin{array}{l}
\angle A D B=\angle A C B+90^{\circ}, \\
A C \cdot B D=A D \cdot B C .
\end{array}
$$
(1) Calculate the ratio $\frac{A B \cdot C D}{A C \cdot B D}$. The key related to the perpendicular tangents at point $C$ is to find the relationship between angles and line segments.
Solve (1) Construct
the circumcircle of $\triangle A B C$, intersecting the extensions of $A D$ and $B D$ at $E$ and $F$ respectively, and connect $B E$, $C E$, $A F$, and $C F$, as shown in the figure.
$$
\begin{aligned}
& \because \angle A D B \\
= & \angle A C B+90^{\circ}, \\
\text { and } & \angle A D B \\
= & \angle A C B+\angle 2 \\
+ & \angle 4
\end{aligned}
$$

and $\angle A D B$
$\therefore \angle 2+\angle 4=90^{\circ}$.
In $\triangle D C E$ and $\triangle D C F$,
$$
\begin{array}{l}
\angle D E C=\angle A B C=\angle D F C=\angle B A C, \\
\therefore \quad \frac{D C}{\sin B}=\frac{D E}{\sin (\angle 1+\angle 5)}, \\
\frac{D C}{\sin A}=\frac{D F}{\sin (\angle 3+\angle 6)} .
\end{array}
$$

Thus, $\frac{D E}{D F}=\frac{\sin A}{\sin B} \cdot \frac{\sin (\angle 1+\angle 5)}{\sin (\angle 3+\angle 6)}$
and $A D \cdot D E=B D \cdot D F$,
and $\square$
$$
\begin{array}{l}
A C \cdot B D=A D \cdot B C, \\
\therefore \quad \frac{D E}{D F}=\frac{B D}{A D}=\frac{B C}{A C}=\frac{\sin A}{\sin B} .
\end{array}
$$

Therefore,
$$
\begin{array}{l}
\sin (\angle 1+\angle 5)=\sin (\angle 3+\angle 6) \text {. } \\
\text { Also, } \angle 1+\angle 5+\angle 3+\angle 6 \\
=\angle 1+\angle 3+\angle A C B \\
=180^{\circ}-\angle A D B+\angle A C B=90^{\circ} \text {, } \\
\therefore \quad \angle 1+\angle 5=\angle 3+\angle 6=45^{\circ} \text {. } \\
\because \angle D B E=\angle 2+\angle 4=90^{\circ} \text {, } \\
\angle B E D=\angle A C B \text {, } \\
\therefore D E=\frac{B D}{\sin C} \text {. } \\
\end{array}
$$

Thus, $\frac{A B \cdot C D}{A C \cdot B D}$
$$
=\frac{A B}{A C} \cdot \frac{1}{B D} \cdot \frac{D E \cdot \sin B}{\sin (\angle 1+\angle 5)}
$$
(Continued from page 13)
$$
\begin{array}{l}
=\frac{\sin C}{\sin B} \cdot \frac{\sin B}{\sin C \cdot \sin (\angle 1+\angle 5)} \\
=\frac{1}{\sin 45^{\circ}}=\sqrt{2} .
\end{array}
$$
(2) Let $M N$ and $P Q$ be the tangents to the circumcircles of $\triangle B D C$ and $\triangle A D C$ at point $C$, respectively, then
$$
\begin{array}{l}
\angle M C D=\angle 4, \angle P C D=\angle 2 . \\
\because \quad \angle 2+\angle 4=90^{\circ}, \\
\therefore \quad \angle M C P=\angle M C D+\angle P C D=90^{\circ},
\end{array}
$$

i.e., $M N \perp P Q$.