Example 8 Let $D$ be a point inside an acute-angled $\triangle ABC$,
$$
\begin{array}{l}
\angle A D B=\angle A C B+90^{\circ}, \text { and } A C \cdot B D=A D \cdot B C . \\
\text { Find the value of } \frac{A B \cdot C D}{A C \cdot B D} \text { . }
\end{array}
$$
(34th IMO)
【Analysis】With $C$ as the origin and $C A$ as the real axis, establish a complex plane, and let $|\overrightarrow{C A}|=r$.

According to the geometric meaning of complex number multiplication and division, the vector $\overrightarrow{D B}$ can be regarded as $\overrightarrow{D A}$ after rotation and scaling. Therefore,
$$
\begin{array}{l}
\overrightarrow{D B}=\overrightarrow{D A} \cdot \frac{|\overrightarrow{D B}|}{|\overrightarrow{D A}|}\left(\cos \left(\angle A C B+90^{\circ}\right)+\right. \\
\left.\quad \mathrm{i} \sin \left(\angle A C B+90^{\circ}\right)\right) \\
=\frac{1}{r}\left(r-z_{D}\right) z_{B} \mathrm{i} . \\
\text { Also, } \overrightarrow{D B}=z_{B}-z_{D} \text {, then } \\
z_{D}=\frac{r z_{B}(1-\mathrm{i})}{r-z_{B} \mathrm{i}} \\
\Rightarrow \frac{A B \cdot C D}{A C \cdot B D}=\frac{\left|z_{B}-r\right|\left|z_{D}\right|}{|r|\left|z_{B}-z_{D}\right|}=|1-\mathrm{i}|=\sqrt{2} .
\end{array}
$$