Let 5 be a point inside an acute triangle $\triangle ABC$ such that $\angle ADB = \angle ACB + 90^{\circ}$, and $AC \cdot BD = AD \cdot BC$. Find the value of $\frac{AB \cdot CD}{AC \cdot BD}$.
Solution: Taking vertex $C$ as the origin and $CA$ as the real axis to establish a complex plane, let $|\overrightarrow{C A}|=r$. According to the geometric meaning of complex number multiplication and division, vector $\overrightarrow{D B}$ can be seen as $\overrightarrow{D A}$ after rotation and scaling. From the given information,
$$
\begin{aligned}
\overrightarrow{D B}= & \overrightarrow{D A} \cdot \frac{|\overrightarrow{D B}|}{|\overrightarrow{D A}|}\left[\cos \left(\angle A C B+90^{\circ}\right)\right. \\
& \left.+\mathrm{i} \sin \left(\angle A C B+90^{\circ}\right)\right] \\
= & \overrightarrow{D A} \cdot \frac{|\overrightarrow{C B}|}{|\overrightarrow{C A}|}(\cos \angle A C B+\mathrm{i} \sin \angle A C B) \mathrm{i} \\
= & \frac{1}{r} \overrightarrow{D A} \cdot \overrightarrow{C B} \cdot \mathrm{i}=\frac{1}{r}\left(r-z_{D}\right) z_{B} \cdot \mathrm{i} .
\end{aligned}
$$

Also, since $\overrightarrow{D B}=z_{B}-z_{D}$, we have
$$
\frac{1}{r}\left(r-z_{D}\right) z_{B} \cdot \mathrm{i}=z_{B}-z_{D} \text {, }
$$

which gives $z_{D}=\frac{r z_{B}(1-\mathrm{i})}{r-z_{B} \mathrm{i}}$.
$$
\begin{array}{l}
\therefore \frac{A B \cdot C D}{A C \cdot B D}=\frac{\left|z_{B}-r\right| \cdot\left|z_{D}\right|}{|r| \cdot\left|z_{D}-z_{B}\right|} \\
\quad=|1-\mathrm{i}|=\sqrt{2} .
\end{array}
$$

In particular, using the geometric meaning of complex number multiplication and division, it is easy to derive the complex number relationship for vectors rotated to special angles. For example, the necessary and sufficient condition for two vectors to be perpendicular is that the quotient of the corresponding complex numbers is a pure imaginary number; the necessary and sufficient condition for two vectors to be parallel or coincident is that the quotient of the corresponding complex numbers is a real number. Using the above conclusions, geometric problems involving perpendicularity, parallelism, and collinearity can be conveniently solved.