Example 3 Let $D$ be a point inside an acute $\triangle A B C$, $\angle A D B = \angle A C B + 90^{\circ}$, and $A C \cdot B D = A D \cdot B C$. Find the value of $\frac{A B \cdot C D}{A C \cdot B D}$.
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\begin{array}{l}
\text{Let } C \text{ be the origin and } \overrightarrow{C A} \text{ be the real axis to establish the complex plane. Let } |\overrightarrow{C A}|=r. \\
\text{According to the geometric meaning of complex number multiplication and division, vector } \overrightarrow{D B} \text{ can be seen as } \overrightarrow{D A} \text{ after rotation and scaling.} \\
\text{From the given,} \\
\overrightarrow{D B}=\overrightarrow{D A} \cdot \frac{|\overrightarrow{D B}|}{|\overrightarrow{D A}|} \mathrm{e}^{\left(\angle A C B+90^{\circ}\right) \mathrm{i}} \\
=\frac{1}{r} \overrightarrow{D A} \cdot \overrightarrow{C B} \mathrm{i}=\frac{1}{r}\left(r-z_{D}\right) z_{B} \mathrm{i}. \\
\text{Also, } \overrightarrow{D B}=z_{B}-z_{D}, \text{ so } \\
\frac{1}{r}\left(r-z_{D}\right) z_{B} \mathrm{i}=z_{B}-z_{D} \Rightarrow 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_{D}-z_{B}\right|}=|1-\mathrm{i}|=\sqrt{2} .
\end{array}
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