Example 3.7.5 (IMO 34) Let $D$ be a point inside an acute-angled $\triangle ABC$ such that $\angle ADB = \angle ACB + 90^{\circ}$, and $AC \cdot BD = AD \cdot BC$. Find the value of $AB \cdot CD : AC \cdot BD$.

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The translation is provided as requested, maintaining the original formatting and structure.
Proof: Taking $C$ as the origin and $C A$ as the real axis to establish a complex plane.
Let $|\overrightarrow{C A}|=r$, and $\overrightarrow{D B}$ can be seen as the result of rotating and scaling $\overrightarrow{D A}$:
$$
\text{Given } \begin{aligned}
\overrightarrow{D B} & =\overrightarrow{D A} \cdot \frac{|\overrightarrow{D B}|}{|\overrightarrow{D A}|}\left[\cos \left(\angle A C B+90^{\circ}\right)+\mathrm{i} \cdot \sin \left(\angle A C B+90^{\circ}\right)\right] \\
& =\overrightarrow{D A} \cdot \frac{|\overrightarrow{C B}|}{|\overrightarrow{C A}|}[\cos \angle A C B+\mathrm{i} \cdot \sin \angle A C B] \cdot \mathrm{i} \\
& =\frac{1}{r} \overrightarrow{D A} \cdot \overrightarrow{C B} \cdot \mathrm{i}=\frac{1}{r}(\overrightarrow{C A}-\overrightarrow{C D}) \cdot \overrightarrow{C B} \mathrm{i} \\
& =\frac{\mathrm{i}}{r}\left(r-z_{D}\right) z_{B}
\end{aligned}
$$

Also, since $\overrightarrow{D B}=z_{B}-z_{D}$, we have $\frac{\mathrm{i}}{r}\left(r-z_{D}\right) z_{B}=z_{B}-z_{D}$.
Solving for $z_{D}$, we get $z_{D}=r z_{B}(1-\mathrm{i}):\left(r-\mathrm{i} z_{B}\right)$,
Therefore, $\frac{A B \cdot C D}{A C \cdot B D}=\frac{\left|z_{B}-r\right|\left|z_{D}\right|}{|r|\left|z_{D}-z_{B}\right|}$, substituting the above expression for $z_{D}$, we get:
$$
=\frac{\left|z_{B}-r\right||1-\mathrm{i}|}{\left|-r \mathrm{i}+z_{B} \mathrm{i}\right|}=|1-\mathrm{i}|=\sqrt{2} .
$$