5. 3. 15 ** Let $D$ be a point inside acute $\triangle A B C$ such that $\angle A D B=\angle A C B+90^{\circ}$, and $A C \cdot B D=A D \cdot B C$. Find the value of $\frac{A B \cdot C D}{A C \cdot B D}$.
Parse the figure, with $C$ as the origin of the complex plane, let $\angle A C B=\theta, \overrightarrow{C A}=1$,
$$
\begin{aligned}
\overrightarrow{D B} & =\frac{|B D|}{|D A|} \cdot \overrightarrow{D A} \cdot \mathrm{e}^{\mathrm{i}\left(\theta+90^{\circ}\right)} \\
& =\frac{|B C|}{|A C|} \cdot\left(1-z_{D}\right) \cdot \mathrm{i} \cdot \mathrm{e}^{\mathrm{i} \theta} \\
& =\overrightarrow{C B} \cdot\left(1-z_{D}\right) \cdot \mathrm{i} \\
& =z_{B}\left(1-z_{D}\right) \cdot \mathrm{i} .
\end{aligned}
$$

Since $\overrightarrow{D B}=z_{B}-z_{D}$, we have $z_{D}=\frac{z_{B}(\mathrm{i}-1)}{z_{B} \mathrm{i}-1}$, thus
$$
\frac{|A B| \cdot|C D|}{|A C| \cdot|B D|}=\frac{\left|z_{B}-1\right|\left|\frac{z_{B} \mathrm{i}-z_{B}}{z_{B} \mathrm{i}-1}\right|}{\left|z_{B}-\frac{z_{B} \mathrm{i}-z_{B}}{z_{B} \mathrm{i}-1}\right|}=\frac{\left|z_{B}\right| \cdot\left|z_{B}-1\right| \cdot|\mathrm{i}-1|}{\left|z_{B}\right| \cdot\left|z_{B}-1\right|}=\sqrt{2} .
$$