7. Let $D$ be a point inside acute $\triangle A B C$ such that $\angle A D B=\angle A C B+90^{\circ}$, and $A C \cdot B D=A D \cdot B C$. Calculate the ratio: $\frac{A B \cdot C D}{A C \cdot B D}$.
7. Establish a complex plane with $C$ as the origin, and let $\angle A C B=\theta$. Denote $\overrightarrow{C A}=1$.
$$
\text { Then } \begin{aligned}
\overrightarrow{D B} & =\frac{|D B|}{|D A|} \cdot \overrightarrow{D A} \mathrm{e}^{\mathrm{i}(\phi \theta+\theta)}=\frac{|B C|}{|A C|}(1-D) \mathrm{ie}^{\mathrm{i} \theta} \\
& =\overrightarrow{C B}(1-D) \mathrm{i}=B(1-D) \mathrm{i} .
\end{aligned}
$$

Also, $\overrightarrow{D B}=B-D$, so $B(1-D) \mathrm{i}=B-D$.
Thus, $D=\frac{B \mathrm{i}-B}{B \mathrm{i}-1}$.
$$
\text { Therefore, } \frac{|A B| \cdot|C D|}{|A C| \cdot|B D|}=\frac{|B-1| \cdot\left|\frac{B \mathrm{i}-B}{B \mathrm{i}-1}\right|}{\left|B-\frac{B \mathrm{i}}{B \mathrm{i}}-\frac{B}{B}\right|}=\frac{|B| \cdot|B-1| \cdot|\mathrm{i}-1|}{|B| \cdot|B-1|}=\sqrt{2} \text {. }
$$