20. (12 points) As shown in Figure 5, let $D$ be a point inside the acute triangle $\triangle A B C$ such that
$$
\begin{array}{l}
\angle A D B \\
=\angle A C B+90^{\circ},
\end{array}
$$

and $A C \cdot B D=A D \cdot B C$.
Find the value of $\frac{A B \cdot C D}{A C \cdot B D}$.
20. Draw perpendiculars from point $D$ to the three sides of $\triangle A B C$, with the feet of the perpendiculars being $E, F, G$. Connect $E F, F G, E G$. Let $R$ be the circumradius of $\triangle A B C$.

Since $A, F, D, E$ are concyclic with $A D$ as the diameter, by the Law of Sines, we have
$\frac{E F}{A D}=\sin A=\frac{B C}{2 R}$.
Thus, $E F=\frac{1}{2 R} A D \cdot B C$.
Similarly, $F G=\frac{1}{2 R} A C \cdot B D$,
$$
G E=\frac{1}{2 R} A B \cdot C D \text {. }
$$

Since $A C \cdot B D=A D \cdot B C$, it follows that $E F=F G$.
From $\angle A D B=\angle A C B+90^{\circ}$, we get
$\angle C A G+\angle E B C=90^{\circ}$.
Since $\angle E F C=\angle C A G, \angle G F C=\angle E B C$, then $\angle E F G=90^{\circ}$.
Therefore, $\triangle E F G$ is an isosceles right triangle.
Thus, $\frac{A B \cdot C D}{A C \cdot B D}=\frac{G E}{F G}=\sqrt{2}$.