G63 (34-2, UK) Let $D$ be a point inside an acute triangle $ABC$ such that $\angle ADB = \angle ACB + 90^\circ$ and $AC \cdot BD = AD \cdot BC$.
(a) Calculate the ratio $\frac{AB \cdot CD}{AC \cdot BD}$;
(b) Prove that the tangents to the circumcircles of $\triangle ACD$ and $\triangle BCD$ at $C$ are perpendicular to each other.
(a) Construct $\triangle C E B \backsim \triangle C D A$. As shown in Figure 4-64, where $\angle C A D=\angle C B E, \angle A C D = \angle B C E$.
From the given conditions and similarity, we know
$$
\frac{B E}{B C}=\frac{A D}{A C}=\frac{B D}{B C}.
$$

Therefore, $B E=B D$.
$$
\begin{aligned}
\angle \triangle D B & =\angle C A D+\angle C B D+\angle A C B \\
& =\angle D B E+\angle A C B \\
& =90^{\circ}+\angle A C B .
\end{aligned}
$$

Thus, $\angle D B E=90^{\circ}$, and $\triangle B D E$ is an isosceles right triangle.
It is also easy to see that $\angle A C B=\angle D C E, \frac{A C}{C D}=\frac{B C}{C E}$, hence $\triangle A C B \backsim \triangle D C E$.
Therefore,
$$
\frac{A B \cdot C D}{A C \cdot B D}=\frac{D E}{B D}=\sqrt{2}.
$$
(b) Suppose the tangents to the circumcircles of $\triangle A C D$ and $\triangle B C D$ at point $C$ are $C F$ and $C G$ (not shown in the figure), such that the angles between the chord and the tangent are $\angle F C D=\angle D B C, \angle G C D=\angle C A D$.
Thus,
$$
\begin{aligned}
\angle F C G & =\angle F C D+\angle G C D \\
& =\angle D B C+\angle C A D \\
& =\angle A D B-\angle A C B \\
& =90^{\circ}.
\end{aligned}
$$

That is, the two tangents are perpendicular.