2. Let $D$ be a point inside acute $\triangle A B C$, $\angle A D B=\angle A C B+90^{\circ}$, and $A C \cdot B D=A D \cdot B C$. Find the value of $\frac{A B \cdot C D}{A C \cdot B D}$.
2. Let $z_{A}=0, z_{B}=1$, so $\arg \frac{z_{D}-1}{z_{D}}=\arg \frac{z_{C}-1}{z_{C}}+\frac{\pi}{2}$, $\left|z_{C}\right| \times\left|z_{D}-1\right|=\left|z_{D}\right| \times\left|z_{C}-1\right|$,

so $\left|\frac{z_{D}-1}{z_{D}}\right|=\left|\frac{z_{C}-1}{z_{C}}\right|$, so $\frac{z_{D}-1}{z_{D}}=\mathrm{i} \times \frac{z_{C}-1}{z_{C}}$, so $(\mathrm{i}-1)\left(z_{C} z_{D}-z_{C}\right)=\mathrm{i}\left(z_{D}-z_{C}\right)$
so $\frac{A B \cdot C D}{A C \cdot B D}=\frac{|1| \times\left|z_{C}-z_{D}\right|}{\left|z_{C}\right| \times\left|z_{D}-1\right|}=\left|\frac{z_{D}-z_{C}}{z_{C} z_{D}-z_{C}}\right|=\left|\frac{\mathrm{i}-1}{\mathrm{i}}\right|=\sqrt{2}$.