II. (20 points) As shown in Figure 1, in the isosceles right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, and $D$ is a moving point on the hypotenuse $AB$. Connect $CD$, and draw $BE \perp CD$, with the foot of the perpendicular being $E$. Find the minimum value of $\frac{S_{\triangle ACD}}{S_{\triangle BCE}}$.
II. Solution 1 Let the length of the legs of the isosceles right triangle be $1$, and $\angle A C D=\alpha$. Then
$$
\angle C B E=\alpha, \angle A D C=\frac{3 \pi}{4}-\alpha, B E=\cos \alpha \text {. }
$$

In $\triangle A C D$, by the Law of Sines, we have
$$
\begin{array}{l}
\frac{C D}{\sin A}=\frac{A C}{\sin \angle A D C} \\
\Rightarrow C D=\frac{\sin \frac{\pi}{4}}{\sin \left(\frac{3 \pi}{4}-\alpha\right)} . \\
\text { Then } \frac{S_{\triangle A C D}}{S_{\triangle B C E}}=\frac{A C \cdot C D \sin \angle A C D}{B C \cdot B E \sin \angle C B E}=\frac{C D}{B E} \\
=\frac{\sin \frac{\pi}{4}}{\sin \left(\frac{3 \pi}{4}-\alpha\right) \cdot \cos \alpha} \\
=\frac{\sqrt{2}}{\sin \frac{3 \pi}{4}+\sin \left(\frac{3 \pi}{4}-2 \alpha\right)} .
\end{array}
$$

Since $2 \alpha \in\left(0, \frac{\pi}{2}\right)$, when $\alpha=\frac{\pi}{8}$, $\frac{S_{\triangle A C D}}{S_{\triangle B C E}}$ achieves its minimum value $\frac{\sqrt{2}}{\frac{\sqrt{2}}{2}+1}=2 \sqrt{2}-2$.
Solution 2 Let $A(1,0), B(0,1), l_{c D}: y=k x$.
Since $B E \perp C D$, we have $l_{B E}: y=-\frac{x}{k}+1$.
Also, $l_{A B}: x+y=1$.
Using the solve command of the TI-Nspire CX CAS graphing calculator
solve $\left(\left\{\begin{array}{l}y=k x, \\ x+y=1,\end{array},\{x, y\}\right)\right.$

and solve $\left(\left\{\begin{array}{l}y=k x, \\ y=-\frac{x}{k}+1,\{x, y\}\end{array}\right)\right.$ to get the solutions of the system of equations.

Thus, the y-coordinate of point $D$ and the x-coordinate of point $E$ are $\frac{k}{k+1}, \frac{k}{k^{2}+1}$, respectively.
Therefore, $\frac{S_{\triangle A C D}}{S_{\triangle B C E}}=\frac{\frac{k}{k+1}}{\frac{k}{k^{2}+1}}=\frac{k^{2}+1}{k+1}$.
Using the command $\left.\mathrm{fMin}\left(\frac{k^{2}+1}{k+1}, k\right) \right\rvert\, k>0$, we get $k=\sqrt{2}-1$.
Thus, when $k=\sqrt{2}-1$, $\frac{S_{\triangle C C D}}{S_{\triangle B C E}}$ achieves its minimum value $2 \sqrt{2}-2$.