One, (Full marks 50 points) Given a right trapezoid $A B C D$ with a point $E$ on the right-angle leg $A B$, such that $\triangle C D E$ is an equilateral triangle, and $\frac{A E}{C D}=\frac{\sqrt{6}}{4}$. Find the value of $\frac{A E}{A B}$.
Solution 1: As shown in Figure 19, draw $D F \perp C E$ at $F$, and through $F$ draw a perpendicular to $B C$ intersecting $B C$ at $G$ and $A D$ at $H$. We get $F$ as the midpoint of $C E$, and $G$ as the midpoint of $B C$, thus
$$
\begin{array}{l}
F G=\frac{1}{2} B E \\
=\frac{1}{2}(A B-A E) .
\end{array}
$$

Then $H F=A B-\frac{1}{2} B E=\frac{1}{2}(A B+A E)$,
$$
\begin{aligned}
B C^{2} & =C E^{2}-B E^{2}=C D^{2}-B E^{2} \\
& =\left(\frac{4}{\sqrt{6}} A E\right)^{2}-(A B-A E)^{2} \\
& =\frac{5}{3} A E^{2}-A B^{2}+2 A B \cdot A E .
\end{aligned}
$$

Also, from $D F \perp C E, H F \perp B C$, we get
$$
\angle E C B=\angle D F H \text {. }
$$
$\therefore \mathrm{Rt} \triangle E C B \sim \mathrm{Rt} \triangle D F H$.
$$
\text { Hence } \frac{B C}{H F}=\frac{C E}{D F}=2 \operatorname{ctg} \angle D C E=\frac{2 \sqrt{3}}{3} \text {. }
$$

Squaring, we get $3 B C^{2}=4 H F$,
which means $5 A E^{2}-3 A B^{2}+6 A B \cdot A E=(4 B+A E)^{2}$.
$$
\therefore A E^{2}+A B \cdot A E-A B^{2}=0
$$

Solving the quadratic equation and taking the positive value, we get $\frac{A E}{A B}=\frac{\sqrt{5}-1}{2}$.
Solution 2: As shown in Figure 20, let $\angle A D E=\alpha$. From the given information,
$$
\sin \alpha=\frac{\sqrt{6}}{4}, \cos \alpha=\frac{\sqrt{10}}{4} \text {. }
$$

Draw $D P \perp B C$ at $P$, then
$$
\begin{array}{l}
D P=A B, \\
\begin{aligned}
\angle P D C & =\angle A D C-90^{\circ} \\
& =\left(\alpha+60^{\circ}\right)-90^{\circ} \\
& =\alpha-30^{\circ} .
\end{aligned}
\end{array}
$$

Thus, $\frac{A B}{C D}=\frac{D P}{C D}=\cos \angle P D C=\cos \left(\alpha-30^{\circ}\right)$,
$$
\therefore \frac{A B}{A E} \cdot \frac{A E}{C D}=\cos \alpha \cos 30^{\circ}+\sin \alpha \sin 30^{\circ} \text {, }
$$

which means
$$
\begin{array}{l}
\frac{A B}{A E} \cdot \frac{\sqrt{6}}{4}=\frac{\sqrt{10}}{4} \times \frac{\sqrt{3}}{2}+\frac{\sqrt{6}}{4} \times \frac{1}{2} . \\
\text { Hence } \frac{A E}{A B}=\frac{2}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{2} .
\end{array}
$$