6. As shown in Figure 2, in the right trapezoid $A B C D$, $\angle B=$ $\angle C=90^{\circ}, A B=B C$, point $E$ is on side $B C$, and makes $\triangle A D E$ an equilateral triangle. Then the ratio of the area of $\triangle A D E$ to the area of trapezoid $A B C D$ is ( ).
(A) $\sqrt{3}-1$
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{2}{3}$
(D) $(\sqrt{3}-1)^{2}$
6. D.

As shown in Figure 6, extend trapezoid $ABCD$ to form square $ABCF$, with side length 1, and let $CD = CE = x$. Then,
$$
BE = FD = 1 - x.
$$

By the Pythagorean theorem, we have $2x^2 = 1 + (1 - x)^2$.
Solving for $x$ gives $x = \sqrt{3} - 1$.
$$
\text{Therefore, } \frac{S_{\triangle ADE}}{S_{\text{trapezoid } ABCD}} = \frac{\frac{\sqrt{3}}{4} \times 2x^2}{\frac{1}{2}(1 + x) \times 1} = (\sqrt{3} - 1)^2.
$$