15. As shown in Figure 3, in rectangle $A B C D$, it is known that $C D=2 B C$, and $E, F$ are on side $A B$, such that $D E, D F$ trisect $\angle A D C$. Then $\frac{S_{\triangle D E F}}{S_{\text {rectangle } A B C D}}=(\quad)$.
(A) $\frac{\sqrt{3}}{6}$
(B) $\frac{\sqrt{6}}{8}$
(C) $\frac{3 \sqrt{3}}{16}$
(D) $\frac{1}{3}$
(E) $\frac{\sqrt{2}}{4}$
15. A.

From the problem, we know that $\angle A D C=90^{\circ}$, $\angle A D E=\angle E D F=\angle F D C=30^{\circ}$.
Thus, $A E=\frac{\sqrt{3}}{3} A D, A F=\sqrt{3} A D$.
Therefore, $E F=\frac{2 \sqrt{3}}{3} A D$.
Since $C D=2 A D$, we have
$$
\frac{S_{\triangle D E F}}{S_{\text {trapezoid } A B C D}}=\frac{\frac{1}{2} \times \frac{2 \sqrt{3}}{3} A D^{2}}{2 A D^{2}}=\frac{\sqrt{3}}{6} \text {. }
$$