7. As shown in the figure, in isosceles trapezoid $A B C D$, $A B / / C D, A B=2 C D, \angle A=60^{\circ}$, and $E$ is a point on the base $A B$, such that $F E=F B=A C$, $F A=A B$.
Then $A E: E B$ equals
(A) $1: 2$;
(B) $1: 3$;
(C) $2: 5$;
(D) $3: 10$.
7. (B)

Let $CD=1$, then $FA=AB=2$, it is easy to prove that $BC=\frac{1}{2} AB=1$,
$$
\begin{array}{c}
\angle ACB=90^{\circ}, FE=FB=AC=\sqrt{3}. \\
\because \quad \triangle ABF \sim \triangle FBE, \\
\therefore \quad \frac{AB}{BF}=\frac{BF}{BE}, \\
\therefore \quad BE=\frac{BF^2}{AB}=\frac{3}{2},
\end{array}
$$

Thus $AE=\frac{1}{2}$, so $AE: EB=1: 3$.