$18 \cdot 114$ As shown in the figure, in isosceles trapezoid $ABCD$, $AB$ $/ / CD, AB=2CD, \angle A=60^{\circ}$, and $E$ is a point on the base $AB$, such that $FE=FB=AC, FA=AB$. Then $AE$ : $EB$ equals
(A) $1: 2$.
(B) $1: 3$.
(C) $2: 5$.
(D) $3: 10$.
[Solution] Let $CD=1$, then $FA=AB=2$. It is easy to prove that $BC=\frac{AB}{2}, \angle ACB=90^{\circ}$. Therefore, $FE=FB=AC=\sqrt{3}$.
Also, $\triangle ABF \sim \triangle FBE$, so $\frac{AB}{BF}=\frac{BF}{BE}, \quad BE=\frac{BF^2}{AB}=\frac{3}{2}$.
Thus, $AE=AB-BE=\frac{1}{2}$.
Hence, $AE: EB=1: 3$.
Therefore, the answer is $(B)$.