$18 \cdot 60$ In the figure, $A E, C F$ are perpendicular to the diagonal $B D$ of rectangle $A B C D$, with $E, F$ being the feet of the perpendiculars. If $B E=1, E F=2$, then the area of rectangle $A B C D$ is
(A) $4 \sqrt{3}$.
(B) $3 \sqrt{5}$.
(C) 6 .
(D) Cannot be determined.
(China Junior High School Mathematics Correspondence Competition, 1989)
[Solution 1] Connect $AC$, intersecting $BD$ at $O$, and let $AE, BC$ intersect at point $G$.
Given that $BE = EO = 1$, then $AE$ is the perpendicular bisector of $BO$.
$$
\text{Also, } \begin{aligned}
\angle BDC & = \angle BGA \\
& = \angle GCA + \angle GAC,
\end{aligned}
$$

and $\angle GCA = \angle DBC$,
$$
\begin{array}{l} 
\angle GAC = \angle GAB = \angle GCA. \\
\therefore \quad \angle BDC = 2 \angle DBC,
\end{array}
$$

Thus,
$$
\begin{array}{c}
\angle DBC = 30^{\circ}. \\
\therefore \quad CD = \frac{BD}{2} = \frac{4}{2} = 2, \quad BC = \sqrt{4^2 - 2^2} = 2 \sqrt{3}.
\end{array}
$$

Therefore, $S_{\square ABCD} = BC \cdot CD = 2 \sqrt{3} \cdot 2 = 4 \sqrt{3}$. Hence, the answer is $(A)$.
[Solution 2] Given $BE = 1, EF = 2$,
we know $FD = 1, ED = 3, \quad BD = 4$.
Since $\triangle AEB \sim \triangle DEA$, we have $AE^2 = BE \cdot ED = 1 \cdot 3$, thus $AE = \sqrt{3}$.
Then $S_{\triangle ABCD} = 2 S_{\triangle ABD} = BD \cdot AE = 4 \sqrt{3}$. Hence, the answer is $(A)$.