$18.75 F$ is any point on side $AD$ of square $ABCD$, $CE \perp CF$ intersects the extension of $AB$ at $E$, if the area of square $ABCD$ is $256$, and the area of $\triangle CEF$ is 200, then the length of $BE$ is
(A) 12.
(B) 14.
(C) 15.
(D) 16.
(E) 20.
(14th American High School Mathematics Examination, 1963)
[Solution] Given $C D=C B, \angle D C F=\angle B C E$, thus $\mathrm{Rt} \triangle C D F \cong \mathrm{Rt} \triangle C B E$, so $C F=C E$.
Then $S_{\triangle C E F}=\frac{1}{2} \cdot C E \cdot C F=\frac{1}{2} C E^{2}=200$,

which means
$$
C E^{2}=400 . \quad S_{\text {square } ABCD }=C B^{2}=256 \text {, }
$$

By the Pythagorean theorem, $B E^{2}=C E^{2}-C B^{2}=400-256=144$, so $B E=12$.
Therefore, the answer is $(A)$.