18.14 As shown in the figure, in quadrilateral $ABCD$, $\angle A, \angle C$ are right angles, points $E, F$ are on $AC$, $DE, BF$ are perpendicular to $AC$. If $AE=3, DE=5$, $CE=7$, then $BF$ is.
(A) 3.6.
(B) 4.
(C) 4.2.
(D) 4.5.
(E) 5.
(41st American High School Mathematics Examination, 1990)
[Solution] It is easy to prove that $\triangle A E D \backsim \triangle B F A$, $\triangle B F C \backsim \triangle C E D$, then
$$
\frac{B F}{A E}=\frac{A F}{E D} \text {, and } \quad \frac{B F}{E C}=\frac{F C}{E D} \text {, }
$$

Let $F C=x$, then $A F=10-x$, so the above two equations become
$$
\frac{B F}{3}=\frac{10-x}{5} \text {, and } \quad \frac{B F}{7}=\frac{x}{5},
$$

Therefore, $\frac{B F}{3}=2-\frac{x}{5}=2-\frac{B F}{7}$,
which means
$$
\frac{B F}{3}+\frac{B F}{7}=2 \text {, }
$$

Solving this, we get $B F=4.2$.
Thus, the answer is $(C)$.