18.33 $F$ is a point on the extension of side $AD$ of square $ABCD$, $BF$ intersects $AC$ at $E$, and intersects $CD$ at $G$. If $EF=32, GF=24$, then $BE$ equals
(A) 4.
(B) 8.
(C) 10.
(D) 12.
(E) 16.
[Solution] Let $B E=x, D G=y$, and $A B=b$.
It is easy to prove that $\triangle B E A \backsim \triangle G E C$. From the diagram, we have $\frac{8}{x}=\frac{b-y}{b}$,
which gives $y=\frac{b(x-8)}{x}$.
Also, $\triangle F D G \backsim \triangle B C G$, so
$$
\frac{24}{x+8}=\frac{y}{b-y} .
$$

Substituting $y=\frac{b(x-8)}{x}$ into the equation and simplifying, we get
$$
x^{2}-64=192 \text {, }
$$

Solving for $x$ yields $x=16$ (the negative value has been discarded).
Therefore, the answer is $(E)$.