18.55 $ABCD$ is a rectangle (as shown in the figure), $P$ is any point on $AB$, $PS \perp BD$, $PR \perp AC$, $AF \perp BD$, $PQ \perp AF$, then $PR + PS$ equals
(A) $PQ$.
(B) $AE$.
(C) $PT + AT$.
(D) $AF$.
(E) $EF$.
[Solution] Connect $P E$. Given that $A B C D$ is a rectangle, then $A E=B E=D E$, let it be $a$.
$$
\text { Also } \begin{aligned}
S_{\triangle A E B} & =\frac{1}{2} \cdot A E \cdot P R+\frac{1}{2} \cdot B E \cdot P S \\
& =\frac{1}{2} a(P R+P S),
\end{aligned}
$$

and $S_{\triangle A E D}=\frac{1}{2} \cdot D E \cdot A F=\frac{1}{2} a A F$,
$$
\begin{array}{ll}
\because & S_{\triangle A E B}=S_{\triangle A E D}, \\
\therefore & P R+P S=A F .
\end{array}
$$

Therefore, the answer is $(D)$.