$ABCD$ is a rectangle (see the accompanying diagram) with $P$ any point on $\overline{AB}$. $\overline{PS} \perp \overline{BD}$ and $\overline{PR} \perp \overline{AC}$. $\overline{AF} \perp \overline{BD}$ and $\overline{PQ} \perp \overline{AF}$. Then $PR + PS$ is equal to:

$\textbf{(A)}\ PQ\qquad  \textbf{(B)}\ AE\qquad  \textbf{(C)}\ PT + AT\qquad  \textbf{(D)}\ AF\qquad  \textbf{(E)}\ EF$
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Since $\overline{\rm PQ}$ and $\overline{\rm BD}$ are both perpendicular to $\overline{\rm AF}$, $\overline{\rm PQ} || \overline{\rm BD}$. Thus, $\angle APQ = \angle ABD$.

Also, $\angle ABD$ = $\angle CAB$ because $ABCD$ is a rectangle. Thus, $\angle APQ = \angle ABD = \angle CAB$.

Since $\angle APQ = \angle CAB$, $\triangle APT$ is isosceles with $PT = AT$.

$\angle ATQ$ and $\angle PTR$ are vertical angles and are congruent. Thus, by HL congruence, $\triangle ATQ \cong \triangle PTR$, so $AQ = PR$.

We also know that $PSFQ$ is a rectangle, since $\overline{\rm PS} \perp \overline{\rm BD}$, $\overline{\rm BF} \perp \overline{\rm AF}$, and $\overline{\rm PQ} \perp \overline{\rm AF}$.

Since $PSFQ$ is a rectangle, $QF = PS$. We also found earlier that $AQ = PR$. Thus, $PR + PS = AQ + QF = AF$.

Our answer is $PR + PS = \fbox{D) AF}$
~ lovesummer