1958 AHSME Problems/Problem 47

Problem

$ABCD$ 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:

$[asy] draw((-2,-1)--(-2,1)--(2,1)--(2,-1)--cycle,dot); draw((-2,-1)--(2,1)--(2,-1)--(-2,1),dot); draw((-2,1)--(-6/5,-3/5),black+linewidth(.75)); draw((6/5,3/5)--(1,1)--(-3/2+1/10,-2/10),black+linewidth(.75)); draw((1,1)--(1-3/5,1-6/5),black+linewidth(.75)); MP("A",(-2,1),NW);MP("B",(2,1),NE);MP("C",(2,-1),SE);MP("D",(-2,-1),SW); MP("Q",(-3/2+1/10,-2/10),W);MP("T",(-2/5,1/5),N);MP("P",(1,1),N); MP("F",(-6/5,-3/5),SE);MP("E",(0,0),S);MP("S",(6/5,3/5),S);MP("R",(1-3/5,1-6/5),S); [/asy]$

$\textbf{(A)}\ PQ\qquad \textbf{(B)}\ AE\qquad \textbf{(C)}\ PT + AT\qquad \textbf{(D)}\ AF\qquad \textbf{(E)}\ EF$

Solution

Step I: 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 $PT = AT$.

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

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

Step IV: 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