# 1974 AHSME Problems/Problem 25

## Problem

In parallelogram $ABCD$ of the accompanying diagram, line $DP$ is drawn bisecting $BC$ at $N$ and meeting $AB$ (extended) at $P$. From vertex $C$, line $CQ$ is drawn bisecting side $AD$ at $M$ and meeting $AB$ (extended) at $Q$. Lines $DP$ and $CQ$ meet at $O$. If the area of parallelogram $ABCD$ is $k$, then the area of the triangle $QPO$ is equal to

$[asy] size((400)); draw((0,0)--(5,0)--(6,3)--(1,3)--cycle); draw((6,3)--(-5,0)--(10,0)--(1,3)); label("A", (0,0), S); label("B", (5,0), S); label("C", (6,3), NE); label("D", (1,3), NW); label("P", (10,0), E); label("Q", (-5,0), W); label("M", (.5,1.5), NW); label("N", (5.65, 1.5), NE); label("O", (3.4,1.75));[/asy]$

$\mathrm{(A)\ } k \qquad \mathrm{(B) \ }\frac{6k}{5} \qquad \mathrm{(C) \ } \frac{9k}{8} \qquad \mathrm{(D) \ } \frac{5k}{4} \qquad \mathrm{(E) \ }2k$

## Solution

Note that $$[QPO]=[QAM]+[PBN]+[AMONB]=[AMONB]+[MDC]+[NCD]$$

$$=[AMONB]+[MDC]+[NOC]+[DOC]=[ABCD]+[DOC]=k+[DOC].$$

Also, note that $DCNM$ is a parallelogram, and so $[DOC]=\frac{1}{4}[DCNM]=\frac{1}{8}[ABCD]=\frac{k}{8}$.

Therefore, $[QPO]=k+\frac{k}{8}=\frac{9k}{8}, \boxed{\text{C}}$.