# Difference between revisions of "1964 AHSME Problems/Problem 22"

## Problem

Given parallelogram $ABCD$ with $E$ the midpoint of diagonal $BD$. Point $E$ is connected to a point $F$ in $DA$ so that $DF=\frac{1}{3}DA$. What is the ratio of the area of $\triangle DFE$ to the area of quadrilateral $ABEF$?

$\textbf{(A)}\ 1:2 \qquad \textbf{(B)}\ 1:3 \qquad \textbf{(C)}\ 1:5 \qquad \textbf{(D)}\ 1:6 \qquad \textbf{(E)}\ 1:7$

## Solution

If it works for a parallelogram $ABCD$, it should also work for a unit square, with $A(0, 0), B(0, 1), C(1, 1), D(1, 0)$. We are given that $E$ is the midpoint of $BD$, so $E(0.5, 0.5)$. If $F$ is on $DA$, then $F(x, 0)$. We note that $DF = 1-x$ and $DA = 1$, so $DF = \frac{1}{3}DA$ means $1-x = \frac{1}{3}$, or $x = \frac{2}{3}$, and hence $F(\frac{2}{3}, 0)$.

We note that $\triangle DFE$ has a base $DF$ that is $\frac{1}{3}$ and an altitude from $E$ to $DF$ that is $\frac{1}{2}$. Therefore, $[DEF] = \frac{1}{2}bh = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{12}$.

Quadrilateral $ABEF$ can be split into $\triangle ABE$ and $\triangle AEF$. The first triangle is $\frac{1}{4}$ of the unit square cut diagonally, so $[ABE] = \frac{1}{4}$. The second triangle has base $AF$ that is $\frac{2}{3}$ and height $E$ to $AF$ that is $\frac{1}{2}$. Therefore, $[AEF] = \frac{1}{2}bh = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{6}$.

The entire quadrilateral $ABEF$ has area $\frac{1}{4} + \frac{1}{6} = \frac{5}{12}$. This is $5$ times larger than the area of $\triangle DFE$, so the ratio is $1:5$, or $\boxed{\textbf{(C)}}$.