# 1965 AHSME Problems/Problem 16

## Problem

Let line $AC$ be perpendicular to line $CE$. Connect $A$ to $D$, the midpoint of $CE$, and connect $E$ to $B$, the midpoint of $AC$. If $AD$ and $EB$ intersect in point $F$, and $\overline{BC} = \overline{CD} = 15$ inches, then the area of triangle $DFE$, in square inches, is:

$\textbf{(A)}\ 50 \qquad \textbf{(B) }\ 50\sqrt {2} \qquad \textbf{(C) }\ 75 \qquad \textbf{(D) }\ \frac{15}{2}\sqrt{105}\qquad \textbf{(E) }\ 100$

## Solution

$[asy] draw((0,0)--(30,0)); dot((0,0)); label("C", (-2,-2)); dot((30,0)); label("E", (32,-2)); dot((15,0)); label("D",(15,-2)); dot((0,15)); label("B",(-2,15)); draw((0,0)--(0,30)--(30,0)); dot((0,30)); label("A",(-2, 32)); dot((15,15)); label("G", (17,17)); draw((0,0)--(15,15)); dot((10,10)); label("F", (10,12)); draw((0,15)--(30,0)); draw((15,0)--(0,30)); markscalefactor=0.25; draw(rightanglemark((0,30),(0,0),(30,0))); label("15",(-2,7.5)); label("15",(-2,22.5)); label("15",(7.5,-2)); label("15",(22.5,-2)); [/asy]$

Draw $\overline{AE}$, as seen in the diagram. From the problem, we know that $\overline{EB}$ and $\overline{AD}$ are medians of $\triangle ACE$. Let $G$ be the midpoint of $\overline{AE}$. Then, $\overline{CG}$ is also a median of $\triangle ACE$, and it goes through $\triangle ACE$'s centroid, $F$. Because medians divide their triangle into $6$ smaller triangles of equal area, we know that $[\triangle DFE]=\frac{1}{6}[\triangle ACE]$. Because $[\triangle ACE]=\frac{1}{2}*(15+15)^2=\frac{900}{2}=450$, $[\triangle DFE]=\frac{450}{6}=75$. Thus, our answer is $\boxed{\textbf{(C) }75}$.