# 1983 AHSME Problems/Problem 28

## Problem 28

Triangle $\triangle ABC$ in the figure has area $10$. Points $D, E$ and $F$, all distinct from $A, B$ and $C$, are on sides $AB, BC$ and $CA$ respectively, and $AD = 2, DB = 3$. If triangle $\triangle ABE$ and quadrilateral $DBEF$ have equal areas, then that area is

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(10,0), C=(8,7), F=7*dir(A--C), E=(10,0)+4*dir(B--C), D=4*dir(A--B); draw(A--B--C--A--E--F--D); pair point=incenter(A,B,C); label("A", A, dir(point--A)); label("B", B, dir(point--B)); label("C", C, dir(point--C)); label("D", D, dir(point--D)); label("E", E, dir(point--E)); label("F", F, dir(point--F)); label("2", (2,0), S); label("3", (7,0), S);[/asy]$

$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ \frac{5}{3}\sqrt{10}\qquad \textbf{(E)}\ \text{not uniquely determined}$

## Solution

Let $G$ be the intersection point of $AE$ and $DF$. Since $[DBEF] = [ABE]$, we have $[DBEG] + [EFG] = [DBEG] + [ADG]$, i.e. $[EFG] = [ADG]$. It therefore follows that $[ADG] + [AGF] = [EFG] + [AGF]$, so $[ADF] = [AFE]$. Now, taking $AF$ as the base of both $\triangle ADF$ and $\triangle AFE$, and using the fact that triangles with the same base and same perpendicular height have the same area, we deduce that the perpendicular distance from $D$ to $AF$ is the same as the perpendicular distance from $E$ to $AF$. This in turn implies that $AF \parallel DE$, and so as $A$, $F$, and $C$ are collinear, $AC \parallel DE$. Thus $\triangle DBE \sim \triangle ABC$, so $\frac{BE}{BC} = \frac{BD}{BA} = \frac{3}{5}$. Since $\triangle ABE$ and $\triangle ABC$ have the same perpendicular height (taking $AB$ as the base), $[ABE] = \frac{3}{5} \cdot [ABC]=\frac{3}{5}\cdot 10 = 6$, and hence the answer is $\boxed{\textbf{(C)}}$.