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 $\fbox{\textbf{C}}$ .

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1983 AHSME Problems/Problem 28

Problem 28

Solution