1980 AHSME Problems/Problem 21

Problem

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E); draw(E--B--A--C--B^^A--D); label("$A$", A, dir(D--A)); label("$B$", B, dir(E--B)); label("$C$", C, dir(0)); label("$D$", D, SE); label("$E$", E, N); label("$F$", F, dir(80));[/asy]$

In triangle $ABC$ , $\measuredangle CBA=72^\circ$ , $E$ is the midpoint of side $AC$ , and $D$ is a point on side $BC$ such that $2BD=DC$ ; $AD$ and $BE$ intersect at $F$ . The ratio of the area of triangle $BDF$ to the area of quadrilateral $FDCE$ is

$\text{(A)} \ \frac 15 \qquad \text{(B)} \ \frac 14 \qquad \text{(C)} \ \frac 13 \qquad \text{(D)}\ \frac{2}{5}\qquad \text{(E)}\ \text{none of these}$

Solution

We can use the principle of same height same area to solve this problem. $\fbox{A}$

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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1980 AHSME Problems/Problem 21

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