# 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 $\fbox{A}$

## See also

 1980 AHSME (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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