Problem
1980 AHSME Problem 21
![[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]](/static/aops/70161822674949b3054b3faa63150a9a6f751e97.png)
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}
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