AHSME 1996 Problem 19

The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon?

$\text{(A)}\ \frac{1}{2} \text{(B)}\ \frac{\sqrt 3}{3} \text{(C)}\ \frac{2}{3} \text{(D)}\ \frac{3}{4} \text{(E)}\ \frac{\sqrt 3}{2}$