Rhombus $ABCD$ is similar to rhombus $BFDE$. The area of rhombus $ABCD$ is 24, and $\angle BAD = 60^\circ$. What is the area of rhombus $BFDE$? $\textrm{(A) } 6 \qquad$ $ \textrm{(B) } 4\sqrt {3} \qquad$ $ \textrm{(C) } 8 \qquad$ $ \textrm{(D) } 9 \qquad$ $ \textrm{(E) } 6\sqrt {3}$
Solution
Since $\angle BAD=60^\circ$ and $AB=AD$, $\triangle ABD$ is equilateral, so diagonal $BD$ equals the side length $s$ of rhombus $ABCD$.
As in the figure, $BFDE$ lies inside $ABCD$, so its $60^\circ$ angles are at $B$ and $D$ and $BD$ is its longer diagonal. In a rhombus with a $60^\circ$ angle the longer diagonal is $\sqrt3$ times the side, so the side of $BFDE$ is $\frac{s}{\sqrt3}$.
The rhombi are similar with ratio $\frac1{\sqrt3}$, so $[BFDE]:[ABCD]=\left(\frac1{\sqrt3}\right)^2=\frac13$.
$\textrm{(A) } 6 \qquad$ $ \textrm{(B) } 4\sqrt {3} \qquad$ $ \textrm{(C) } 8 \qquad$ $ \textrm{(D) } 9 \qquad$ $ \textrm{(E) } 6\sqrt {3}$ 
