2006 AMC 12B 第 13 题

Since $\angle BAD = 60^\circ,$ triangle $ABD$ is equilateral, so $BD$ equals a side

If the diagonals meet at $O,$ then $\triangle ABO$ is a $30^\circ$–$60^\circ$–$90^\circ$ triangle; compare the long and short diagonals

Let the diagonals of $ABCD$ meet at $O.$ They bisect each other at right angles, and since $\angle BAD=60^\circ,$ triangle $ABO$ is a $30^\circ$–$60^\circ$–$90^\circ$ triangle. Hence the half-diagonals satisfy $AO=\sqrt3\,BO.$

The segment $BD$ is the short diagonal of $ABCD$ and the long diagonal of the similar rhombus $BFDE.$ Thus the smaller-to-larger length ratio is $\frac{BD}{AC}=\frac{BO}{AO}=\frac{1}{\sqrt3}.$ Areas scale by the square of this ratio, so the area of $BFDE$ is $24\cdot\dfrac13=8.$

Thus, the correct answer is C.