HMMT November

First, note that $B D$ is the long diagonal of $A B C D$, and $A C$ is the long diagonal of $E C F A$. Because the ratio of the areas of similar figures is equal to the square of the ratio of their side lengths, we know that the ratio of the area of $E C F A$ to the area of $A B C D$ is equal to the ratio $\frac{A C^{2}}{B D^{2}}$. Using law of cosines on triangle $A B D$, we have $B D^{2}=A D^{2}+A B^{2}-2(A D)(A B) \cos \left(120^{\circ}\right)=2^{2}+7^{2}-2(2)(7)\left(-\frac{1}{2}\right)=67$. Using law of cosines on triangle $A B C$, we have $A C^{2}=A B^{2}+B C^{2}-2(A B)(B C) \cos \left(60^{\circ}\right)=7^{2}+2^{2}-2(7)(2)\left(\frac{1}{2}\right)=39$. Finally, $\frac{A C^{2}}{B D^{2}}=\frac{39}{67}$.