HMMT-Feb 2014 geo, Problem 6

In quadrilateral $A B C D$, we have $A B=5, B C=6, C D=5, D A=4$, and $\angle A B C=90^{\circ}$. Let $A C$ and $B D$ meet at $E$. Compute $\frac{B E}{E D}$.

Solution: We find that $A C=\sqrt{61}$, and applying the law of cosines to triangle $A C D$ tells us that $\angle A D C=120$. Then $\frac{B E}{E D}$ is the ratio of the areas of triangles $A B C$ and $A D C$, which is $\frac{(5)(6)}{(4)(5) \frac{\sqrt{3}}{2}}=\sqrt{3}$. $\fbox{\sqrt{3}}$.

Answer: \sqrt{3}