Question 88
The rhombus $AKLM$ is inscribed in the triangle $ABC$, so that point $K$ is on $\overline{AB}$, point $L$ is on $\overline{BC}$ and point $M$ is on $\overline{CA}$. If the rhombus has side of length $2\sqrt{2}$, the area of triangle $LMC$ is $3$, and the area of triangle $KLB$ is $4$, prove that $\angle BAC = 60^\circ$. (Mea Bombardelli)