Let $ABCD$ be a rectangle with sides $AB,BC,CD$ and $DA$. Let $K,L$ be the midpoints of the sides $BC,DA$ respectivily. The perpendicular from $B$ to $AK$ hits $CL$ at $M$. Find
$$\frac{[ABKM]}{[ABCL]}$$
Let the entire area of the rectangle be $A$ (in the sequel a lot of areas will be just written since it is extremely easy to prove them). Then $[ABKM] = [ABK] + [AKM] = \frac{A}{4} + \frac{1}{2}[AKCL] = \frac{A}{4} + \frac{A}{4} = \frac{A}{2}$. Also $[ABCL] = \frac{3A}{4}$ and thus the required ratio is $2/3$.