Areas in a rectangle Suppose $P,Q, R$, and $S$ are the midpoints of the sides $AB, BC, CD$, and $DA$, respectively of rectangle $ABCD$. If the area of the rectangle is $\delta$, then prove that the area of the figure bounded by the straight lines $AQ, BR, CS$, and $DP$ is $\frac{\delta}{5}$.
I began by imposing a coordinate system but couldn't find a way to relate the given area with the area of the rectangle.
How should I begin?

Without loss of generality, make $ABCD$ a square with $A(0,0)\ B(0,2) \ C(2,2)\ D(2,0)$.
Using the lines $y = -2x + 2,y= -2x+4, y=x/2,y= x/2 + 1$. Two of these intersections are $(\frac{2}{5}, \frac{6}{5}), (\frac{4}{5},\frac{2}{5})$. The distance between them is a side of the small square and is $\frac{2}{\sqrt{5}}$. The area of the small square is $\frac{4}{5}$, divided by the total area of $4$ gives you a $\frac{1}{5}$ area. Transformations to a rectangle preserve the areas because everything is proportional.
