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?
 A: This problem deserves a proof without words:

A: 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.
A: see this picture.
$
this is the picture of problem.
we know the medians in a triangle are ... in $ G $ !!!
$ \dfrac{GM}{GA'}=\dfrac{1}{3} $ ( these points are not the points of our problem! )
so $ BM=MN=ND $
also $ SA=SD $ and $ SD' \parallel AA' $
so $ D'D=D'A $ 
thus $ S_{A'B'C'D'} = 2.S_{A'MND'} = 2*3.S_{DD'N} $
so you need $ S_{DD'N} $
if you need the compelte solution say please!

A: Assume $A$ at origin and vectors $\vec B$ and $\vec C$. Now by writing a point on a line,say, $BC$, you can proceed by : $\vec B+\lambda(\vec{C}-\vec{B})$. Similarly, you can do it for other lines and find points of intersection in terms of $\vec B,\vec C$. Now you have all points, I assume know what to do... 
