Let the rectangle be ABCD with length L and width W,
with midpoints of sides E,F,G,H. We are to show that
EFGH is a rhombus.
We may then place the figure on a graph like this with
the coordinates as shown:
We us the distance formula to show that all the sides of EFGH have
the same length:
Since EF = FG = GH = HE, EFGH is a rhombus.
I didn't bother simplifying the length of the sides of rhombus EFGH,
but if you want to, it's
Edwin