Question 1064274
<pre><b><font size=4>

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:

{{{drawing(400,800/3,-2,10,-2,6,

line(-3,0,12,0),line(0,-3,0,8), 

red(line(0,0,8,0),line(8,0,8,4), line(8,4,0,4), line(0,4,0,0)),

green(line(4,0,0,2),line(0,2,4,4), line(4,4,8,2), line(8,2,4,0)),

locate(.1,0,"A(0,0)"), locate(3,0,E(matrix(1,3,L/2,",",0))),
locate(7.3,0,"B(L,0)"), locate(8.1,2.5,F(matrix(1,3,L,",",W/2))),
locate(7.3,4.5,"C(L,W)"), locate(3,5.1,G(matrix(1,3,L/2,",",W))),
locate(.1,4.5,"D(0,W)"),locate(-1.9,2.5,H(matrix(1,3,0,",",W/2))),
locate(10,0,x),locate(.1,6,y) 

   )}}}

We us the distance formula to show that all the sides of EFGH have
the same length:

{{{matrix(1,5,EF,""="",sqrt((L-L/2)^2+(W/2-0)^2),""="",sqrt((L/2)^2+(W/2)^2))}}}

{{{matrix(1,7,FG,""="",sqrt((L/2-L)^2+(W-W/2)^2),""="",sqrt((-L/2)^2+(W/2)^2),""="",sqrt((L/2)^2+(W/2)^2))}}}

{{{matrix(1,7,GH,""="",sqrt((0-L/2)^2+(W/2-W)^2),""="",sqrt((-L/2)^2+(-W/2)^2),""="",sqrt((L/2)^2+(W/2)^2))}}}

{{{matrix(1,5,HE,""="",sqrt((L/2-0)^2+(W/2-0)^2),""="",sqrt((L/2)^2+(W/2)^2))}}}

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

{{{matrix(1,7,sqrt((L/2)^2+(W/2)^2),""="",sqrt(L^2/4+W^2/4),""="",sqrt((L^2+W^2)/4),""="",sqrt(L^2+W^2)/2
)}}}

Edwin</pre></b></font>