Shearing a parallelogram back to a rectangle doesn't change the relative sizes of any of these areas.
Compressing the rectangle to a square doesn't change the relative sizes of any of these areas.
So we just need to prove this for a square. For further convenience
A=(0,6)
B=(0,0)
C=(6,0)
D=(6,6)
P=(3p,0)
so that
K=(p,2)
L=(p+2,4)
M=(p+4,2)
BCMK is a trapezoid so
[BCMK] = .5*2*(6+4) = 10
[AKL] + [DLM] = [AKMD]  [KLM]  [ADL]
= .5*4*(6+4)  .5*2*4  .5*6*2
= 20  4  6 = 10
This completes the proof. There are other correspondences between areas as can be seen from the above. The last two triangles areas also sum to
[ABK] + [DCM]
=.5*6*p + .5*6*(2p)
=3p + 63p = 6 = [ADL]

Posted by Jer
on 20150529 09:00:05 