Dear Alex,
I recommend two proofs for your variant.
W = intersection of DR and AB

1. Proof used my variant 1)
As my variant:
(XNCP) - (AMXQ) = 2*(BXD) = (XBYD) = (XRWB) = (XRSM)
From this:
(XNCP) = (AMXQ) + (XRSM) = (ASRQ)

2. Proof used Euclid I.43
A line from W parallel with BC intersects NQ at E, CD at F.
K is intersection of RS and CD
RE = BW - NR = XR - NR = XN there for two paralellograms CNXP and FERK are congruent.
By Euclid I.43: (ASRQ) = (FERK) = (XNCP)

Also please check to see some typo mistakes in your new page:
http://www.cut-the-knot.org/Curriculum/Geometry/EuclidI43Extended.shtml

In the first line:
"Through point X inside the parallelogram ABCD draw lines MP||CD and NQ||AB."
I think MP is parallel with BC.

At the end of the page:
"This means (and follows from the fact) that the point of intersection of RS and UV (not shown) lies on the line AX, the diagonal of parallelogram ASXR."
I think there is one typo mistake because ASXR is not a parallelogram.

Best regards,
Bui Quang Tuan