> I think there is a simpler dissection solution to the problem. > Using the figure you posted as reference, we observe that DCG > is congruent to EMF. Hence area(DEFG) = area(DEMC). > Since length(AB) = length(EM), it follows > > area(ABCD) = area(DEMC), > > therefore S2 = S1. > So we need only the additional point M for the proof.
I agree that nothing else is required for the *proof*.
The other points and lines on my drawing are just the edges of the 4 necessary pieces (blue, green, yellow, pink) in there successive positions, when dissecting : rectangle ABCD -> parallelogram EMCD -> rectangle DEFG by 1) translation of triangle DAE -> CBM 2) then translation of triangle EFM -> DGC (of course the same dissection might be in reverse order DEFG -> ABCD)
PS. I finally found my old login at Drexel, so I can answer directly here. There seems to be a problem with crossfeed of Drexel to usenet. (geometry.puzzles is *declared* as a public usenet group. This seeems then to be false...)