### Proof

We overlap the square to the rectangle and trace straight lines FC, ED and BG. Our main objective is to prove the congruence of the pairs of triangles GDJ and BHE, and DCH and JFE.
As we are supposing that the base of the rectangle is less than the double of the base of the square, straight line ED intercepts the side of the square in the interior of the rectangle.
The triangles on which we want to show congruences, are all right triangles. Visually we have in the figure different parallelograms, which prove the congruences between legs of the triangles. Thus it is enough to show that in fact these are parallelograms.

From the equality of areas we have AB*AD = AE*AG, and therefore AD/AG=AE/AB. From this we conclude that straight lines BG and ED are parallel segments, and therefore BEJG and GBHD are parallelograms, using the proportional sides here with the characterization of parallelogram in terms of parallel opposing sides.
Also we have that FJDC is parallelogram. In fact: since BEJG is a parallelogram, we have congruent sides EB and JG, and as EB is congruent to the FI it follows, from transitivity, that FI is congruent to the JG. We have then FJ congruent to the IG, and consequently, congruent to CD, from which it allows that FJDC also is parallelogram, using here the characterization of parallelogram in terms of pair of parallel and congruentes opposing sides.
Since FJDC is a parallelogram we conclude that straight lines ED and FC are parallel segments, and therefore EFCH also is parallelogram, since it too has a pair of parallel opposing sides.
From the existence of these four parallelograms we conclude that the segments are congruent:
* GD and BH, BE and GJ, and therefore triangles GDJ and BHE are congruent ;
* DC and JF, CH and FE, and therefore triangles DCH and JFE are congruent . Return to Rectangle transformed to Square