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Demonstration that the rectangle ABCD is scissors congruent to the square
AEFG.

We overlap the square to the rectangle and trace straight lines FC, ED
and BG. Our main objective is to demonstate 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 twice
the base of the square, straight line ED intercepts the side of the square
in the interior of the rectangle.

The triangles which we want to show congruent, are all of the right
triangles. Visually we have in the figure diverse parallelograms, which
go in guaranteeing congruences to them between legs of the triangles. Thus
it is enough to show that these parallelograms exist.
Of 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
here the characterization of parallelogram in terms of parallel opposite
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 FI it follows, from transitivity, that FI is congruent
to JG. We have then FJ congruent to IG, and consequently,
is congruent to CD, which allows one to conclude that **FJDC also
is parallelogram **, using here the characterization of parallelogram
in terms of pair of parallel and congruent opposite sides. |

Since FJDC is a parallelogram we conclude that straight lines ED and FC are
parallel segments, and therefore **EFCH also is parallelogram **having pairs
of parallel opposite 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 DCH
and JFE are congruent triangles.