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.
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.
||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.
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 .
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