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. |