[math]\overline{BC}\cong\overline{AD}[/math] and [math]\overline{FC}\cong\overline{ED}[/math], because the opposite sides of parallelograms are congruent.[br][br]Also, [math]\overline{BA}\cong\overline{CD}[/math] and [math]\overline{FE}\cong\overline{CD}[/math], because the opposite sides of parallelograms are congruent.[br][br]By the transitive property of congruence, [math]\overline{BA}\cong\overline{FE}[/math][br][br]Therefore, by the SSS Theorem of Congruence, [math]\triangle{FBC}\cong\triangle{EAD}[/math].

Since [math]\triangle{FBC}\cong\triangle{EAD}[/math], they have the same area. But then we can subtract the red triangle from both triangles and we are left with the green trapezoids having the same area.

Since the green trapezoids have the same area, we can add the same blue triangle to both and we will get equal areas again, but each green trapezoid combined with the blue triangle gives one of our desired parallelograms.

The two green triangle have the same area, so adding each to the blue trapezoid will result in two figures with equal area, which are the two parallelograms we are interested in.