Trapezoid Riddle III

If we draw the extending lines, parallel to segments $AD$ and $DC$ , to meet at point $G$ , as shown above, we will create a parallelogram $AGCD$ . Then if we draw $BF$ parallel to $AE$ , it is also obvious that $ABFE$ is another parallelogram, thus making the areas of triangles $ABE$ = $BFE$ because $BE$ is the diagonal of the parallelogram.

Thereby, since $E$ is the midpoint of $AD$ , the segments $BF = HC$ and $BI = IC$ because of parallel rules. Moreover, since $\angle BIF = \angle HIC$ for the intersection angles, the triangles $BIF$ and $HIC$ are congruent, thus having the same area.

As a result, we can virtually demonstrate the area equality as two bisected parallelograms, as shown below: