Nina Guindilla explained it:
If A, B, C, D are the vertices of the quadrilateral and E, F, G, H are its side midpoints, and if O is the origin, then the vector
EF = OF − OE = (OC+OB)/2 − (OA+OB)/2 =
= OC/2 − OA/2 =
= (OC+OD)/2 − (OA+OD)/2 = OG − OH = HG
That is, EF and HG are equipollent and EFGH is a parallelogram.
Pepe Chapuzas did it differently:
AC and BF are the diagonals of the quadrilateral and we have the following pairs of similar triangles:
BEF and BAC
DAC and DHG
AEH and ABD
CBD and CFGSo, we have the parallel segments
EF || AC || HG
EH || BD || FGSo, EFGH is a parallelogram.
The area of the parallelogram is obviously half the area of the quadrilateral if this is convex. If not, the following drawing helps to understand it: