1261. Side midpoints...

    Varignon's theorem states that the side midpoints of any quadrilateral are vertices of a parallelogram, and the area of this parallelogram is half the area of that quadrilateral...

    Nina Guindilla explained it:

     Dear Teacher:

    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:

    Dear Teacher:

    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 CFG

    So, we have the parallel segments

EF || AC || HG

EH || BD || FG

    So, 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:

    Furthermore, if the initial quadrilateral is a kite, a dart or a rhombus then the parallelogram is a rectangle. If the initial quadrilateral is an isosceles trapezoid or a rectangle then the parallelogram is a rhombus... If the quadrilateral is a square then the parallelogram is another square...