martes, 28 de febrero de 2017

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...

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