If E, F, G and H are respectively the midpoints of the sides of the parallelogram ABCD, show that area of E F G H is equal to area of half ABCD

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Solution

$A H = \frac{A D}{2}, B F = \frac{B C}{2}$
$A B C D$ is a parallelogram
Parallelogram $F H A B$ and $△ F H E$ are on the same base and between the same parallel lines
So $\frac{1}{2} A r (F H A B) = A r (△ F H E)$
Similarly $\frac{1}{2} A r (F H D C) = A r (△ F H G)$
Adding these two equations
$A r (△ F H E) + A r (△ F H G) = \frac{1}{2} [A r (F H A B) + A r (F H D C)]$
$A r (F G H E) = \frac{1}{2} A r (A B C D)$

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If E, F, G and H are respectively the midpoints of the sides of the parallelogram ABCD, show that area of E F G H is equal to area of half ABCD

AH=AD2,BF=BC2ABCD is a parallelogramParallelogram FHAB and △FHE are on the same base and between the same parallel lines So 12Ar(FHAB)=Ar(△FHE)Similarly 12Ar(FHDC)=Ar(△FHG)Adding these two equationsAr(△FHE)+Ar(△FHG)=12[Ar(FHAB)+Ar(FHDC)]Ar(FGHE)=12Ar(ABCD)