If E, F G and H are respectively themidpoints of the sides AB, BC, CD and ADof a parallelogram ABCD, show thatar(EFGH) 1ar(ABCD)2=.
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Let us join HF
In parallelogram ABCD,
AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)
AB = CD (Opposite sides of a parallelogram are equal)
AD = BC
And AH || BF
AH = BF and AH || BF (H and F are the mid-points of AD and BC)
Therefore, ABFH is a parallelogram
Since,
ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF
Therefore,
Area of triangle HEF = × Area (ABFH) (i)
Similarly,
It can also be proved that
Area of triangle HGF = × Area (HDCF) (ii)
On adding (i) and (ii), we get
Area of triangle HEF + Area of triangle HGF = Area (ABFH) + Area (HDCF)
Area (EFGH)= [Area (ABFH) + Area (HDCF)]
Area (EFGH) = Area (ABCD)
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