Question

If E,F,G and H are the mid-points of the area of a parallelogram ABCD, show that are(EFGH)=1/2 ar(ABCD).

Answer 1

given,
ABCD a parallelogram in which
E,F,G,H, are mid points of AB,BC,CD,AD.
Then to prove,
ar(EFGH) = ar(ABCD)
Construction, Draw HF parallel to AB and CD
Proof:
AB is parallel and equal to HF . Therefore, ABFH is a parallelogram
Since, triangle EFH and parallelogram ABFH lies on the same base HF and between same parallels AB and HF
Therefore, ar(EFH) = 1/2 ar (ABFH) ( 1)

Now, DC is parallel and equal to HF. Therefore,DCFH is a parallelogram
Since, triangle GFH and parallelogram DCFH lies on the same base HF and between same parallels DC and HF.
Therefore, ar(GFH) = 1/2 ar (DCFH) (2)
From 1 and 2 we get,
ar(EFH) + ar (GFH) = 1/2 ar (ABFH) + 1/2 ar (DCFH)
ar (EFGH) = 1/2 ( ar(ABFH) + ar(DCFH))
ar (EFGH) = 1/2 ( ar(ABCD))
ar (EFGH) = 1/2 ar (ABCD)