Data: E, F, G and H are mid-points of the sides of a parallelogram ABCD,
To Prove: area (EFGH) = \(\frac{1}{2}\)area (ABCD)
Construction: HF is joined.
Proof:
Now, AD = BC and AD || BC
∴ 2AH = 2BF
∴ AH = BF and AH || BF
∴ AHFB is a parallelogram.
Similarly, HDCF is a parallelogram.
Now, DEHF and Quadrilateral AHFD are on base HF and in between HF || AB.