Given that □ABCD is a parallelogram.
E, F, G and H are the midpoints of the sides. Join E, G.
Now ΔEFG and □EBCG he on the same base EG and between the same parallels
EG // BC. ∴ ΔEFG = 1/2 □EBCG ……………(1)
Similarly,
ΔEHG = 1/2 □EGDA …………….(2)
Adding (1) and (2);
ΔEFG + ΔEHG = 1/2 □EBCG + 1/2 □EGDA
□EFGH = 1/2 [□EBCG +□ EGDA]
□EFGH = 1/2 [□ABCD]
Hence proved.