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

Given: E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD


Construction: Let us join HF.

To prove: ar(EFGH) = \(\frac{1}{2} \) ar(ABCD)

Proof: In parallelogram ABCD,
AD = BC and AD || BC
( Opposite sides of a parallelogram are equal and parallel)
Multiplying both sides by \(\frac{1}{2} \) we get,
\(\Rightarrow \) \(\frac{1}{2} \) x AD = \(\frac{1}{2} \) x BC
\(\Rightarrow \) AH = BF and AH || BF
\(\therefore \) ABFH is a parallelogram.

Since, \(\triangle{HEF}\) and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,
\(\therefore \) Area (\(\triangle{HEF}\)) = \(\frac{1}{2} \) Area (ABFH) ...(i)
Similarly, it can be proved that,

Area (\(\triangle{HGF}\)) = \(\Rightarrow \) Area (HDCF) ...(ii)
On adding equations (i) and (ii), we get,
\(\Rightarrow \) Area (\(\triangle{HEF}\)) + Area (\(\triangle{HGF}\)) = \(\frac{1}{2} \) Area (ABFH) + \(\frac{1}{2} \) Area (HDCF)
\(\Rightarrow \) Area [(\(\triangle{HEF}\)) + (\(\triangle{HGF}\))] = [Area (ABFH) + Area (HDCF)]
\(\therefore\) Area (EFGH) = \(\frac{1}{2} \) Area (ABCD)
Hence, proved.