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# 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,
( 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.