If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that ar (EFGH) = 1/2ar (ABCD)
Let us join HF.
In parallelogram ABCD,
AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)
AB = CD (Opposite sides of a parallelogram are equal)
⇒ 1/2AD = 1/2BC and AH || BF
⇒ AH = BF and AH || BF (∵ H and F are the mid-points of AD and BC)
Therefore, ABFH is a parallelogram.
Since ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,
∴ Area (ΔHEF) = 1/2Area (ABFH) ... (1)
Similarly, it can be proved that
Area (ΔHGF) = 1/2Area (HDCF) ... (2)
On adding equations (1) and (2), we obtain
Area(ΔHEF) + Area(ΔHGF) = 1/2Area (ABFH) + 1/2Area (HDCF)
= 1/2[Area (ABFH) + Area (HDCF)]
⇒ Area(EFGH) = 1/2Area(ABCD)
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- Theorem: Parallelograms on the Same Base and Between the Same Parallels.