If E,F,G, and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = $\frac{1}{2}$ ar (ABCD).

If a triangle and parallelogram are on the same base and between the same parallel lines, then the area of the triangle will be half of the area of a parallelogram.

We can observe in the figure shown below that triangle GHF and parallelogram DHFC are on the same base and between parallel lines.

Similarly, triangle EHF and Parallelogram AHFB are on the same base and between parallel lines. By using these results, we can find areas and add them to get the required result.

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)

⇒ 1/2 AD = 1/2 BC and AH || BF

⇒ AH = BF and AH || BF (Since H and F are the mid-points of AD and BC)

Therefore, ABFH is a parallelogram.

Since ∆HEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,

∴ Area (ΔHEF) = 1/2 Area (ABFH)   .....(1)

Similarly, Area (ΔHGF) = 1/2 Area (HDCF)     .....(2)

On adding Equations (1) and (2), we obtain

Area (ΔHEF) + Area (ΔHGF) = 1/2 Area (ABFH) + 1/2 Area (HDCF) ...... (3)

From the figure, we can see that,

Area (ΔHEF) + Area (ΔHGF) = Area (EFGH)  ....... (4)

1/2 Area (ABFH) + 1/2 Area (HDCF) = 1/2 Area (ABCD)   ..... (5)

From (3) & (4) and (5) we have,

Area (EFGH) = 1/2 Area (ABCD)