Question

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

Answer 1

In parallelogram ABCD, let us join HF.
AD = BC and AD|| BC (Opposite sides of a parallelogram are equal and parallel)
AB = CD (Opposite sides of a parallelogram are equal)
$\Rightarrow \frac{1}{2}AD=\frac{1}{2}BCandAH||BF$
$\Rightarrow$ AH = BF and AH || BF (since H and F are the mid-points of AD and BC)
Therefore, ABFH is a parallelogram.
$\Delta HEF$ and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,
$\therefore \left(\Delta HEF\right)=\frac{1}{2}ar\left(ABFH\right)...\left(1\right)$

Similarly , it can be proved that;
$\left(\Delta HGF\right)=\frac{1}{2}ar\left(HDCF\right)....\left(2\right)$

On adding equations (1) and (2), we obtain;
$ar\left(\Delta HEF\right)+ar\left(\Delta HGF\right)=\frac{1}{2}ar\left(ABFH\right)+\frac{1}{2}ar\left(HDCF\right)$ $ar\left(EFGH\right)=\frac{1}{2}\left[ar\right(ABFH)+ar(HDCF\left)\right]$
$\Rightarrow ar\left(EFGH\right)=\frac{1}{2}ar\left(ABCD\right)$