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

(N/A) Let us join $E$ and $G$.
If a triangle and a parallelogram are on the same base and between the same parallels,then the area of the triangle is equal to half the area of the parallelogram.
Since $E$ and $G$ are the mid-points of $AB$ and $CD$ respectively,$EG$ is parallel to $BC$ and $AD$.
Also,$\text{ar}(\text{parallelogram } EBCG) = \text{ar}(\text{parallelogram } AEGD) = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD) \dots (1)$
Now,$\Delta EFG$ and parallelogram $EBCG$ are on the same base $EG$,and between the same parallels $EG$ and $BC$.
Therefore,$\text{ar}(\Delta EFG) = \frac{1}{2} \text{ar}(\text{parallelogram } EBCG) \dots (2)$
Similarly,$\text{ar}(\Delta EHG) = \frac{1}{2} \text{ar}(\text{parallelogram } AEGD) \dots (3)$
Adding $(2)$ and $(3)$,we get:
$\text{ar}(\Delta EFG) + \text{ar}(\Delta EHG) = \frac{1}{2} [\text{ar}(\text{parallelogram } EBCG) + \text{ar}(\text{parallelogram } AEGD)]$
$\Rightarrow \text{ar}(EFGH) = \frac{1}{2} [\text{ar}(\text{parallelogram } ABCD)]$
Thus,$\text{ar}(EFGH) = \frac{1}{2} \text{ar}(ABCD)$.