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

(N/A) In parallelogram $ABCD$,$E, F, G$ and $H$ are the midpoints of $AB, BC, CD$ and $DA$ respectively. Draw $GE$.
In parallelogram $ABCD$,$AB \parallel CD$ and $AB = CD$.
$\therefore BE \parallel CG$ and $BE = (\frac{1}{2} AB) = CG = (\frac{1}{2} CD)$.
$\therefore$ Quadrilateral $EBCG$ is a parallelogram.
$\therefore GE \parallel BC$.
Now,$\Delta EFG$ and parallelogram $EBCG$ are on the same base $GE$ and between the same parallels $GE$ and $BC$.
$\therefore ar(EFG) = \frac{1}{2} ar(EBCG)$ ... $(1)$
Similarly,$\Delta EHG$ and parallelogram $AEGD$ are on the same base $GE$ and between the same parallels $GE$ and $AD$.
$\therefore ar(EHG) = \frac{1}{2} ar(AEGD)$ ... $(2)$
Adding $(1)$ and $(2)$,
$ar(EFG) + ar(EHG) = \frac{1}{2} ar(EBCG) + \frac{1}{2} ar(AEGD)$
$\therefore ar(EFGH) = \frac{1}{2} [ar(EBCG) + ar(AEGD)]$
$\therefore ar(EFGH) = \frac{1}{2} ar(ABCD)$