Question

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

Given:
$E,F,G$ and $H$ are respectively the mid-points of the sides of a parallelogram $ABCD$.

To Prove:
$ar\left(EFGH\right)=\frac{1}{2}ar\left(ABCD\right)$

Construction:
$H$ and $F$ are joined.

Proof:
$AD\parallel BC$ and $AD=BC$ (Opposite sides of a parallelogram)
$\Rightarrow \frac{1}{2}AD=\frac{1}{2}BC$

Also,
$AH\parallel BF$ and and $DH\parallel CF$

$\Rightarrow AH=BF$ and $DH=CF$ $\mid$ $H$ and $F$ are mid points
Thus,

$ABFH$ and $HFCD$ are parallelograms.

Now,

$△EFH$ and ||gm $ABFH$ lie on the same base $FH$ and between the same parallel lines $AB$ and $HF$.
$\therefore$ Area of $EFH=\frac{1}{2}ar\left(ABFH\right)$ --- (i)
Also,

Area of $GHF=\frac{1}{2}ar\left(HFCD\right)$ --- (ii)

Adding (i) and (ii),
Area of $△EFH+$ area of $△GHF$ $=\frac{1}{2}ar\left(ABFH\right)+\frac{1}{2}ar\left(HFCD\right)$
$\Rightarrow$ Area of $EFGH=$ Area of $ABFH$
$\Rightarrow ar\left(EFGH\right)=\frac{1}{2}ar\left(ABCD\right)$