Question

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

Answer 1

$ABCD$ a parallelogram in which $E,F,G,H$ are mid points of $AB,BC,CD,AD$

Construction: Draw $HF$ parallel to $AB$ and $C$

$AB$ is parallel and equal to $HF$ . Therefore, $ABFH$ is a parallelogram

Since, $△EFH$ and parallelogram $ABFH$ lies on the same base $HF$ and between same parallels $AB$ and $HF$

$\therefore area\left(\Delta EFH\right)=\frac{1}{2}area\left(ABHF\right)$...............(1)

Now, $DC$ is parallel and equal to $HF$. Therefore, $DCFH$ is a parallelogram.

Since, triangle $GFH$ and parallelogram $DCFH$ lies on the same base

$HF$ and between same parallels $DC$ and $HF$.

$\therefore area\left(\Delta GFH\right)=\frac{1}{2}area\left(DCHF\right)$...................(2)

From $1$ and $2$ we get,

$area\left(\Delta EFH\right)+area\left(\Delta GFH\right)=\frac{1}{2}area\left(ABFH\right)+\frac{1}{2}area\left(DCFH\right)$

$\Rightarrow area\left(EFGH\right)=\frac{1}{2}\left(area\right(ABFH)+area(DCFH\left)\right)$

$\Rightarrow area\left(EFGH\right)=\frac{1}{2}area\left(ABCD\right)$