Question

$E,F,G$ and $H$ are the midpoints of the sides of a parallelogram $ABCD.$ SHow that area of quadrilateral $EFGH$ is half of the area of parallelogram $ABCD.$

Answer 1

Also given that ,
$E,F,G$ and $H$ are respectively the midpoints of the sides $AB,BC,CD$ and $AD$ of a ||gm $ABCD.$
To prove: $ar\left(EFGH\right)=\frac{1}{2}ar\left(ABCD\right).$
Proof:
Join $FH,$ such that $FH\parallel AB\parallel CD.$
Since, $FH\parallel AB$ and $AH\parallel BF.$
So, $ABFH$ is a parallelogram.
$△EFH$ and parallelogram $ABFH$ are on same base $FH$ and between the same parallel lines.
Hence, $ar\left(△EFH\right)=\frac{1}{2}ar\left(parallelogramABFH\right).....\left(1\right).$
Since, $FH\parallel CD$ and $DH\parallel FC.$
$DCFH$ is a parallelogram.
we Know that,
$△FGH$ and parallelogram $DCFH$ are on same base $FH$ and between the same parallel lines.
$ar\left(△FGH\right)=\frac{1}{2}ar\left(parallelogramDCFH\right).....\left(2\right).$
Adding equation (1) and (2), we get
$ar\left(△EFH\right)+ar\left(△FGH\right)=\frac{1}{2}ar\left(parallelogramABFH\right)+\frac{1}{2}ar\left(parallelogramDCFH\right).$
$ar\left(△EFH\right)+ar\left(△FGH\right)=\frac{1}{2}\left[ar\right(parallelogramABFH)+\frac{1}{2}ar(parallelogramDCFH\left)\right].$
$ar\left(EFGH\right)=\frac{1}{2}ar\left(ABCD\right).$
Hence proved.