Question

Prove that the line segments joining the mid-point of a pair of opposite sides of a parallelogram divided it into two equal $//g{m}^{\prime }s$.

Answer 1

Let us consider $ABCD$ be a parallelogram in which $E$ and $F$ are mid-points of $AB$ and CD. Join EF.
To prove: ar $(\parallel AEFD)=\mathrm{ar}(\parallel EBCF)$
Let us construct DG $\perp AG$ and let $DG=h$ where, $h$ is the altitude on side $AB$.
Proof: $\mathrm{ar}(\parallel ABCD)=AB\times h$
$\mathrm{ar}(\parallel AEFD)=AE\times h$
$=1/2AB\times h\dots \dots \left(1\right)\left[\text{since,}E\text{is the mid-point of}AB\right]$
$\mathrm{ar}(\parallel EBCF)=EF\times h$
$=1/2AB\times h\dots \dots \left(2\right)\left[\text{since,}E\text{is the mid-point of}AB\right]$
From ( 1 ) and ( 2 ) $\mathrm{ar}(\parallel ABFD)=\mathrm{ar}(\parallel EBCF)$
Hence proved.