Question

In the quadrilateral (1) given below, $AB||DC$, $E$ and $F$ are mid point of $AD$ and $BD$ respectively. Prove that
$EG=\frac{1}{2}(AB+DC)$

Answer 1

Given:- $AB\parallel DC,E$ and $F$ are mid-points of $AD$ and $BD$ respectively.

To prove:- $EG=\frac{1}{2}(AB+DC)$

Proof:-

In $△ABD$,

$DF=BF(\because \text{F is the mid-point of BD})$

Also, $E$ is the mid-point of $AD\left(\text{Given}\right)$

Therefore,

$EF\parallel AB$ and $EF=\frac{1}{2}AB.....\left(1\right)$

$\Rightarrow EG\parallel CD(\because AB\parallel CD)$

Now,

$F$ is the mid-point of $BD$ and $FG\parallel DC$

$\therefore G$ is the mid-point of $BC$

$\Rightarrow FG=\frac{1}{2}CD.....\left(2\right)$

Adding equation $\left(1\right)\&\left(2\right)$, we have

$EF+FG=\frac{1}{2}AB+\frac{1}{2}DC$

$\Rightarrow EG=\frac{1}{2}(AB+CD)$

Hence proved.