Question

$ABCD$ is a quadrilateral $E$ and $F$ are mid points of $DC$ and $AB$ respective. Prove that area of area of EFHG=$\frac{1}{2}$ABCD

Answer 1

REF.Image.

Join AC and BD,

In $△ADC$ ,G is the mid point of AD and

E is the mid point of DC. so by mid pt. theorem

$GE\parallel AC$ and $GE=\frac{1}{2}AC...\left(1\right)$

Similarly, $FH\parallel AC$ and $FH=\frac{1}{2}AC...\left(2\right)$

By (1) & (2),

$GE\parallel FH$ and $GE=FH$

As we know, opposite sides of quad. are equal

and parallel, so EHFG is a parallelogram

and (area of EHFG) = $\frac{1}{2}$(area of ABCD)

from above definition.