Question

$\text{In a triangle ABC, E is the mid-point of median AD. Show that ar(BED)}=\frac{1}{4}ar\left(ABC\right).$

Answer 1

$ar\left(BED\right)=\frac{1}{2}BD\times DE$
Since, E is the mid-point of AD, AE = DE
Since, AD is the median on side BC of triangle ABC,
BD = DC,
$DE=\frac{1}{2}AD—\left(i\right)$
$BD=\frac{1}{2}BC—\left(ii\right)$
From (i) and (ii), we get,
$ar\left(BED\right)=\frac{1}{2}\times \frac{1}{2}\times BC\times \frac{1}{2}AD$
$\Rightarrow ar\left(BED\right)=\frac{1}{2}\times \frac{1}{2}ar\left(ABC\right)$
$\Rightarrow ar\left(BED\right)=\frac{1}{2}\times \frac{1}{2}ar\left(ABC\right)$
$\Rightarrow ar\left(BED\right)=\frac{1}{4}ar\left(ABC\right)$