Question

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

Answer 1

Given that:- $ABC$ is a triangle with $AD$ as mdian, i.e., $BD=CD$ and $E$ is the mid-point of $AD$, i.e., $AE=DE$

To prove:- $ar\left(△BED\right)=\frac{1}{4}ar\left(△ABC\right)$

Proof:-

As we know that median divides a triangle into wo triangles of equal area.

$\therefore ar\left(△ABD\right)=ar\left(△ACD\right)$

$\Rightarrow ar\left(△ABD\right)=\frac{1}{2}ar\left(△ABC\right).....\left(1\right)$

Now in $△ABD$

$BE$ is the median $[\because \text{E is mid-point of AD}]$

$\therefore ar\left(△BED\right)=ar\left(△BEA\right)$

$\Rightarrow ar\left(△BED\right)=\frac{1}{2}ar\left(△ABD\right)$

$\Rightarrow ar\left(△BED\right)=\frac{1}{4}ar\left(△ABC\right)[\because ar\left(△ABD\right)=\frac{1}{2}ar\left(△ABC\right)]$

Hence proved.