Question

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

Answer 1

Let $ABC$ be a triangle and $AD$ is the median of $\Delta ABC$

$E$ is the mid point of $AD$.

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

In $\Delta ABC,$

$ar\left(ABD\right)=ar\left(ACD\right)$ ___________ (1)

In $\Delta ABD,BE$ is the median

$ar\left(ABE\right)=ar\left(BED\right)$ __________ (2)

Now, $ar\left(ABD\right)=ar\left(BED\right)$

$=2.ar\left(BED\right)$______ (3)

$ar\left(ABC\right)=ar\left(ABD\right)+ar\left(ACD\right)$

$ar\left(ABD\right)=2.ar\left(ABD\right)$ using (1)

$ar\left(ABC\right)=\mathrm{2.2.}ar\left(BED\right)$ - using (2)

$ar\left(ABC\right)=4.ar\left(BED\right)$

$\therefore ar\left(BED\right)=\frac{1}{4}ar\left(ABC\right)$

Hence it is proved.