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

Given:

In a triangle $\mathrm{ABC}, \mathrm{E}$ is the mid-point of median AD.

To do:

We have to show that $\operatorname{ar}(\mathrm{BED})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC})$.
Solution:

We know that,

A median divides a triangle into two triangles of equal areas.

This implies,

$\operatorname{ar}(\triangle \mathrm{ABD})=\operatorname{ar}(\triangle \mathrm{ADC})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABC})$............(i)

In $\triangle \mathrm{ABD}$

$\mathrm{BE}$ is the median.

This implies,

$\operatorname{ar}(\triangle \mathrm{BED})=\operatorname{ar}(\triangle \mathrm{BAE})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABD})$

$\Rightarrow \operatorname{ar}(\triangle \mathrm{BED})= \frac{1}{2}[\frac{1}{2}ar(\triangle \mathrm{ABC})]$       [From (i)]

$\operatorname{ar}(\triangle \mathrm{BED})=\frac{1}{4} \operatorname{ar}(\triangle \mathrm{ABC})$

Hence proved.

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Updated on: 10-Oct-2022

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