In figure below, $D$ and $ \mathrm{E} $ are two points on $ \mathrm{BC} $ such that $ \mathrm{BD}=\mathrm{DE}=\mathrm{EC} $. Show that $ \operatorname{ar}(\mathrm{ABD})=\operatorname{ar}(\mathrm{ADE})=\operatorname{ar}(\mathrm{AEC}) $.
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Given:

$D$ and \( \mathrm{E} \) are two points on \( \mathrm{BC} \) such that \( \mathrm{BD}=\mathrm{DE}=\mathrm{EC} \).

To do:

We have to show that \( \operatorname{ar}(\mathrm{ABD})=\operatorname{ar}(\mathrm{ADE})=\operatorname{ar}(\mathrm{AEC}) \).

Solution:

In $\triangle ABE$

$BD=DE$

This implies,

$AD$ is the median.

We know that,

The median of a triangle divides it into two parts of equal areas.

This implies,

$ar(\triangle ABD) = ar(\triangle AED)$.........(i)

In $\triangle ADC$,

$DE=EC$

$AE$ is the median

This implies,

$ar(\triangle ADE) = ar(\triangle AEC)$..........(ii)

From (i) and (ii), we get,

$ar(\triangle ABD) = ar(\triangle ADE) = ar(\triangle AEC)$

Hence proved.

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Updated on: 10-Oct-2022

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