A point $D$ is taken on the side $BC$ of a $\triangle ABC$ such that $BD = 2DC$. Prove that $ar(\triangle ABD) = 2ar(\triangle ADC)$.


Given:

A point $D$ is taken on the side $BC$ of a $\triangle ABC$ such that $BD = 2DC$. 

To do:

We have to prove that $ar(\triangle ABD) = 2ar(\triangle ADC)$.

Solution:

Draw $AL \perp BC$.


$\operatorname{area}(\triangle \mathrm{ABD})=\frac{1}{2} \times$ base $\times$ height

$=\frac{1}{2} \mathrm{BD} \times \mathrm{AL}$

$\operatorname{area}(\triangle \mathrm{ADC})=\frac{1}{2} \mathrm{DC} \times \mathrm{AL}$

$\mathrm{BD}=2 \mathrm{DC}$

$\Rightarrow \mathrm{DC}=\frac{1}{2} \mathrm{BD}$

Therefore,

$\operatorname{area}(\triangle \mathrm{ADC})=\frac{1}{2}\times(\frac{1}{2} \mathrm{BD}) \times \mathrm{AL}$

$=\frac{1}{2}[\frac{1}{2} \mathrm{BD} \times \mathrm{AL}]$

$=\frac{1}{2}[\operatorname{ar}(\Delta \mathrm{ABD})]$

$\operatorname{ar}(\Delta \mathrm{ADC})=2 \mathrm{area}(\triangle \mathrm{ADC})$

Hence proved.

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Updated on: 10-Oct-2022

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