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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