Question
The base $BC$ of $\triangle\text{ABC}$ is divided at $D$ such that $\text{BD}=\frac{1}{2}\text{DC}.$ Prove that $\text{ar}(\triangle\text{ABD})=\frac{1}{3}\times\text{ar}(\triangle\text{ABC}).$
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Answer
Given: $D$ is a point on $BC$ of $\triangle\text{ABC},$ such that $\text{BD}=\frac{1}{2}\text{DC}$
To prove: $\text{ar}(\triangle\text{ABD})=\frac{1}{3}\times\text{ar}(\triangle\text{ABC}).$
Construction: Draw $\text{AL}\perp\text{BC}.$
Proof: In $\triangle\text{ABC},$
we have: $BC = BD + DC \Rightarrow BD + 2BD = 3 \times BD$
Now, we have: $\text{ar}(\triangle\text{ABD})=\frac{1}{2}\times\text{BD}\times\text{AL}$
$\text{ar}(\triangle\text{ABC})=\frac{1}{2}\times\text{BC}\times\text{AL}$
$\Rightarrow\ \text{ar}(\triangle\text{ABC})=\frac{1}{2}\times3\text{BD}\times\text{AL}$
$\Rightarrow\ 3\times\Big(\frac{1}{2}\times\text{BD}\times\text{AL}\Big)$
$\Rightarrow\ \text{ar}(\triangle\text{ABC})=3\times\text{ar}(\triangle\text{ABD})$
$\therefore\ \text{ar}(\triangle\text{ABD})=\frac{1}{3}\text{ar}(\triangle\text{ABC})$
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