In a , if =60^ , prove that (a + b + c) = 3ca

Question

In a $\triangle\text{ABC},$ if $\angle\text{B}=60^{\circ},$ prove that (a + b + c) = 3ca

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Answer

Given, $\angle\text{B}=60^{\circ}$
We know that, $\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$
$\Rightarrow\cos60^{\circ}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$
$\Rightarrow\frac{1}{2}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$ $\Big(\because\cos60^{\circ}=\frac{1}{2}\Big)$
$\Rightarrow\text{ac}=\text{a}^2+\text{c}^2-\text{b}^2$
$\Rightarrow3\text{ac}-2\text{ac}=\text{a}^2+\text{c}^2-\text{b}^2$
$\Rightarrow3\text{ac}=\text{a}^2+\text{c}^2-\text{b}^2+2\text{ac}$
$\Rightarrow3\text{ac}=\text{a}^2+\text{c}^2+2\text{ac}-\text{b}^2$
$\Rightarrow3\text{ac}=(\text{a + c})^2-\text{b}^2$
$\Rightarrow3\text{ac}=(\text{a + c + b})\text{(a + c}-\text{b})$
$\Rightarrow3\text{ac}=(\text{a + b + c})(\text{a}-\text{b + c})$
Hence proved.

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