in triangle ABC ANGLE B = 60 prove that (a+ b+ c ) ( a - b + c ) = 3ca

Hi Ashutosh,
Please find below the solution to the asked query:

$$\begin{gathered} \angle \mathrm{B}=60 \\ \mathrm{We} \mathrm{know} \mathrm{that}: \\ \cos \left(\frac{\mathrm{B}}{2}\right)=\sqrt{\frac{\mathrm{s}\left(\mathrm{s}-\mathrm{b}\right)}{\mathrm{ac}}} \\ \Rightarrow \cos \left(\frac{60}{2}\right)=\sqrt{\frac{\mathrm{s}\left(\mathrm{s}-\mathrm{b}\right)}{\mathrm{ac}}} \\ \Rightarrow \cos \left(30\right)=\sqrt{\frac{\mathrm{s}\left(\mathrm{s}-\mathrm{b}\right)}{\mathrm{ac}}} \\ \Rightarrow \frac{\sqrt{3}}{2}=\sqrt{\frac{\mathrm{s}\left(\mathrm{s}-\mathrm{b}\right)}{\mathrm{ac}}} \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides} \mathrm{we} \mathrm{get} \\ \frac{3}{4}=\frac{\mathrm{s}\left(\mathrm{s}-\mathrm{b}\right)}{\mathrm{ac}} \\ \Rightarrow 4\mathrm{s}\left(\mathrm{s}-\mathrm{b}\right)=3\mathrm{ac} \\ \Rightarrow 2\mathrm{s}\left(2\mathrm{s}-2\mathrm{b}\right)=3\mathrm{ac} \\ \mathrm{We} \mathrm{know} \mathrm{that} 2\mathrm{s}=\mathrm{a}+\mathrm{b}+\mathrm{c} \\ \Rightarrow \left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{a}+\mathrm{b}+\mathrm{c}-2\mathrm{b}\right)=3\mathrm{ac} \\ \mathbf{\Rightarrow }\left(\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}\right)\left(\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{+}\mathbf{c}\right)\mathbf{=}\mathbf{3}\mathbf{ac}\mathbf{ }\left(\mathbf{Hence}\mathbf{ }\mathbf{proved}\right) \end{gathered}$$

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