Question

In $∆ABC,\mathrm{if}\angle B=60°,$ prove that $\left(a+b+c\right)\left(a-b+c\right)=3ca$.

Answer 1

$$\begin{gathered} \mathrm{Given},\angle B=60° \\ \mathrm{We}\mathrm{know}\mathrm{that},\cos B=\frac{a^2+c^2-b^2}{2ac} \\ \Rightarrow \cos 60°=\frac{a^2+c^2-b^2}{2ac} \\ \Rightarrow \frac{1}{2}=\frac{a^2+c^2-b^2}{2ac}\left(\because \cos 60°=\frac{1}{2}\right) \\ \Rightarrow ac=a^2+c^2-b^2 \\ \Rightarrow 3ac-2ac=a^2+c^2-b^2 \\ \Rightarrow 3ac=a^2+c^2-b^2+2ac \\ \Rightarrow 3ac=a^2+c^2+2ac-b^2 \\ \Rightarrow 3ac={\left(a+c\right)}^2-b^2 \\ \Rightarrow 3ac=\left(a+c+b\right)\left(a+c-b\right) \\ \Rightarrow 3ac=\left(a+b+c\right)\left(a-b+c\right) \end{gathered}$$

Hence proved.