In ΔABC,

Question:

In $\Delta A B C$, if $\angle B=60^{\circ}$, prove that $(a+b+c)(a-b+c)=3 c a$.

Solution:

Given, $\angle B=60^{\circ}$

We know that, $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$

$\Rightarrow \cos 60^{\circ}=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$

$\Rightarrow \frac{1}{2}=\frac{a^{2}+c^{2}-b^{2}}{2 a c} \quad\left(\because \cos 60^{\circ}=\frac{1}{2}\right)$

$\Rightarrow a c=a^{2}+c^{2}-b^{2}$

$\Rightarrow 3 a c-2 a c=a^{2}+c^{2}-b^{2}$

$\Rightarrow 3 a c=a^{2}+c^{2}+2 a c-b^{2}$

$\Rightarrow 3 a c=(a+c)^{2}-b^{2}$

$\Rightarrow 3 a c=(a+c+b)(a+c-b)$

$\Rightarrow 3 a c=(a+b+c)(a-b+c)$

hence proved.