In a Δ ABC, if ∠B=60∘, prove that (a+b+c)(a−b+c)=3ca
Now, It is given that :(a+b+c)(a−b+c)=3ca⇒ a2−ab+ac+ab−b2+bc+ac−bc+c2=3ca⇒ a2−b2+c2+ac=3ca⇒ a2−b2+c2=2ca⇒ a2−b2−2ca=b2 ....(i)We have, ∠B=60∘cos B=12 ⇒a2+b2−b22ac=12⇒ a2+c2−b2=ac⇒ a2+c2−ac=b2 ....(ii)(a+b+c)(a−b+c)=3caa2−ab+ac+ab−b2+bc+ac−bc+c2=3ac=a2+c2−b2+2ac−3ac=0a2+c2−ac=b2 ....(iii)from (i)~(ii)~and~(iii)If B=60∘ and (a+b+c)=3ca hence proved.