Question

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

Answer 1

$\text{Now, It is given that :}(a+b+c)(a-b+c)=3ca\Rightarrow a^2-ab+ac+ab-b^2+bc+ac-bc+c^2=3ca\Rightarrow a^2-b^2+c^2+ac=3ca\Rightarrow a^2-b^2+c^2=2ca\Rightarrow a^2-b^2-2ca=b^2....\left(i\right)\text{We have,}\angle B={60}^{\circ }cosB=\frac{1}{2}\Rightarrow \frac{a^2+b^2-b^2}{2ac}=\frac{1}{2}\Rightarrow a^2+c^2-b^2=ac\Rightarrow a^2+c^2-ac=b^2....\left(ii\right)(a+b+c)(a-b+c)=3ca{a}^2-ab+ac+ab-b^2+bc+ac-bc+c^2=3ac=a^2+c^2-b^2+2ac-3ac=0a^2+c^2-ac=b^2....\left(iii\right)\text{from (i)~(ii)~and~(iii)}IfB={60}^{\circ }and(a+b+c)=3ca\text{hence proved.}$