Question

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

Answer 1

Given, $\angle B={60}^o$, then $\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{1}{2}$ [cosine rule $\cos {60}^o=1/2$]

$\Rightarrow a^2+c^2-b^2=ac$ ........ $\left(1\right)$

Now, $LHS=(a+b+c)(a-b+c)$

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

$=a^2+c^2+2ac-b^2$

$=ac+2ac$

$=3ac=RHS$ [from $\left(1\right)$]

Hence proved