Question

In a $\Delta ABC$, if $\angle B={60}^0$,prove that $\left(a+b+c\right)\left(a-b-c\right)=3ca$

Answer 1

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

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

$\frac{1}{2}\times 2ac=a^2+c^2-b^2$

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

Adding $2ac$ on both sides.

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

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

$\therefore 3ac=(a+c-b)(a+c+b)$.