If  $\left(a+b+c\right)\left(b+c-a\right)=kbc$  then 

 Given   $\left(a+b+c\right)\left(b+c-a\right)=kbc$

$2s2\left(s-a\right)=kbc$

$\Rightarrow \frac{s\left(s-a\right)}{bc}=\frac{k}{4}$$\Rightarrow$${\cos }^2\frac{A}{2}=\frac{k}{4}$

$0\le \frac{k}{4}\le 1$ $\Rightarrow$ $0\le k\le 4$

$k=0\Rightarrow {\cos }^2\frac{A}{2}=0\Rightarrow \frac{A}{2}={90}^0\Rightarrow A={180}^0$

$k=0$  does not exists

$k=4\Rightarrow {\cos }^2\frac{A}{2}=\frac{A}{2}=0\Rightarrow A=0$

 $k=4$  does not exists