If $ c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ prove that it is Isosceles Triangle 
In a $\triangle  ABC\;,$ If $\displaystyle c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ Then how can we prove that $\triangle ABC$
is an Isoceles $\triangle.$

$\bf{My\; Try::}$ Using $\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\;,$ We get
$$k\sin C\left[k\sin A+k\sin B\right]\cdot \cos \frac{B}{2} = k\sin B\left[k\sin A+k\sin C\right]\cdot \cos \frac{C}{2} $$
So we get $$\sin C\left[\sin \left(\frac{A+B}{2}\right)\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\sin \left(\frac{A+C}{2}\right)\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
Now Using $A+B+C=\pi\;,$ We get $\displaystyle \frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$ and $\displaystyle \frac{A+C}{2}=\frac{\pi}{2}-\frac{B}{2}$
So we get $$\sin C\left[\cos \frac{C}{2}\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\cos \frac{B}{2}\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
So we get $$\sin C\cdot \cos \left(\frac{A-B}{2}\right)=\sin B\cdot \cos \left(\frac{A-C}{2}\right)$$
Now if we put $B=C\;,$ Then these two are equal.
My question is how can we prove it.
Help me, Thanks

we have $$\frac{cos\left(\frac{B}{2}\right)}{cos\left(\frac{C}{2}\right)}=\frac{ab+bc}{ac+bc}$$
Applying Componendo and Dividendo we get
$$\frac{cos\left(\frac{B}{2}\right)-cos\left(\frac{C}{2}\right)}{cos\left(\frac{B}{2}\right)+cos\left(\frac{C}{2}\right)}=\frac{a(b-c)}{ab+ac+bc}$$ $\implies$
$$-tan\left(\frac{B+C}{2}\right)tan\left(\frac{B-C}{2}\right)=\frac{a(b-c)}{ab+ac+bc} \tag{1}$$
Now $$tan\left(\frac{B+C}{2}\right)=cot\left(\frac{A}{2}\right)$$
and by Napier's Rule $$tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}cot\left(\frac{A}{2}\right)$$
so $(1)$ becomes
$$\frac{-(b-c)cot^2\left(\frac{A}{2}\right)}{b+c}=\frac{a(b-c)}{ab+ac+bc} $$ $\implies$
$$(b-c) \times \left(\frac{a}{ab+bc+ac}+\frac{cot^2\left(\frac{A}{2}\right)}{b+c}\right)=0 $$ which is possible only if $$b=c$$
