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
 A: using the sine rule in the form:
$$
b = 4R\sin\frac{B}2\cos\frac{B}2 \\
c = 4R\sin\frac{C}2\cos\frac{C}2
$$
we obtain:
$$
(a+c)\sin\frac{B}2 = (a+b)\sin\frac{C}2
$$
which, together with the original relation:
$$
c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}
$$
gives
$$
b \tan\frac{B}2 = c\tan\frac{C}2
$$
but the function $x\tan\frac{x}2$ is monotonic on $[0,\pi)$ hence the result
A: 
Let's use the angle bisector theorem on $C$, we get
$$\frac{b+a}{c}=\frac{a}{EB}.$$
By the angle bisector theorem on $B$, we get
$$\frac{c+a}{b}=\frac{a}{CD}.$$
Now from the law of sines
$$EB=\frac{a\sin{C/2}}{\sin y}$$
and
$$CD=\frac{a\sin{B/2}}{\sin x}.$$
Putting all together we find
$$\frac{(b+a)\sin {C/2}}{c \sin y}=\frac{(c+a)\sin {B/2}}{b \sin x}.$$
Plugging in the equation given by the problem, we find
$$\sin x=\sin y$$
thus $x=y$ and $ABC$ is isosceles.
A: with $\cos(\beta/2)=\sqrt{\frac{s(s-b)}{ac}}$ and $\cos(\gamma/2)=\sqrt{\frac{s(s-b)}{ab}}$ and $s=(a+b+c)/2$ we get after squaring
$$-1/2\,bc \left( b-c \right)  \left( {a}^{3}+{a}^{2}b+{a}^{2}c+3\,abc+{
b}^{2}c+b{c}^{2} \right) 
=0$$ thus $b=c$
A: $$\sin C\cos\dfrac{A-B}2=\sin\dfrac C2\cdot2\sin\dfrac{A+B}2\cos\dfrac{A-B}2 =\sin\dfrac C2\left(\sin\dfrac A2+\sin\dfrac B2\right)$$
So,
$$\sin C\cdot \cos \left(\frac{A-B}{2}\right)=\sin B\cdot \cos \left(\frac{A-C}{2}\right)$$
$$\iff\sin\dfrac C2\left(\sin\dfrac A2+\sin\dfrac B2\right)=\sin\dfrac B2\left(\sin\dfrac A2+\sin\dfrac C2\right)$$
As $\sin\dfrac A2\ne0,\sin\dfrac C2=\sin\dfrac B2$
Can you take it from here?
A: 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 $$b=c$$
