$\triangle ABC$ has an angle $\angle A=60^{\circ}$. Also $AB=c$, $BC=a$, $AC=b$ and $2\cos B-1=\frac{a+b}{a+c}$. Find all the other angles. 
$\triangle ABC$ has an angle $\angle A=60^{\circ}$. Also $AB=c$, $BC=a$, $AC=b$ and $2\cos B-1=\frac{a+b}{a+c}$. Find all the other angles.

And we can't use calculus, logarithms, limits or something else advanced. Of course, the $\cos$ and $\sin$ theorems are allowed.
The first thing I tried when I saw the problem was using the $\cos$ theorem to get that $a^2=b^2+c^2-bc$.
I don't think substituting $\cos B$ with an expression in terms of $a,b,c$ would help us find the angles.
There's probably a simple idea here that I can't see. So some observations would be great. Thanks.

 A: From the law of sine we derive $b=2a\sin(B)/\sqrt{3}$ and $c=2a\sin(C)/\sqrt{3}$.  Plug that in $2(a+c)\cos(B)=2a+b+c$ to get
$$\sqrt{3}\cos(B)+2\sin(C)\cos(B)=\sqrt{3}+\sin(B)+\sin(C).$$
Now express $\cos(C)$ and $\sin(C)$ in terms of $\sin(B)$ and $\cos(B)$.  Substituting $\cos(B)=2t/(1+t^2)$ and $\sin(t)=2t/(1+t^2)$ yields a polynomial of degree four.  Only one zero qualifies $B$ to be an angle in a triangle and that is ... 
$$B=30^{\circ}\text{, hence }C=90^{\circ}\text{!}$$
Edit (a lot easier): remembering that $\sqrt{3}=2\sin(60)=2\sin(B+C)$ the first displayed equation becomes
$$\sqrt{3}\cos(B)=2\sin(B)\cos(C)+\sin(B)+\sin(C).$$
By expressing this equation solely through $C$ we arrive in
$$\cos(C)\bigl(3\cos(C)+\sin(C)+\sqrt 3\bigr)=0.$$
Now clearly $C=90$ whereas the second factor has no zeroes in $[0,120]$.
A: This isn't a full solution but you know this:
$$2\cos B-1=\frac{a+b}{a+c}$$
We also know:
$$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}.$$
You found that:
$$a^2 = b^2 + c^2 - bc.$$
And finally:
$$B + C = 120^{\circ}.$$
It seems like that's enough equations to solve for what you need?
A: Brute-force solution:
$$\frac{a+b}{a+c}=2\cos B-1=\frac{a^2+c^2-b^2-ac}{ac}=\frac{b^2+c^2-bc+c^2-b^2-ac}{ac}=\frac{2c-a-b}{a}$$
$$
0=a(a+b)+(a+c)(a+b-2c)=2a^2+2ab-ac+bc-2c^2=2(b^2+c^2-bc)+2ab-ac+bc-2c^2=(a+b)(2b-c)
$$
Therefore, $c=2b$, and consequently $a=\sqrt{3}b$. The angles follow easily.
