<s>[INST] $\triangle ABC$ 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.

 [/INST] 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.
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