Triangle angles. 
For $\vartriangle$ABC it is given that $$\frac1{b+c}+\frac1{a+c}=\frac3{a+b+c}$$ Find the measure of angle $C$.

This is a "challenge problem" in my precalculus book that I was assigned. How do I find an angle from side lengths like this? I have tried everything I can. I think I may need to employ the law of cosines or sines. Thanks.
 A: Multiply both sides by $a+b+c$ to get $$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}=3$$
We get 
$$\frac{a}{b+c}+\frac{b}{a+c}=1$$
$$a(a+c)+b(b+c)=(a+c)(b+c)$$
$$a^2+ac+b^2+bc=ab+ac+bc+c^2$$
$$a^2+b^2=ab+c^2$$
$$a^2-ab+b^2=c^2$$
By the cosine law, we have $a^2-2\cos(C)ab+b^2=c^2$. Hence $\cos(C)=\frac12$, so $C=60^\circ$.
A: Short answer: 
According to the problem, the solution is unique, so any triple of values that satisfies the equation provides a solution. We immediately see that $a=b=c=1$ is a solution, hence the angle is 60 degrees.
Medium answer.
What count are the ratios between sides. So we can assume that $c=1$.
The equation is symmetric in $a$ and $b$, and we know it's unique, so as $a$ varies, $b$ varies. We can try to see if this infinite family of solutions has an intersection with isoscele triangles, so we put $b=a$ and we solve
$$\frac{1}{a+1}+\frac{1}{a+1}=\frac{3}{2a+1}$$
finding $a=b=c=1$. So the angle is 60 deg.
Notice that the fun thing is that $k\cdot (1,1,1)$ is not the unique solution. They are indeed infinite. Example, $a=15$, $b=8$, $c=13$.
A: Hint:
Simplify the equation to $$a^2 + b^2 - ab  = c^2$$
And use cosine law.
