If $\dfrac {1}{a+b} +\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ $ If $\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ$.
My Attempt 
$$\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$$
$$\dfrac {a+2b+c}{(a+b)(b+c)}=\dfrac {3}{a+b+c}$$
$$a^2-ac-b^2+c^2=0$$.
How to prove further? 

You're almost there. You had
$$ \frac{a+2b+c}{(a+b)(b+c)} = \frac{3}{a+b+c} $$
which implies
$$ a^2 + ab + ac + 2ab + 2b^2 + 2bc + ac + bc + c^2 = 3ab + 3b^2 + 3ac + 3bc $$
so
$$ a^2 + c^2 - ac = b^2 $$
Now recall the cosine rule $b^2 = a^2 + c^2 - 2ac\cos(\angle B)$. This means that $\angle B = \arccos(\frac{1}{2}) = 60^\circ$. 
