<s>[INST] 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? 
 [/INST] In a triangle,
$b^2=c^2+a^2−2ca \cos B$ .....(1)
From question,
$(a+2b+c)(a+b+c) = 3(a+b)(b+c)$
$a^2+ab+ac+2ab+2b^2+2bc+ca+cb+c^2=3ab+3ac+3b^2+3bc$
$a^2+c^2-ca=3b^2−2b^2$
$a^2+b^2-ca=b^2$ .....(2)
Comparing equation (1) and (2),
$a^2+b^2−2ca\cos B=a^2+b^2–ca$
$\cos B = \frac 12$
$B = 60°$
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