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? 
 A: By the cosine rule, $\cos B = \frac{1}{2}$ iff $b^2=a^2+c^2-ac$. That's the result you need to try to obtain by rearrangement.
A: from your condition we get $$a^2+c^2-ac=b^2$$ apllying the theorem of cosines we get
$$a^2+c^2-ac=a^2+c^2-2ac\cos(\beta)$$
thus we get $$\cos(\beta)=\frac{1}{2}$$
A: 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 $\angle B = \arccos(\frac{1}{2}) = 60^\circ$. 
A: 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),
$a^2+b^2−2ca\cos B=a^2+b^2–ca$
$\cos B = \frac 12$
$B = 60°$
A: 
Here's a way to prove it without using trigonometry. For sake of discussion, in the pictured triangle assume $∠B$ is lower left corner and side $a$ is the bottom (side $c$ is left and $b$ is right). Now look at the three cases where $∠C$ is $<90°$, $=90°$, and $>90°$. 
The easy case is $∠C =90°$. In this case $c=2a$ and $b=\sqrt{3}a$. Plug those values into $a^2−ac−b^2+c^2=0$ to get $a^2 - 2a^2 - 3a^2 +4a^2 = 0$ which confirms the equality.
Take the case  $∠C <90°$. In this case let $h$ be the height of the triangle from $a$ up to $∠A$. Let $d$ be the length of the line segment from the lower left corner to where the height intercepts $a$. And let $e=a-d$.
Calculate the area of the triangle using Heron's formula $A=\sqrt{s(s-a)(s-b)(s-c)}$ where $S=(a+b+c)/2$. Since $A=ah/2$ calculate $$h=2A/a=\frac{\sqrt{2(c^2a^2+c^2b^2+b^2a^2)+a^4+b^4+c^4}}{2a}$$ Knowing $h$ and $c$ allows you to calculate $d$ by Pythagorean. Then calculate $e=a-d$. Knowing $e$ and $h$ allows you to calculate $b=\sqrt{a^2-ac+c^2}$. 
Plugging that into your equation gives $a-ac-(a^2-ac+c^2)+c^2=0$, again confirming the equality.
I'll leave the third case as an exercise.
