Angle in a triangle whose side lengths satisfy $\dfrac{3}{a+b+c}=\dfrac{1}{a+b}+\dfrac{1}{a+c}$ Suppose that we have a triangle $ABC$ such that
$$\dfrac{3}{a+b+c}=\dfrac{1}{a+b}+\dfrac{1}{a+c}$$
What is the angle $\widehat{BAC}$ between the sides of length $b$ and $c$?
If $p$ is the perimeter, then $3(p-c)(p-b)=p(p-b+p-c)=p(p+a)$ if I'm not mistaken.
I have made only very little progress, so I would appreciate some hints. Thank you!

Multiplying the both sides by $(a+b+c)(a+b)(a+c)$ gives
$$a^2+bc=b^2+c^2$$
Now consider the law of cosines.
