A geometry problem by Guy Fox

$\dfrac{3}{a+b+c} =\dfrac{1}{a+b}+\dfrac{1}{a+c}$

After some algebraic manipulations we get:

$a^{2}=b^{2}+c^{2}-b\cdot c$

Which resembles the cosine law

$a^{2}=b^{2}+c^{2}-2\cdot b\cdot c\cdot \cos α$

$-b \cdot c = -2\cdot b\cdot c\cdot \cos α$

$α = 60°$