If $\dfrac{a}{b+c} = \dfrac{b}{c+a} = \dfrac{c}{a+b}$ and $a+b+c \ne 0$, then by the sum rule each ratio equals—

Adding numerators and denominators by the sum rule gives $\dfrac{a+b+c}{(b+c)+(c+a)+(a+b)}=\dfrac{a+b+c}{2(a+b+c)}$. Since $a+b+c\ne 0$, this simplifies to $\dfrac{1}{2}$, which therefore equals each individual ratio.