If $$\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=k$$ then value of $$k$$ is.

$$b+c=\dfrac{a}{k}\rightarrow1$$

$$c+a=\dfrac{b}{k}\rightarrow2$$

$$a+b=\dfrac{c}{k}\rightarrow3$$

Adding eq. 1, eq. 2 and eq. 3

$$2(a+b+c)=\dfrac{a+b+c}{k}$$

$$k=\dfrac{1}{2}$$

Subtracing eq. 2 from eq. 1

$$a-b=\dfrac{b-a}{k}$$

$$k=-1$$

Thus, k can be either $$k=\dfrac{1}{2}$$ or $$k=-1$$

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