If $a,b,c,x$ are real numbers, such that $abc\ne 0$ and :
$$\dfrac{xb+(1-x)c}{a}=\dfrac{xc+(1-x)a}{b}=\dfrac{xa+(1-x)b}{c}$$ then prove that either $a+b+c=0$ or $a=b=c$
My Approach:
$$\dfrac{xb+(1-x)c}{a}=\dfrac{xc+(1-x)a}{b}=\dfrac{xa+(1-x)b}{c}=k\ \text{(suppose)}$$
assuming $a+b+c\ne 0$:
$$k=\dfrac{xb+(1-x)c+xc+(1-x)a+xa+(1-x)b}{a+b+c}=1$$ $$\text{Thus}\ x=\dfrac{a-c}{b-c}=\dfrac{b-a}{c-a}=\dfrac{c-b}{a-b}$$
From the first two equalities, we have $(a-c)^2=(b-a)(c-b)\Rightarrow a^2+b^2+c^2=ab+bc+ca$
$a(a-c)+b(b-a)+c(c-b)=0\Rightarrow (a-c)(a-bx-cx)=0$
Either $a=c$ and since the function is symmetric, $a=b=c$, or $x=\dfrac{a}{b+c}$
Suppose $x=\dfrac{a}{b+c}$, since the function is symmetric:
$$x=\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{1}{2}\Rightarrow 2a=b+c$$
Now $2a=b+c\ \text{and}\ 2b=c+a \Rightarrow 2(a+b)=(a+b)+2c\Rightarrow b=2c-a\Rightarrow 2c-a=c-2a\Rightarrow a=c$
Thus $x=\dfrac{a}{b+c}$ again gives $a=b=c$
How can I find the case $a+b+c=0$. Please check my approach and give suggestions as to what I must add to my approach to make it better and find the missing solution.
THANKS