If$\displaystyle \frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}$Prove$\displaystyle (b+c)x+(c+a)y+(a+b)z=0$
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Here is my hint: x/(b-c) = y/(c-a) = (x+y)/(b-c+c-a) = (x+y)/(b-a) , since (x+y)/(b-a) = z/(a-b) => x+y = -z
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