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Question

# If $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}.$ then show that x(b $-$ c) + y (c $-$ a) + z (a $-$ b) = 0.

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Solution

## $\mathrm{Let}:\phantom{\rule{0ex}{0ex}}\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}=k\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{we}\mathrm{have}:\phantom{\rule{0ex}{0ex}}x=\left(b+c-a\right)k,y=\left(c+a-b\right)kandz=\left(a+b-c\right)k\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{LHS}=x\left(b-c\right)+y\left(c-a\right)+z\left(a-b\right)\phantom{\rule{0ex}{0ex}}=\left(b+c-a\right)\left(b-c\right)k+\left(c+a-b\right)\left(c-a\right)k+\left(a+b-c\right)\left(a-b\right)k\phantom{\rule{0ex}{0ex}}=k\left({b}^{2}+bc-ba-bc-{c}^{2}+ac+{c}^{2}-ca+ac-{a}^{2}-bc+ba+{a}^{2}-ab+ba-{b}^{2}-ca+cb\right)\phantom{\rule{0ex}{0ex}}=k\left(0\right)\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\therefore x\left(b-c\right)+y\left(c-a\right)+z\left(a-b\right)=0\phantom{\rule{0ex}{0ex}}$

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