If $\frac{a y - b x}{c} = \frac{c x - a z}{b} = \frac{b z - c y}{a}$ $(ay-bx)/c=(cx-az)/b=(bz-cy)/a$ , then how to prove that $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ $x/a=y/b=z/c$ ?

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If $\frac{a y - b x}{c} = \frac{c x - a z}{b} = \frac{b z - c y}{a}$ $(ay-bx)/c=(cx-az)/b=(bz-cy)/a$ , then how to prove that $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ $x/a=y/b=z/c$ ?

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Answer 1 · 2017-07-06T15:39:50

$\frac{a y - b x}{c} = \frac{c x - a z}{b} = \frac{b z - c y}{a}$ $(ay-bx)/c=(cx-az)/b=(bz-cy)/a$

$each = \frac{a c y - b c x}{c^{2}} = \frac{b c x - a b z}{b^{2}} = \frac{a b z - c a y}{a^{2}}$ $"each"= (acy-bcx)/c^2=(bcx-abz)/b^2=(abz-cay)/a^2$

By addendo we get

$each = \frac{(a c y - b c x) + (b c x - a b z) + (a b z - c a y)}{a^{2} + b^{2} + c^{2}} = 0$ $"each"=( (acy-bcx)+(bcx-abz)+(abz-cay))/(a^2+b^2+c^2)=0$

So $\frac{a y - b x}{c} = 0 \Rightarrow \frac{x}{a} = \frac{y}{b}$ $(ay-bx)/c=0=>x/a=y/b$

$\frac{c x - a z}{b} = 0 \Rightarrow \frac{x}{a} = \frac{z}{c}$ $(cx-az)/b=0=>x/a=z/c$

Hence

$\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ $x/a=y/b=z/c$

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If $\frac{a y - b x}{c} = \frac{c x - a z}{b} = \frac{b z - c y}{a}$ $(ay-bx)/c=(cx-az)/b=(bz-cy)/a$ , then how to prove that $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ $x/a=y/b=z/c$ ?

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If $\frac{a y - b x}{c} = \frac{c x - a z}{b} = \frac{b z - c y}{a}$ $(ay-bx)/c=(cx-az)/b=(bz-cy)/a$ , then how to prove that $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ $x/a=y/b=z/c$ ?