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If $\frac{b+c-a}{y+z-x} = \frac{c+a-b}{z+x-y} = \frac{a+b-c}{x+y-z}$, prove that each ratio is equal to $\frac{a}{x} = \frac{b}{y} = \frac{c}{z}$.

My attempt:

By addendo,

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

Now,

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

⇒ $\frac{b+c-a}{a+b+c} = \frac{y+z-x}{x+y+z}$

By dividendo,

⇒ $\frac{(b+c-a)-(a+b+c)}{a+b+c} = \frac{(y+z-x)-(x+y+z)}{x+y+z}$

⇒ $\frac{-2a}{a+b+c} = \frac{-2x}{x+y+z}$

⇒ $\frac{a}{x} = \frac{a+b+c}{x+y+z}$

Similarly,

$\frac{c+a-b}{z+x-y} = \frac{a+b+c}{x+y+z} ⇒ \frac{b}{y} = \frac{a+b+c}{x+y+z}$ and $\frac{a+b-c}{x+y-z} = \frac{a+b+c}{x+y+z} ⇒ \frac{c}{z} = \frac{a+b+c}{x+y+z}$

Hence each given ratio = $\frac{a}{x} = \frac{b}{y} = \frac{c}{z}$

I have proved this but i am not satisfied with this method. Is there any elegant and short method to prove this?

MrAP
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1 Answers1

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You can do it simplier.

You have $$\frac{b+c-a}{y+z-x} = \frac{c+a-b}{z+x-y} = \frac{a+b-c}{x+y-z} = \frac{(b+c-a)+(c+a-b)}{(y+z-x)+(z+x-y)} = \frac{2c}{2z} = \frac{c}{z}.$$

Similarly, you have $\frac{b}{y} = \frac{a}{x} = \frac{b+c-a}{y+z-x} = \frac{c}{z}$

GAVD
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  • can you help me in proving this result by a method similar to the method used in the answer for this question:http://math.stackexchange.com/questions/2144602/difficulty-in-proving-a-result? – MrAP Feb 18 '17 at 17:26