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Math Help - Ratio and Proportion

  1. #1
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    Ratio and Proportion

    If \frac{x}{(b+c-a)}=\frac{y}{c+a-b}=\frac{z}{a+b-c} Prove that (b-c)x+(c-a)y+(a-b)z=0

    If we add the numerator parts and denominator parts we get : \frac{x+y+z}{a+b+c} Can we do it this way... please help further.

    Or : multiplying numerator and denominator with (b-c) of : \frac{x}{(b+c-a)} = \frac{(b-c)x}{(b-c)(b+c-a)} similar way the other...
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  2. #2
    Junior Member Nehushtan's Avatar
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    Re: Ratio and proportion

    (b-c)x+(c-a)y+(a-b)z

    =\ \left[\frac{(b-c)(b+c-a)}{a+b-c}+\frac{(c-a)(c+a-b)}{a+b-c}+(a-b)\right]z

    =\ \frac{b^2-c^2-ab+ca+c^2-a^2-bc+ab+a^2-b^2-ca+bc}{a+b-c}\cdot z

    =\ 0
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