# Math Help - Ratio and Proportion

1. ## 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...

2. ## Re: Ratio and proportion

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

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

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

$=\ 0$