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 March 13th, 2013, 09:31 AM #1 Senior Member   Joined: Feb 2013 Posts: 114 Thanks: 0 Ratio and Proportion If $\frac{x}{(b-c)(b+c-2a)}= \frac{y}{(c-a)(c+a-2b)}=\frac{z}{(a-b)(a+b-2c)}$, prove that x+y+z=0 Can we proceed like this : Let $\frac{x}{(b-c)(b+c-2a)}= \frac{y}{(c-a)(c+a-2b)}=\frac{z}{(a-b)(a+b-2c)}$ = $\frac{Adding ~the ~antecedents}{Adding ~the~ consequents}$ we get : $\frac{x+y+z}{a^2+b^2+c^2}$ Please guide further .....
 March 13th, 2013, 10:10 AM #2 Member   Joined: Mar 2013 Posts: 90 Thanks: 0 Re: Ratio and proportion $x+y+z$ $=\ \left[\frac{(b-c)(b+c-2a)}{(a-b)(a+b-2c)}+\frac{(c-a)(c+a-2b)}{(a-b)(a+b-2c)}+1\right]z$ $=\ \frac{b^2-c^2-2ab+2ca+c^2-a^2-2bc+2ab+a^2-b^2-2ca+2bc}{(a-b)(a+b-2c)}\cdot z$ $=\ 0$
 March 13th, 2013, 10:22 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,065 Thanks: 1621 If each fraction is equal to n, x + y + z = ((b-c)(b+c-2a) + (c-a)(c+a-2b) + (a-b)(a+b-2c))n              = ((b-c)(b-a) + (b-c)(c-a) + (c-a)(c-b) + (c-a)(a-b) + (a-b)(a-c) + (a-b)(b-c))n              = 0.

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