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If x/(b+c-a)=y/(c+a-b)=z/(a+b-c);prove that x(by+cz-ax)=y(cz+ax-by)=z(ax+by-cz)
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If, `x/(b+c-a)=y/(c+a-b) =z/(a+b-c)`
`(b+c-a)/x=(c+a-b)y=(a+b-c)/z = k`
`b+c-a = kx`
`c+a-b = ky`
`a+b-c = kz`
`c = k/2(x+y)`
`b = k/2(x+z)`
`a = k/2(y+z)`
Multiplying above three equations by z,y and x respectively.
`cz = k/2(x+y)z`
`by = k/2(x+z)y`
`ax = k/2(y+z)x`
Let `k/2 = 1/p` , then,
`xz+yz = pcz`
`xy+yz = pby`
`xy+xz = pax`
`pby+pcz-pax = xy+yz+xz+yz-xy-xz`
`pby+pcz-pax = yz`
`by+cz-ax = (yz)/p`
Therefore, multiplying by x on both sides,
`x(by+cz-ax) = (xyz)/p`
Similarly, you can get.
`y(cz+ax-by) = (xyz)/p`
`z(ax+by-cz) = (xyz)/p`
Therefore, by combining above three,
`x(by+cz-ax) =y(cz+ax-by) =z(ax+by-cz) =(xyz)/p`
`x(by+cz-ax) =y(cz+ax-by) =z(ax+by-cz) `
Posted by thilina-g on May 24, 2012 at 1:08 PM (Answer #1)
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