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) If, `x/(b+c-a)=y/(c+a-b) =z/(a+b-c)`

Then,

`(b+c-a)/x=(c+a-b)y=(a+b-c)/z = k`

Then,

`b+c-a = kx`

`c+a-b = ky`

`a+b-c = kz`

This gives,

`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...

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If, `x/(b+c-a)=y/(c+a-b) =z/(a+b-c)`

Then,

`(b+c-a)/x=(c+a-b)y=(a+b-c)/z = k`

Then,

`b+c-a = kx`

`c+a-b = ky`

`a+b-c = kz`

This gives,

`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,

Rearranging gives,

`xz+yz = pcz`

`xy+yz = pby`

`xy+xz = pax`

Therefore,

`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`

Therefore,

`x(by+cz-ax) =y(cz+ax-by) =z(ax+by-cz) `

pby+pcz-pax =

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