1.

If `(by+cz)/(b^2 + c^2) = (cz + ax)/(c^2 + a^2) = (ax + by)/(a^2 + b^2)` then prove that each ratio is equal to `x/a = y/b = z/c`.

Answer» We are given,
`(by+cz)/(b^2+c^2)=(cz+ax)/(c^2+a^2)=(ax+by)/(a^2+b^2)->Eq(1)`
Also,`a/b=c/d=e/f`, can be written as `(a+c+e)/(b+d+f)` So, Eq(1) becomes
`(by+cz)/(b^2+c^2)=(cz+ax)/(c^2+a^2)=(ax+by)/(a^2+b^2)=(ax+by+cz)/(a^2+b^2+c^2)`
Taking,
`(by+cz)/(b^2+c^2)=(ax+by+cz)/(a^2+b^2+c^2)`
`=>(by+cz)(a^2+b^2+c^2))=(ax+by+cz)(b^2+c^2)`
`=>a^2(by+cz)+(b^2+c^2)(by+cz) = ax(b^2+c^2)+(b^2+c^2)(by+cz)`
`=>(by+cz)/(b^2+c^2)=x/a-> Eq(2)`
Similarly, we can prove that
`(cz+ax)/(c^2+a^2) = y/b->Eq(3)`
and
`(ax+by)/(a^2+b^2) = z/c->Eq(4)`
From `Eq(1),Eq(2),Eq(3),and Eq(4)`, we can conclude that,
`x/a=y/b=z/c`


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