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Question Number 194695 by horsebrand11 last updated on 13/Jul/23

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Answered by som(math1967) last updated on 13/Jul/23

let (x/(a+b−c))=(y/(b+c−a))=(z/(c+a−b))=k x=k(a+b−c) y=k(b+c−a) z=k(c+a−b) (a−b)x+(b−c)y+(c−a)z =(a−b)(a+b−c)k+(b−c)(b+c−a)k (c−a)(c+a−b)k =k×(a^2 −b^2 −ca+bc+b^2 −c^2 −ab+ca c^2 −a^2 −bc+ab) =k×0=0

letxa+b−c=yb+c−a=zc+a−b=kx=k(a+b−c)y=k(b+c−a)z=k(c+a−b)(a−b)x+(b−c)y+(c−a)z=(a−b)(a+b−c)k+(b−c)(b+c−a)k(c−a)(c+a−b)k=k×(a2−b2−ca+bc+b2−c2−ab+cac2−a2−bc+ab)=k×0=0

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