Question

If x/b+c=y/c+a=z/a+b, let us show that a/y+z-x=b/z+x-y=c/x+y-z​

Answer 1

Answer:

If x/b+c=y/c+a=z/a+b, then how do you prove that (b-c) x+(c-a) y+(a-b) z=0?

Write xb+c=yc+a=za+b=k.

Sort of “unreducing” each fraction in a useful manner, we could write

x(b−c)(b+c)(b−c)=y(c−a)(c+a)(c−a)=z(a−b)(a+b)(a−b)=k.

Componendo-Dividendo says that the fraction formed by the sum of the tops over the sum of the bottoms is still equal to k. That is

(b−c)x+(c−a)y+(a−b)zb2−c2+c2−a2+a2−b2=k.

Therefore,

(b−c)x+(c−a)y+(a−b)z=k(b2−c2+c2−a2+a2−b2)

and so

(b−c)x+(c−a)y+(a−b)z=0

Quick, smart and stylish.

Since all the terms are equal it must be equal to some constant value. So say

x/(b+c) = y/(c+a) = z/(a+b) =p

equating separately we get

x=p(b+c)

y=p(c+a)

z=p(a+b)

Now consider the equation

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

k=p(b-c)(b+c)+p(c-a)(c+a)+p(a-b)(a+b)

k=p(b²-c²+c²-a²+a²-b²)

k=p(0)

k=0

So as considered

(b-c)x +(c-a)y+(a-b)z=0