If x(b+c)=y(c+a)=z(a+b), then prove that a÷(y+z-x)=b÷(z+x-y)=c÷(x+y-z)

Let us see:

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

=> x = ka+kb-kc —————————→ 1

and y = kb+kc-ka —————————→ 2

and z = kc+ka-kb —————————-> 3

add 1 and 3 above to get z+x = 2ka => (z+x)/a = 2k ————————-> 4

add 1 and 2 above to get x+y = 2kb => (x+y)/b = 2k ————————→ 5

add 2 and 3 above to get y+z = 2kc => (y+z)/c = 2k —————————→ 6

from 4, 5 and 6 above, we have (z+x)/a = (x+y)/b = (y+z)/c =====> PROVED