If `(x+y)/(3a-b) = (y+z)/(3b-c) = (z+x)/(3c-a)`, then show that `(x+y+z)/(a+b+c)= (ax + by+cz)/(a^(2)+b^(2)+c^(2))`

Explanation

`(x + y)/(3a - b) = (y+z)/(3b - c) = (z+x)/(3c - a) = (x+ y + y + z + z + x)/(3a - b + 3b - c + 3c - a) ` [by addendo process]
` = (2(x+y+z))/(2(a+b+c)) = (x + y+z)/(a+b +c)`
` :. " each ratio " = (x+y+z)/(a + b + c ) ` ............... (1)
Again, ` (x + y)/(3a - b) = (y+z)/(3b - c) = (z+x)/(3c - a) = (x+y - y - z + z + x)/(3a - b - 3b + c + 3c - a)` [by addendo process]
` = (2x)/(2a - 4b + 4c) = x/(a -2b + 2c)` ltbr. ` = (ax)/(a^(2) - 2ab + 2ca) ` ................(2)
Similarly , ` (x+y)/(3a - b) = (y + z)/(3b - c) = (z+x)/(3c -a) = (by)/(2ab + b^(2) - 2bc)` ......................(3)
and ` (x+y)/(3a-b) = (y+z)/(3b-c) = (z+x)/(3c -a) = (cz)/(c^(2) - 2ca + 2bc)`.....................(4)
Now appliying on (2), (3) and (4) we get ,
`(x + y)/(3a -b) = (y+z)/(3b-c) = (z+x)/(3c-a) = (ax + by +cz)/(a^(2) - 2ab + 2ca + 2ab + b^(2) - 2bc + c^(2) - 2ca + 2bc)`
` = (ax + by + cz)/(a^(2)+b^(2)+c^(2))`................(5)
So, from (1) and (5) we get, `(x + y+z)/(a + b+c) = (ax+by+cz)/(a^(2)+b^(2)+c^(2))`