If ` x^(2) : (by + cz) =y^(2) : (cz + ax) = z^(2) : (ax + by) = 1`, then show that ` a/(a+x) + b/(b+y) + c/(c+z) = 1`.

Explanation

`x^(2) : (by + cz) = y^(2) : (cz+ax) = z^(2) : (ax + by) = 1`.
or, ` x^(2)/(by + cz) = y^(2)/(cz + ax) = z^(2)/(ax + by) = 1`.
` :. X^(2)/(by+cz) = 1 rArr x/(by+cz) = 1/x` [Dividing by x ]
` rArr (ax)/(by + cz) = a/x ` [Multiplying by a]
`rArr (ax)/(ax + by + cz) = a/(a + x) ` [by componendo process]
` :. (ax)/(ax + by + cz) = a/(a + x) ` ...............(1)
Similarly , it can be proved that ` (by)/(ax + by + cz) = b/(b +y) ` ....................(2)
and `(cz)/(ax + by + cz) = c/(c+z) ` ..................(3)
Now , adding (1), (2) and (3) we get,
` (ax)/(ax + by+cz) + (by)/(ax+by+cz) + (cz)/(ax + by+cz) = a/(a+x) + b/(b+y) + c/(c+z)`
or, `(ax+by+cz)/(ax+by+cz) = a/(a+x) + b/(b+y) + c/(c+z)`
or, ` 1 = a/(a + x) + b/(b + y) + c/(c +z)`
` :. a/(a +x) + b/(b=y) + c/(c+z) = 1`