If ` a/(y+z) = b/(z + x) = c/(x+y)`, then prove that ` (a(b-c))/(y^(2)-z^(2)) = (b(c-a))/(z^(2)-x^(2)) = (c(a-b))/(x^(2)-y^(2))`.

Explanation

`a/(y + z) = b/(z + x) = c/(x+ y) " or . " a/(y + z) = (b-c)/(z + x - x - y)` [by addendo process]
` = (b-c)/(z-y)`
` :. a/(y+z) = (b-c)/(z-y) ` …………..(1)
Similarly, it can be proved that ` b/(z + x) = (c-a)/(x - z) ` …………..(2)
and, ` c/(x + y) = (a - b)/(y - x) ` ..............(3)
Now, expressions on the LHS (1) , (2) and (3) are equal (Given)
` :. ` All the 6 expressions of (1), (2) and (3) of both the sides are equal.
`:. a/(y+z) xx ( b -c)/(z - y) = b/(z + x) xx (c -a)/(x+y) xx (a-b)/(y-x)`.
or, `(a(b-c))/(-(y+z)(y-z)) = (b(c-a))/(-(z+x)(z-x)) = (c(a-b))/(-(x+y)(x-y))`
or, `(a(b-c))/(y^(2)-z^(2)) = (b(c-a))/(z^(2)-x^(2)) = (c(a-b))/(x^(2)-y^(2))`.
` :. (a(b-c))/(y^(2)-z^(2)) = (b(c-a))/(z^(2)-x^(2)) = (c(a-b))/(x^(2) - y^(2))`.