If ` b/(a+b) = (a+c-b)/(b+c-a) = (a+b+c)/(2a+b+2c)("where " a + b + c ne 0)`, then prove that ` a/2 = b/3 = c/4`.

Explanation

`b/(a+b) = (a+c-b)/(b+c-a) = (a+b+c)/(2a+b+2c)`
or, ` (2b)/(2(a+b)) = (a+c-b)/(b+c-a) = (2(a+b+c))/(2(2a+b+2c))`
`= (2b+a+c-b+2a+2b+2c)/(2a+2b+b+c-a+4a+2b+4c)` (by addendo process)
` = (3a+3b+3c)/(5a+5b+5c)`
` = (3(a+b+c))/(5(a+b+c)) = 3/5 [:' a + b + c ne 0]`
` :. b/(a + b) = 3/5 " or, " 5b + 3a + 3b " or, " 5b - 3b = 3a`.
or, `2b = 3a " or, " a/b = 2/3 " or, " a/2 = b/2 ` ..............(1)
Again, `(a + c-b)/(b+c-a) = 3/5 " or, " 5a + 5c - 5b = 3b + 3c - 3a`
or, ` 5a + 3a - 5b - 3b = 3c - 5c`
or, ` 8a - 8b = - 2c " or, " 8a - 12a = - 2c [ :' 2b = 3a]`
or, ` - 4a = - 2c " or, " a/(-2) = c/(-4) " or , " a/(-2) = c/(-4) " or, " a/2 = c/4 ` ..................(2)
From (1) and (2) we get , ` a/2 = b/3 = c/4`.
Hence ` a/2 = b/3 = c/4`