If ` x/(b + c -a) = y/(c + a - b) = z/(a + b -c)`, then show that ` (b -c) x + (c-a) y + (a -b) z = 0` .

Explanation

`x/(b+c-a) = y/(c+a-b) = z/(a +b -c) = k " (let)" [k ne 0]`
` :. x = k (b + c - a) , " or, " (b -c) x = k (b -c)(b + c -a)`
or, ` (b -c) x = k (b^(2) - c^(2)) - ak (b -c) ` …………………..(1)
` y = k (c + a - b) , " or , " (c -a) y = k ( c -a) (c + a - b)`
or, ` (c -a) y = k (c^(2) -a^(2)) - bk (c -a) ` ........................ (2)
` z = k (a + b -c), " or, " (a - b) z = k (a - b) (a +b - c)`
or, ` (a - b) z = k (a^(2) - b^(2)) - ck (a - b)` ........................(3)
` :. ` adding (1) , (2) and (3) we get,
` (b -c) x + (c - a) y + (a - b) z = k (b^(2) - c^(2) + c^(2) - a^(2) + a^(2)-b^(2)) - k (ab - ac + bc - ab + ca - bc)`
` = k xx 0 - k xx 0`
` = 0 - 0 = 0`
` :. (b - c) x + (c -a) y + (a - b) z = 0`