If $\dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c},$ prove that each ratio is equal to

$\dfrac{x + y + z}{a + b + c}.$

Let, $\dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} = k.$

∴ x = k(b + c - a), y = k(c + a - b), z = k(a + b - c).

Putting values of x, y and z in $\dfrac{x + y + z}{a + b + c}$ we get,

$\dfrac{k(b + c - a) + k(c + a - b) + k(a + b - c)}{a + b + c} \\[1em] = \dfrac{kb + kc - \cancel{ak} + \cancel{kc} + \cancel{ak} - \cancel{bk} + ak + \cancel{bk} - \cancel{kc}}{a + b + c} \\[1em] = \dfrac{k(a + b + c)}{(a + b + c)} \\[1em] = k.$

Since, the value of all ratios = k, hence, each ratio = $\dfrac{x + y + z}{a + b + c}.$