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

Also, show that (b − c)x + (c − a)y + (a − b)z = 0.

Given,

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

Let the common value of the given ratios be k.

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

Therefore,

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

Adding x,y and z, we get:

⇒ x + y + z = k[(b + c − a) + (c + a − b) + (a + b − c)]

⇒ x + y + z = k(a + b + c)

⇒ k = $\dfrac{x + y + z}{a + b + c}$

Therefore,

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

Given,

(b − c)x + (c − a)y + (a − b)z = 0.

Substituting value of x, y, z in L.H.S of above equation, we get :

⇒ (b − c)[k(b + c - a)] + (c − a)[ k(c + a - b)] + (a − b)[k(a + b - c)]

⇒ k[(b − c)(b + c - a) + (c − a)(c + a - b) + (a − b)(a + b - c)]

⇒ k[(b² - c²) - a(b - c) + (c² - a²) - b(c - a) + (a² - b²) - c(a - b)]

⇒ k[b² - c² - ab + ac + c² - a² - bc + ab + a² - b² - ca + bc]

⇒ k[b² - b² + c² - c² + a² - a² - ab + ab + ac - ac - bc + bc]

⇒ k(0)

⇒ 0.

Hence, proved that $\dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} = \dfrac{x + y + z}{a + b + c}$ and (b − c)x + (c − a)y + (a − b)z = 0.