Question

If $$\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}$$, prove that each ratio is equal to $$\left(\frac{x + y + z}{a + b + c}\right)$$. Also, show that $$(b - c)x + (c - a)y + (a - b)z = 0$$.

\[ \begin{array}{l} [\textbf{Hint :}\ \text{Each ratio} = \dfrac{x + y + z}{(b + c - a) + (c + a - b) + (a + b - c)} = \dfrac{x + y + z}{a + b + c}. \\[16pt] \text{Let each ratio be equal to } k\text{. Then,} \\[8pt] x = (b + c - a)k,\ y = (c + a - b)k \ \text{and} \ z = (a + b - c)k.] \end{array} \]

Answer 1

Given: $$\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}$$

To prove: Each ratio $$= \frac{x + y + z}{a + b + c}$$ and $$(b - c)x + (c - a)y + (a - b)z = 0$$

Proof:

1. By the property of equal ratios, each ratio $$= \frac{\text{Sum of antecedents}}{\text{Sum of consequents}}$$.
2. $$\text{Each ratio} = \frac{x + y + z}{(b + c - a) + (c + a - b) + (a + b - c)} = \frac{x + y + z}{a + b + c}$$
3. Let each ratio be equal to $$k$$. Then $$x = k(b + c - a)$$, $$y = k(c + a - b)$$, and $$z = k(a + b - c)$$.
4. $$(b - c)x + (c - a)y + (a - b)z = k[(b - c)(b + c - a) + (c - a)(c + a - b) + (a - b)(a + b - c)]$$
5. $$(b - c)(b + c - a) = (b^2 - c^2) - a(b - c)$$
6. $$(c - a)(c + a - b) = (c^2 - a^2) - b(c - a)$$
7. $$(a - b)(a + b - c) = (a^2 - b^2) - c(a - b)$$
8. Adding these gives $$[(b^2 - c^2) + (c^2 - a^2) + (a^2 - b^2)] - [ab - ac + bc - ab + ca - bc] = 0 - 0 = 0$$
9. Therefore, $$(b - c)x + (c - a)y + (a - b)z = k \times 0 = 0$$

Hence proved.