Question

If $$\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$$, prove that $$\frac{ax - by}{(a + b)(x - y)} + \frac{by - cz}{(b + c)(y - z)} + \frac{cz - ax}{(c + a)(z - x)} = 3$$.

[Hint : Use k-method in each.]

Answer 1

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

To prove: $$\frac{ax - by}{(a + b)(x - y)} + \frac{by - cz}{(b + c)(y - z)} + \frac{cz - ax}{(c + a)(z - x)} = 3$$

Proof:

1. Let $$\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k$$, then $$x = ka, y = kb, z = kc$$
2. $$\frac{ax - by}{(a + b)(x - y)} = \frac{a(ka) - b(kb)}{(a + b)(ka - kb)} = \frac{k(a^2 - b^2)}{k(a + b)(a - b)} = \frac{(a - b)(a + b)}{(a + b)(a - b)} = 1$$
3. Similarly, $$\frac{by - cz}{(b + c)(y - z)} = 1$$
4. And $$\frac{cz - ax}{(c + a)(z - x)} = 1$$
5. $$\text{L.H.S.} = 1 + 1 + 1 = 3 = \text{R.H.S.}$$

Hence proved.