If $\dfrac{x + y}{ax + by} = \dfrac{y + z}{ay + bz} = \dfrac{z + x}{az + bx}$, prove that each of these ratio is equal to $\dfrac{2}{a + b},$ unless x + y + z = 0.

We know that if $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f}$, then each ratio

$= \dfrac{a + c + e}{b + d + f} = \dfrac{\text{sum of antecedents}}{\text{sum of consequents}}$

$\therefore \dfrac{x + y}{ax + by} = \dfrac{y + z}{ay + bz} = \dfrac{z + x}{az + bx} \\[1em] = \dfrac{x + y + y + z + z + x}{ax + by + ay + bz + az + bx} \\[1em] = \dfrac{2(x + y + z)}{a(x + y + z) + b(x + y + z)} \\[1em] = \dfrac{2(x + y + z)}{(a + b)(x + y + z)} \\[1em] = \dfrac{2}{a + b}.$

Hence, proved that,

$\dfrac{x + y}{ax + by} = \dfrac{y + z}{ay + bz} = \dfrac{z + x}{az + bx} = \dfrac{2}{a + b}.$