If a : b = c : d, prove that :

(i) 5a + 7b : 5a - 7b = 5c + 7d : 5c - 7d

(ii) (9a + 13b)(9c - 13d) = (9c + 13d)(9a - 13b)

(iii) xa + yb : xc + yd = b : d

(i) Given,

⇒ a : b = c : d

$\therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em]$

Multiplying both sides by $\dfrac{5}{7}$:

$\Rightarrow \dfrac{5a}{7b} = \dfrac{5c}{7d}$

Applying componendo and dividendo:

$\Rightarrow \dfrac{5a + 7b}{5a - 7b} = \dfrac{5c + 7d}{5c - 7d}$

Hence, proved that 5a + 7b : 5a - 7b = 5c + 7d : 5c - 7d.

(ii) Given,

a : b = c : d

$\therefore \dfrac{a}{b} = \dfrac{c}{d}$

Multiplying both sides by $\dfrac{9}{13}$:

$\Rightarrow \dfrac{9a}{13b} = \dfrac{9c}{13d}$

Applying componendo and dividendo:

$\Rightarrow \dfrac{9a + 13b}{9a - 13b} = \dfrac{9c + 13d}{9c - 13d}$

On cross-multiplication:

⇒ (9a + 13b)(9c - 13d) = (9c + 13d)(9a - 13b).

Hence, proved that (9a + 13b)(9c - 13d) = (9c + 13d)(9a - 13b).

(iii) Given,

a : b = c : d

$\therefore \dfrac{a}{b} = \dfrac{c}{d}$

Multiplying both sides by $\dfrac{x}{y}$:

$\Rightarrow \dfrac{xa}{yb} = \dfrac{xc}{yd}$

Applying componendo: $\Rightarrow \dfrac{xa + yb}{yb} = \dfrac{xc + yd}{yd}$

On cross-multiplication:

$\Rightarrow \dfrac{xa + yb}{xc + yd} = \dfrac{yb}{yd} \\[1em] \Rightarrow \dfrac{xa + yb}{xc + yd} = \dfrac{b}{d}.$

Hence, proved that xa + yb : xc + yd = b : d.