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

$\begin{matrix} \text{(i)} & \dfrac{2a + 5b}{2a - 5b} = \dfrac{2c + 5d}{2c - 5d}. \\[0.5em] \text{(ii)} & \dfrac{5a + 11b}{5c + 11d} = \dfrac{5a - 11b}{5c - 11d}. \\[0.5em] \text{(iii)} & (2a + 3b)(2c - 3d) \ & = (2a - 3b)(2c + 3d). \\ \text{(iv)} & (la + mb) : (lc + md) \ & : : (la - mb) : (lc - md). \end{matrix}$

(i) Given, a : b : : c : d,

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em]$

Multiplying the equation by $\dfrac{2}{5}$,

$\Rightarrow \dfrac{2a}{5b} = \dfrac{2c}{5d} \\[0.5em]$

By componendo and dividendo,

$\Rightarrow \dfrac{2a + 5b}{2a - 5b} = \dfrac{2c + 5d}{2c - 5d} \\[0.5em]$

Hence, proved that $\dfrac{2a + 5b}{2a - 5b} = \dfrac{2c + 5d}{2c - 5d}.$

(ii) Given, a : b : : c : d,

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em]$

On multiplying the equation by $\dfrac{5}{11}$,

$\Rightarrow \dfrac{5a}{11b} = \dfrac{5c}{11d} \\[0.5em]$

By componendo and dividendo,

$\Rightarrow \dfrac{5a + 11b}{5a - 11b} = \dfrac{5c + 11d}{5c - 11d} \\[0.5em]$

By alternendo,

$\Rightarrow \dfrac{5a + 11b}{5c + 11d} = \dfrac{5a - 11b}{5c - 11d}$

Hence, proved that $\dfrac{5a + 11b}{5c + 11d} = \dfrac{5a - 11b}{5c - 11d}.$

(iii) Given, a : b : : c : d,

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em]$

On multiplying equation by $\dfrac{2}{3}$,

$\Rightarrow \dfrac{2a}{3b} = \dfrac{2c}{3d} \\[0.5em]$

By componendo and dividendo,

$\Rightarrow \dfrac{2a + 3b}{2a - 3b} = \dfrac{2c + 3d}{2c - 3d} \\[0.5em]$

On cross multiplication,

$\Rightarrow (2a + 3b)(2c - 3d) = (2c + 3d)(2a - 3b).$

Hence, proved that (2a + 3b)(2c - 3d) = (2a - 3b)(2c + 3d).

(iv) Given, a : b : : c : d,

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em]$

On multiplying the equation by $\dfrac{l}{m}$,

$\Rightarrow \dfrac{la}{mb} = \dfrac{lc}{md} \\[0.5em]$

By componendo and dividendo,

$\Rightarrow \dfrac{la + mb}{la - mb} = \dfrac{lc + md}{lc - md} \\[0.5em]$

By alternendo,

$\Rightarrow \dfrac{la + mb}{lc + md} = \dfrac{la - mb}{lc - md} \\[0.5em] \Rightarrow (la + mb) : (lc + md) : : (la - mb) : (lc - md).$

Hence, proved that (la + mb) : (lc + md) : : (la - mb) : (lc - md).