If (11a² + 13b²)(11c² - 13d²) = (11a² - 13b²)(11c² + 13d²), prove that a : b : : c : d.

Given, (11a² + 13b²)(11c² - 13d²) = (11a² - 13b²)(11c² + 13d²).

On cross-multiplication,

$\Rightarrow \dfrac{11a^2 + 13b^2}{11a^2 - 13b^2} = \dfrac{11c^2 + 13d^2}{11c^2 - 13d^2} \\[0.5em]$

By componendo and dividendo,

$\Rightarrow \dfrac{11a^2 + 13b^2 + 11a^2 - 13b^2}{11a^2 + 13b^2 - 11a^2 + 13b^2} = \dfrac{11c^2 + 13d^2 + 11c^2 - 13d^2}{11c^2 + 13d^2 - 11c^2 + 13d^2} \\[0.5em] \Rightarrow \dfrac{22a^2}{26b^2} = \dfrac{22c^2}{26d^2} \\[0.5em]$

On dividing the equation by $\dfrac{22}{26}$,

$\Rightarrow \dfrac{a^2}{b^2} = \dfrac{c^2}{d^2} \\[0.5em] \Rightarrow \sqrt{\big(\dfrac{a^2}{b^2}\big)} = \sqrt{\big(\dfrac{c^2}{d^2}\big)} \\[0.5em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em] \Rightarrow a : b : : c : d.$

Hence, proved that a, b, c, d are in proportion.