Question

)If (5a+ 9b) (5c-9d) = (5a — 9b) (5c + 9d), using properties of
proportion, prove that a : b = c : d.

Answer 1

Basic Concept Used :-

$\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d} \: then$

$\bf \: 1. \: \: \dfrac{a}{c} = \dfrac{b}{d} \: \: \: \: \{ \: alternendo \}$

$\bf \: 2. \: \: \dfrac{a + b}{b} = \dfrac{c + d}{d} \: \: \{componendo \}$

$\bf \: 3. \: \: \dfrac{a - b}{b} = \dfrac{c - d}{d} \: \: \{dividendo \}$

$\bf \: 4. \: \: \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d} \: \: \{componendo \: and \: dividendo\}$

$\bf \: 5. \: \: \dfrac{b}{a} = \dfrac{d}{c} \: \: \: \: \{ \: invertendo \}$

$\bf \: 6. \: \: ad = bc$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:(5a + 9b)(5c - 9d) = (5a - 9b)(5c + 9d)$

can be rewritten as

$\rm :\longmapsto\:\dfrac{5a + 9b}{5a - 9b} = \dfrac{5c + 9d}{5c - 9d}$

Apply Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{5a + \cancel{9b} + 5a - \cancel{9b}}{\cancel{5a} + 9b - \cancel{5a} + 9b} = \dfrac{5c + \cancel{9d} + 5c - \cancel{9d}}{\cancel{5c} + 9d - \cancel{5c} + 9d}$

$\rm :\longmapsto\:\dfrac{\cancel{10} \: a}{\cancel{18} \: b} = \dfrac{\cancel{10} \: c}{\cancel{18} \: d}$

$\rm :\implies\:\dfrac{a}{b} = \dfrac{c}{d}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$