If $(p a + q b) : (p c + q d) :: (p a - q b) : (p c - q d)$ , prove that $a : b :: c : d .$

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Solution

given

$(p a + q b) : (p c + q d) :: (p a - q b) : (p c - q d)$

$\Rightarrow \frac{p a + q b}{p c + q d} = \frac{p a - q b}{p c - q d}$

$\frac{p a + q b}{p a - q b} = \frac{p c + q d}{p c - q d}$

now using compodendo and dividendo rule [a/b=c/d implies (a+b)/(a-b)=(c+d)/(c-d)] we get

$\frac{(p a + q b) + (p a - q b)}{(p a + q b) - (p a - q b)} = \frac{(p c + q d) + (p c - q d)}{(p c + q d) - (p c - q d)}$

$\frac{2 p a}{2 q b} = \frac{2 p c}{2 q d} \Rightarrow \frac{a}{b} = \frac{c}{d}$

therefore $a : b :: c : d$

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