Question

If $a,b,c$ are in continued proportion, then prove that
$\frac{b}{b+c}=\frac{a-b}{a-c}$

Answer 1

Given, $a,b,c$ are in contained proportion

$\Rightarrow \frac{a}{b}=\frac{b}{c}$

$\Rightarrow ac-b^2=0$

$\Rightarrow ac-b^2+ab-bc=ab-bc$

$\Rightarrow c(a-b)+b(a-b)=b(a-c)$

$\Rightarrow (c+b)(a-b)=b$