Question

If $\frac{8a-5b}{8c-5d}=\frac{8a+5b}{8c+5d}$, then show that $\frac{a}{b}=\frac{c}{d}$

Answer 1

Given, $\frac{8a-5b}{8c-5d}=\frac{8a+5b}{8c+5d}$

$\Rightarrow \frac{8a-5b}{8a+5b}=\frac{8c-5d}{8c+5d}$ [Applying alternendo] ( 1 Mark)

$\frac{8a-5b+8a+5b}{8a-5b-8a-5b}=\frac{8c-5d+8c+5d}{8c-5d-8c-5d}$ [Applying componendo and dividendo] ( 1 Mark)

$\frac{16a}{-10b}=\frac{16c}{-10d}$

$\frac{a}{b}=\frac{c}{d}$ ( 1 Mark)

Hence, it is proved that $\frac{a}{b}=\frac{c}{d}$