Đặt $\frac{a}{b}$ =$\frac{c}{d}$ =k
⇒ $\left \{ {{a=bk} \atop {c=dk}} \right.$
Ta có:
$\frac{a}{a-b}$ =$\frac{bk}{bk-b}$ =$\frac{bk}{b(k-1)}$ =$\frac{k}{k-1}$
$\frac{c}{c-d}$ =$\frac{dk}{dk-d}$ =$\frac{dk}{d(k-1)}$ =$\frac{k}{k-1}$
⇒$\frac{a}{a-b}$ = $\frac{c}{c-d}$