Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\begin{cases}a=b.k\\c=d.k\end{cases}\)
Ta có:
\(\frac{a+b}{a-b}=\frac{b.k+b}{b.k-b}=\frac{b.\left(k+1\right)}{b.\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)
\(\frac{c+d}{c-d}=\frac{d.k+d}{d.k-d}=\frac{d.\left(k+1\right)}{d.\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\)
Từ (1) và (2) => \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)