Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=k\Rightarrow a=ck,b=dk\)
Có: \(\dfrac{a-c}{a+c}=\dfrac{ck-c}{ck+c}=\dfrac{c\left(k-1\right)}{c\left(k+1\right)}=\dfrac{k-1}{k+1}\left(1\right)\)
\(\dfrac{b-d}{b+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)
Từ (1), (2) `=> đpcm`