Đặt :\(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\) Ta có :\(\dfrac{a+b}{c+d}=\dfrac{bk+b}{dk+d}=\dfrac{b.\left(k+1\right)}{d.\left(k+1\right)}=\dfrac{b}{d}\left(1\right)\) \(\dfrac{a-b}{c-d}=\dfrac{bk-b}{dk-d}=\dfrac{b.\left(k-1\right)}{d.\left(k-1\right)}=\dfrac{b}{d}\left(2\right)\) Từ (1) và (2) suy ra : \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\) (đpcm)