Đặt:
$\dfrac{a}{b}=\dfrac{c}{d}=k$
Suy ra:
`a=bk; c=dk`
Có:
$\dfrac{a-b}{c-d}=\dfrac{bk-b}{dk-d}=\dfrac{b(k-1)}{d(k-1)}=\dfrac{b}{d}$ (1)
$\dfrac{a+b}{c+d}=\dfrac{bk+b}{dk+d}=\dfrac{b(k+1)}{d(k+1)}=\dfrac{b}{d}$ (2)
Từ `(1); (2)` suy ra $\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}$