Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\) (tính chất tỉ lệ thức)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\) \(\left(k\ne0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
Ta có:
\(\dfrac{a-2c}{b-2d}=\dfrac{ck-2c}{dk-2d}=\dfrac{c\times\left(k-2\right)}{d\times\left(k-2\right)}=\dfrac{c}{d}\) \(\left(1\right)\)
\(\dfrac{a+2c}{b+2d}=\dfrac{ck+2c}{dk+2d}=\dfrac{c\times\left(k+2\right)}{d\times\left(k+2\right)}=\dfrac{c}{d}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{a-2c}{b-2d}=\dfrac{a+2c}{b+2d}\)
Vậy \(\dfrac{a-2c}{b-2d}=\dfrac{a+2c}{b+2d}\) \(\left(đpct\right)\).
