Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Khi đó:
\(\frac{2a-3c}{2b-3d}=\frac{2bk-3dk}{2b-3d}=\frac{k\left(2b-3d\right)}{2b-3d}=k\)
\(\frac{2a+3c}{2a+3d}=\frac{2bk+3dk}{2a+3d}=\frac{k\left(2a+3d\right)}{2a+3d}=k\)
Vậy \(\frac{2a-3c}{2b-3d}=\frac{2a+3c}{2a+3d}=k\)
Ta có đpcm