Ta có: \(\frac{a}{b}=\frac{c}{d}\)
\(\frac{2a-3b}{5b}=\frac{2c-3d}{5d}\\ \Leftrightarrow\frac{d}{b}=\frac{2c-3d}{2a-3b}\)
Áp dụng TCDTSBN ta có:
\(\frac{d}{b}=\frac{2c-3d}{2a-3b}=\frac{d+2c-3d}{b+2a-3b}=\frac{2c-2d}{2a-2b}=\frac{c-d}{a-b}\)
\(\Rightarrow\frac{d}{b}=\frac{c-d}{a-b}\)
Áp dụng TCDTSBN ta có:
\(\frac{d}{b}=\frac{c-d}{a-b}=\frac{d+c-d}{b+a-b}=\frac{c}{a}\\ \Rightarrow\frac{d}{b}=\frac{c}{a}\\ \Leftrightarrow\frac{a}{b}=\frac{c}{d}\left(Theo.gt\right)\\ \Rightarrow dpcm\)