Ta có: $\frac{a}{b}=\frac{c}{d}=t\implies a=bt;c=dt$.Khi đó: $\frac{2a+3b}{3a-4b}=\frac{2bt+3b}{3bt-4b}=\frac{2t+3}{3t-4}(1)$.
Và: $\frac{2c+3d}{3c-4d}=\frac{2dt+3d}{3dt-4d}=\frac{2t+3}{3t-4}(2)$.
Từ $(1),(2)\implies$\frac{2a+3b}{3a-4b}=$\frac{2c+3d}{3c-4d} (đpcm)$