Lời giải:
Ta có:
$\frac{a}{b}=\frac{c}{d}=\frac{4c}{4d}=\frac{a+4c}{b+4d}$ (theo TCDTSBN)
$\frac{a}{b}=\frac{c}{d}=\frac{2a}{2b}=\frac{3c}{3d}=\frac{2a-3c}{2b-3d}$ (theo TCDTSBN)
$\Rightarrow \frac{a+4c}{b+4d}=\frac{2a-3c}{2b-3d}$
$\Rightarrow (a+4c)(2b-3d)=(2a-3c)(b+4d)$ (đpcm)