Ta đặt: \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có: \(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\) (1)
\(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\) (2)
Từ (1) và (2): \(\frac{a}{3a+b}=\frac{b}{3c+d}\)