解：过点$C$作$CF⊥x$轴于点$F$，过点$D$作$DE⊥x$轴于点$E$
由题意设点$C$的坐标为$(a$，$\frac {4}{a} )$，点$D$的坐标为$(b$，$\frac {4}{b} )$
则点$A $的坐标为$(\frac {ma}{4}$，$\frac {4}{a} )$，
点$B$的坐标为$(\frac {mb}{4}$，$\frac {4}{b} )$
∴$AC= \frac {ma}{4} -a$，$BD=\frac {mb}{4}-b $
∵$BD=3AC$
∴$\frac {mb}{4} -b=3(\frac {ma}{4}-a)$
整理，得$m(3a-b)=4(3a-b)$
又$m>4$
∴$3a-b=0$，即$b=3a$
∴点$D$的坐标为$(3a$，$\frac {4}{3a} )$
则$CF= \frac {4}{a}$，$DE= \frac {4}{3a}$，$EF=3a-a=2a$
∵$S_{△OCF}=S_{△ODE}= \frac {1}{2} ×4=2$
∴$S_{△OCD}=S_{△OCF}+S_{梯形CFED}-S_{△ODE}=S_{梯形CFED}$
又$S_{梯形CFED}= \frac {1}{2} (DE+CF)×EF$
$= \frac {1}{2}×(\frac {4}{3a}+\frac {4}{a} )×2a=\frac {16}{3}$
∴$△OCD$的面积为$ \frac {16}{3} $
