解：过点$D$作$DF//BE，$交$AC$于点$F$
∵$DF//BE$
∴$\frac {CF}{CE}=\frac {CD}{BC}=\frac {1}{2}$
∴点$F $为$CE$的中点
∵$DF//BC$
∴$\frac {AO}{AD}=\frac {AE}{AF}$
$ (1)$∵$\frac {AE}{AC}=\frac {1}{2}$
∴$\frac {AE}{AF}=\frac {2}{3}$
∴$\frac {AO}{AD}=\frac {2}{3}$
$ (2)$∵$\frac {AE}{AC}=\frac {1}{3}$
∴$\frac {AE}{AF}=\frac {1}{2}$
∴$\frac {AO}{AD}=\frac {1}{2}$
$ (3)$∵$\frac {AE}{AC}=\frac {1}{4}$
∴$\frac {AE}{AF}=\frac {2}{5}$
∴$\frac {AO}{AD}=\frac {2}{5}$
$ (4)$∵$\frac {AE}{AC}=\frac {1}{n+1}$
∴$\frac {AE}{AF}=\frac {2}{n+2}$
∴$\frac {AO}{AD}=\frac {2}{n+2}$
