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