过点E作EF‖BC,交AD于点F.则有:
OF∶OD = EF∶BD = EF∶CD = AF∶AD = AE∶AC = 1∶(1+n) .
OF∶FD = OF∶(OF+OD) = 1∶(1+1+n) = 1∶(2+n) ;
FD∶AD = (AD-AF)∶AD = (1+n-1)∶(1+n) = n∶(1+n) ;
OF∶AD = (OF∶FD)·(FD∶AD) = n∶(2+n)(1+n) ;
AO∶AD = AF∶AD + OF∶AD = 1∶(1+n) + n∶(2+n)(1+n) = 2∶(2+n) .