Question

在三角形ABC中,D为BC边的中点,E为AC边上任意一点,BE交AD于点O,且AE/EC=1/n,求AO/OD

Answer 1

过点E作EG//BC,交AD于点G ,
∵AE/EC=1/n ,AC=AE+EC
∴AE/AC=1/(n+1)
∵EG//BC ,
∴EG/DC=AE/AC
∴EG/DC=AE/AC=1/(n+1)
∴EG/DC=1/(n+1)
∵BD=DC
∴EG/BD=1/(n+1)
又∵EG//BC ,EG/BD=1/(n+1)
∴EG/BD=OG/OD

∴OG/OD=1/(n+1)
∴OG=OD/(n+1).....................................(1)
∵EG//BC ,AE/EC=1/n
∴AG/DG=AE/EC
∴AG/DG=1/n
∵AG=OA-OG ,DG=OD+OG
∴(OA-OG)/(OD+OG)=1/n......................(2)
把(1)代入(2)得
OD+[OD/(n+1)]=n[OA-OD/(n+1)]
OD+[2OD/(n+1)]=nOA
(n+1)OD+2OD=n(n+1)OA
(n+3)OD=n(n+1)OA
OA/OD=(n+3)/(n^2+n)