延长AD交⊙D于另一点E,连接EF
设AC＝BC＝BD＝a,则AB＝2a,AE＝6a
易得△AFH∽△AEF
∴AF/AE＝AH/AF,得：2a/6a＝AH/2a,AH＝2/3·a
∴AH/AB＝1/3