(I)证明:∵DE2=EF•EC,∠DEF公用, ∴△DEF∽△CED, ∴∠EDF=∠C. 又∵弦CD∥AP,∴∠P=∠C, ∴∠EDF=∠P,∠DEF=∠PEA ∴△EDF∽△EPA. ∴=,∴EA•ED=EF•EP. 又∵EA•ED=CE•EB, ∴CE•EB=EF•EP; (II)∵DE2=EF•EC,DE=3,EF=2. ∴32=2EC,∴CE=. ∵CE:BE=3:2,∴BE=3. 由(I)可知:CE•EB=EF•EP,∴×3=2EP,解得EP=, ∴BP=EP-EB=-3=. ∵PA是⊙O的切线,∴PA2=PB•PC, ∴PA2=×(+),解得PA=. |