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解：将AB与CD的交点设为O，设∠BCP＝∠1, ∠DCP＝∠2, ∠BAP＝∠3, ∠DAP＝∠4∵AP平分∠DAB∴∠3＝∠4∴∠DAB＝2∠3∵CP平分∠BCD∴∠1＝∠2∴∠BCD＝2∠1∵∠BOD＝∠B+∠BCD＝∠B+2∠1, ∠BOD＝∠D+∠DAB＝∠D+2∠3∴∠B+2∠1＝∠D+2∠3∴∠1-∠3＝（∠D-∠B）/2∵∠BNP＝∠B+∠1, ∠BNP＝∠P+∠3∴∠B+∠1＝∠P+∠3∴∠1-∠3＝∠P-∠B∴∠P-∠B＝（∠D-∠B）/2∴∠P＝（∠D-∠B）/2+∠B＝（∠B+∠D）/2∵∠B＝30, ∠D＝40∴∠P＝（30+40）/2＝35°这是我之前做... |
