证明：过点p作PD⊥OA，PE⊥OB，，∴∠NDP=∠MEP=90°∵∠OMP+∠ONP=180°，∠OMP+∠PME=180°∴∠ONP=∠PME．∵OP平分∠AOB，PD⊥OA，PE⊥OB，∴PD=PE．在△PDN和△PEM中∠PDN＝∠PEM∠PND＝∠PMEPD＝PE，∴△PDN≌...