证明：连结PA交DN于E,PD'交DC'于F
在△ADP与△DCN中
∵AD=DC,∠ADP=∠DCN=90°,DP=CN
∴△ADP≌△DCN
∠APD=∠DNC
∠AED=∠APD+∠PDN=∠DNC+∠PDN=90°
∴PA⊥DN
又∵AA'⊥面ADN,DN∈面ADN
∴AA'⊥DN
∵DN⊥AA',DN⊥PA,AA'∩PA于A
∴DN⊥面PAA'
而PA'∈面PAA'
∴DN⊥PA'
在△D'DP与△DCM中
∵D'D=DC,∠D'DP=∠DCM=90°,DP=CM
∴△D'DP≌△DCM
∠D'PD=∠DMC
∠D'FD=∠D'PD+∠PDM=∠DMC+∠PDM=90°
∴PD'⊥DM
又∵A'D'⊥面D'DM,DM∈面D'DM
∴A'D'⊥DM
∵DM⊥A'D',DM⊥PD',A'D'∩PD'于D'
∴DM⊥面PA'D'
而PA'∈面PA'D'
∴DM⊥PA'
∵PA'⊥DN,PA'⊥DM,DN∩DM于D
∴PA'⊥面DMN