求解一道高二数学平面与平面垂直的判定题
1个回答

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

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