M,N,P分别是正方体ABCD—A1B1C1D1的棱CC1,BC,DC的点,求证:A1P垂直平面DMN

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  • 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