如图,四棱锥P-ABCD的底面是正方形,PD⊥底面ABCD,点E在棱PB上,(1)求证平面AEC

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  • ((1)证明:∵四边形ABCD是正方形ABCD,∴AC⊥DB.

    ∵PD⊥面ABCD,AC⊂面ABCD,

    ∴PD⊥AC,

    ∵PD∩PB=P,

    ∴AC⊥面PDB,

    ∵AC⊂面AEC,

    ∴平面AEC⊥平面PDB;

    设AC∩BD=O,则AO⊥BD

    ∵AO⊥PD,BD∩PD=D,

    ∴AO⊥面PDE,

    ∵AO=1,VA-PED=

    1

    3

    •AO•S△PDE=

    1

    3

    ,

    ∴S△PDE=1

    在直角三角形ADB中,DB=PD=2,则PB=

    2

    ∴Rt△PDB中斜边PB的高h=

    2

    ,

    1

    2

    •h•PE=1,

    ∴PE=

    2

    ,

    PE

    EB

    =1

    即E为PB的中点.