((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.