The bottom P_ABCD of the quadrangular pyramid is an ABCD square, the vertical plane of PD ABCD, e is a point on PB, and 2BE = EP；;

( 1)。 Prove AC vertical de; (2)。 If PC=√2? BC finds the cosine of dihedral angle e-ac-p.

Solution: (1). Proof: ∵PD⊥ plane ABCD, AC? Aircraft ABCD, ∴ AC ⊥ PD; Connect BD, ABCD is square,

∴ac⊥bd； Therefore, AC⊥ aircraft pdb；; De? Plane pdb aircraft

(2)。 Let the intersection of two diagonal lines of the bottom square ABCD be O. Connect PO and EO; ∵PD⊥ plane ABCD, PD=PD,

AD=CD, ∴RT△PDA≌RT△PDC, ∴AP=CP, that is, △PAC is an isosceles triangle; O is the midpoint of AC, so po

⊥ac； It turns out that AC⊥ aircraft PDB, EO? Planar PDB, ∴ EO ⊥ AC; So ∠POE is the flatness of dihedral angle e-AC-p.

Face angle.

∫PC =√2？ BC=√2？ cd,∴∠pcd=45； Let PD=CD= 1, then BD = √ 2; PB？ =PD？ +BD？ = 1+2=3;

PE =(2/3)PB =(2/3)√3； Po. =PD？ +DO？ = 1+ 1/2=3/2; BE =( 1/3)PB =( 1/3)√3； BO = BD/2 =√2/2；

cos∠PBD = BD/PB =√(2/3)；

EO？ = Bo? +BE？ -2BO？ Bekos ∠ PBD =1/2+(1/3)-2× (√ 2/2) × (√ 3/3 )× √ (2/3) =1/6;

So cos∠POE=(PO? +EO？ -Sports? )/(2po×EO)=(3/2+ 1/6-4/3)/[2×√( 3/2)×√( 1/6)]= 1/3。