解:(Ⅰ) ∵ PA = PD = 1 ,PD = 2 , ∴ PA2 + AD2 = PD2, 即:PA ⊥ AD 又PA ⊥CD , AD , CD 相交于点D, ∴ PA ⊥平面ABCD (Ⅱ)过E作EG∥PA 交AD于G,从而EG ⊥平面ABCD, 且AG = 2GD , EG = PA = , 连接BD交AC于O, 过G作GH∥OD ,交AC于H,连接EH. ∵GH ⊥AC , ∴EH ⊥AC , ∴∠ EHG为二面角D-AC-E的平面角. ∴tan∠EHG = = . ∴二面角D-AC-E的平面角的余弦值为- (Ⅲ)以AB , AD , PA为x轴、y轴、z轴建立空间直角坐标系. 则A(0 ,0, 0),B(1,0,0) ,C(1,1,0),P(0,0,1),E(0 , ), = (1,1,0), = (0 , ) 设平面AEC的法向量= (x, y,z) , 则 , 即:, 令y = 1 , 则 = (- 1,1, - 2 ) 假设侧棱PC上存在一点F, 且= , (0 ≤ ≤ 1), 使得:BF∥平面AEC, 则· = 0. 又因为:= + = (0 ,1,0)+ (-,-,)= (-,1-,), ∴· =+ 1- - 2 = 0 , ∴ = , 所以存在PC的中点F, 使得BF∥平面AEC.
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