![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/35a85edf8db1cb13943db6a1de54564e92584b28?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:(1)如图,连AC、BD,则由PA⊥底面ABCD,得平面PAC⊥底面ABCD于AC,
又由底面ABCD为菱形可得BD⊥AC于O,∴DO⊥平面PAC.
连OE,则OE为DE在平面PAC上的射影,∴∠DEO即为DE与平面PAC所成的角.
由E为PC的中点可得
EO=PA=
.
又由菱形的性质可得,在Rt△AOD中,
∠ADO=60°,AD=1,∴
DO=.
∴在Rt△DEO中,
tan∠DEO==,
∴∠DEO=30°.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4afbfbedab64034f4b9a48b4acc379310a551d28?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
(2)设AC∩BD=O,过O作OM⊥PC于M,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/9f2f070828381f30b14e071baa014c086e06f029?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
则由PA⊥底面ABCD可得
平面PAC⊥底面ABCD于AC.
又BD⊥AC,BD?底面ABCD,
∴BD⊥平面PAC,
∴BD⊥PC,
而由OM?平面PAC且OM⊥PC
可得PC⊥平面MBD.
故在线段PC上是存在一点M,使PC⊥平面MBD成立.
此时OM∥AE,且OM=
AE=
PC=