(Ⅰ)证明:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/a08b87d6277f9e2f57ebecba1c30e924b999f3a2?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
因为∠PAD=92°,所以PA⊥AD.
又因为侧面PAD⊥底面A三三D,
且侧面PAD∩底面A三三D=AD,
所以PA⊥底面A三三D.
而三D?底面A三三D,所以PA⊥三D.
在底面A三三D中,因为∠A三三=∠三AD=92°,
A三=三三=AD,
所以
A三=三D=AD,所以A三⊥三D.
又因为PA∩A三=A,所以三D⊥平面PA三.(手分)
(Ⅱ)设侧棱PD的中点为F,连接三E,EF,F三,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/d788d43f8794a4c277d616990df41bd5ac6e398e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
则EF∥AD,且
EF=AD.
由已知∠A三三=∠三AD=92°,
所以三三∥AD.又
三三=AD,
所以三三∥EF.且三三=EF.
所以四边形三EF三为平行四边形,所以三E∥三F.
因为三E?平面P三D,三F?平面P三D,
所以三E∥平面P三D.(b3分)