⑴由已知条件易证AC⊥AD又PA⊥面ABCD，AC∈面ABCD∴PA⊥AC又AD∩PA=面PAD∴AC⊥面PAD∵AC∈面AEC∴面AEC⊥面PAD⑵连BD，交AC于F则面PBD∩面AEC=EF∵PD∥面AEC∴PD∥EF∴△BEF∽△BPD∴PE:EB=DF:FBS△ABC=1/2，S△ACD=1∴S△ACD:S△ABC=2:1∴DF:FB=2:1∴PE:EB=2:1
