1.连AC
∵∠ABC=∠BAD=90°
AB=BC=AD/2
∴CD⊥AC
∵PA⊥面ABCD
∴PA⊥CD
∴CD⊥面PAC
又CD∈面PCD
∴面PCD⊥面PAC
2.延长DA至F,使AF=AB,连PF
则∠BPF为异面直线AE与PB所成的角,设为α
设AB=1
则BF=√2,PB=PF=2
cosα=(PB²+PF²-BF²)/(2·PB·PF)=6/8=3/4
1.连AC
∵∠ABC=∠BAD=90°
AB=BC=AD/2
∴CD⊥AC
∵PA⊥面ABCD
∴PA⊥CD
∴CD⊥面PAC
又CD∈面PCD
∴面PCD⊥面PAC
2.延长DA至F,使AF=AB,连PF
则∠BPF为异面直线AE与PB所成的角,设为α
设AB=1
则BF=√2,PB=PF=2
cosα=(PB²+PF²-BF²)/(2·PB·PF)=6/8=3/4