在a-bcd中,bc⊥ad,∠acb=∠bcd=45° ,∠acd=60°,求面abc⊥面bcd

过点A作AE⊥CD,垂足为E,作AF⊥BC,垂足为 F
在Rt△ADE中,∠ADC=45°
∵AD=2√2
∴AE=ED=2
∵AC平分∠BCD=60°
∴∠ACD=∠ACB=30°,AE=AF=2
在Rt△ACE和Rt△ACF中
FC=2√3,CE=2√3
在Rt△AFB中
∵AB=2√2,AF=2
∴BF=2
则四边形ABCD的面积
=S△ACD+S△ABC
=S△ACD+S△AFC-S△ABF
=1/2×（CE+ED)×AE+1/2×AF×FC-1/2×AF×BF
=1/2×（2+2√3）×2+1/2×2×2√3-1/2×2×2
=2+2√3+2√3-2
=4√3请采纳回答