连AC,BD交于点M,ME为平面ACC1A1与平面BDE的交线,设A1C过平面DEB交于点F则F必在交线ME上
(平面ACD与直线BD)
∵AC⊥BD,AA1⊥平面ABCD=>AA1⊥BD
AA1∩AC=A
∴BD⊥平面AA1C ∴BD⊥A1C
(平面ACC1A1内,RT△MCE与RT△AA1C)
CE:AC=1:2√2=CM:AA1=√2:4
所以△MCE∽△AA1C=>∠CME=∠AA1C ∠MEC=∠CME
所以∠CFM=∠CAA1=90°即A1C⊥ME
BD∩ME=M所以A1C⊥平面BDE