证明：（1）Because(B)：A1C1=B1C1=2 and D1是A1B1的中点
So(s):C1D1⊥A1B1
又B:A1ABB1⊥平面A1B1C1
S:C1D1⊥A1ABB1
又B:AB1包含于A1ABB1
S:AB1⊥C1D1
(2)连接BD1
B:C1D1⊥A1ABB1故C1D1⊥BD1.\1\
且C1D1⊥AB1.\2\
且C1D1//CD
S:CD⊥A1ABB1
S:CD⊥AB1.\3\
又B:异面直线AB1和C1B互相垂直（即AB1⊥C1B）.\4\
由\2\和\4\知AB1⊥平面BC1D1
S:AB1⊥BD1 又因为A1D//BD1
S:AB1⊥A1D.\5\
由\3\和\5\即得AB1⊥面A1CD
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