(2014?贵州二模)已知直三棱柱ABC-A1B1C1中,∠ACB=π2.AC=CB=AA1=2,E为BB1的中点,D在
(2014?贵州二模)已知直三棱柱ABC-A1B1C1中,∠ACB=π2.AC=CB=AA1=2,E为BB1的中点,D在AB上,且∠A1DE
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(2014?贵州二模)已知直三棱柱ABC-A
1B
1C
1中,∠ACB=[π/2].AC=CB=AA
1=2,E为BB
1的中点,D在AB上,且∠A
1DE=[π/2].
(Ⅰ)求证:CD⊥面ABB
1A
1;
(Ⅱ)求二面角D-A
1C-A的正弦值.