(1)
易证
AA1⊥底面
CC1⊥底面
AA1与CC1共面
AA1C1C⊥底面于AC
又AC⊥BD
BD⊥AA1C1C
BD⊥A1C
(2)
AA1C1C交BA1D于A1E
AA1C1C交BC1D于C1E
故A1BD与BC1D交角是A1EC1
直角三角形ADC中
AD=2,DC=2sqrt(3),AC=4
ED⊥AC于E,AE=1,EC=3
A1E=sqrt(3+1)=2
C1E=sqrt(3+9)=2sqrt(3)
A1C1=AC=4
A1C^2=A1E^2+C1E^2-2A1E*C1E*cos(∠A1EC)
16=4+12-4sqrt(6)*cos(∠A1EC)
cos(∠A1EC)=0
∠A1EC=π/2(直角)
(3)
AD=AB=2
AE⊥BD
BAD是等腰三角形
从而BCE是等腰三角形
易知∠ACB=π/6
过B做BM‖AD交CD于M
易知∠MBC=π/6,BM⊥MC1
BM=3
BC1=sqrt(15)
cos(AD与BC1交角)=cos(MBC1)=MB/C1B=sqrt(15)/5
AD与BC1交角=aercos(sqrt(15)/5)
