(1)证明:在Rt△DAC中,AD=AC,∴∠ADC=45° 同理:∠A1DC1=45°,∴∠CDC1=90° ∴DC1⊥DC,DC1⊥BD ∵DC∩BD=D ∴DC1⊥面BCD ∵BC⊂面BCD ∴DC1⊥BC (2)∵DC1⊥BC,CC1⊥BC,DC1∩CC1=C1,∴BC⊥面ACC1A1, ∵AC⊂面ACC1A1,∴BC⊥AC 取A1B1的中点O,过点O作OH⊥BD于点H,连接C1O,OH ∵A1C1=B1C1,∴C1O⊥A1B1, ∵面A1B1C1⊥面A1BD,面A1B1C1∩面A1BD=A1B1, ∴C1O⊥面A1BD 而BD⊂面A1BD ∴BD⊥C1O, ∵OH⊥BD,C1O∩OH=O, ∴BD⊥面C1OH∴C1H⊥BD,∴点H与点D重合且∠C1DO是二面角A1-BD-C1的平面角 设AC=a,则C1O=,C1D=a=2C1O, ∴sin∠C1DO= ∴∠C1DO=30° 即二面角A1-BD-C1的大小为30°
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