(1)证明DC1⊥BC,只需证明DC1⊥面BCD,即证明DC1⊥DC,DC1⊥BD;
(2)证明BC⊥面ACC1A1,可得BC⊥AC取A1B1的中点O,过点O作OH⊥BD于点H,连接C1O,C1H,可得点H与点D重合且∠C1DO是二面角A1-BD-C1的平面角,由此可求二面角A1-BD-C1的大小.
(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)【解析】
∵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,,
∴sin∠C1DO=
∴∠C1DO=30°
即二面角A1-BD-C1的大小为30°