(2012海南数学)如图,直三棱柱ABC﹣A1B1C1中,AC=BC=1/2AA1,D是棱AA1的中点,DC1⊥BD

（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
∵OH⊥BD,∴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= 2 a 2 ,C1D= 2 a=2C1O,
∴sin∠C1DO=1 2∴∠C1DO=30°
即二面角A1-BD-C1的大小为30°
向量法解（1）
DC1=CD+CC1=CC1+CA+AD=CC1+CA+1/2AA1
DC1*BC=(CC1+CA+1/2AA1)*BC=(CC1+C1D)*BC=CC1*BC+C1D*BC =0+0=0
所以DC1⊥BC