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如图,在三棱台ABC

解:(I)∵侧棱AA1⊥平面ABC,AB?平面ABC,∴AA1⊥AB,又∵∠BAC=90°∴AB⊥AC,AA1∩AC=A,从而AB⊥平面AA1C1C…(4分)(II)由(I)可知AB⊥平面AA1C1C,C1C?平面AA1C1C,∴C1C⊥AB又∵C1C⊥BC1并且AB∩BC1=B,∴C1C⊥平面ABC1…(8分)(III)连接A1B,∵AC∥A1C1∴...

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