证明：连接MO．∵DB⊥A1A，DB⊥AC，A1A∩AC=A，∴DB⊥平面A1ACC1．又A1O⊂平面A1ACC1，∴A1O⊥DB．在矩形A1ACC1中，tan∠AA1O=22，tan∠MOC=22，∴∠AA1O=∠MOC，则∠A1OA+∠MOC=90°．∴A1O⊥OM．∵OM∩DB=O，∴A1O...