摘要:面∴AD⊥CC1. ------6(Ⅱ)延长B1A1与BM交于N, 连结C1N. ∵AM=MA1, ∴NA1=A1B1. ∵A1B1=A1C1, ∴A1C1= A1N=A1B1. ∴C1N⊥C1B1. ------9∵截面N B1C1⊥侧面BB1C1C,

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