(Ⅰ)求证:CB1⊥平面ABC1;
(Ⅱ)求证:MN∥平面ABC1.
侧面BB1C1C⊥底面ABC,且侧面BB1C1C∩底面ABC=BC,
∵∠ABC=90°,即AB⊥BC,
∴AB⊥平面BB1C1
∵CB1⊂平面BB1C1C