(Ⅰ)证明AA1⊥BC,只需证明BC⊥平面OO1A1A,取BC,B1C1的中点为点O,O1,连接AO,OO1,A1O,A1O1,即可证得;
(Ⅱ)延长A1O1到D,使O1D=OA,则可得AD∥OO1,AD=OO1,可证OO1⊥面A1B1C1,从而AD⊥面A1B1C1,即可求AA1的长;
(Ⅲ)证明∠AOA1是二面角A-BC-A1的平面角,在直角△OAA1中,利用余弦定理,可求二面角A-BC-A1的余弦值.
(Ⅰ)证明:取BC,B1C1的中点为点O,O1,连接AO,OO1,A1O,A1O1,
∵AB=AC,∴AO⊥BC
∵平面ABC⊥平面BB1C1C,平面ABC∩平面BB1C1C=BC
∴AO⊥平面BB1C1C
同理A1O1⊥平面BB1C1C,∴AO∥A1O1,∴A、O、A1、O1共面
∵OO1⊥BC,AO⊥BC,OO1∩AO=O,∴BC⊥平面OO1A1A
∵AA1⊂平面OO1A1A,∴AA1⊥BC;
(Ⅱ)【解析】
延长A1O1到D,使O1D=OA,则∵O1D∥OA,∴AD∥OO1,AD=OO1,
∵OO1⊥BC,平面A1B1C1⊥平面BB1C1C,平面A1B1C1∩平面BB1C1C=B1C1,
∴OO1⊥面A1B1C1,
∵AD∥OO1,
∴AD⊥面A1B1C1,
∵AD=BB1=4,A1D=A1O1+O1D=2+1=3
∴AA1==5;
(Ⅲ)【解析】
∵AO⊥BC,A1O⊥BC,∴∠AOA1是二面角A-BC-A1的平面角
在直角△OO1A1中,A1O=
在直角△OAA1中,cos∠AOA1=-
∴二面角A-BC-A1的余弦值为-.