在三棱柱ABC-A1B1C1中,AB=AC=2AA1,∠BAA1=∠CAA1=60°,D,E分别是AB,A1C中点,求证：BB1垂直于平面A1BC

设AA1=1,AB=AC=2,在三角形A1AB中,〈A1AB=60°,根据余弦定理,A1B=√3,AA1^2+A1B^2=4,AB^2=4,根据勾股逆定理,△A1AB是RT△,〈AA1B=〈B1BA1=90°,则BB1⊥A1B,同理,〈AA1C=90°,AA1⊥A1C,BB1//AA1,则BB1⊥A1C,A1C∩A1B=A...