Question

如图,正方体ABCD-A1B1C1D1中,E是AA1中点 求证：平面C1BD⊥平面BDE

Answer 1

设正方体边长为2,取BD中点为F,连接EF、C1F、C1E.在Rt△EAF中,可知AE=1,AF=根号2,则得到EF=根号3；在Rt△CC1F中,可知CC1=2,CF=根号2,则得到C1F=根号6；在Rt△A1C1E中,可知A1E=1,A1C1=2根号2,则得到C1E=3；因为EF²...