Question

已知在正方体ABCD-A1B1C1D1中,O是下底面ABCD的中心,E是CC1中点,证明：A1O⊥平面EBD

Answer 1

令正方体边长为a,则DO=（√2/2）a,A1O=√（A1A^2+AO^2）=（√6/2）a,A1D=√（A1A^2+AD^2）=√2a=√（A1O^2+DO^2）,所以A1O⊥DB,

同理可得A1E^2=A1O^2+OE^2,所以A1O⊥OE,

所以A1O⊥平面EBD