图15
求证:A1O⊥平面GBD.
证明:
又∵A1O2=A1A2+AO2=a2+(a)2=a2,OG2=OC2+CG2=(a)2+()2=a2,
A1G2=A1C12+C1G2=(2a)2+()2=a2,
∴A1O2+OG2=A1G2.
∴A1O⊥OG.又BD∩OG=O,∴A1O⊥平面GBD.
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