连接AD1和AB1以及A1D和A1B根据正方体的性质D1C1⊥A1D,A1D⊥AD1则A1D⊥△C1D1A则有A1D⊥AC1又F,G为AD,AA1中点FG‖A1D所以FG⊥AC1同理GE⊥AC1所以AC1⊥△EFG
∵∴⊥∵ABCD是正方形∴BD⊥AC∵BD⊥CC1∴BD⊥面ACC1∴BD⊥AC1∵EF//BD∴EF⊥AC1同理可证EG⊥AC1∴AC1⊥面EFG