ABC-A1B1C1是正三棱柱,且AB=AA1,∴ABBA1是正方形,∴AB1⊥A1B.
∵AC=B1C1,CD=C1D,∠ACD=∠B1C1D=90°,∴△ACD≌△B1C1D,∴AD=B1D.
令AB1与A1B的交点为E,显然有:AE=B1E,而AD=B1D,∴AB1⊥DE.
由AB1⊥A1B,AB1⊥DE,A1B∩DE=E,∴AB1⊥面A1BD,∴BD⊥AB1.
ABC-A1B1C1是正三棱柱,且AB=AA1,∴ABBA1是正方形,∴AB1⊥A1B.
∵AC=B1C1,CD=C1D,∠ACD=∠B1C1D=90°,∴△ACD≌△B1C1D,∴AD=B1D.
令AB1与A1B的交点为E,显然有:AE=B1E,而AD=B1D,∴AB1⊥DE.
由AB1⊥A1B,AB1⊥DE,A1B∩DE=E,∴AB1⊥面A1BD,∴BD⊥AB1.