因为A1C1∥AC,AC在面AB1C内
所以A1C1∥面AB1C
C1到AB1C的距离为A1到AB1C的距离
过A1作A1M⊥AB1于M
由于AC⊥AB,AC⊥AA1
AC⊥面AA1B1B
A1M⊥AC,又A1M⊥AB1
得A1M⊥面AB1C
因为AB=BB1=1,AB1=√2,A1M=√2/2
因为A1C1∥AC,AC在面AB1C内
所以A1C1∥面AB1C
C1到AB1C的距离为A1到AB1C的距离
过A1作A1M⊥AB1于M
由于AC⊥AB,AC⊥AA1
AC⊥面AA1B1B
A1M⊥AC,又A1M⊥AB1
得A1M⊥面AB1C
因为AB=BB1=1,AB1=√2,A1M=√2/2