Question

已知正方体ABCD-A’B’C’D，求证：

AC’⊥B’ C
AC’⊥平面CB’D’

Answer 1

AB⊥面BCC'B'==>AB⊥B'C;
正方形对角线垂直，BC'⊥B'C;
故 面ABC'⊥B'C;
从而 AC'⊥B'C.

AA'⊥面A'B'C'D'==>AA'⊥B'D';
正方形对角线垂直，A'C'⊥B'D';
故 面AA'C'⊥B'D';
从而 AC'⊥B'D'.
于是 AC'⊥面CB'D'.