已知正方体ABCD-A’B’C’D,求证:AC’⊥B’ CAC’⊥平面CB’D’

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已知正方体ABCD-A’B’C’D,求证:
AC’⊥B’ C
AC’⊥平面CB’D’

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'.