Given a rectangular parallelepiped ABCDA1B1C1D1 Prove that edge A1D1 is perpendicular to plane D1DCC1.

In a rectangular parallelepiped, all opposite faces are parallel to each other in pairs. If we designate the vertices of the front face clockwise as ABCD, then we see the following picture:

1. Edge A1D1 is located on the opposite face to ABCD. In this case, A1D1 is parallel to AD, which in turn is perpendicular to DC of the D1DCC1 face.

2. Therefore A1D1 is perpendicular to D1C1 of the face of D1DCC1. That is A1D1 and this face itself, and hence the plane in which the D1DCC1 face is located.




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