Given a parallelepiped ABCDA1B1C1D1, the base of which is the rhombus ABCD, and the lateral edge
| May 4, 2021 | education
Given a parallelepiped ABCDA1B1C1D1, the base of which is the rhombus ABCD, and the lateral edge is perpendicular to the plane of the base. Prove that the diagonal B1D of the parallelepiped is perpendicular to the diagonal AC of its base.
Since the side edges of the parallelepiped are perpendicular to the plane of its base, the parallelepiped is rectangular. Then the projection of the В1D diagonal onto the plane of the parallelepiped will be the ВD diagonal at the base of the parallelepiped.
AВСD is a rhombus, then its diagonals intersect at right angles. Since the projection of the ВD is perpendicular to the AС, then the inclined V1D is perpendicular to the AС, which was required to be proved.
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