The answer to the second question on page 58 of compulsory mathematics in senior one.

By connecting D 1B 1, triangles A 1B 1D 1 and C 1B 1D 1 can be obtained, and because n, m, e, It is easy to get MF/= a1d1by connecting MF. Because a1d1= ad, MF/= AD, the quadrilateral ADFM is a parallelogram, so AM//DE, because EF is contained in the plane EFBD, so NM// plane EFBD, similarly AM/.

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