If FB‖AC is the diagonal of a square, then the quadrilateral ACBF is a parallelogram.
The connection of MN is MP+PN.
Since CM=BN=a,
NP‖AF and MP‖AD can be obtained,
Therefore, the plane composed of MNP is parallel to the plane composed of ADF.
That is, the AFD of MN‖ plane is proved.
(2)NP=(√2)a/2,PM= 1-(√2)a/2
mn=√[(np)^2+(mp)^2]=√{[(√2)a/2]^2+[ 1-(√2)a/2)]^2}
mn=√[a^2/2+ 1-(√2)a+a^2/2]=√[a^2+ 1-(√2)a]