In the cube abcd-a'b'c'd ', find the size of dihedral angle c' - b'd '- A

In the cube abcd-a'b'c'd ', find the size of dihedral angle c' - b'd '- A

Link a'c ', b'd', intersect o '. Link ab', AC ', Ao'
B'O'⊥A'C',
If o 'is the midpoint of B'd', then Ao '⊥ B'd',
be

In the cube abcd-a1b1c1d1, e is the midpoint of edge Aa1. Find the dihedral angle of plane eb1c and plane ABCD

As shown in the figure, if the square abb1a1 is extended one time to the left, then ch is the intersection of the plane eb1c and the plane ABCD, make BF ⊥ ch, then B1F ⊥ CH (three vertical line), ∠ b1fb is the plane angle of the dihedral angle, and let AB = 1, then ch = √ 5. BF = BC × BH / CH = 2 / √ 5. Tan ∠ b1fb = BB1 / BF = √ 5 / 2

It is known that in the cube abcd-a1b1c1d1, e and F are the midpoint of Aa1 and CC1 respectively

Prove that two faces are parallel! Then it must be proved that there are two intersecting lines in one face which are parallel to the other face respectively! Then the two faces are parallel!
Drawing first (omitted)
Because cube AC1 (cube can be represented by volume diagonal)
So aa1b1b / / dd1c
And because EF is the midpoint of Aa1 and CC1
So DF / / b1e
So EB1 / / face DBF
Because dd1 is parallel and equal to BB1
So the quadrilateral bb1d is a parallelogram!
So dB / / d1b1
So b1d1 / / DBF
And because b1d1 and b1e intersect at B1
So face BDF / / face b1d1e

If e is the midpoint of BC, then the vector AE * is the vector CD=
In space, if the edge length of oabc is 1, then the vector OC * AB =?

(1) E is the midpoint of BC
2 vector AE = vector AB + vector AC
2 vector AE. Vector CD
=(vector AB + vector AC). (vector Ad vector AC)
=Vector ab. vector ad - vector ab. vector AC + vector AC. vector ad-ac & # 178;
=1*1*cos60°-1*1*cos60°+1*1*cos60°-1*1
=1/2-1
=-1/2
Ψ vector AE * vector CD = - 1 / 4
(2) Vector OC * vector ab
=Vector OC. (vector ob vector OA)
=Vector OC. Vector ob vector OC. Vector OA
=1*1*cos60°-1*1*cos60°
=0