As shown in the figure, the edge length of cube abcd-a'b'c'd 'is a, e and F are the midpoint of edge AB and BC respectively The angle between a'd and BC ' The angle between BC 'and ef

Use the space vector
Let Da DC DD 'be the X, y, Z axis
a’（1,0,1）d（0,0,0）
Vector a'd = (- 1,0, - 1)
Similarly, the vector BC '= (- 1,0,1)
Cos angle = 0 / root sign 2 = 0, so the angle is 90 degrees. The second problem is to do the same

In the cube abcd-a1b1c1d1, ab = 2, point E is the midpoint of AD, and point F is on CD. If EF ∥ plane ab1c, then the length of the line EF is equal to that of the line=

∩ EF on the bottom surface ABCD ∩ surface ab1c = AC EF ∩ surface ab1c
∴EF∥AC
∵ e is the midpoint of AD, AC = 2 √ 2
∴EF=AC/2=√2

In cube ABCD, e and F are the midpoint of BC and CD respectively, and G is the midpoint of EF. Now they fold along AE, AF and EF to make B, C and D coincide,
The overlapped point is p, so there are several pairs of perpendicular edges in the six edges of tetrahedral a-efp

There are six pairs of edges perpendicular to each other, and three pairs are obtained directly from the method: (PA, PE), (PE, PF), (PF, PA). Furthermore, because PA is perpendicular to the plane PEF, PE is perpendicular to the plane PAF, PF is perpendicular to the plane PAE, (two intersecting lines perpendicular to the plane are perpendicular to this plane)

Known cube ABCD_ In a1b1c1d1, point E is the middle point of dd1. Verification: plane a1bd ‖ plane cb1d1

Proof: ∵ A1B ∥ CD1
A1B is not on plane cb1d1, CD1 is on plane cb1d1
{A1B} plane cb1d1
The same as plane cb1d1
A1B and BD are all on plane a1bd
A1B to BD at point B
Plane a1bd plane cb1d1
Get proof
I wish you success in your study

As shown in the figure, the edge length of cube abcd-a'b'c'd 'is a, e and F are the midpoint of edge AB and BC respectively The angle between a'd and BC ' The angle between BC 'and ef