In the cube ABCD-A 'B' C'd ', if the midpoint of BC is e and the midpoint of CD is f, then the angle between ad' and EF is

In the cube ABCD-A 'B' C'd ', if the midpoint of BC is e and the midpoint of CD is f, then the angle between ad' and EF is

E→M,F→N,'→1
Connect BD, BC1, DC1, Mn ‖ BD, AD1 ‖ BC1,
The angle between AD1 and Mn is dbc1
The triangle dbc1 is an equilateral triangle
∴∠DBC1=60°

In the known parallelepiped abcd-a1b1c1d1, the bottom surface is a square and the side edge Aa1 is B,

Through a, make AE ⊥ A1B1 intersect A1B1 to e, AF ⊥ a1d1 intersect a1d1 to F, then make Ao ⊥ plane a1b1c1d1 intersect plane a1b1c1d1 to o, connect EO, fo. ∵ Aa1 = Aa1, ∵ aa1e = aa1f = 60 °, ∵ aea1 = ∵ afa1 = 90 °, ∵ AE = AF. ∵ O is the projection of a on plane a1b1c1d1, ∵ EO, fo are ae

What is the analogy between parallelogram ABCD AC ^ 2 + BD ^ 2 = 2Ab ^ 2 and parallelogram abcd-a'b'c'd '
AC ^ 2 + BD ^ 2 = 2Ab ^ 2 to parallelepiped abcd-a'b'c'd'a'c ^ 2 + b'd ^ 2 + c'a ^ 2 + d'a ^ 2 =?

The formula of parallelogram you wrote is wrong: AC & sup2; + BD & sup2; = 2Ab & sup2; + 2CD & sup2;
In the parallelogram ABCD, AC and BD are the diagonals of the parallelogram ABCD, then the sum of squares of the four sides is equal to the sum of squares of the diagonals
In the straight parallel hexahedron abcd-a'b'c'd ', the sum of squares of four diagonals is equal to the sum of squares of each side (a'c & sup2; + b'd2 + c'a & sup2; + d'b & sup2; = 4AB & sup2; + 4CD & sup2; + 4a'a & sup2;)

Why is the vector AB + ad = AC?

Let vector AB = vector a, vector ad = vector B,
For the parallelogram ABCD, the parallelogram rule of vector addition is as follows:
Vector AB + vector ad = vector AC
The triangle rule of vector addition is
Since vector ad = vector BC
Vector AB + vector ad = vector AB + vector BC = vector AC