In the cube abcd-a1b1c1d1, e and F are the midpoint of edge BC and c1d1 respectively （2）AC⊥EF (3) Plane aa1c1c ⊥ plane ab1d1

In the cube abcd-a1b1c1d1, e and F are the midpoint of edge BC and c1d1 respectively （2）AC⊥EF (3) Plane aa1c1c ⊥ plane ab1d1

(1) Take the midpoint h of b1c1 to connect FH and GH
∵ E and F are the midpoint of BC and c1d1, respectively
∴FH∥B1D1 HE∥BB1
‖ FH ‖ plane bdd1b1 he ‖ plane bdd1b1
Plane fhe plane bdd1b1
‖ EF ‖ plane bdd1b1
(2) Take point G of CD to connect Ge
∴EG∥BD ∴GE⊥AC
It was also found in the cube abcd-a1b1c1d1
∴HE⊥AC
Ge hand over he to e
Plane fceg
∴AC⊥FE
(3) In the cube abcd-a1b1c1d1
A1C1⊥B1D1 AA1⊥B1D1
Aa1 to a1c1 to A1
{b1d1 ⊥ plane aa1c
B1d1 is also contained in plane ab1d1
Plane ab1d1 ⊥ plane aa1c

In the cube abcd-a1b1c1d1, e and F are the midpoint of edge BC and c1d1 respectively

Let G be the midpoint of b1c1, with eg ‖ BB1, GF ‖ b1d1, EFG ‖ bdd1b1, EF ∈ EFG,
The plane bdd1b1

A right triangle, the ratio of its acute angle degree is 4:1, how many degrees is the difference between the two acute angles?
It's a practical problem in life

90 × (4-1) / (4 + 1) = 54 degrees