In the cube abcd-a1b1c1d1, e, F and G are the midpoint of A1B1, b1c1 and B1B respectively

In the cube abcd-a1b1c1d1, e, F and G are the midpoint of A1B1, b1c1 and B1B respectively

In the cube abcd-a1b1c1d1, there are several diagonals of the plane 60 degrees to AD1

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A points out two, D1 points out two
They have two parallel lines on the opposite side

In the cube abcd-a1b1c1d1, the number of diagonal lines of the face at 60 ° angle to AD1 is ()

AD1 is on the plane addiai. The surface 60 degrees from AD1 is a quadrilateral. Therefore, there are two diagonals

In rectangle abcd-a1b1c1d1, ad = Aa1 = 1, ab = 2, point e moves on edge ab
（1）D1E⊥A1D；
(2) When e is the midpoint of AB, calculate the distance from point e to plane acd1;
(2) The dihedral angle d1-ec-d is 45 ° when the AE value is 0

It is proved that: (1) in rectangle aa1d, ad = Aa1 = 1
Then the rectangle aa1d is a square
So a1d ⊥ AD1
And ab ⊥ plane aa1d, a1d in plane aa1d
Then ab ⊥ a1d
Because AB and AD1 are two intersecting lines in the plane abc1d1
Therefore, from the judgment theorem of perpendicularity of lines and planes, we can get the following results:
A1d ⊥ plane abc1d1
Because d1e is in plane abc1d1
So d1e ⊥ a1d
(2) Set the distance from point e to plane acd1 as D
Then, the volume of d1-ace is: 1
V＝1/3 *d*(S_ △ACD1)=1/3 *DD1*(S_ △ACE)
That is, d = dd1 * (s)_ △ACE)/(S_ △ACD1)=(S_ △ACE)/(S_ △ACD1) （×）
The following is the area of △ acd1
Make ad point O, connect Co
In the cuboid, ad = Aa1 = 1, ab = 2
Diagonals AC = CD1 = √ 5, AD1 = √ 2
Then Ao = od = √ 2 / 2, so:
Co ⊥ AD1 and from Pythagorean theorem: CO = √ (5-1 / 2) = (3 √ 2) / 2
So s_ △ACD1=1/2 *AD1*CO=1/2 *√2*(3√2)/2=3/2
S again_ △ACE＝1/2 *BC*AE=1/2 *1*1=1/2
From the above formula (×), it can be concluded that:
d=(S_ △ACE)/(S_ △ACD1) =(1/2)/(3/2)=1/3
That is, when e is the midpoint of AB, the distance from point e to plane acd1 is 1 / 3
(3) Through point d make DP ⊥ CE, foot drop is p, connect d1p, De
Because d1d ⊥ plane ABCD, the projection of d1p in plane ABCD is DP
And in the plane ABCD, DP ⊥ CE
From the three perpendicular theorem, we can get: d1p ⊥ CE
Therefore, dpd1 is the plane angle of dihedral angle d1-ec-d
The dihedral angle d1-ec-d is 45 degrees, i.e. dpd1 = 45 degrees
It is easy to know that △ dd1p is an isosceles right triangle
So DP = dd1 = ad = 1
And De is the common edge of RT △ DAE and RT △ DPE
So RT △ DAE ≌ RT △ DPE (HL)
Then AE = PE
Let AE = PE = a
Then be = ab-ae = 2-A
From Pythagorean theorem: CE = √ (be & # 178; + BC & # 178;) = √ [(2-A) &# 178; + 1]
In RT △ DPC, CD = 2, DP = 1, then CP = √ 3
Because CE = CP + PE
So √ [(2-A) &# 178; + 1] = √ 3 + a
That is, (2-A) & # 178; + 1 = (√ 3 + a) & # 178;
4-4a+a²+1=3+2√3*a+a²
2(2+√3)a=2
The solution is a = 2 - √ 3
So when AE is 2 - √ 3, the dihedral angle d1-ec-d is 45 degrees