In the straight triangular prism abc-a1b1c1, ad ⊥ plane a1bc, its perpendicular foot D falls on the straight line A1B, P is the midpoint of AC, (1) Verification: B1C ‖ plane a1bp (2) Verification: BC ⊥ A1B

In the straight triangular prism abc-a1b1c1, ad ⊥ plane a1bc, its perpendicular foot D falls on the straight line A1B, P is the midpoint of AC, (1) Verification: B1C ‖ plane a1bp (2) Verification: BC ⊥ A1B

prove:
(1) Connect Ab1 to A1B to Q, connect PQ
The quadrilateral aa1b1b is a parallelogram
Q is the midpoint of Ab1 and P is the midpoint of AC,
And PQ is in plane a1bp, B1C is not in plane a1bp
Ψ B1C ‖ plane a1bp
(2) ∵ the triangular prism abc-a1b1c1 is a straight triangular prism
⊥ A1A ⊥ plane ABC, and BC is in plane a1bc
∴A1A⊥BC
∵ ad ⊥ plane a1bc and BC in plane a1bc
∴AD⊥BC
And A1A is in plane a1ab, ad is in plane a1ab, and A1A ∩ ad = a
And A1B is in plane a1ab
∴BC⊥A1B

In the triangular prism abc-a1b1c1, Aa1 ⊥ BC, ∠ a1ac = 60 degrees, A1A = AC = BC = 1. A1B = √ 2 (1) prove: plane a1bc
In the triangular prism abc-a1b1c1, Aa1 ⊥ BC, ∠ a1ac = 60 degrees, A1A = AC = BC = 1. A1B = √ 2 (1) prove: (1) plane a1bc ⊥ plane acc1a1 (2) if D is the midpoint of AB, prove: B1C is parallel to plane a1cd
As shown in figure (1), e and F are the middle points of AB and AC on ABC side of right triangle respectively. ∠ B = 90 degrees. Along EF, the triangle ABC is folded into the Riel free intersection a1-ef-b as shown in figure (2). If M is the middle point of line A1C, it is proved that the line FM is parallel to plane a1eb (2) plane a1fc ⊥ plane a1bc

1
(1) . proof: connect A1C
In △ aa1c:
∵∠A1AC=60°,AC=A1C=1
The equilateral triangle is aa1c
∴A1C=1
In △ a1bc:
∵A1C²+BC²=A1B²
Ψ△ a1bc is a right triangle, and ∠ a1cb = 90 degree
∴BC⊥A1C
And ∩ BC ⊥ Aa1, and Aa1 ∩ A1C = A1
Ψ BC ⊥ plane aa1c
Plane a1bc
Plane a1bc ⊥ plane aa1c
(2) Prove: connect B1C and BC1 to e, let the midpoint of A1C be f, connect EF and de
∵ the quadrilateral bb1c is a planar quadrilateral
E is the midpoint of B1C
And ∵ f is the midpoint of A1C
Ψ EF is the median line of △ a1b1c
∴EF//A1B1,EF=(1/2)A1B1
And ∵ A1B1 / / AB, A1B1 = AB, D is the midpoint of ab
Ψ EF / / BD, and EF = BD
The BDFE is a parallelogram
That is, DF / / BC1
The plane a1cd
｜ BC1 / / plane a1cd
2
(1) . proof: connect Aa1
∵ F, M is the midpoint of AC, A1C
Ψ MF is the median of △ aa1c
∴MF//AA1
Aa1 and a1eb
Ψ FM / / plane a1eb
(2) . certification:
∵ e, f are the midpoint of AB, AC
∴EF//BC
∴EF⊥AB
∵ the figure is folded by EF
And ef ⊥ be
And ∩ AE ∩ be = E
Ψ EF ⊥ plane aa1b
Ψ BC ⊥ plane aa1b
∵ Aa1 &; plane aa1b
∴BC⊥AA1
∵AA1//MF
∴BC⊥MF
∵ f is the midpoint of AC
That is to say, △ a1fc is an isosceles triangle
M is the midpoint of A1C
∴MF⊥A1C
That is MF ⊥ A1C, MF ⊥ BC, A1C ∩ BC = C
Ψ MF ⊥ plane a1bc
The plane a1fc
Plane a1fc ⊥ plane a1bc

In the regular triangular prism abc-a1b1c1, if AB = 2 and a & nbsp; A1 = 1, then the distance from point a to plane a1bc is ()
A. 34B. 32C. 334D. 3

Let the distance from point a to plane a1bc be h, then the volume of the triangular pyramid va1 − ABC is va1 − ABC = VA − a1bc, that is, 13s △ ABC · Aa1 = 13s △ a1bc · h ｛ 13 · 3 · 1 = 13 · 2 · h ｛ H = 32