It is known that PA is perpendicular to the plane of rectangle ABCD, and M is the midpoint of ab. if PD and ABCD form an angle of 45 degrees, it is proved that plane PCM is perpendicular to plane PCD

It is known that PA is perpendicular to the plane of rectangle ABCD, and M is the midpoint of ab. if PD and ABCD form an angle of 45 degrees, it is proved that plane PCM is perpendicular to plane PCD

PD~ABCD=45 then PA=AD
AB=2b,AD=a
PM*2=MC*2=a*2+b*2
PC midpoint Q, MQ * 2 = a * 2 / 2
DM*2=a*2+b*2
DQ * 2 = a * 2 / 2 + b * 2 (from solving triangle PCD)
OK!

The distance between a point P in square ABCD and three points ABC is 1:2:3. Calculate the angle APB and degree

Conclusion: APB is 135 ° proof: point B is the vertex of point B, and the point B is the vertex of the point B, and the point B is the vertex of the point B, and the point P is located at the Q point; PA = a; PA = a; Pb = 2A, PC = 3A, PC = 3A; all of the \ʊ APB is 135 ° proof: point B is the vertex of the point B, and the point B is the vertex of B, and the point B is the vertex B, and the point B and Ba is the point P falls at the Q point; PA = a; PA = a; Pb = a; Pb = b-bpc; point B is the point; PA = a; PA = a; QB = a; PQ = 2 {2A; PQ = 2 {2A; QA = 2A; PA = 2A; PA; PA = PA = PA = PA = pa = a; PA = a; PA = a; PA = a; PQ = 2; PQ = 2 ; ∴∠QPA=90°；∴∠APB=∠QPA+∠QPB=90+45=135°

P is the point of ABCD outside the parallelogram, O is the key point of PA

Take the midpoint e of BD and connect QE
Because Q and E are the midpoint, QE is the median line of the triangular APC
So QE is parallel to PC
Because QE belongs to bdq, PC does not belong to bdq
So PC is parallel to plane bdq

P is a point out of the plane of the parallelogram ABCD. Find a point E on PC to make the PA ‖ face be, and give the proof

Let me give you some ideas, because it is a surface bed and ABCD is a parallelogram, then we can establish a space rectangular coordinate system with the midpoint o in BD, and the PA vector is parallel to the OE vector

Given the plane of the PA vertical parallelogram ABCD, if PC ⊥ BD, the parallelogram ABCD must be______ ．

According to the meaning of the question, draw a figure as shown in the figure, ∩ PC = P. ∩ BD ⊥ PAC ⊂ AC ⊂ PAC ⊂ AC ⊥ BD ⊂ ABCD, PC ⊂ ABCD, PA ∩ PC = P. ∩ BD ⊥ PAC ⊂ AC ⊂ BD ⊥ ABCD is parallelogram ⊂ parallelogram ABCD must be diamond

As shown in the figure, the bottom surface is a parallelogram pyramid p-abcd, the point E is on PD, and PE: ed = 2:1. Q: is there a point F on the edge PC, which makes BF ‖ plane AEC? Prove your conclusion

Let BD ∩ AC = O. connect BF, MF, BM, OE. ∩ PE: ed = 2:1, f is the midpoint of PC, e is the midpoint of MD, ∩ MF ∥ EC, BM ∥ OE. ∩ MF ⊄ plane AEC, CE ⊂ plane AEC, B

It is known that P -- ABCD is a quadrangular pyramid whose base is a parallelogram, point E is on PD, and PE: ed = 2:1. Is there a point F on edge PC, which makes BF parallel to plane AEC

Existence
When point f satisfies PF: FC = 2:1, BF ‖ is on the AEC
Extend Fe to h such that HF = ab
∵PE:ED=2:1,PF:FC=2:1,∴EF‖DC,
And ∵ ab ∥ DC, ∵ ab ∥ EF,
That is ab ‖ HF, ab = HF,
The quadrilateral abfh is a parallelogram
∴BF‖AH
And ∵ ah ∈ plane AEC,
Ψ BF ‖ in plane AEC

In the p-abcd pyramid with a parallelogram at the bottom, if the point E is on the PD and PE: ed = 2:1, is there a point F on the edge PC to make BF parallel to the plane AEC

Existence
AC and BD intersect at O and connect with EO
Take the midpoint m of PE and the midpoint n of PC to connect BM, Mn and Nb
Mn / / EC in △ PEC
In △ DBM, EO / / BM
So plane AEC / / plane BMN
So BN / / plane AEC
So we just need to take point d to the midpoint of PC, (F, n coincide)
BF is parallel to plane AEC

There is a point outside the square ABCD with side length a, so that PA ⊥ plane ABCD, PA = A. find the size of dihedral angle p-bc-a
Who's going to teach me --
It should be done with cosine theorem

Let p-bc-a be the size of dihedral angle@
Cos @ = area of triangle ABC / area of triangle PBC
Because the triangle ABC is the projection of the triangle PBC on the ABCD plane
Area of triangle ABC = (1 / 2) * 1 * 1 = 1 / 2
The area of triangle PBC = (1 / 2) * BC * Pb = (1 / 2) * 1 * radical 2
= (radical 2) / 2
So we get: cos @ = 1 / (root 2)
@= 45 degrees

In the pyramid p-abcd, it is known that PA is perpendicular to the ground ABCD, and the bottom ABCD is rectangular, then the following conclusion is wrong: plane ABC is perpendicular to plane PCD
Why? What is the angle perpendicular to the intersection line?

In order for two planes to be perpendicular to each other, a line in one plane must be perpendicular to the two intersecting lines in the other plane at the same time
There is no such condition in plane ABC and plane PCD