To solve the math problem, it is known that the vertical plane ABCD E F of the PA is the midpoint of

1) Proof: Take the midpoint G of the PC and connect eg and fg

f is the midpoint of pd, fg cd and fg=cd

And ae cd and ae = cd, ea gf and ea = gf, so the quadrilateral egfa is a parallelogram, thus eg af

also af plane pec, eg plane pec, af plane pec

2) Proof of: PA planar ABCD, AD is the projection of PD on planar ABCD.

CD AD, CD PD, PDA are the planar angles of the dihedral angle P-CD-B.

adp = 45°, then af pd

and af cd, pd cd=d, af planar pcd

By (1), EG AF, EG planar PCD

Whereas EG Plane PEC, Plane PEC Plane PCD

This is a geometry problem, there should be a diagram, please give it up.

Can you tell me if ABCD is an irregular figure or a parallelogram, or something else...

As an auxiliary line, this tetrahedron can be supplemented into a cuboid, by setting the other side of the cuboid as padh, according to the line plane perpendicular and line line perpendicular angle of pa, pada is the face angle of p-cd-b, 45 degrees, and pa perpendicular to ad, pa=ad, that is, padh is square, ah is perpendicular to pd, and ah is perpendicular to cd; AH perpendicular to the surface PCD passes E as EM in the surface ABH and parallel to the intersection of AH and PC at the point N, then EN is perpendicular to the surface PCD, and EN belongs to the surface PEC, so the surface PEC perpendicular plane PCD is obtained.

The quadrilateral abcd is a rectangle, ab cd, ab=cd, and e and f are the midpoints of the side ab and cd respectively, and there is no huaizhou df=be, and ab cd, and the quadrilateral debf is a parallelogram, a bright bridge.

de=bf．

This method is slightly complicated, and it is only for those who ask about the cave model to choose by themselves.

Proof: Take the midpoint of cd and set it to g to connect AG FG.

f g are the midpoints of pd cd, fg is parallel and equal to pc, and pc is included in the surface pce, fg is not included in the surface pcefg, and the surface is pce

and e g are ab cd midpoint quadrilateral, and aecg is a parallelogram.

ag ec and ag do not contain traces of the surface pce ag face pce and fg ag=g

So the face AFG face PCE

And AF is included in the surface AFG

Parallel Plane PCE of Neutrian Slow AF

As a PC, the middle reed carrying ruler is set to g to connect fg ag, because ae=1 2ab=1 2dc=fg (there is a parallel on the equal sign, which means parallel and equal).

So the quadrilateral AEGF is a parallelogram.

So AF collapses eg

Because of the AF surface PCE

EG belongs to the surface PCE

So af plane PCE

Proof: Let the midpoint of AB be H, and connect DH to the point M, because E and F are the midpoints of the ABCD side of the square BC and Cd respectively, so.

Bae CBF, CBF+ AEB 90, so BF is perpendicular to AE, and DH is perpendicular to AE

So MH is parallel to BP, H is the midpoint of AB, so M is the midpoint of AP, so AD=PD

Proof: Extend the intersection of BF and AD at point G

E is the midpoint of BC.

be＝bc/2

f is the midpoint of the cd.

cf＝df＝cd/2

bc＝cdbe＝cf

ab＝bc，∠abc＝∠bcd＝90

abe≌△bcf

bae＝∠cbf

bae+∠aeb＝90

cbf+∠aeb＝90

bpe＝90

apg＝∠bpe＝90

bfc＝∠gfd，∠bcd＝∠gdc＝90∴△gdf≌△bcf

dg＝bcdg＝ad

d is the midpoint of ag.

AD PD (Right Triangle Midline Property).

My method may not be the easiest way to do it, so take a look.

Establish a planar Cartesian coordinate system with d as the coordinate origin, so that the coordinates of ABCEF can be represented.

Then we find the intersection point p of the straight line AE and the straight line bf, and then use the distance formula to calculate AD=PD

I couldn't figure out how to use the proof of geometric relations for a while.

Let ab=a, so bc= a, df= 2a, ad=2a

Because af +df = ad

So the angle dfa=90 degrees, so df af

Because the PA is perpendicular to the planar ABCD, the PA DF because the AF line segment belongs to the planar PAF

So df plane paf

Proof is complete.

Right? DF doesn't seem to be perpendicular to PAF.

If ad=ab, the result is true.

Solution: Quadrilateral ABCD is rectangular.

bad=∠d=90°

∠afb=∠1+∠3=90°

and d= afb=90°

1) Connect EF, BE=1 2PA, BE BABE=AF

The quadrilateral ABEF is a parallelogram.

ab∥=ef

Again cd = ab

cd∥=ef

Quadrilateral CDFE is a parallelogram.

CE=DFCE belongs to surface PEC, DF does not belong to surface PECDF surface PEC

2) PE= 2, EF=AB=1 EB=1 (Pythagorean) Are you a freshman or a junior in high school, if you are a junior in high school, this problem is very good to do with vectors.