Then FG//CD, FG= 1/2CD.

FG//AE，FG=AE

So AEGF is a parallelogram.

AF//EG, because EG is in plane PEC, af//plane PEC.

(2) The bottom of P-ABCD is rectangular, while PA⊥ plane ABCD,

PA⊥CD, so CD⊥ surface pad, CD⊥PD, CD⊥AF.

Therefore, the angle PDA is the plane angle of dihedral angle P-CD-B.

That is to say, the angle PDA = 45 degrees, so PA = AD.

ec^2=bc^2+be^2 pe^2=pa^2+ae^2

Therefore: PE=EC, and EG⊥PC is obtained.

It has been proved above: AF//EG, CD⊥AF, so EG⊥CD.

So the surface PCD of EG⊥, EG is in the plane PEC.

So aircraft PEC⊥ aircraft PCD.