It is proved that the quadrilateral ABCD as shown in figure 1 and ∵ is a square, and P and C coincide.

∴OB=OP,∠BOC=∠BOG=90。

∵PF⊥BG,∠PFB=90，

∴∠GBO=90 -∠BGO，

∠EPO=90 -∠BGO，

∴∠GBO=∠EPO，

In delta swamp and delta slope,

∠GBO =∠ Osi

OB=OC

∠BOG=∠COE

∴△BOG≌△POE.

∴OE=OG，

∫∠EOG = 90 degrees,

∴ Rotate the line segment oe 90 clockwise around the O point to get OG.

And ∵OB=OP, ∠ pob = 90,

∴ Rotate the line segment op 90 clockwise around the O point to get OB.

∴△BOG can be obtained by rotating △POE 90 degrees clockwise around the O point.

(2) As shown in Figure 2, let PM∑AC pass through BG in M and BO in N,

∴∠PNE=∠BOC=90，∠BPN=∠OCB，

∠∠obc =∠OCB = 45 ,∴∠nbp=∠npb，

∴NB=NP.

∠∠MBN = 90-∠BMN∠NPE = 90-∠BMN，

∴∠MBN=∠NPE，

In the △BMN and △ pens.

∠MBN=∠NPE

NB=NP

∠MNB=∠ENP

∴△BMN≌△PEN，

∴BM=PE.

∫∠BPE = 1/2

∠ACB∠BPN =∠ACB，

∴∠BPF=∠MPF.

∵PF⊥BM,∴∠BFP=∠MFP=90。

Also in △BPF and △MPF

BPF =∠ MPF

PF=PF

∠BFP=∠MFP

∴△BPF≌△MPF，

∴BF=MF, namely BF= 1/2?

BM，

∴BF= 1/2PE, namely BF /PE= 1/2.

.

(3) As shown in Figure 2, the intersection P is PM∑AC, BG is M, and BO is N,

∴∠BPN=∠BCA，

∫∠BPE = 1/2∠BCA，

∴∠BPF=∠MPF，

∵PF⊥BG，

∴∠BFP=∠MFP，

At △BFP and △MFP.

∠BFP=∠MFP

PF=PF

∠BPF =∠ MPF

∴△BFP≌△MFP(ASA)，

∴BF=FM，

That is BF= 1/2.

BM，

The quadrilateral ABCD is a diamond,

∴DB⊥AC，

∫PM∑AC，

∴∠BPN=∠ACB=α,∠PNE=∠BOC=90，

∴∠BNM=90

∫∠PFM = 90 degrees,

∴∠MBN+∠BMN=90，∠MPF+∠BMN=90，

∴∠MBN=∠NPE，

∠∠BNM =∠ENP，

∴△BMN∽△PEN.

∴BM/？ PE=BN？ /PN

,

∫tanα= BN/PN = BM/？ PE

=2BF/？ PE

∴BF/？ PE

= 1/2 tanα。