∴△BCD is an isosceles right triangle.

∴BD=CD.

∠ DBF = 90-∠ BFD, ∠ DCA = 90-∠ EFC and ∠BFD=∠EFC,

∴∠DBF=∠DCA.

In Rt△DFB and Rt△DAC,

∠∠DBF =∠DCA BD = CD∠BDF =∠ADC

∴Rt△DFB≌Rt△DAC(ASA).

∴bf=ac；

(2) proof: ∫ divided into equal parts ∠ABC,

∴∠ABE=∠CBE.

In Rt△BEA and Rt△BEC.

∠ABE=∠CBEBE=BE∠BEA=∠BEC，

∴Rt△BEA≌Rt△BEC(ASA).

∴CE=AE= 12AC.

From (1), BF=AC,

∴ce= 12ac= 12bf；

(3) Proof: ∠ ABC = 45, CD is perpendicular to AB and D, then CD = BD.

H is the midpoint of BC, and then DH⊥BC (isosceles triangle "three lines in one")

If CG is connected, BG=CG, ∠ GCB = ∠ GBC =12 ∠ ABC =12× 45 = 22.5, ∠ EGC = 45.

And ∫ perpendicular to AC, so ∠ EGC = ∠ ECG = 45, CE = Ge.

∵△GEC is a right triangle,

∴CE2+GE2=CG2，

∫DH vertically divides BC,

∴BG=CG，

∴ce2+ge2=cg2=bg2； Namely 2CE2=BG2, bg2 = 2ce,

∴BG>CE.