The quadrilateral ABHG is a rectangle,
∴AB=GH，
∵ quadrilateral ABCD is a square,
∴AB=BC，
∴BC=GH，
∵EG⊥BF，
∴∠CBF+∠BEG=∠HGE+∠BEG=90，
∴∠CBF=∠HGE，
In △BCF and △GHE, ∠ CBF = ∠ HGEBC = GH ∠ C = ∠ GeH = 90,
∴△BCF≌△GHE(ASA)，
∴ge=bf；
(2) Proof: As shown in the figure, rotate △BCF 90 degrees counterclockwise around point B to get △BAH, and then connect it with HF.
Then △BFH is an isosceles right triangle,
∴HF=2BF，
From the nature of rotation, BH=BF,
BF = GE，
∴BH=GE，
∵EG⊥BF,∠HBF=90，
∴BH∥GE，
∴ Quadrilateral BEGH is a parallelogram,
∴GH=BE，
At △GHF, fg+GH > HF,
∴fg+be>2bf；
(3) Solution: As shown in the figure, GH=BE, HF=2BF, which can be obtained as shown in Equation (2).
At △GHF, HF+GH > FG,
∴2BF+BE>FG.