(1)AB∨CD can be found according to the parallelism of the opposite sides of a rectangle, then ∠BAC=∠FCO can be found according to the parallelism of two straight lines and the equal internal angles, and then the congruence of △AOE and △COF can be proved by "angular edge", and then it can be proved according to congruent triangles's.

(2) Connecting OB, we can get BO⊥EF according to the properties of isosceles triangle with three lines in one, OA=OB according to the properties of rectangle, BAC = ∠ ABO according to the properties of equilateral and equilateral angles, and ABO = 30 according to the internal angle of triangle and theorem formula.

Answer:

(1) proves that ABCD,

∴∠BAC=∠FCO，

At △AOE and △COF,

{∠BAC=∠FCO

{∠AOE=∠COF，

{AE=CF

∴△AOE≌△COF(AAS)，

∴OE=OF.

(2) Solution: As shown in the figure, connect OB,

BE = BF，OE=OF，

∴BO⊥EF，

∴, ∠ BEF+∠ ABO = 90 in Rt△BEO,

According to the fact that the median line on the hypotenuse of a right triangle is equal to half of the hypotenuse, OA=OB=OC,

∴∠BAC=∠ABO，

∫∠BEF = 2∠BAC，

That is 2 < BAC+< BAC = 90,

The solution is ∠ BAC = 30,

∫BC = 2√3，

∴AC=2BC=4√3，

∴ab=√(ac^2-bc^2)=[(4√3)^2-(2√3)^2]=6.