The quadrilateral OBFC is a parallelogram,

∴BM=CM,OM=MF

OA = AE，OM=MF

∴AM∥EF,AM=EF/2

∫∠BAC = 90，BM=CM

∴AM=BC/2 ,BD⊥CD，

∴EF⊥BC,EF=BC

Because the midline on the base of the isosceles triangle coincides with the height on the base, so

When △ABC is an equilateral triangle, EF⊥BC and ef = BC can be proved in the same way;

When △ABC is an isosceles triangle, EF⊥BC and ef = BC can be proved in the same way;

Because the median line on the hypotenuse of a right triangle is equal to half of the hypotenuse, so

When △ABC is a right triangle, ef = BC.

When △ABC is an arbitrary triangle, none of the above conclusions are valid.