Connecting OB, we can get ∠ OAB = ∠ ABO = 25 according to the definition of angle bisector, ∠ AB = AC = ∠ ACB = 65 according to the properties of isosceles triangle, and OA=OB according to the properties of perpendicular bisector, then ∠ OBA = ∠ OAB. According to the nature of isosceles triangle, OA bisects BC vertically, so BO=OC, so ∠ 1 = ∠ 2 = 40, and then according to the nature of folding, EO=EC, so ∠ 2 = ∠ 3 = 40, and then ∠ OEC is calculated according to the triangle internal angle sum theorem.

Solution: link OB,

∠∠ ∠BAC = 50, and the bisector of∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠

∴∠OAB=∠ABO=25，

AB = AC，∠BAC=50，

∴∠ABC=∠ACB=65，

∫OD vertically divides AB,

∴OA=OB，

∴∠OBA=∠OAB=25，

∴∠ 1=65 -25 =40 ,

AB = AC, OA average score ∠BAC,

∴OA vertically divides BC,

∴BO=OC，

∴∠ 1=∠2=40 ,

Point c is folded along EF and coincides with point o,

∴EO=EC，

∴∠2=∠3=40 ,

∴∠oec= 180-40-40 = 100。

So the answer is 100.

I hope it helps you.