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解:如图,连接OB、OC,
∵∠BAC=54°,AO为∠BAC的平分线,
∴∠BAO=
∠BAC=
×54°=27°,
又∵AB=AC,
∴∠ABC=
(180°-∠BAC)=
(180°-54°)=63°,
∵DO是AB的垂直平分线,
∴OA=OB,
∴∠ABO=∠BAO=27°,
∴∠OBC=∠ABC-∠ABO=63°-27°=36°,
∵AO为∠BAC的平分线,AB=AC,
∴△AOB≌△AOC(SAS),
∴OB=OC,
∴点O在BC的垂直平分线上,
又∵DO是AB的垂直平分线,
∴点O是△ABC的外心,
∴∠OCB=∠OBC=36°,
∵将∠C沿EF(E在BC上,F在AC上)折叠,点C与点O恰好重合,
∴OE=CE,
∴∠COE=∠OCB=36°,
在△OCE中,∠OEC=180°-∠COE-∠OCB=180°-36°-36°=108°.
故答案为:108.