注意A,B关于y轴对称,AD平分∠OAC,CE // BD是整个题目的前提.
设∠CEA = x,则∠DBO = x (同位角相等).
于是∠DAO = x (∵△ADB等腰),∠CAO = 2x (∵AD平分∠OAC).
(2)延长EO至G,使EG = EC,连CG.
∵EG = EC,∠CEG = x,∴∠CGE = 90°-x/2.
∵CF平分∠ACE,∠ACE = 180°-3x (△ACE内角和),∴∠ACF = 90°-3x/2.
∴∠CFA = 180°-2x-(90°-3x/2) = 90°-x/2 = ∠CGE.
于是△GCF等腰,有GO = OF.故CE = GE = GO+OE = OF+OE.
(3)CN // x轴.
在AE上取点H使AH = AC,连NH,NE.
由AN平分∠OAC,可证△ACN ≌ △AHN (SAS),得CN = HN,∠ACN = ∠AHN.
∵MN是CE垂直平分线,∴EN = CN = HN,∴∠NHE = ∠NEH.
∴∠ACN = ∠AHN = 180°-∠NHE = 180°-∠NEH.
于是∠CNE = 360°-∠ACN-∠NEA-∠CAE = 180°-2x.
又∵NC = NE,∴∠NCE = x = ∠CEA.故CN // x轴 (内错角相等).
因为思维受到圆的知识的影响,可能(3)问证得比较麻烦.