1. 如图,四边形$ABCD$内接于圆,点$E在\overset{\frown}{CD}$上,若$\overset{\frown}{AB} = \overset{\frown}{AD}$,$\overset{\frown}{BC} = \overset{\frown}{CE} = \overset{\frown}{ED}$,$∠BCD = 105^{\circ}$,则$∠CDE$为
25
$^{\circ}$。
答案:
25
2. 如图,点$A$,$B$,$C$,$D$,$E都是\odot O$上的点,$\overset{\frown}{AC} = \overset{\frown}{AE}$,$∠D = 128^{\circ}$,则$∠B = $

116
$^{\circ}$。
答案:
116
3. 如图,在$\triangle ABC$中,$AB > AC$,$∠CAB的外角平分线交\triangle ABC的外接圆于点E$,过点$E作EF⊥AB$,垂足为$F$。求证:$AB - AC = 2AF$。

答案:
证明:如答图,在AB上取点D使FD = AF,连接ED并延长交圆于点G,连接BG;
∵ EF⊥AB,
∴ ∠EAD = ∠EDA = ∠BDG.
又
∵ ∠EAD = ∠G,
∴ ∠BDG = ∠G,
∴ BD = BG.
又
∵ ∠BAC = 180° - 2∠EAD = 180° - (∠EAD + ∠EDA) = ∠AEG,
∴ $\overset{\frown}{BGC} = \overset{\frown}{ACG}$,
∴ $\overset{\frown}{AC} = \overset{\frown}{BG}$,
∴ AC = BG,
∴ AC = BD,
∴ AB - AC = AB - BD = AD = 2AF.
证明:如答图,在AB上取点D使FD = AF,连接ED并延长交圆于点G,连接BG;
∵ EF⊥AB,
∴ ∠EAD = ∠EDA = ∠BDG.
又
∵ ∠EAD = ∠G,
∴ ∠BDG = ∠G,
∴ BD = BG.
又
∵ ∠BAC = 180° - 2∠EAD = 180° - (∠EAD + ∠EDA) = ∠AEG,
∴ $\overset{\frown}{BGC} = \overset{\frown}{ACG}$,
∴ $\overset{\frown}{AC} = \overset{\frown}{BG}$,
∴ AC = BG,
∴ AC = BD,
∴ AB - AC = AB - BD = AD = 2AF.
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