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1. 如图,$\triangle ABC内接于\odot O$,$DE$,$FG是\odot O$的弦,$AB = DE$,$FG = AC$. 下列结论：①$DE + FG = BC$；②$\overset{\frown}{DE}+\overset{\frown}{FG}= \overset{\frown}{BC}$；③$∠DOE + ∠FOG = ∠BOC$；④$∠DEO + ∠FGO = ∠BAC$. 其中所有正确结论的序号是(
A.①②③④
B.②③
C.②④
D.②③④
D
)A.①②③④
B.②③
C.②④
D.②③④
答案：
D
2. 如图,在$\odot O$中,$AB$是直径,点$C$,$D$,$E$在圆上,$AC = 2$,$AD = 6$,$AE = 8$,$AB = 10$. 下列结论：①$\overset{\frown}{AD}= \overset{\frown}{CE}$；②$\overset{\frown}{AE}= \overset{\frown}{BD}$；③$\overset{\frown}{AD}+\overset{\frown}{DE}= \overset{\frown}{AE}$；④$\overset{\frown}{AD}+\overset{\frown}{AE}= \overset{\frown}{ADB}$. 其中正确结论的个数是(
A.4 个
B.3 个
C.2 个
D.1 个
B
)A.4 个
B.3 个
C.2 个
D.1 个
答案：
B
3. 如图,$\overset{\frown}{AB}的度数为90^{\circ}$,点$C$,$D将\overset{\frown}{AB}$三等分,弦$AB与半径OC$,$OD分别交于点E$,$F$,求证：$AE = CD = FB$.
答案：
证明:如答图,连接 BD.
∵∠AOB = 90°,点 D,C 将$\widehat {AB}$三等分,
∴∠1 = ∠2 = ∠3 = 30°,CD = BD.
又
∵OA = OB,
∴∠OAE = ∠OBF = 45°.
在△OAE与△OBF中,$\begin{cases}∠1 = ∠3\\OA = OB\\∠OAE = ∠OBF\end{cases}$
∴△OAE≌△OBF(ASA),
∴AE = BF.
又
∵OB = OD,∠3 = 30°,
∴∠5 = $\frac{1}{2}$×(180° - 30°) = 75°,
而∠4 = ∠3 + ∠OBF = 30° + 45° = 75°,
∴BF = BD.而CD = DB,AE = BF,
∴AE = DC = BF.
证明:如答图,连接 BD.
∵∠AOB = 90°,点 D,C 将$\widehat {AB}$三等分,
∴∠1 = ∠2 = ∠3 = 30°,CD = BD.
又
∵OA = OB,
∴∠OAE = ∠OBF = 45°.
在△OAE与△OBF中,$\begin{cases}∠1 = ∠3\\OA = OB\\∠OAE = ∠OBF\end{cases}$
∴△OAE≌△OBF(ASA),
∴AE = BF.
又
∵OB = OD,∠3 = 30°,
∴∠5 = $\frac{1}{2}$×(180° - 30°) = 75°,
而∠4 = ∠3 + ∠OBF = 30° + 45° = 75°,
∴BF = BD.而CD = DB,AE = BF,
∴AE = DC = BF.
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