《2025年精彩练习就练这一本九年级数学全一册浙教版评议教辅》

9. 如图,以平行四边形ABCD的顶点A为圆心,AB为半径作⊙A,分别交AD,BC于点E,F,延长BA交⊙A于点G,求证：$\overset{\frown}{GE}= \overset{\frown}{EF}$.

答案：
证明:如图,连结AF,$\because AB=AF$,$\therefore \angle ABF=\angle AFB$.$\because$四边形ABCD是平行四边形,$\therefore AD// BC$,$\therefore \angle DAF=\angle AFB$,$\angle GAE=\angle ABF$,$\therefore \angle GAE=\angle EAF$,$\therefore GE=EF$.

10. 如图,C是⊙O直径AB上的一点,过点C作弦DE,使CD= CO,若$\overset{\frown}{AD}$所对圆心角的度数为40°,则$\overset{\frown}{BE}$所对圆心角的度数为(D)

A.40°
B.80°
C.150°
D.120°

答案： D

11. 已知⊙O的半径为2,△ABC内接于⊙O,$\overset{\frown}{AB}$,$\overset{\frown}{BC}$,$\overset{\frown}{AC}$的长之比为3∶2∶3,则BC的长为$2\sqrt{2}$.

答案： $2\sqrt{2}$

12. 如图,点C在⊙O上,CD⊥OA于点D,CE⊥OB于点E,OD= OE. 求证：C是$\overset{\frown}{AB}$的中点.

答案：
证明:如图,连结OC.$\because CD\perp OA$于点D,CE$\perp OB$于点E,$\therefore \angle CDO=\angle CEO=90^\circ$.$\because OC=OC$,$OD=OE$,$\therefore \triangle COD\cong \triangle COE(HL)$,$\therefore \angle AOC=\angle BOC$,$\therefore \overset{\frown}{AC}=\overset{\frown}{BC}$,$\therefore$C是$\overset{\frown}{AB}$的中点.

13. 如图,AB,BC,AC都是⊙O的弦,且∠AOB= ∠BOC. 求证：(1)∠BAC= ∠BCA. (2)∠ABO= ∠CBO.

答案： 证明:
(1)$\because \angle AOB=\angle BOC$,$\therefore \overset{\frown}{AB}=\overset{\frown}{BC}$,$BA=BC$,$\therefore \angle BAC=\angle BCA$.
(2)$\because OA=OB$,$\therefore \angle ABO=\angle BAO=\frac{1}{2}(180^\circ -\angle AOB)$.同理,$\angle CBO=\angle BCO=\frac{1}{2}(180^\circ -\angle BOC)$.又$\because \angle AOB=\angle BOC$,$\therefore \angle ABO=\angle CBO$.