8. 如图,在$\odot O$中,$\overset{\frown}{AB}=2\overset{\frown}{CD}$,则下列结论正确的是().

A.$AB＞2CD$
B.$AB=2CD$
C.$AB＜2CD$
D.以上都不正确

(第8题图)
(第9题图)

8. C 解析 如图,取$\overset{\frown}{AB}$的中点E,连接AE,BE.
$\because$在$\odot O$中,$\overset{\frown}{AB}=2\overset{\frown}{CD}$,
$\therefore\overset{\frown}{AE}=\overset{\frown}{BE}=\overset{\frown}{CD}$,
$\therefore AE=BE=CD$.
$\because AE+BE＞AB,\therefore 2CD＞AB$.故选C.

9.如图，AB，CD是$\odot O$的弦，且$AB=CD$，若$∠ BOD=84°$，则$∠ ACO$的度数为（）.

A.$42°$
B.$44°$
C.$46°$
D.$48°$

9. D 解析 如图,连接OA.$\because AB=CD$,
$\therefore\overset{\frown}{AB}=\overset{\frown}{CD}$,
$\therefore\overset{\frown}{AB}-\overset{\frown}{AD}=\overset{\frown}{CD}-\overset{\frown}{AD}$,
$\therefore\overset{\frown}{AC}=\overset{\frown}{BD}$,
$\therefore∠ AOC=∠ BOD=84°$.
$\because OA=OC$,
$\therefore ∠ ACO = ∠ CAO = \frac{1}{2} ×（180° -∠ AOC)=\frac{1}{2}×(180°-84°)=48°$,故选D.

10. 如图，AB为$\odot O$的直径，D是$\overset{\frown}{AC}$的中点，过点D作$DE ⊥ AB$于点E，延长DE交$\odot O$于点F。若$AE = 3$，$\odot O$的直径为15，则AC的长为().

A.10
B.13
C.12
D.11

10. C 解析 如图,连接OF.$\because DE⊥ AB$,$AB$为$\odot O$的直径,
$\therefore DE=EF$,$\overset{\frown}{AD}=\overset{\frown}{AF}$.
$\because D$为$\overset{\frown}{AC}$的中点,
$\therefore\overset{\frown}{AD}=\overset{\frown}{DC}$,$\therefore\overset{\frown}{AC}=\overset{\frown}{DF}$,$\therefore AC=DF$.
$\because\odot O$的直径为15,$\therefore OF=OA=\frac{15}{2}$.
$\because AE=3$,$\therefore OE=OA-AE=\frac{9}{2}$.
在$\mathrm{Rt}△ OEF$中,由勾股定理,得$EF=\sqrt{OF^2-OE^2}=\sqrt{(\frac{15}{2})^2-(\frac{9}{2})^2}=6$,
$\therefore DE=EF=6$,
$\therefore AC=DF=DE+EF=6+6=12$.

11. 如图，A，B是$\odot O$上的两点，$∠ AOB = 120°$，C是$\overset{\frown}{AB}$的中点，连接OA，AC，BC，OB.
(1)试判断四边形OACB的形状，并说明理由；
(2)延长OA至点P，使得$AP=OA$，连接PC，若$\odot O$的半径为2，求PC的长.

解 (1)四边形OACB是菱形.
理由:连接OC,如图所示.$\because∠ AOB=120°$,$C$是$\overset{\frown}{AB}$的中点,
$\therefore\overset{\frown}{AC}=\overset{\frown}{BC}$,
$\therefore∠ AOC=∠ BOC=\frac{1}{2}∠ AOB=60°$.
$\because OA=OC=OB$,
$\therefore△ AOC$与$△ BOC$都是等边三角形,
$\therefore AC=OA=OC=OB=BC$,
$\therefore$四边形OACB是菱形.
(2)$\because AP=OA$,$AC=OA$,$\therefore AP=AC$,
$\therefore∠ P=∠ ACP=\frac{1}{2}∠ OAC=30°$.
$\therefore∠ OCP=90°$.
$\because\odot O$的半径为2,$\therefore OC=2$,$OP=4$,
$\therefore PC=\sqrt{OP^2-OC^2}=2\sqrt{3}$.

12. 如图,在$\odot O$中,C,D是直径AB上的两点,且$AC=BD$,$MC ⊥ AB$,$ND ⊥ AB$,点M,N在$\odot O$上.
(1)求证:$\overset{\frown}{AM}=\overset{\frown}{BN}$.
(2)若C,D分别为OA,OB的中点,则$\overset{\frown}{AM}=\overset{\frown}{MN}=\overset{\frown}{BN}$成立吗?请说明理由.

(1)证明 如图,连接OM,ON.
$\because OA=OB$,$AC=BD$,
$\therefore OA - AC = OB - BD$,$\therefore OC=OD$.
$\because MC⊥ AB$,$ND⊥ AB$,
$\therefore∠ OCM=∠ ODN=90°$.
又$OM=ON$,$\therefore△ OCM≌△ ODN$,
$\therefore∠ AOM=∠ BON$,$\therefore\overset{\frown}{AM}=\overset{\frown}{BN}$.
(2)解 成立.理由如下:
如图,连接AM.
$\because C$为$OA$的中点,$MC⊥ AB$,
$\therefore MA=MO$.
$\because OM=OA$,$\therefore MA=OM=OA$,
$\therefore△ AOM$为等边三角形,$\therefore∠ AOM=60°$.
同理可得$∠ BON=60°$,
$\therefore∠ MON=180°-∠ AOM-∠ BON=60°$,
$\therefore\overset{\frown}{AM}=\overset{\frown}{MN}=\overset{\frown}{BN}$.