8 如图, A B 是 $\odot O$ 的直径, C 为 $\overgroup{BD}$ 的中点, C F 为 $\odot O$ 的弦, 且 $C F ⊥ A B$, 垂足为 E, 连接 B D, 交CF 于点 G, 连接 C D, A D, B F.
(1) 求证: $△ B F G ≌ △ C D G$;
(2) 若 $A D=B E=2$, 求 B F 的长.

8. (1) $\because C$是$\overset{\frown}{BD}$的中点,$\therefore \overset{\frown}{CD}=\overset{\frown}{BC}$.$\because AB$是$\odot O$的直径,且$CF⊥ AB$,$\therefore \overset{\frown}{BC}=\overset{\frown}{BF}$,$\therefore \overset{\frown}{CD}=\overset{\frown}{BF}$,$\therefore CD=BF$.$\because \overset{\frown}{BC}=\overset{\frown}{BC}$,$\therefore ∠ BFG=∠ CDG$.在$△ BFG$和$△ CDG$中,$\begin{cases} ∠ FGB=∠ DGC, \\ ∠ BFG=∠ CDG, \\ BF=CD, \end{cases}$$\therefore △ BFG≌△ CDG(\mathrm{AAS})$ (2) 连接$OF$,设$\odot O$的半径为$r$.$\because AB$是$\odot O$的直径,$\therefore ∠ ADB=90°$,$\therefore$在$\mathrm{Rt}△ ADB$中,$BD^2=AB^2-AD^2$,即$BD^2=(2r)^2-2^2$.$\because CF⊥ AB$,$\therefore CF=2EF$.在$\mathrm{Rt}△ OEF$中,$OF^2=OE^2+EF^2$,即$EF^2=r^2-(r-2)^2$.$\because \overset{\frown}{CD}=\overset{\frown}{BC}=\overset{\frown}{BF}$,$\therefore \overset{\frown}{BD}=\overset{\frown}{CF}$,$\therefore BD=CF$,$\therefore BD^2=CF^2=(2EF)^2=4EF^2$,即$(2r)^2-2^2=4[r^2-(r-2)^2]$,解得$r_1=1$(不合题意,舍去),$r_2=3$.$\therefore$在$\mathrm{Rt}△ EFB$中,$BF^2=EF^2+BE^2=3^2-(3-2)^2+2^2=12$,$\therefore BF=2\sqrt{3}$

9 如图, A B 为 $\odot O$ 的直径, 直线 C D 与 $\odot O$ 相切于点 C, 连接 A C. 若 $∠ A C D=50°$, 则 $∠ B A C$ 的度数为（）



A.$30°$
B.$40°$
C.$50°$
D.$60°$

9. B

10 [2024 广州]如图,在$\odot O$中,弦$AB$的长为$4\sqrt{3}$,点$C$在$\odot O$上,$OC ⊥ AB$,$∠ ABC=30^{ \circ }$,$\odot O$所在的平面内有一点$P$.若$OP=5$,则点$P$与$\odot O$的位置关系是.

10. 点$P$在$\odot O$外

11 如图,在扇形$ABC$中,$CD ⊥ AB$,垂足为$D$,$\odot E$是$△ ACD$的内切圆,连接$AE$,$BE$,则$∠ AEB$的度数为.

11. $135°$

12 [2025 山东]如图,在$△ OAB$中,点$A$在$\odot O$上,边$OB$交$\odot O$于点$C$,$AD ⊥ OB$于点$D$,$AC$是$∠ BAD$的平分线.
(1) 求证:$AB$为$\odot O$的切线;
(2) 若$\odot O$的半径为$2$,$∠ AOB=45^{ \circ }$,求$CB$的长.

12. (1) $\because AD⊥ OB$于点$D$,$\therefore ∠ ADB=90°$.$\because AC$是$∠ BAD$的平分线,$\therefore ∠ DAC=∠ BAC$.$\because OA=OC$,$\therefore ∠ OAC=∠ OCA$.$\because ∠ OAC=∠ OAD+∠ DAC=∠ OAD+∠ BAC$,$∠ OCA=∠ B+∠ BAC$,$\therefore ∠ OAD=∠ B$,$\therefore ∠ OAB=∠ OAD+∠ BAD=∠ B+∠ BAD=180°-90°=90°$,即$AB⊥ OA$.$\because OA$是$\odot O$的半径,$\therefore AB$为$\odot O$的切线 (2) $\because ∠ OAB=90°$,$∠ AOB=45°$,$\therefore ∠ B=∠ AOB=45°$,$\therefore AB=OA$.$\because \odot O$的半径为2,$\therefore AB=OA=OC=2$,$\therefore$在$\mathrm{Rt}△ OAB$中,$OB=\sqrt{AB^2+OA^2}=2\sqrt{2}$,$\therefore CB=OB-OC=2\sqrt{2}-2$,$\therefore CB$的长是$2\sqrt{2}-2$