三、解答题(解答应写出文字说明、证明过程或演算步骤)(共52分)
15.(6分)在$\mathrm{Rt}△ ABC$中，$AB=10\ \mathrm{cm}$，$BC=6\ \mathrm{cm}$，$AC=8\ \mathrm{cm}$，求以点$C$为圆心、$r$为半径的$\odot C$与直线$AB$的位置关系.
(1)$r=4\ \mathrm{cm}$.
(2)$r=4.8\ \mathrm{cm}$.
(3)$r=6\ \mathrm{cm}$.

15. 解:过点$C$作$CD ⊥ AB$于点$D$,则$CD=\frac{AC · BC}{AB}=4.8\ \mathrm{cm}$. (1)$\because$ 当$r=4\ \mathrm{cm}$时，$CD＞r$，$\therefore \odot C$与直线$AB$相离. (2)$\because$ 当$r=4.8\ \mathrm{cm}$时，$CD=r$，$\therefore \odot C$与直线$AB$相切. (3)$\because$ 当$r=6\ \mathrm{cm}$时，$CD＜r$，$\therefore \odot C$与直线$AB$相交.

16.(8分)如图，△ABC的顶点A，C在⊙O上，圆心O在边AB上，∠ACB=120°，BC与⊙O相切于点C，连接OC.
(1)求∠ACO的度数.
(2)求证:AC=BC.

16. (1) 解: $\because BC$与$\odot O$相切于点$C$，$\therefore OC ⊥ BC$，$\therefore ∠ OCB=90°$，$\therefore ∠ ACO=∠ ACB-∠ OCB=120°-90°=30°$. (2) 证明: $\because OA=OC$，$\therefore ∠ A=∠ ACO=30°$，$\therefore ∠ B=180°-∠ A-∠ ACB=180°-30°-120°=30°$，$\therefore ∠ A=∠ B$，$\therefore AC=BC$.

17. (7分)如图,△ABC的内切圆⊙O与BC,CA,AB分别相切于点D,E,F,且AB=18 cm, BC=28 cm, CA=26 cm,求AF,BD,CE的长.

17. 解: $\because △ ABC$的内切圆$\odot O$与$BC,CA,AB$分别相切于点$D,E,F$，$\therefore AE=AF$，$BF=BD$，$CE=CD$. 设$AE=AF=x\ \mathrm{cm}$，则$CE=CD=(26-x)\ \mathrm{cm}$，$BD=BF=(18-x)\ \mathrm{cm}$.$\because BC=28\ \mathrm{cm}$，$\therefore (18-x)+(26-x)=28$，解得$x=8$，$\therefore AF=8\ \mathrm{cm}$，$BD=10\ \mathrm{cm}$，$CE=18\ \mathrm{cm}$.

18.(8分)如图,在△OAB中,点A在⊙O上,边OB交⊙O于点C,AD⊥OB于点D,AC是∠BAD的平分线.
(1)求证:AB为⊙O的切线.
(2)若⊙O的半径为2,∠AOB=45°,求CB的长.

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