《2025年通城学典课时作业本九年级数学上册苏科版江苏专版》

7. (2023·牡丹江)如图,A、B、C为$\odot O$上的三个点,$∠AOB=4∠BOC$.若$∠ACB=60^{\circ }$,则$∠BAC$的度数为 (C)

A.$20^{\circ }$
B.$18^{\circ }$
C.$15^{\circ }$
D.$12^{\circ }$

答案： 7. C

8. (2024·菏泽)如图,$△ABC$是$\odot O$的内接三角形.若$OA// CB,∠ACB=25^{\circ }$,则$∠CAB=$40$^{\circ }$.

答案： 8. 40

9. 如图,AB、CD是$\odot O$的弦,延长AB、CD相交于点P.若$∠P=30^{\circ },∠AOC=80^{\circ }$,则$\widehat {BD}$的度数是$20^{\circ}$.

答案： 9. $20^{\circ}$

10. (转化与化归思想)(2023·鞍山)如图,AC、BC为$\odot O$的两条弦,D、G分别为AC、BC的中点,$\odot O$的半径为2.若$∠C=45^{\circ }$,则DG的长为$\sqrt{2}$.

答案： 10. $ \sqrt{2} $ 解析：连接 $ AO $、$ BO $、$ AB $。根据圆周角定理，得 $ \angle AOB = 2 \angle C = 90^{\circ} $。在 $ Rt \triangle AOB $ 中，根据勾股定理，可得 $ AB = 2 \sqrt{2} $。在 $ \triangle CAB $ 中，根据三角形的中位线定理，可得 $ DG = \frac{1}{2} AB = \sqrt{2} $。

11. 如图,在四边形ABCD中,$AD=BC,∠B=∠D$,AD不平行于BC,过点C作$CE// AD$,交$△ABC$的外接圆$\odot O$于点E,连接AE.
(1) 求证:四边形AECD为平行四边形;
(2) 连接CO,求证:CO平分$∠BCE$.

答案：
11.
(1) $ \because \overset{\frown}{AC} = \overset{\frown}{AC} $，$ \therefore \angle B = \angle E $。$ \because \angle B = \angle D $，$ \therefore \angle E = \angle D $。$ \because CE // AD $，$ \therefore \angle D + \angle ECD = 180^{\circ} $，$ \therefore \angle E + \angle ECD = 180^{\circ} $，$ \therefore AE // CD $，$ \therefore $ 四边形 $ AECD $ 为平行四边形
(2) 如图，连接 $ OE $、$ OB $。$ \because $ 四边形 $ AECD $ 为平行四边形，$ \therefore AD = EC $。$ \because AD = BC $，$ \therefore EC = BC $。又 $ \because OC = OC $，$ OE = OB $，$ \therefore \triangle COE \cong \triangle COB $，$ \therefore \angle OCE = \angle OCB $，即 $ CO $ 平分 $ \angle BCE $

12. (2023·温州)如图,四边形ABCD的四个顶点均在$\odot O$上,$BC// AD,AC⊥BD$.若$∠AOD=120^{\circ },AD=\sqrt {3}$,求:
(1)$∠CAO$的度数;
(2) BC的长.

答案：
12.
(1) 如图，设 $ AC $、$ BD $ 交于点 $ E $。$ \because AC \perp BD $，$ \therefore \angle AED = 90^{\circ} $。$ \because BC // AD $，$ \therefore \angle DBC = \angle ADB $。$ \because \overset{\frown}{CD} = \overset{\frown}{CD} $，$ \therefore \angle DBC = \angle DAC $，$ \therefore \angle ADB = \angle DAC $，$ \therefore $ 在 $ Rt \triangle AED $ 中，$ \angle ADB = \angle DAC = 45^{\circ} $。$ \because OA = OD $，$ \therefore \angle OAD = \angle ODA $。$ \because $ 在 $ \triangle OAD $ 中，$ \angle AOD = 120^{\circ} $，$ \therefore \angle OAD = 30^{\circ} $，$ \therefore \angle CAO = \angle DAC - \angle OAD = 15^{\circ} $
(2) 如图，连接 $ OB $、$ OC $，过点 $ O $ 作 $ OH \perp AD $，垂足为 $ H $。$ \because OA = OD $，$ OH \perp AD $，$ \therefore AH = \frac{1}{2} AD = \frac{\sqrt{3}}{2} $。$ \because $ 在 $ Rt \triangle OHA $ 中，$ \angle OAH = 30^{\circ} $，$ \therefore $ 易得 $ OH = \frac{1}{2} OA $。在 $ Rt \triangle OHA $ 中，由勾股定理，得 $ OH^{2} + AH^{2} = OA^{2} $，$ \therefore \left( \frac{1}{2} OA \right)^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2} = OA^{2} $，解得 $ OA = 1 $（负值舍去）。$ \because \overset{\frown}{CD} = \overset{\frown}{CD} $，$ \therefore \angle COD = 2 \angle DAC = 90^{\circ} $。同理，得 $ \angle AOB = 90^{\circ} $。$ \because \angle AOD = 120^{\circ} $，$ \therefore \angle BOC = 360^{\circ} - 90^{\circ} - 90^{\circ} - 120^{\circ} = 60^{\circ} $。$ \because OB = OC $，$ \therefore \triangle OBC $ 是等边三角形，$ \therefore BC = OB $。$ \because OB = OA = 1 $，$ \therefore BC = 1 $