《2025年高效精练九年级数学上册苏科版》

1. 如图,AB是$\odot O$的直径,点C是$\odot O$上一点,连接AC,BC,则$∠C$的度数是 (B)



A.$60^{\circ }$
B.$90^{\circ }$
C.$120^{\circ }$
D.$150^{\circ }$

答案： B

2. (2023·广东)如图,AB是$\odot O$的直径,$∠BAC=50^{\circ }$,则$∠D=$ (B)
A.$20^{\circ }$
B.$40^{\circ }$
C.$50^{\circ }$
D.$80^{\circ }$

答案： B

3. (2023·湖北)如图,在$\odot O$中,直径AB与弦CD相交于点P,连接AC,AD,BD,若$∠C=20^{\circ },∠BPC=70^{\circ }$,则$∠ADC=$ (D)

A.$70^{\circ }$
B.$60^{\circ }$
C.$50^{\circ }$
D.$40^{\circ }$

答案： D

4. 如图,AB是半圆的直径,C,D是半圆上的两点,$∠ADC=106^{\circ }$,则$∠CAB$等于 (C)
A.$10^{\circ }$
B.$14^{\circ }$
C.$16^{\circ }$
D.$26^{\circ }$

答案： C

5. (2023·常州)如图,AD是$\odot O$的直径,$△ABC$是$\odot O$的内接三角形.若$∠DAC=∠ABC,AC=4$,则$\odot O$的直径$AD=$$4\sqrt{2}$.

答案： $4\sqrt{2}$

6. 如图,OA是$\odot O$的半径,以OA为直径的$\odot C$与$\odot O$的弦AB相交于点D.点D是AB的中点吗? 为什么?

点D是AB的中点. 理由: 连接OD, 可得$OD \perp AB$, 在$\odot O$中, 根据垂径定理, 得$AD = DB$.

答案： 点D是AB的中点. 理由: 连接OD, 可得$OD \perp AB$, 在$\odot O$中, 根据垂径定理, 得$AD = DB$.

7. 如图,AB是半圆O的直径,C,D是半圆O上的两点,且$OD// BC$,OD与AC交于点E.
(1)若$∠B=70^{\circ }$,求$∠CAD$的度数;35°
(2)若$AB=4,AC=3$,求DE的长.$2 - \frac{\sqrt{7}}{2}$

答案：
(1) $\because OD // BC$, $\angle B = 70^{\circ}$, $\therefore \angle DOA = \angle B = 70^{\circ}$. 又$\because OA = OD$, $\therefore \angle DAO = \angle ADO = 55^{\circ}$. $\because AB$是直径, $\therefore \angle ACB = 90^{\circ}$, $\therefore \angle CAB = 90^{\circ} - \angle B = 20^{\circ}$, $\therefore \angle CAD = \angle DAO - \angle CAB = 35^{\circ}$.
(2) 在$Rt\triangle ACB$中, $BC = \sqrt{AB^{2} - AC^{2}} = \sqrt{7}$. $\because OE \perp AC$, $\therefore AE = EC$, 又$\because OA = OB$, $\therefore OE = \frac{1}{2}BC = \frac{\sqrt{7}}{2}$. 又$\because OD = \frac{1}{2}AB = 2$, $\therefore DE = OD - OE = 2 - \frac{\sqrt{7}}{2}$.

8. (2024·安徽)如图,$\odot O$是$△ABC$的外接圆,D是直径AB上一点,$∠ACD$的平分线交AB于点E,交$\odot O$于另一点F,$FA=FE$.
(1)求证:$CD⊥AB;$
(2)设$FM⊥AB$,垂足为M,若$OM=OE=1$,求AC的长.

(1) $\because FA = FE$, $\therefore \angle FAE = \angle AEF$. $\because \angle FAE$与$\angle BCE$都是$\overset{\frown}{BF}$所对的圆周角, $\therefore \angle FAE = \angle BCE$. $\because \angle AEF = \angle CEB$, $\therefore \angle CEB = \angle BCE$. $\because CE$平分$\angle ACD$, $\therefore \angle ACE = \angle DCE$. $\because AB$是直径, $\therefore \angle ACB = 90^{\circ}$, $\therefore \angle CEB + \angle DCE = \angle BCE + \angle ACE = \angle ACB = 90^{\circ}$, $\therefore \angle CDE = 90^{\circ}$, $\therefore CD \perp AB$;
(2) $4\sqrt{2}$

答案：
(1) $\because FA = FE$, $\therefore \angle FAE = \angle AEF$. $\because \angle FAE$与$\angle BCE$都是$\overset{\frown}{BF}$所对的圆周角, $\therefore \angle FAE = \angle BCE$. $\because \angle AEF = \angle CEB$, $\therefore \angle CEB = \angle BCE$. $\because CE$平分$\angle ACD$, $\therefore \angle ACE = \angle DCE$. $\because AB$是直径, $\therefore \angle ACB = 90^{\circ}$, $\therefore \angle CEB + \angle DCE = \angle BCE + \angle ACE = \angle ACB = 90^{\circ}$, $\therefore \angle CDE = 90^{\circ}$, $\therefore CD \perp AB$;
(2) 由
(1)知, $\angle BEC = \angle BCE$, $\therefore BE = BC$. $\because AF = EF$, $FM \perp AB$, $\therefore MA = ME = 2$, $AE = 4$, $\therefore$圆的半径$OA = OB = AE - OE = 3$, $\therefore BC = BE = OB - OE = 2$, 在$\triangle ABC$中, $AB = 6$, $BC = 2$, $\angle ACB = 90^{\circ}$, $\therefore AC = \sqrt{AB^{2} - BC^{2}} = \sqrt{6^{2} - 2^{2}} = 4\sqrt{2}$.