《2025年高效精练九年级数学上册苏科版》

7. 已知:$\odot O$的两条弦AB,CD相交于点M.
(1) 如图1,若$AB=CD$,连接AD.求证:$∠A=∠D$.
(2) 如图2,若$AB⊥CD$,在$\widehat {BD}$上取一点E,使$\widehat {BE}=\widehat {BC}$,AE交CD于点F,连AD,DE.若$∠E=72^{\circ }$,求$∠BAE$的大小.

答案：

(1) $ \because AB = CD $，$ \therefore \overset{\frown}{AB} = \overset{\frown}{CD} $，$ \therefore \overset{\frown}{AB} - \overset{\frown}{BC} = \overset{\frown}{CD} - \overset{\frown}{BC} $，$ \therefore \overset{\frown}{AC} = \overset{\frown}{BD} $，$ \therefore \angle A = \angle D $；
(2) 连接 $ CA $，$ \because \angle E = 72^{\circ} $，$ \therefore \angle E = \angle ACF = 72^{\circ} $。$ \because AB \perp CD $，$ \therefore \angle AMC = 90^{\circ} $，$ \therefore \angle CAM = 90^{\circ} - \angle ACM = 18^{\circ} $。$ \because \overset{\frown}{BE} = \overset{\frown}{BC} $，$ \therefore \angle BAE = \angle CAM = 18^{\circ} $，$ \therefore \angle BAE $ 的度数为 $ 18^{\circ} $。

8. 如图,AB,CD是$\odot O$的弦,延长AB,CD相交于点P.已知$∠P=30^{\circ },∠AOC=80^{\circ }$,则$\widehat {BD}$的度数是 (C)

A. $30^{\circ }$
B. $25^{\circ }$
C. $20^{\circ }$
D. $10^{\circ }$

答案： C

9. 如图,已知$AB=AC=AD,∠CBD=2∠BDC,∠BAC=44^{\circ }$,则$∠CAD$的度数为 (B)

A. $68^{\circ }$
B. $88^{\circ }$
C. $90^{\circ }$
D. $112^{\circ }$

答案： B

10. (2023·郴州)如图,某博览会上有一圆形展示区,在其圆形边缘的点P处安装了一台监视器,它的监控角度是$55^{\circ }$,为了监控整个展区,最少需要在圆形边缘上共安装这样的监视器______4台.

答案： 4

11. (2023·武汉)如图,OA,OB,OC都是$\odot O$的半径,$∠ACB=2∠BAC$.
(1) 求证:$∠AOB=2∠BOC$;
(2) 若$AB=4,BC=\sqrt {5}$,求$\odot O$的半径.

答案：

(1) $ \because \angle ACB = \frac{1}{2} \angle AOB $，$ \angle BAC = \frac{1}{2} \angle BOC $，$ \angle ACB = 2 \angle BAC $，$ \therefore \angle AOB = 2 \angle BOC $；
(2) 过点 $ O $ 作半径 $ OD \perp AB $ 于点 $ E $，连接 $ DB $，$ \therefore AE = BE $。$ \because \angle AOB = 2 \angle BOC $，$ \angle DOB = \frac{1}{2} \angle AOB $，$ \therefore \angle DOB = \angle BOC $，$ \therefore BD = BC $。$ \because AB = 4 $，$ BC = \sqrt{5} $，$ \therefore BE = 2 $，$ DB = \sqrt{5} $，在 $ \text{Rt} \triangle BDE $ 中，$ \angle DEB = 90^{\circ} $，$ \therefore DE = \sqrt{BD^{2} - BE^{2}} = 1 $，在 $ \text{Rt} \triangle BOE $ 中，$ \angle OEB = 90^{\circ} $，$ OB^{2} = (OB - 1)^{2} + 2^{2} $，解得 $ OB = \frac{5}{2} $，即 $ \odot O $ 的半径是 $ \frac{5}{2} $。