1. (2024·长沙)如图,在$\odot O$中,弦$AB$的长为8,圆心$O$到$AB$的距离$OE=4$,则$\odot O$的半径为
( )
A. 4
B. $4\sqrt{2}$
C. 5
D. $5\sqrt{2}$

B

2. 如图，AB是$\odot O$的直径，$OD⊥ AC$于点D，DO的延长线交$\odot O$于点E，连接BC.若$AC=4\sqrt{2}$，$DE=4$，则BC的长是______．

2

3. (2024·启东期中)如图，AB是$\odot O$的弦，C，D为直线AB上两点，且$OC=OD$，求证：$AC=BD$.

解: 如图，作$OH\perp AB$于点$H，$则$AH = BH.$$\because OC = OD，$$OH\perp AB，$$\therefore CH=DH.$$\therefore CH - AH = DH - BH，$即$AC = BD$ ;

考点二 圆心（周）角定理及其推论

4. (2024·泰安)如图，AB是$\odot O$的直径，C，D是$\odot O$上两点，BA平分$∠ CBD$.若$∠ AOD=50°$，则$∠ A$的度数为 （ ）
A. $65°$
B. $55°$
C. $50°$
D. $75°$

A

5. (2024•滨州)如图,四边形ABCD内接于$\odot O$.若四边形OABC是菱形,则$∠D=$______°.

60

6. 如图,在$△ ABC$中,$AB=AC$,以$AB$为直径作$\odot O$,交$BC$边于点$D$,交$CA$的延长线于点$E$,连接$AD$,$DE$.
(1) 求证:$BD=CD$;
(2) 若$AB=5$,$DE=4$,求$AD$的长.

(1)证明:$\because AB$是$\odot O$的直径，$\therefore \angle ADB = 90^{\circ}，$即$AD\perp BD.$又$\because AB = AC，$$\therefore BD = CD$ (2)解:$\because AB = AC，$$\therefore \angle B = \angle C.$$\because \angle B = \angle E，$$\therefore \angle E = \angle C.$$\therefore CD = DE = 4.$由 (1)，知$AD\perp BC.$$\because CD = 4，$$AC = AB = 5，$$\therefore$在$Rt\triangle ACD$中，由勾股定理，得$AD=\sqrt{AC^{2}-CD^{2}}=\sqrt{5^{2}-4^{2}} = 3$