1. [2026镇江扬中市一中月考]如图,点B在$\odot A$上,点C在$\odot A$外,以下条件不能判定BC是$\odot A$切线的是 ()

A.$∠ A=50°, ∠ C=40°$
B.$∠ B - ∠ C = ∠ A$
C.$AB^2 + BC^2 = AC^2$
D.$\odot A$与AC的交点是AC的中点

1 D 对于A,$\because ∠A=50°,∠C=40°,\therefore ∠B=180°-∠A-∠C=90°,\therefore BC⊥AB,\because$ 点B在$\odot A$上,$\therefore AB$是$\odot A$的半径,$\therefore BC$是$\odot A$的切线,故A不符合题意;对于B,$\because ∠B-∠C=∠A,\therefore ∠B=∠A+∠C,$又$\because ∠A+∠B+∠C=180°,\therefore ∠B=90°,\therefore BC⊥AB,\because AB$是$\odot A$的半径,$\therefore BC$是$\odot A$的切线,故B不符合题意;对于C,$\because AB^2+BC^2=AC^2,\therefore △ABC$是直角三角形,$∠B=90°,\therefore BC⊥AB,$又$\because AB$是$\odot A$的半径,$\therefore BC$是$\odot A$的切线,故C不符合题意;对于D,当$\odot A$与AC的交点是AC的中点时,不能判定BC是$\odot A$的切线,故D符合题意.

2. 教材例题变式 如图，$△ ABC$是$\odot O$的内接三角形，AB是直径，过点A作直线EF.添加下列一个条件：①$AB ⊥ EF$；②$∠ ACB = ∠ FAB$；③$∠ ABC = ∠ EAC$；④$∠ EAC = ∠ BAC$.其中不能证明EF是$\odot O$切线的是.(填序号)

2 ④ $\because AB$是$\odot O$的直径,$\therefore ∠ACB=90°.$若添加$AB⊥EF$,则EF是$\odot O$的切线;若添加$∠ACB=∠FAB$,则$∠FAB=90°$,$\therefore EF$是$\odot O$的切线;若添加$∠ABC=∠EAC$,$\because ∠ABC+∠BAC=90°,\therefore ∠EAC+∠BAC=90°$,即$EF⊥AB,\therefore EF$是$\odot O$的切线,故添加条件①②③都能证明EF是$\odot O$的切线.易知添加④不能证明EF是$\odot O$的切线.

3. [2025 南京建邺区期中] 如图,在$△ ABC$中,$AB = AC$,以$AB$为直径作$\odot O$交$BC$于点$D$,过点$D$作$DE ⊥ AC$,垂足为$E$. 求证:$DE$为$\odot O$的切线.

3 证明:如图,连接OD.
$\because OD=OB,\therefore ∠ODB=∠B.$
$\because AB=AC,\therefore ∠C=∠B,$
$\therefore ∠ODB=∠C,\therefore OD// AC.$
$\because DE⊥AC,\therefore ∠ODE=∠DEC=90°.$
$\because OD$为$\odot O$的半径,且$DE⊥OD,$
$\therefore DE$为$\odot O$的切线.

4. [2024山西中考]如图，已知$△ ABC$，以AB为直径的$\odot O$交BC于点D，与AC相切于点A，连接OD.若$∠ AOD=80°$，则$∠ C$的度数为 （）

A.$30°$
B.$40°$
C.$45°$
D.$50°$

4 D $\because ∠AOD=80°,∠B,∠AOD$分别是弧AD所对的圆周角和圆心角,$\therefore ∠B=\frac{1}{2}∠AOD=40°. \because \odot O$与AC相切于点A,$\therefore ∠BAC=90°,\therefore ∠C=90°-∠B=50°.$

5. [2025福建中考]如图,PA与$\odot O$相切于点A,PO的延长线交$\odot O$于点C,$AB// PC$,且交$\odot O$于点B.若$∠P=30^{\circ }$,则$∠BCP$的度数为（）

A.$30^{\circ }$
B.$45^{\circ }$
C.$60^{\circ }$
D.$75^{\circ }$

5 C 如图,连接OA,OB,$\because PA$与$\odot O$相切于点A,$\therefore OA⊥PA,\therefore ∠AOP=90°-∠P=90°-30°=60°. \because AB// PC,$
$\therefore ∠OAB=∠AOP=60°. \because OA=OB,\therefore △AOB$为等边三角形,$\therefore ∠AOB=60°,\therefore ∠BOC=180°-∠AOP-∠AOB=60°.$
$\because OB=OC,\therefore △BOC$为等边三角形,$\therefore ∠BCP=60°.$

6. [2025无锡梁溪区期中] 如图，$\odot O$是$\mathrm{Rt}△ ABC$的外接圆，$∠ ACB=90°$，D是$\overset{\frown}{BC}$上的一点，且$\overset{\frown}{CD}=\overset{\frown}{CA}$，连接AD交BC于点F，过点A作$\odot O$的切线AE交BC的延长线于点E. 求证：$CF=CE$.

6 证明:$\because ∠ACB=90°,$
$\therefore ∠CAB+∠B=90°,∠ACE=90°=∠ACB.$
$\because AE$是$\odot O$的切线,$\therefore ∠BAE=90°,$
$\therefore ∠EAC+∠CAB=90°,\therefore ∠EAC=∠B.$
$\because \overset{\frown}{CD}=\overset{\frown}{CA},\therefore ∠CAD=∠B,$
$\therefore ∠EAC=∠CAD.$
$\because AC=AC,\therefore △ACE≌ △ACF,$
$\therefore CE=CF.$

7. [2026 常州河海实验学校月考] 如图，AB是$\odot O$的直径，C是$\odot O$上一点，过点C的切线交AB的延长线于点D，连接AC，BC.
(1)求证：$∠ BAC = ∠ BCD$.
(2)若$OA = 2$，$CD = 2\sqrt{3}$，求BD的长.

7 (1)证明:如图,连接OC.
$\because CD$是$\odot O$的切线,$\therefore OC⊥CD,$
$\therefore ∠OCD=90°$,即$∠OCB+∠DCB=90°.$
$\because AB$是$\odot O$的直径,
$\therefore ∠ACB=90°$,即$∠OCB+∠ACO=90°,$
$\therefore ∠ACO=∠BCD.$
$\because AO=CO,\therefore ∠ACO=∠OAC.$
$\therefore ∠BAC=∠BCD.$
(2)解:$\because OA=2,\therefore OC=OB=OA=2.$
设$BD=x$,则$OD=2+x.$
$\because ∠OCD=90°,\therefore OC^2+CD^2=OD^2.$
$\therefore 2^2+(2\sqrt{3})^2=(2+x)^2$,解得$x=2$或$x=-6$(舍去).
$\therefore BD=2.$