7. 如图，PA，PB是$\odot O$的切线，A，B为切点，过点A作$AC // PB$交$\odot O$于点C，连接BC，若$∠ P = α$，则$∠ PBC$的度数为 （）

A.$90° + \frac{1}{2}α$
B.$90° - \frac{1}{2}α$
C.$180° - α$
D.$180° - \frac{1}{2}α$

7.A

8.(2025·玄武区二模)如图,在四边形ABCD中,AD//BC,经过A,B,C三点的⊙O与AD相切于点A,与CD交于点E,连接BE.若∠ABC=58°,则∠BEC的度数为.

8.$64°$

9.如图,等边三角形ABC的边长为4,$\odot C$的半径为$\sqrt{3}$,P为AB边上一动点,过点P作$\odot C$的切线PQ,切点为Q,则PQ长度的最小值为________.

9.3

10.(2025·海门市二模)如图,AB为$\odot O$的直径,C为$\odot O$上一点,BD和过点C的切线互相垂直,垂足为D,DB的延长线交$\odot O$于点E,连接BC,CE.
(1)求证:BC平分$∠ ABD$;
(2)若$AB=5,CE=4$,求BC的长.

10.(1)证明:连接OC,如答图①,
由题意,得$OC⊥ CD,ED⊥ CD$,
$\therefore ∠ OCD=∠ EDG=90°$,
$\therefore OC// ED,\therefore ∠ OCB=∠ CBD$.
又$\because OB=OC,\therefore ∠ OCB=∠ OBC$,
$\therefore ∠ CBD=∠ OBC,\therefore BC$平分$∠ ABD$.
(2)解:连接CA,AE,如答图②,
则$∠ CAE+∠ CBE=180°,∠ CBE+∠ CBD=180°$,
$\therefore ∠ CAE=∠ CBD$.
又$\because ∠ ABC=∠ AEC,∠ CBD=∠ OBC,\therefore ∠ CAE=∠ AEC,\therefore AC=CE=4$.
$\because ∠ ACB=90°,\therefore BC=\sqrt{AB^2-AC^2}=\sqrt{5^2-4^2}=3$.

11.(2025·滨湖区一模)如图,AB为$\odot O$的直径,点C在$\odot O$上,$∠ ACB$的平分线交$\odot O$于点D,过点D作$DE// AB$,交CB的延长线于点E.
(1)求证:ED是$\odot O$的切线;
(2)若$AC=12$,$BC=5$,求CD的长.

11.(1)证明:连接OD,如答图,
$\because CD$是$∠ ACB$的平分线,$\therefore ∠ ACD=∠ BCD$,
$\therefore ∠ AOD=∠ BOD$.
$\because AB$为$\odot O$的直径,
$\therefore ∠ AOD=∠ BOD=\frac{1}{2}× 180°=90°$,
$\therefore OD⊥ AB$.
$\because DE// AB,\therefore OD⊥ DE$.
$\because OD$为$\odot O$的半径,$\therefore ED$是$\odot O$的切线.
(2)解:如答图,连接BD,过点B作$BH⊥ CD$于点H.
$\because AB$为$\odot O$的直径,
$\therefore ∠ ACB=90°,∠ ADB=90°$.
$\because AC=12,BC=5$,
$\therefore AB=\sqrt{AC^2+BC^2}=\sqrt{12^2+5^2}=13$.
$\because ∠ ACB$的平分线CD交$\odot O$于点D,
$\therefore ∠ ACD=∠ BCD,\therefore \overset{\frown}{AD}=\overset{\frown}{BD}$,
$\therefore AD=BD=\frac{\sqrt{2}}{2}AB=\frac{13\sqrt{2}}{2}$.
$\because ∠ BCD=\frac{1}{2}∠ ACB=45°$,
$\therefore BH=CH=\frac{\sqrt{2}}{2}BC=\frac{5\sqrt{2}}{2}$,
$\therefore DH=\sqrt{BD^2-BH^2}=\sqrt{(\frac{13\sqrt{2}}{2})^2-(\frac{5\sqrt{2}}{2})^2}=6\sqrt{2}$,
$\therefore CD=CH+DH=\frac{5\sqrt{2}}{2}+6\sqrt{2}=\frac{17\sqrt{2}}{2}$.