8.(2025·扬州)如图,点A,B,C在⊙O上,∠BAC=50°,则∠OBC=

40°
.答案
8.40°
9.(2025·宿迁沭阳县二模)如图,在$\odot O$中,半径$OA$,$OB$互相垂直,点$C$在劣弧$AB$上.若$∠ ABC=19°$,则$∠ BAC$的度数是

26
°.答案
9.26
10.(2025·南京)如图,已知扇形AOB,D为圆弧上一点,且$\overset{\frown}{ADB}$的度数为$260°$,若P为扇形内一点,则$∠ APB$的取值范围是

$50°<∠ APB<100°$
.答案
10.$50°<∠ APB<100°$
11.(2025·宿迁宿城区期末)如图,直径AB,CD的夹角∠AOC=60°,P为弧BC上的一个动点(不与点B,C重合).PM,PN分别垂直于CD,AB,垂足分别为M,N.若$\odot O$的半径长为2,则MN的长为

$\sqrt{3}$
.答案
11.$\sqrt{3}$
12.如图,点A,B,C为⊙O上的三个点,且△ABC为等边三角形,P为BC上一点.求证:PA=PB+PC.

答案
12.证明:如答图,在PA上截取PD=PB,连接BD.
$\because △ ABC$是等边三角形,
$\therefore ∠ ABC=∠ ACB=60°,AB=BC$,
$\therefore ∠ BPA=∠ ACB=60°,∠ APC=∠ ABC=60°$,
$\therefore ∠ BPC=120°$.
又$\because PD=PB,\therefore △ PBD$是等边三角形,
$\therefore ∠ BDP=60°,BD=PB,\therefore ∠ BDA=120°$.
又$\because ∠ BAP=∠ BCP,\therefore △ ABD≌△ CBP(\mathrm{AAS})$,
$\therefore AD=PC,\therefore PA=PD+AD=PB+PC$.
13.如图,$\odot O$中两条互相垂直的弦AB,CD交于点P,AB经过点O,E是AC的中点,连接OE,EP,延长EP交BD于点F。
(1)若$AB=10$,$OE=\sqrt{10}$,求AC的长;
(2)求证:$EF⊥ BD$。

(1)若$AB=10$,$OE=\sqrt{10}$,求AC的长;
(2)求证:$EF⊥ BD$。
答案
13.(1)解:$\because E$是AC的中点,$\therefore OE⊥ AC,AC=2AE$.
$\because AB=10,\therefore OA=\frac{1}{2}AB=5$.
在$\mathrm{Rt}△ AOE$中,$OE=\sqrt{10}$,
$\therefore AE=\sqrt{OA^2-OE^2}=\sqrt{5^2-(\sqrt{10})^2}=\sqrt{15}$,
$\therefore AC=2AE=2\sqrt{15}$,$\therefore AC$的长为$2\sqrt{15}$.
(2)证明:$\because AB⊥ CD,\therefore ∠ APC=∠ BPD=90°$,
$\therefore ∠ DPF+∠ BPF=90°$.
$\because E$是AC的中点,$\therefore EP=EC=\frac{1}{2}AC$,
$\therefore ∠ EPC=∠ C$.
$\because ∠ EPC=∠ DPF,∠ B=∠ C,\therefore ∠ DPF=∠ B$,
$\therefore ∠ B+∠ BPF=90°$,
$\therefore ∠ BFP=180°-(∠ B+∠ BPF)=90°,\therefore EF⊥ BD$.
$\because AB=10,\therefore OA=\frac{1}{2}AB=5$.
在$\mathrm{Rt}△ AOE$中,$OE=\sqrt{10}$,
$\therefore AE=\sqrt{OA^2-OE^2}=\sqrt{5^2-(\sqrt{10})^2}=\sqrt{15}$,
$\therefore AC=2AE=2\sqrt{15}$,$\therefore AC$的长为$2\sqrt{15}$.
(2)证明:$\because AB⊥ CD,\therefore ∠ APC=∠ BPD=90°$,
$\therefore ∠ DPF+∠ BPF=90°$.
$\because E$是AC的中点,$\therefore EP=EC=\frac{1}{2}AC$,
$\therefore ∠ EPC=∠ C$.
$\because ∠ EPC=∠ DPF,∠ B=∠ C,\therefore ∠ DPF=∠ B$,
$\therefore ∠ B+∠ BPF=90°$,
$\therefore ∠ BFP=180°-(∠ B+∠ BPF)=90°,\therefore EF⊥ BD$.
登录