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8.如图，A,B,C,D,E都是圆O上的点，$\overset{\frown}{AC}=\overset{\frown}{AE}$，∠B=116°，则∠D的度数为
128°
。答案
8.$128°$
9.(2025·宿迁宿城新区教学共同体期末)如图,四边形ABCD是$\odot O$的内接四边形,$\odot O$的半径为5,$∠ B=135°$,则弦AC的长为
$5\sqrt{2}$
.答案
9.$5\sqrt{2}$
10.如图，在△ABC中，$BC=4\sqrt{2}$，$∠ BAC=135°$，求△ABC的外接圆$\odot O$的半径长.
答案
10.解:在$BC$的下侧的$\odot O$上取点$D$,连接$BD,CD,BO,CO$,
$\because ∠BAC=135°, \therefore ∠D=180°-∠BAC=45°$,
$\therefore ∠BOC=2∠D=90°, \therefore OB^2+OC^2=BC^2$.
$\because OB=OC,BC=4\sqrt{2}, \therefore 2OB^2=(4\sqrt{2})^2$,
$\therefore OB=4, \therefore △ABC$的外接圆$\odot O$的半径长为4.
11.如图,在$△ ABC$中,$AB=AC$,以$AB$为直径的$\odot O$交$BC$于点$D$,交$AC$于点$E$,连接$BE$.
(1)若$AB=6$,$CD=2$,求$CE$的长;
(2)当$∠ A$为锐角时,判断$∠ BAC$与$∠ CBE$的关系,并证明你的结论.
(1)若$AB=6$,$CD=2$,求$CE$的长;
(2)当$∠ A$为锐角时,判断$∠ BAC$与$∠ CBE$的关系,并证明你的结论.
答案
11.解:连接$AD$.如答图.
(1)$\because AB$为$\odot O$的直径,$\therefore AD⊥BC,BE⊥AC$.
$\because AB=AC=6, \therefore BD=CD=2, \therefore BC=4$.
由勾股定理,得$BE^2=AB^2-AE^2,BE^2=BC^2-CE^2$,
即$AB^2-AE^2=BC^2-CE^2$.
设$CE=x$,则$6^2-(6-x)^2=4^2-x^2$,
解得$x=\frac{4}{3}$,即$CE=\frac{4}{3}$.
(2)$∠BAC=2∠CBE$.证明如下:
$\because AB$为$\odot O$的直径,
$\therefore ∠ADC=∠ADB=90°,∠BEA=90°$,
$\therefore ∠CAD+∠C=90°,∠CBE+∠C=90°$,
$\therefore ∠CAD=∠CBE$.
$\because AB=AC, \therefore ∠BAD=∠CAD$,
$\therefore ∠BAC=2∠CAD=2∠CBE$.
12.(2025·宿迁宿城区期中)如图,在$\odot O$的内接四边形ABCD中,AB=AD,∠BCD=110°,点E在$\overset{\frown}{AD}$上.
(1)∠BAD=
(2)求∠AED的度数.
(1)∠BAD=
70
°;(2)求∠AED的度数.
答案
12.(1)70
(2)解:连接$BD$,如答图.
$\because AB=AD,∠BAD=70°, \therefore ∠ABD=55°$.
$\because$ 四边形$ABDE$内接于$\odot O$,
$\therefore ∠ABD+∠AED=180°$,
$\therefore ∠AED=180°-∠ABD=125°$.
13.如图，四边形ABCD内接于$\odot O$，$∠ ABC=60°$，对角线DB平分$∠ ADC$。
(1)求证：$△ ABC$是等边三角形；
(2)过点B作$BE // CD$交DA的延长线于点E，若$AD=2$，$DC=3$，求$△ BDE$的面积。
(1)求证：$△ ABC$是等边三角形；
(2)过点B作$BE // CD$交DA的延长线于点E，若$AD=2$，$DC=3$，求$△ BDE$的面积。
答案
13.(1)证明:$\because$ 四边形$ABCD$内接于$\odot O$,
$\therefore ∠ABC+∠ADC=180°$.
$\because ∠ABC=60°, \therefore ∠ADC=120°$.
$\because DB$平分$∠ADC, \therefore ∠ADB=∠CDB=60°$,
$\therefore ∠ACB=∠ADB=60°,∠BAC=∠CDB=60°$,
$\therefore ∠ABC=∠BCA=∠BAC, \therefore △ABC$是等边三角形.
(2)解:$\because BE// CD, \therefore ∠EBD=∠BDC$.
$\because ∠ADB=∠CDB=60°, \therefore ∠EBD=∠EDB=60°$,
$\therefore △BDE$是等边三角形.
又$\because △ABC$为等边三角形,
$\therefore ∠EBD=∠ABC=60°, \therefore ∠ABE=∠CBD$.
在$△ ABE$和$△ CBD$中,$\begin{cases} BE=BD, \\ ∠ABE=∠CBD, \\ AB=CB, \end{cases}$
$\therefore △ ABE≌ △ CBD(\mathrm{SAS})$,
$\therefore AE=CD=3, \therefore DE=AE+AD=5$,
$\therefore △ BDE$的面积为$\frac{1}{2}×5×\frac{\sqrt{3}}{2}×5=\frac{25\sqrt{3}}{4}$.
$\therefore ∠ABC+∠ADC=180°$.
$\because ∠ABC=60°, \therefore ∠ADC=120°$.
$\because DB$平分$∠ADC, \therefore ∠ADB=∠CDB=60°$,
$\therefore ∠ACB=∠ADB=60°,∠BAC=∠CDB=60°$,
$\therefore ∠ABC=∠BCA=∠BAC, \therefore △ABC$是等边三角形.
(2)解:$\because BE// CD, \therefore ∠EBD=∠BDC$.
$\because ∠ADB=∠CDB=60°, \therefore ∠EBD=∠EDB=60°$,
$\therefore △BDE$是等边三角形.
又$\because △ABC$为等边三角形,
$\therefore ∠EBD=∠ABC=60°, \therefore ∠ABE=∠CBD$.
在$△ ABE$和$△ CBD$中,$\begin{cases} BE=BD, \\ ∠ABE=∠CBD, \\ AB=CB, \end{cases}$
$\therefore △ ABE≌ △ CBD(\mathrm{SAS})$,
$\therefore AE=CD=3, \therefore DE=AE+AD=5$,
$\therefore △ BDE$的面积为$\frac{1}{2}×5×\frac{\sqrt{3}}{2}×5=\frac{25\sqrt{3}}{4}$.
