1 . 如图所示,在⊙O中,相交于点E的两弦AB,CD的中点分别是M,N,直线MO与直线CD相交于点F.证明
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/26/cdcc2cdf-e541-42c8-bbce-48c05d3a473d.png?resizew=240)
(1)∠MEN+∠NOM=180°;
(2)FE·FN=FM·FO.
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/26/cdcc2cdf-e541-42c8-bbce-48c05d3a473d.png?resizew=240)
(1)∠MEN+∠NOM=180°;
(2)FE·FN=FM·FO.
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2 . 如图所示,PA与圆O相切于点A,不过圆心O的割线PCB与直径AE相交于D点.已知∠BPA=30°,AD=2,PC=1,则圆O的半径等于________ .
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758200027602944/1817081830203392/STEM/91b436abd3534b12b4bbcc4893490293.png?resizew=87)
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3 . 如图所示,AB是半圆O的直径,点C在半圆上,CD⊥AB于点D,且AD=3DB,设∠COD=θ,则tan2
=__ .
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a16d92d3ccdf0ecfaa43eb8ed62ae5a4.png)
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758200027602944/1817081830195200/STEM/027e6e38fed14b0d94131d2fd983e44c.png?resizew=107)
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4 . 如图所示,⊙O是△ABC的外接圆,AD是⊙O的直径,连接CD,若⊙O的半径r=
,AC=2,则cos B的值是( )
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758200027602944/1817081830154240/STEM/c15e7f13ca2b468897843317de4c17b0.png?resizew=84)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a4b8503f4706b8321e4e79a87eadea84.png)
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758200027602944/1817081830154240/STEM/c15e7f13ca2b468897843317de4c17b0.png?resizew=84)
A.![]() | B.![]() |
C.![]() | D.![]() |
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5 . 如图,已知Rt△ABC,∠ABC=90°,D是AC的中点,⊙O经过A,B,D三点,CB的延长线交⊙O于点E,过点E作⊙O的切线,交AC的延长线于点F.在满足上述条件的情况下,当∠CAB的大小变化时,图形也随着改变,但在这个变化过程中,有些线段总保持着相等的关系.
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758848846274560/1815497844531200/STEM/0591bcb2e32041fca413dbc5eaf15a7c.png?resizew=104)
(1)连接图中已标明字母的某两点,得到一条新线段与线段CE相等,并说明理由;
(2)若CF=CD,求sin F的值.
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758848846274560/1815497844531200/STEM/0591bcb2e32041fca413dbc5eaf15a7c.png?resizew=104)
(1)连接图中已标明字母的某两点,得到一条新线段与线段CE相等,并说明理由;
(2)若CF=CD,求sin F的值.
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6 . 如图,已知AB切⊙O于B,BC是⊙O的直径,AC交⊙O于D,DE是⊙O的切线,CE⊥DE于E,DE=3,CE=4,求AB的长.
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758848846274560/1815497844514816/STEM/be063f0aebf649ecad15635038670570.png?resizew=83)
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7 . 如图,已知AB为⊙O的弦,CD切⊙O于P,AC⊥CD于C,BD⊥DC于D,PQ⊥AB于Q.
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758848846274560/1815497844498432/STEM/ebb5e43b97724ba79ce98d2a7ee094a6.png?resizew=91)
求证:PQ2=AC·BD.
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758848846274560/1815497844498432/STEM/ebb5e43b97724ba79ce98d2a7ee094a6.png?resizew=91)
求证:PQ2=AC·BD.
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8 . 如图,在△ABC中,∠C=90°,AC=8 cm,AB=10 cm,点P由C出发以每秒2 cm的速度沿线段CA向点A运动(不运动至A点),⊙O的圆心在BP上,且⊙O分别与AB、AC相切,当点P运动2 s时,⊙O的半径是( )
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758848846274560/1815497844424704/STEM/6f215c156a5e4376b8b1fb61b4dcfa95.png?resizew=139)
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758848846274560/1815497844424704/STEM/6f215c156a5e4376b8b1fb61b4dcfa95.png?resizew=139)
A.![]() | B.![]() | C.![]() | D.2 cm |
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9 . 如图,四边形ABCD为圆内接四边形,AC为BD的垂直平分线,∠ACB=60°,AB=a,则CD等于( )
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758848846274560/1815497844408320/STEM/7aedc0bac21d4ba08abc492881e89365.png?resizew=91)
![](https://img.xkw.com/dksih/QBM/2017/8/14/1758848846274560/1815497844408320/STEM/7aedc0bac21d4ba08abc492881e89365.png?resizew=91)
A.![]() | B.![]() | C.![]() | D.![]() |
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10 . 如图,AB是⊙O的直径,C为半圆上一点,CD⊥AB于D,若BC=3,AC=4,则AD∶CD∶BD等于( )
A.4∶6∶3 | B.6∶4∶3 |
C.4∶4∶3 | D.16∶12∶9 |
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