如图,在△BCE中,点A是边BE上一点,以AB为直径的圆O与CE相切于点D,
,点F为OC与圆O的交点,连接AF.
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/15/55b37827-9b64-4907-8033-ce90504edd45.png?resizew=201)
(1)求证:CB是圆O的切线.
(2)若
,图中阴影部分面积为
,求圆O的直径AB.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9204fa555e4c2945323c6c49116ccfa4.png)
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/15/55b37827-9b64-4907-8033-ce90504edd45.png?resizew=201)
(1)求证:CB是圆O的切线.
(2)若
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9685ab781f942a3e28b2f39778515185.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/05bce74fba12850cbda6af301f3e3933.png)
更新时间:2022-05-07 15:58:58
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【推荐1】如图,AC⊥BC,DC⊥EC, AC=BC, DC=EC, AE与BD交于点F.
![](https://img.xkw.com/dksih/QBM/editorImg/2023/2/27/3901c380-1558-4eba-b6d3-8722bb107ac8.png?resizew=171)
(1)求证: AE=BD;
(2)求∠DFE的度数.
![](https://img.xkw.com/dksih/QBM/editorImg/2023/2/27/3901c380-1558-4eba-b6d3-8722bb107ac8.png?resizew=171)
(1)求证: AE=BD;
(2)求∠DFE的度数.
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【推荐2】(1)【感知】:如图1,点P是
角平分线上一点,过点
作
于点
,
于点
,证明
(不需要证明 )
(2)【探究】如图2,在
中,
,
是
的平分线,点
在
边上,![](https://staticzujuan.xkw.com/quesimg/Upload/formula/337c5999cee92a096abaf0e9f5eeede7.png)
①证明:
;
②请判断
,
,
三条线段之间的数量关系,并说明理由.
(3)【拓展】如图3,
的外角
的平分线
与内角
的平分线
交于点
,若
,请直接写出
的度数.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6d7b2fe01a33c4825f9974ed9663a99c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dad2a36927223bd70f426ba06aea4b45.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4d1e98c3ba2bb2199c1985e4d3728639.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c5db41a1f31d6baee7c69990811edb9f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9bb69be70ba0babb236757648695faca.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8455657dde27aabe6adb7b188e031c11.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6ed66431681da1db8f7cb0f40cd19201.png)
(2)【探究】如图2,在
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b3c65edad25ddd666cdce0d7e5afefc9.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7e3262fc038bbec5e7c8cc47df08bef7.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/03902478df1a55bc99703210bccab910.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7cbce11aa19b8bd2bf6ee5a834e005de.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2a30f3a8b673cc28bd90c50cf1a35281.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/60ef95894ceebaf236170e8832dcf7e3.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/337c5999cee92a096abaf0e9f5eeede7.png)
①证明:
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ce25e95729c34d26e482512d36dd89b1.png)
②请判断
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f52a58fbaf4fea03567e88a9f0f6e37e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/68a83fdd2ba72a2dba0b6b10bb3e06b9.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4eedae8d316c76e3d0b451256de03fb9.png)
(3)【拓展】如图3,
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b3c65edad25ddd666cdce0d7e5afefc9.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/bdb8eca20ce2c918ea4034ea15210c7f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/63a253c7fdf589ee3dece13d5b5b5732.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d39b8d91afc34e4a9b0fdbb6bafb9087.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2cdba1337ec85fa9722cb4b320a82ae6.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dad2a36927223bd70f426ba06aea4b45.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/84494e057cb7bc129257c4cb962678ee.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2e800c6fa5152f2d250adf572550a1ca.png)
![](https://img.xkw.com/dksih/QBM/editorImg/2023/1/19/eed3aef8-ea5a-4c8f-b29c-9224a09aa827.png?resizew=497)
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【推荐3】【发现】(1)如图1,在△ABC中,AD是∠BAC的平分线,对于以下结论:
①AD是△ABC的中线;②S△ABD:S△ACD=AB:AC;③AB:AC=BD:DC,
其中正确的是 (只填序号)
【探究】(2)请你选择(1)中正确的一个选项,简述理由
【应用】(3)如图2,△ABC的三个内角的角平分线相交于点O,且AB=40,BC=48,AC=32,则SABO:S△BCO:S△ACO= : :
【拓展】(4)在(1)中的条件下,过点D作DE⊥AB于点E,DF⊥AB于点F,连接EF,求证:AD垂直平分EF.
①AD是△ABC的中线;②S△ABD:S△ACD=AB:AC;③AB:AC=BD:DC,
其中正确的是 (只填序号)
【探究】(2)请你选择(1)中正确的一个选项,简述理由
【应用】(3)如图2,△ABC的三个内角的角平分线相交于点O,且AB=40,BC=48,AC=32,则SABO:S△BCO:S△ACO= : :
【拓展】(4)在(1)中的条件下,过点D作DE⊥AB于点E,DF⊥AB于点F,连接EF,求证:AD垂直平分EF.
![](https://img.xkw.com/dksih/QBM/2019/1/10/2115709008912384/2120484639842304/STEM/c6fe6ef9c2154d13988b5978501a3981.png?resizew=394)
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【推荐1】在△ABC中,∠C=90°.
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/28/0820bfd1-b07e-4fac-ba8b-6dab19c3c3d2.png?resizew=527)
(1)如图①,点O在斜边AB上,以点O为圆心,OB长为半径的圆交AB于点D,交BC于点E,与边AC相切于点F,求证:∠1=∠2;
(2)在图②中作⊙M,使它满足以下条件:①圆心在边AB上;②经过点B;③与边AC相切;
(尺规作图,只保留作图痕迹,不要求写出作法)
(3)在(2)问条件下,若∠A=30°,⊙M的半径为2,求线段BC的长.
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/28/0820bfd1-b07e-4fac-ba8b-6dab19c3c3d2.png?resizew=527)
(1)如图①,点O在斜边AB上,以点O为圆心,OB长为半径的圆交AB于点D,交BC于点E,与边AC相切于点F,求证:∠1=∠2;
(2)在图②中作⊙M,使它满足以下条件:①圆心在边AB上;②经过点B;③与边AC相切;
(尺规作图,只保留作图痕迹,不要求写出作法)
(3)在(2)问条件下,若∠A=30°,⊙M的半径为2,求线段BC的长.
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【推荐2】如图,
与⊙O相切于点A,过点A作
于点C,交⊙O于点D,连接
交直径
的延长线于点E.
是⊙O的切线;
(2)若⊙O的半径为6,
,求
的长.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/bd33764ff4efddfe11a98a609753715c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0a807cc9a44fe75aa93afbdbeecbbf99.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e0629ce42392a7fe9be21d25c39c3e64.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f52a58fbaf4fea03567e88a9f0f6e37e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e0629ce42392a7fe9be21d25c39c3e64.png)
(2)若⊙O的半径为6,
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/49437f474e5805688dff21ded2d1fd7c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e0629ce42392a7fe9be21d25c39c3e64.png)
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【推荐1】已知:四边形
是
的内接四边形,
,连接
.
![](https://img.xkw.com/dksih/QBM/editorImg/2023/2/26/1cf0436b-9546-4512-9dc4-eee9f9ee4430.png?resizew=476)
(1)如图1,求证:
;
(2)如图2,连接
交
于点
.点
在
上,且
,过点
作
交
于点
,求证:
;
(3)如图3,在(2)的条件下,
是
的直径,点
在
上,若
,
,求线段
的长.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/411b38a18046fea8e9fab1f9f9b80a5f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3d97cdc586744d208b6f69c9813af977.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2e735a28578ba191da6d4f3b0f8e8729.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/60ef95894ceebaf236170e8832dcf7e3.png)
![](https://img.xkw.com/dksih/QBM/editorImg/2023/2/26/1cf0436b-9546-4512-9dc4-eee9f9ee4430.png?resizew=476)
(1)如图1,求证:
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e294887ab065cedf03b77f49292bc6a9.png)
(2)如图2,连接
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d40b319212a7e7528b053e1c7097e966.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/60ef95894ceebaf236170e8832dcf7e3.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2a30f3a8b673cc28bd90c50cf1a35281.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a0ed1ec316bc54c37c4286c208f55667.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/60ef95894ceebaf236170e8832dcf7e3.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/db3b0b5010813ccc388a41ba74f3b932.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a0ed1ec316bc54c37c4286c208f55667.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1fa5088fedf996af3d7c50a39bdaae73.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9d78abbad68bbbf12af10cd40ef4c353.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/895dc3dc3a6606ff487a4c4863e18509.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b880e9fe882e72e8e341646ca8ef1070.png)
(3)如图3,在(2)的条件下,
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/60ef95894ceebaf236170e8832dcf7e3.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3d97cdc586744d208b6f69c9813af977.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a0ed1ec316bc54c37c4286c208f55667.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/828628c0876b45381c9a0edeb0fec236.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7c339b0e700b7eef33972441df06faa6.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/da1d81a96bb1286a600694169a99dff5.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f52a58fbaf4fea03567e88a9f0f6e37e.png)
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【推荐2】在古代,智慧的劳动人民已经会使用“石磨”,其原理为在磨盘的边缘连接一个固定长度的“连杆”,推动“连杆”带动磨盘转动,将粮食磨碎,物理学上称这种动力传输工具为“曲线连杆机构”.小明受此启发设计了一个“双连杆机构”,设计图如图1,两个固定长度的“连杆”AP,BP的连接点P在
上,当点P在
上转动时,带动点A,B分别在射线OM,ON上滑动,
当AP与
相切时,点B恰好落在
上,如图2,请仅就图2的情形解答下列问题.
![](https://img.xkw.com/dksih/QBM/editorImg/2022/10/21/29e023cf-a95e-468b-82c4-39d701db7454.png?resizew=514)
(1)求证:
;
(2)若线段AO与
交与点C,AC=
,
,求
的半径
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3d97cdc586744d208b6f69c9813af977.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3d97cdc586744d208b6f69c9813af977.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/bff67929abd32bd40176de0c93770593.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3d97cdc586744d208b6f69c9813af977.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3d97cdc586744d208b6f69c9813af977.png)
![](https://img.xkw.com/dksih/QBM/editorImg/2022/10/21/29e023cf-a95e-468b-82c4-39d701db7454.png?resizew=514)
(1)求证:
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e89c90275323c4cb261043f6ac30a7ec.png)
(2)若线段AO与
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3d97cdc586744d208b6f69c9813af977.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5991e9ec7666f533a528a4173c58f0ff.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ed363b59b947b37d867c0c192a9f119a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3d97cdc586744d208b6f69c9813af977.png)
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