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1、NO.1 双基巧整合NO.2 典例悟内涵NO.3 演练练理考场场工具工具选修4-1 几何证明选讲栏目导引栏目导引1圆周角定理(1)圆周角定理 圆上一条弧所对的圆周角等于它所对的圆心角的(2)圆心角定理 圆心角的度数等于推论1 同弧或等弧所对的圆周角；同圆或等圆中，相等的圆周角所对的弧也推论2 半圆(或直径)所对的圆周角是；90的圆周角所对的弦是一半它所对弧的度数相等相等直角直径Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with A
2、spose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011
3、Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.NO.1 双基巧整合NO.2 典例悟内涵NO.3 演练练理考场场工具工具选修4-1 几何证明选讲栏目导引栏目导引2圆内接四边形的性质与判定定理(1)性质定理1 圆的内接四边形的对角 定理2 圆内接四边形的外角等于它的内角的(2)判定判定定理 如果一个四边形的对角互补，那么这个四边形的四个顶点 推论 如果四边形的一个外角等于它的内角的对角，那么这个四边形的四个顶点互补对角共圆共圆Evaluation only.Evaluation only. Created with Aspose.Slides
4、for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NE
5、T 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.NO.1 双基巧整合NO.2 典例悟内涵NO.3 演练练理考场场工具工具选修4-1 几何证明选讲栏目导引栏目导引3圆的切线的性质及判定定理(1)性质性质定理 圆的切线垂直于经过 切点的推论1 经过圆 心且垂直于切线的直线必过 推论2 经过 切点且垂直于切线的直线必过(2)判定定理 经过 半径的外端并且垂直于这条半径的直线是圆的4弦切角的性质定理 弦切角等于它所夹的弧所对的半径切点圆心切线圆周角Ev
6、aluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5
7、 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.NO.1 双基巧整合NO.2 典例悟内涵NO.3 演练练理考场场工具工具选修4-1 几何证明选讲栏目导引栏目导引5与圆有关的比例线段(1)相交弦定理 圆内的两条相交弦，被交点分成的两条线段长的相等(2)割线定理 从圆外一点引圆的两条割线，这一点到每条割线与圆的交点的两条线段长的 相等
8、(3)切割线定理 从圆外一点引圆的切线和割线，切线长 是这点到割线与圆交点的两条线段长的(4)切线长 定理 从圆外一点引圆的两条切线，它们的切线长 相等，圆心和这一点的连线 平分两条切线的 积积比例中项夹角Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Co
9、pyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.NO.1 双基巧整合NO.2 典例悟内涵NO.3 演练练理考场场工具工具选
10、修4-1 几何证明选讲栏目导引栏目导引证明多点共圆，当它们在一条线段同侧时 ，可证它们对 此线段张角相等，也可以证明它们与某一定点距离相等；如两点在一条线段异侧，则证 明它们与线段两端点连成的凸四边形对角互补如图，已知AP是O的切线，P为切点，AC是O的割线，与O交于B、C两点，圆心O在PAC的内部，点M是BC的中点(1)证明：A，P，O，M四点共圆；(2)求OAMAPM的大小Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with A
11、spose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011
12、Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.NO.1 双基巧整合NO.2 典例悟内涵NO.3 演练练理考场场工具工具选修4-1 几何证明选讲栏目导引栏目导引解析： (1)证明：连接OP，OM，因为AP与O相切于点P，所以OPAP，因为M是O的弦BC的中点，所以OMBC，于是OPAOMA180.由圆心O在PAC的内部，可知四边形APOM的对角互补，所以A，P，O，M四点共圆(2)由(1)得A，P，O，M四点共圆，所以OAMOPM，由(1)得OPAP，由圆心O在PAC的内部，可知OPMAPM90，所以OAMAPM90.Evaluation o
13、nly.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Pr
14、ofile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.NO.1 双基巧整合NO.2 典例悟内涵NO.3 演练练理考场场工具工具选修4-1 几何证明选讲栏目导引栏目导引【变式训练】 1.如图，已知ABC的两条角平分线AD和CE相交于H， B 60， F在 AC上，且AE AF.(1)证明：B、 D、 H、 E四点共圆；(2)证明：CE平分DEF.Evalua
15、tion only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Cli
16、ent Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.NO.1 双基巧整合NO.2 典例悟内涵NO.3 演练练理考场场工具工具选修4-1 几何证明选讲栏目导引栏目导引证明： (1)在ABC中，因为B60，所以BACBCA120.因为AD、CE是角平分线，所以HACHCA60，故AHC120.于是EHDAHC120.因为EBDEHD180，所以B、D、H、E四点共圆Evaluation only.Evaluation o
