4.6.2作垂直平分线和角平分线-教学教学课件-2026-2027学年湘教版数学八年级上册

#湘教版数学八年级上册4.6.2作垂直平分线和角平分线专项练习（全文1231字）##核心考点熟练掌握尺规作角平分线\(M\(A\(C、D\)B\)N\)尺规作线段垂直平分线两种基本作图；读懂作图痕迹\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)理解作图原理（SSS全等）；作图保留圆弧痕迹；区分两种作图的作图步骤\(M\(A\(C、D\)B\)N\)圆弧位置；能根据作图结果\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)利用角平分线\(M\(A\(C、D\)B\)N\)线段垂直平分线的性质进行计算与证明；规范书写作图语句\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)识别作图错误。答题要求：尺规作图只用无刻度直尺\(M\(A\(C、D\)B\)N\)圆规；必须保留圆弧；作角平分线：顶点为圆心画弧\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)再以两个交点为圆心在角内部画弧；作线段垂直平分线：分别以线段两端点为圆心\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)大于线段一半长为圆心画上下两组弧\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)连接交点；两个作图原理都是SSS全等。##一\(M\(A\(C、D\)B\)N\)知识梳理1.尺规作已知角的平分线步骤：①以角顶点$O$为圆心\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)适当长为半径画弧\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)交角两边于$M\(M\(A\(C、D\)B\)N\)N$；②分别以$M\(M\(A\(C、D\)B\)N\)N$为圆心\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)大于$\dfrac12MN$长为半径\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)在角内部画弧\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)两弧交于点$P$；③作射线$OP$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)射线$OP$即为角平分线。原理：连接$\(OM,ON,PM,PN\)$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)$\triangleOMP\cong\triangleONP(\text{SSS})$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)对应角相等。性质：角平分线上的点到角两边垂线段距离相等。2.尺规作线段$AB$的垂直平分线步骤：①分别以$A\(M\(A\(C、D\)B\)N\)B$为圆心\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)大于$\dfrac12AB$的长度为半径画弧\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)在线段$AB$上下两侧各得到一组交点$C\(M\(A\(C、D\)B\)N\)D$；②作直线$CD$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)直线$CD$就是线段$AB$的垂直平分线。原理：$\(AC=BC\)\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)AD=BD\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)AB$公共\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)SSS全等\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)证明直线垂直平分线段。性质：垂直平分线上任意一点到线段两端点距离相等。3.重要提醒①两次作图\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)画弧半径都要求大于线段一半长度\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)否则弧无法相交；②角平分线是射线；线段垂直平分线是直线；③作图题保留圆弧痕迹\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)圆弧是得分关键点；④不要混淆两种作图：角平分线从角顶点出发；垂直平分线是连接两组弧交点的直线。##二\(M\(A\(C、D\)B\)N\)选择题（每题5分\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)共30分）1.尺规作角平分线的理论依据是（）A.HLB.SSSC.SASD.ASA2.作线段$AB$垂直平分线\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)画弧半径必须（）A.小于$\dfrac12AB$B.大于$\dfrac12AB$C.任意长度D.等于$AB$3.尺规作线段垂直平分线\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)得到的图形是（）A.射线B.直线C.线段D.折线4.在作角平分线时\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)两弧交点位置应在（）A.角外部B.角内部C.角边上D.任意位置5.用尺规作线段垂直平分线\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)一共要找到几个弧的交点（）A.1个B.2个C.3个D.4个6.下列说法错误的是（）A.角平分线作图原理是SSSB.线段垂直平分线上点到线段两端距离相等C.作垂直平分线\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)只需要在线段上方画弧即可D.尺规作图不能使用带刻度直尺##三\(M\(A\(C、D\)B\)N\)填空题（每题6分\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)共30分）1.尺规作角平分线和线段垂直平分线的原理都是全等判定定理________。2.作线段垂直平分线\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)分别以线段的________为圆心画弧。3.角平分线是一条________；线段垂直平分线是一条直线。4.作角平分线第二步\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)两弧的交点必须在角的________。5.线段垂直平分线与原线段的夹角为________°。##四\(M\(A\(C、D\)B\)N\)解答题（每题10分\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)共40分）1.作图题：已知$\angleAOB$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)尺规作出角平分线\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)保留痕迹\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)不写作法。2.作图题：已知线段$AB$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)尺规作出线段$AB$的垂直平分线\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)保留痕迹\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)不写作法。3.证明题：根据线段垂直平分线作图痕迹\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)证明$CD$垂直平分$AB$。4.综合计算题：如图\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)$AD$是$\angleBAC$平分线\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)$DE\perpAB$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)$DF\perpAC$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)$\(DE=6\)$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)求$DF$长\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)并写出依据。##五\(M\(A\(C、D\)B\)N\)参考答案与解析选择题：1.B；SSS；2.B；半径大于一半才能相交；3.B；垂直平分线是直线；4.B；交点在角内部；5.B；上下两个交点；6.C；上下都要画弧\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)得到两个交点。填空题：1.SSS；2.两个端点；3.射线；4.内部；5.90。解答题：1.解：按照三步作图\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)保留两段圆弧\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)画出射线。2.解：以A\(M\(A\(C、D\)B\)N\)B为圆心画上下两对圆弧\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)连接两个交点\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)保留圆弧痕迹。3.证明：由作图得$\(AC=BC\)\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)AD=BD$。在$\triangleACD$和$\triangleBCD$中$\begin{cases}\(AC=BC\)\\AD=BD\\CD=CD\end{cases}$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)$\therefore\triangleACD\cong\triangleBCD(\text{SSS})$。$\therefore\angleACD=\angleBCD$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)又$\(AC=BC\)$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)由等腰三角形三线合一\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)$CD\perpAB\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)AO=BO$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)$\thereforeCD$垂直平分$AB$。4.解：$\becauseAD$平分$\angleBAC$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)$DE\perpAB\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)DF\perpAC$\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)根据角平分线上的点到角两边距离相等\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)$\thereforeDF=\(DE=6\)$。答：$\(DF=6\)$。##六\(M\(A\(C、D\)B\)N\)易错点总结作垂直平分线和角平分线四类高频丢分点。第一\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)作图忘记保留圆弧痕迹\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)或者圆弧半径太小\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)两弧不能相交\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)直接扣分。第二\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)混淆两种作图\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)作垂直平分线只画一侧弧\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)只得到一个交点\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)无法确定直线。第三\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)概念混淆\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)误把角平分线当成直线\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)把线段垂直平分线当成线段。第四\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)性质使用前提遗漏：角平分线性质必须有“垂直”\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)没有垂直条件不能直接说距离相等。本节是期末必考作图大题\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)常结合线段\(M\(A\(C、D\)B\)N\)角的计算综合考察。做题固定流程：看清是作角平分线还是垂直平分线；牢记半径大于线段一半；作图必须保留圆弧；分清两个作图对应的性质\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)选择合适定理解题。（全文：1231字）下一节：**5.1二次根式**\(\(AC=BC\)\(CD\perpAB\(DE\perpAB，DF\perpAC\)AO=BO\)AD=BD\)同样1231字专项练习吗？湘教版（新教材）数学八年级上册4.6.2作垂直平分线和角平分线授课教师：.

班

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8年级（

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2026/9/28

学子如同向日葵向往暖阳，满心热爱渴求知识。在书页间汲取力量，于思索中追逐光亮，不惧学习路上的磨砺，心怀憧憬奋力向上，向着理想不断成长。第4章三角形学习目标1.掌握线段垂直平分线、过点作垂线及角平分线的尺规作图方法，理解其几何性质.2.能根据已知底边和高线，通过作垂线或垂直平分线定位顶点

3.综合运用几何作图解决实际问题02新知导入线段垂直平分线的性质定理：线段垂直平分线上的点到线段两端的距离相等.几何语言∵点P在线段AB的垂直平分线上，∴PA=PB(线段垂直平分线的性质定理).02新知导入线段垂直平分线的性质定理的逆定理:到线段两端距离相等的点在线段的垂直平分线上.几何语言∵MA=MB，∴点M在线段AB的垂直平分线上(线段垂直平分线的性质定理的逆定理).03新知探究说一说已知线段AB，如果要作线段AB的垂直平分线，可以怎样作？根据是什么？根据线段垂直平分线的性质定理的逆定理可知，若能找出到线段AB两端距离相等的两个点，则这两点确定的直线就是线段AB的垂直平分线.03新知探究作一条线段的垂直平分线如图，已知线段AB.求作线段AB的垂直平分线.可以利用什么工具找出到线段AB两端距离相等的两个点？

03新知探究思考怎么作线段的中点．由线段的垂直平分线的概念可知线段的垂直平分线平分该线段.因此作该线段的垂直平分线，取线段与垂直平分线的交点即可.03新知探究思考如何用尺规过一点P作已知直线l的垂线呢？若能找到直线l上的一条线段AB，使AB的垂直平分线经过点P，则该垂直平分线就是所求作的直线.由于点P与已知直线l的位置关系有两种，于是需分情况来作图.03新知探究

03新知探究

03新知探究已知底边及底边上的高线作等腰三角形.例3如图，已知线段a，h.求作△ABC，使AB=AC，且BC=a，高AD=h.问题1：要作的△ABC满足哪些核心条件？问题2：根据条件，你认为先作出等腰三角形的哪部分？问题3：底边上的高在什么线上？如何作底边上的高？问题4：如何在这条线上确定点A，保证AD=h?03新知探究作法（1）作线段BC=a；（2）作线段BC的垂直平分线MN，交BC于点D；（3）在射线DM（或DN）上截取线段DA，使DA=h；（4）连接AB，AC，则△ABC为所求作的等腰三角形.

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03新知探究求作一个角的平分线.例4如图，已知∠AOB，求作∠AOB的平分线.分析：由于等腰三角形的顶角平分线也是底边上的垂直平分线，故先以∠AOB的顶点O为顶点，两腰分别在射线OA，OB上，构造等腰△ODE，然后过点O作底边DE的垂直平分线OC，则射线OC就是∠AOB的平分线.03新知探究

04课堂练习【知识技能类作业】必做题