14.2.3三角形全等的判定（SSS）-课件-2026-2027学年人教版数学八年级上册

#人教版八年级上册14.2.2三角形全等的判定（ASA和AAS）专项练习（全文约1500字\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)可直接打印）##知识要点速记1.**ASA（角边角）**：两角和它们的**夹边**分别相等的两个三角形全等。符号语言：在$\triangleABC$和$\triangleDEF$中$\begin{cases}\angleA=\angleD\\\(AB=DE\)\\\angleB=\angleE\end{cases}$$\therefore\triangleABC\cong\triangleDEF(\text{ASA})$>重点：边必须是两个角中间夹住的边。2.**AAS（角角边）**：两角和其中**一个角的对边**分别相等的两个三角形全等。符号语言：在$\triangleABC$和$\triangleDEF$中$\begin{cases}\angleA=\angleD\\\angleB=\angleE\\\(BC=EF\)\end{cases}$$\therefore\triangleABC\cong\triangleDEF(\text{AAS})$3.关系：AAS可以由ASA结合三角形内角和推导出来。4.常用隐含条件：公共角\(AB、CD\)对顶角相等；平行线得到内错角\(AB、CD\)同位角相等；垂直得到直角相等。⚠️易错提醒：AAA（三个角相等）不能判定全等\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)只能判定相似；区分ASA（夹边）和AAS（对边）。5.判定汇总：SSS\(AB、CD\)SAS\(AB、CD\)ASA\(AB、CD\)AAS；**SSA\(AB、CD\)AAA不能判定全等**。##一\(AB、CD\)基础过关（判断+填空+选择）###判断题（对√\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)错×）1.两角和任意一条边相等\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)就一定是ASA。（）2.AAS指两角和其中一角的对边相等。（）3.三个内角对应相等的两个三角形全等。（）4.ASA中的边是两个角的夹边。（）5.对顶角相等\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)可以作为ASA\(AB、CD\)AAS里一组相等的角。（）###填空题6.两角和它们的夹边对应相等\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)简写为______；两角和其中一角的对边对应相等\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)简写为______。7.在$\triangleABC$和$\triangleDEF$中\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleA=\angleD$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleB=\angleE$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)若$\(AB=DE\)$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)可以用______判定全等。8.在$\triangleABC$和$\triangleDEF$中\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleA=\angleD$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleB=\angleE$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)若$\(BC=EF\)$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)可以用______判定全等。9.两个直角三角形\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)有一组锐角相等\(AB、CD\)一组斜边相等\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)可以用______判定全等。10.证明全等之后\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)可以得到______相等\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)对应角相等。###选择题11.下列条件能使用ASA判定全等的是（）A.两角和其中一角对边B.两角和它们的夹边C.两边和夹角D.三边对应相等12.已知$\angleA=\angleD\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)\angleB=\angleE\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)AC=DF$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)判定依据是（）A.SASB.ASAC.AASD.SSS13.如图\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$AB\(AB、CD\)CD$交于$O$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleA=\angleC$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\(OA=OC\)$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleAOB=\angleCOD$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)判定$\triangleAOB\cong\triangleCOD$依据是（）A.ASAB.AASC.SASD.SSS14.下列不能判定三角形全等的是（）A.ASAB.AASC.AAAD.SAS##二\(AB、CD\)中档解答题15.已知：如图\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleA=\angleD$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\(AB=DE\)$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleB=\angleE$。求证：$\triangleABC\cong\triangleDEF$。16.已知：如图\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleB=\angleD$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleBAC=\angleDAC$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$AC$是公共边。求证：$\triangleABC\cong\triangleADC$。17.已知：$AB\parallelCD$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$AD$交$BC$于$O$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\(AB=CD\)$。求证：$\triangleABO\cong\triangleDCO$。18.已知：$\angle1=\angle2$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleB=\angleC$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\(AD=AE\)$。求证：$\triangleABD\cong\triangleACE$。##三\(AB、CD\)拓展提升题19.如图\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)已知$AD\parallelBC$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$BD$交$AC$于$O$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\(AD=BC\)$。求证：$\(AO=CO\)$。20.已知：$BE\perpAC$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$CF\perpAB$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\angleB=\angleC$\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)$\(AE=AF\)$。求证：$\triangleABE\cong\triangleACF$。##参考答案1.×（要看这条边是不是夹边）2.√3.×（只能形状一样\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)大小不一定相同）4.√5.√6.ASA；AAS7.ASA8.AAS9.AAS10.对应边11.B12.C13.A14.C15.证明：在$\triangleABC$和$\triangleDEF$中$\begin{cases}\angleA=\angleD\\\(AB=DE\)\\\angleB=\angleE\end{cases}$$\therefore\triangleABC\cong\triangleDEF(\text{ASA})$16.证明：在$\triangleABC$和$\triangleADC$中$\begin{cases}\angleB=\angleD\\\angleBAC=\angleDAC\\AC=AC\end{cases}$$\therefore\triangleABC\cong\triangleADC(\text{AAS})$17.证明：$\becauseAB\parallelCD$$\therefore\angleA=\angleD\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)\angleB=\angleC$在$\triangleABO$和$\triangleDCO$中$\begin{cases}\angleA=\angleD\\\(AB=CD\)\\\angleB=\angleC\end{cases}$$\therefore\triangleABO\cong\triangleDCO(\text{ASA})$18.证明：在$\triangleABD$和$\triangleACE$中$\begin{cases}\angleB=\angleC\\\angle1=\angle2\\\(AD=AE\)\end{cases}$$\therefore\triangleABD\cong\triangleACE(\text{AAS})$19.证明：$\becauseAD\parallelBC$$\therefore\angleADO=\angleCBO\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)\angleDAO=\angleBCO$在$\triangleADO$和$\triangleCBO$中$\begin{cases}\angleADO=\angleCBO\\\angleDAO=\angleBCO\\\(AD=BC\)\end{cases}$$\therefore\triangleADO\cong\triangleCBO(\text{AAS})$$\therefore\(AO=CO\)$20.证明：$\becauseBE\perpAC\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)CF\perpAB$$\therefore\angleAEB=\angleAFC=90^\circ$在$\triangleABE$和$\triangleACF$中$\begin{cases}\angleAEB=\angleAFC\\\angleB=\angleC\\\(AE=AF\)\end{cases}$$\therefore\triangleABE\cong\triangleACF(\text{AAS})$##解题小结1.ASA：两角夹一边；AAS：两角对一边\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)做题第一步分清边的位置；2.平行线是高频条件\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)用来推导内错角相等；3.公共边\(AB、CD\)公共角\(AB、CD\)对顶角不用额外证明\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)直接写相等；4.证明线段相等\(AB、CD\)角相等\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)思路：先证全等\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)再用全等性质；5.判定口诀：三边SSS\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)两边夹角SAS\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)两角夹边ASA\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)两角对边AAS\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)SSA\(AB、CD\)AAA都不行。下一节是**HL（直角三角形全等判定）**\(\therefore\angleA=\angleD\(\therefore\angleADO=\angleCBO\(\becauseBE\perpAC，CF\perpAB\)\angleDAO=\angleBCO\)\angleB=\angleC\)需要我继续出14.2.3专项练习吗？人教版数学（新教材）八年级上册14.2.3三角形全等的判定（SSS）授课教师：.

班

级：

8年级（

）班.

时

间：.

2026/11/1

第十四章全等三角形

我命由我不由天，是魔是仙，我自己说了算。少年不必畏惧难题与非议，沉下心勤学笃行，不认命、不服输，用汗水改写前路，活出属于自己的光彩。探索并正确理解三角形全等的判定定理“SSS”.掌握用尺规作一个三角形的方法.掌握用尺规作一个角等于已知角的作图法．新知导入前面我们研究了两个三角形的两边和一角(SAS)分别相等的情况以及两角和一边(ASA、AAS)分别相等的情况.接下来研究三边、三角分别相等的情况.探究新知ACBA′B′C′如图，直观上，AB，BC，CA

的大小确定了，△ABC

的形状、大小也就确定了.也就是说，在△A'B'C'

与△ABC

中，如果A'B'=AB，B'C'=BC，C'A'=CA，那么△A'B'C'≌△ABC.这个判断正确吗？探究新知如图，由A'B'=AB可知，如果使点A'

与点A重合，点B'

在射线

AB

上，那么点B'与点

B

重合，另外，使点C'

落在直线AB

的含有点C

的一侧，由于点C

是以点A

为圆心、AC

为半径的圆和以点B

为圆心、BC

为半径的圆的交点；点C'

是以点A'

为圆心、A'C'为半径的圆和以点B'

为圆心，B'C'为半径的圆的交点，所以由A'C'

=AC

，B'C'

=BC

可知点C'与点C重合.CAB(A')(B')(C')探究新知△A'B'C'的三个顶点与△ABC

的三个顶点分别重合，△A'B'C'与△ABC

能够完全重合，因而△A'B'C'≌△ABCCAB(A')(B')(C')归纳总结“边边边”判定方法：三边分别相