14.1 全等三角形及其性质-课件-2026-2027学年人教版数学八年级上册

#人教版八年级上册14.1全等三角形及其性质专项练习（全文约1500字\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)可直接打印）##知识要点速记1.**全等形**：能够完全重合的两个图形叫做全等形。全等形形状相同\(AD、A'D'\)大小相等。2.**全等三角形定义**：能够完全重合的两个三角形叫做全等三角形。3.**对应元素**-重合的顶点：**对应顶点**-重合的边：**对应边**-重合的角：**对应角**4.**表示方法**：$\triangleABC\cong\triangleDEF$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)书写时**对应顶点写在对应位置**\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)方便直接找对应边\(AD、A'D'\)对应角。5.**全等三角形的性质（核心）**-全等三角形的**对应边相等**；-全等三角形的**对应角相等**；-延伸性质：周长相等\(AD、A'D'\)面积相等；对应中线\(AD、A'D'\)对应高\(AD、A'D'\)对应角平分线相等。⚠️易错提醒：周长相等或者面积相等的两个三角形\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)**不一定全等**；找对应边角口诀：公共边是对应边\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)公共角是对应角\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)对顶角是对应角\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)最长边对最长边\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)最大角对最大角。##一\(AD、A'D'\)基础过关（判断+填空+选择）###判断题（对√\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)错×）1.面积相等的两个三角形一定全等。（）2.全等三角形的周长一定相等。（）3.$\triangleABC\cong\triangleDEF$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)说明$\angleA$一定对应$\angleF$。（）4.全等三角形的对应角相等。（）5.两个等边三角形一定是全等三角形。（）###填空题6.能够完全重合的两个三角形叫做________。7.若$\triangleABC\cong\triangleMNP$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)则对应边$\(AB=\)$______\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)对应角$\angleC=$______。8.已知$\triangleABC\cong\triangleDEF$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\(AB=\)6$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)则$\(DE=\)$______。9.若$\triangleABC\cong\triangleDEF$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleA=52^\circ$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleE=48^\circ$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)则$\angleF=$______$^\circ$。10.全等三角形的对应中线\(AD、A'D'\)对应高\(AD、A'D'\)对应角平分线都________。###选择题11.已知$\triangleABC\cong\triangleA'B'C'$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleB$的对应角是（）A.$\angleA'$B.$\angleB'$C.$\angleC'$D.无法确定12.下列说法正确的是（）A.两个长方形一定全等B.两个正方形一定全等C.能够完全重合的两个三角形全等D.所有等腰三角形全等13.$\triangleABC\cong\triangleDEF$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)若$\(BC=8\)$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)则下列线段长等于8的是（）A.$DE$B.$EF$C.$DF$D.$AB$14.如图\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\triangleABC\cong\triangleADC$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleB=70^\circ$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)则$\angleD=$（）A.$70^\circ$B.$110^\circ$C.$20^\circ$D.$90^\circ$##二\(AD、A'D'\)中档解答题15.已知$\triangleABC\cong\triangleDEF$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\(AB=\)5$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\(BC=7\)$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\(AC=9\)$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)求$\triangleDEF$的各边长以及周长。16.如图\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\triangleABC\cong\triangleADE$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleB=30^\circ$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleC=80^\circ$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)求$\angleDAE$的度数。17.已知$\triangleABC\cong\triangleFED$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleA=60^\circ$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleB=45^\circ$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)求$\angleD$的度数。18.若$\triangleABC\cong\triangleDEF$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\triangleABC$的周长为24\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\(DE=\)7$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\(EF=8\)$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)求$DF$的长。##三\(AD、A'D'\)拓展提升题19.如图\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\triangleABC\cong\triangleDEC$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)点B\(AD、A'D'\)C\(AD、A'D'\)E在同一直线上\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleA=25^\circ$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleB=40^\circ$。（1）求$\angleD$和$\angleDEC$；（2）求$\angleACB$的度数。20.已知$\triangleABC\cong\triangleA'B'C'$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$AD\(AD、A'D'\)A'D'$分别是两个三角形对应边上的高\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)求证：$AD=A'D'$。##参考答案1.×面积相等不一定能重合2.√3.×要看顶点顺序4.√5.×边长不一定相等6.全等三角形7.$MN$；$\angleP$8.$6$9.$80$10.相等11.B12.C13.B14.A15.解：$\because\triangleABC\cong\triangleDEF$$\therefore\(DE=\)\(AB=\)5\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)EF=\(BC=7\)\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)DF=\(AC=9\)$周长$\(=5+7+9=21\)$答：$\(DE=\)5\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)EF=7\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)\(DF=9\)$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)周长为21。16.解：在$\triangleABC$中\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleBAC=180^\circ-\angleB-\angleC=180^\circ-30^\circ-80^\circ=70^\circ$$\because\triangleABC\cong\triangleADE$$\therefore\angleDAE=\angleBAC=70^\circ$答：$\angleDAE=70^\circ$。17.解：$\because\triangleABC\cong\triangleFED$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\angleD$对应$\angleB$$\therefore\angleD=\angleB=45^\circ$答：$\angleD=45^\circ$。18.解：全等三角形周长相等\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\triangleDEF$周长=24$DF=24-DE-EF=24-7-8=9$答：$\(DF=9\)$。19.解：（1）$\because\triangleABC\cong\triangleDEC$$\therefore\angleD=\angleA=25^\circ\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)\angleDEC=\angleB=40^\circ$（2）$\angleACB=180^\circ-\angleA-\angleB=180^\circ-25^\circ-40^\circ=115^\circ$答：$\angleD=25^\circ\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)\angleDEC=40^\circ\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)\angleACB=115^\circ$。20.证明：$\because\triangleABC\cong\triangleA'B'C'$$\therefore\(AB=\)A'B'\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)\angleB=\angleB'$$\becauseAD\perpBC\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)A'D'\perpB'C'$\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)$\therefore\angleADB=\angleA'D'B'=90^\circ$在$\triangleABD$和$\triangleA'B'D'$中$\begin{cases}\angleADB=\angleA'D'B'\\\angleB=\angleB'\\\(AB=\)A'B'\end{cases}$$\therefore\triangleABD\cong\triangleA'B'D'(AAS)$$\thereforeAD=A'D'$。##解题小结1.做题第一准则：**按式子顶点顺序找对应边\(AD、A'D'\)对应角**\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)这是本节最容易丢分的地方；2.性质两条主线：对应边相等\(AD、A'D'\)对应角相等；衍生出周长\(AD、A'D'\)面积\(AD、A'D'\)对应线段相等；3.区分：全等$\Rightarrow$面积\(AD、A'D'\)周长相等；反过来\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)面积/周长相等$\nRightarrow$全等；4.综合题经常结合内角和\(AD、A'D'\)平角一起求角度。下一节是**SSS边边边判定**\(\therefore\angleD=\angleA=25^\circ\(\thereforeAB=A'B'\(\becauseAD\perpBC，A'D'\perpB'C'\)\angleB=\angleB'\)\angleDEC=\angleB=40^\circ\)要不要继续出14.2.1专项练习？人教版数学（新教材）八年级上册14.1全等三角形及其性质授课教师：.

班

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8年级（

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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。熟记全等形及全等三角形的概念;能够正确找出全等三角形的对应边、对应角.熟练掌握全等三角形的性质，并能灵活运用全等三角形的性质解决相应的几何问题.初步帮助学生建立平移、翻折、旋转三种图形变化与全等形的关系.你能再举出一些类似的例子吗？归纳总结形状、大小相同的图形放在一起能够完全重合.能够完全重合的两个图形叫作全等形.探究新知能够完全重合的两个图形叫做全等形.两个三角形完全重合，又叫什么呢？能够完全重合的两个三角形叫做全等三角形.EDFABC探究新知【思考】在图（1）中，将△ABC沿直线BC平移，得到△DEF.在图（2）中，将△ABC沿直线BC翻折180°，得到△DBC.在图（3）中，将△ABC绕点A旋转，得到△ADE.各图中的两个三角形全等吗？ABCABC（3）D（2）ABCDEDEF（1）探究新知ABCDEF（1）平移