5.3 直角三角形全等的判定 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5.3直角三角形全等的判定专项练习##知识点回顾###HL定理（斜边\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)直角边定理）**斜边和一条直角边对应相等的两个直角三角形全等。**几何语言：在$Rt\triangleABC$和$Rt\triangleA'B'C'$中\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\begin{cases}AB=A'B'（斜边）\\AC=A'C'（一条直角边）\end{cases}$$\thereforeRt\triangleABC\congRt\triangleA'B'C'(\text{HL})$✅补充：直角三角形也可以用普通全等判定：$\boldsymbol{SSS\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)SAS\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)ASA\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)AAS}$>注意：HL**只适用于直角三角形**\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)普通三角形不能用HL。✅解题要点1.使用HL前\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)必须先写明两个三角形是**直角三角形**；2.HL条件：**斜边+一条直角边**对应相等\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)不是两条直角边；两条直角边用SAS；3.证明书写格式：先标注直角\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)再列出斜边\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)直角边两组相等条件。满分：100分时间：40分钟##一\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)选择题（每题4分\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)共20分）1.判定两个直角三角形全等特有的方法是（）A.SSSB.SASC.AASD.HL2.在$Rt\triangleABC$与$Rt\triangleDEF$中\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\angleC=\angleF=90^\circ$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)斜边$\(AB=DE\)$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)直角边$\(AC=DF\)$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)则判定全等依据是（）A.HLB.ASAC.SSSD.SAS3.下列条件\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)不能判定两个直角三角形全等的是（）A.一条直角边和一个锐角对应相等B.斜边和一条直角边对应相等C.两个锐角对应相等D.斜边和一个锐角对应相等4.两个直角三角形\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)两条直角边对应相等\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)判定全等依据是（）A.HLB.SASC.AASD.ASA5.关于HL说法正确的是（）A.任意三角形都可以用HL判定全等B.HL需要两条直角边对应相等C.HL只适用于直角三角形\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)斜边和一条直角边对应相等D.两个直角三角形\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)斜边相等就一定全等##二\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)填空题（每题4分\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)共20分）1.直角三角形特有的全等判定定理是________。2.HL定理：________和一条直角边对应相等的两个直角三角形全等。3.证明两个直角三角形全等\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)除HL外\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)还可以用________\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)SAS\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)ASA\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)AAS。4.在$Rt\triangleABC\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)Rt\triangleDEF$中\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\angleC=\angleF=90^\circ$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)若$\(AB=DE\)$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\(BC=EF\)$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)则依据________全等。5.两个直角三角形只有锐角相等\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)只能证明它们________\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)不能证明全等。##三\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)基础解答题（每题8分\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)共32分）1.已知：$\angleC=\angleD=90^\circ$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\(AC=AD\)$。求证：$Rt\triangleACB\congRt\triangleADB$。2.判断对错\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)并说明理由（1）两个直角三角形\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)斜边相等\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)一条直角边相等\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)可用HL证全等。（2）HL可以用来判定任意两个三角形全等。3.已知：$AC\perpBC\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)AD\perpBD$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)垂足分别是$C\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)D$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\(AC=BD\)$。求证：$Rt\triangleABC\congRt\triangleBAD$。4.已知：$Rt\triangleABC$和$Rt\triangleDEF$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\angleA=\angleD=90^\circ$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\(BC=EF\)\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)\(AB=DE\)$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)求证两三角形全等。##四\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)拓展提升题（每题14分\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)共28分）1.已知：$CD\perpAB\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)BE\perpAC$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)垂足分别为$D\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)E$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\(BD=CE\)$。求证：$Rt\triangleBDC\congRt\triangleCEB$。2.已知：$AD$是$\triangleABC$的高\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\(BD=CD\)$。求证：$\(AB=AC\)$。---#参考答案与解析##一\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)选择题1.D2.A3.C4.B5.C##二\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)填空题1.**HL（斜边\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)直角边定理）**2.**斜边**3.**SSS**4.**HL**5.**相似**##三\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)基础解答题1.证明：$\because\angleC=\angleD=90^\circ$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\therefore\triangleACB\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)\triangleADB$是直角三角形在$Rt\triangleACB$和$Rt\triangleADB$中$\begin{cases}AB=AB（公共斜边）\\\(AC=AD\)（已知）\end{cases}$$\thereforeRt\triangleACB\congRt\triangleADB(\text{HL})$2.（1）√；HL定理内容\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)斜边+一条直角边对应相等\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)两直角三角形全等。（2）×；HL**只适用于直角三角形**\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)普通三角形不能使用。3.证明：$\becauseAC\perpBC\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)AD\perpBD$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\therefore\angleC=\angleD=90^\circ$$\therefore\triangleABC\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)\triangleBAD$是直角三角形在$Rt\triangleABC$和$Rt\triangleBAD$中$\begin{cases}AB=BA（公共斜边）\\\(AC=BD\)（已知）\end{cases}$$\thereforeRt\triangleABC\congRt\triangleBAD(\text{HL})$4.证明：$\because\angleA=\angleD=90^\circ$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\therefore\triangleABC\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)\triangleDEF$是直角三角形在$Rt\triangleABC$和$Rt\triangleDEF$中$\begin{cases}\(BC=EF\)（斜边）\\\(AB=DE\)（直角边）\end{cases}$$\thereforeRt\triangleABC\congRt\triangleDEF(\text{HL})$##四\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)拓展提升题1.证明：$\becauseCD\perpAB\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)BE\perpAC$\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\therefore\angleCDB=\angleBEC=90^\circ$$\therefore\triangleBDC\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)\triangleCEB$是直角三角形在$Rt\triangleBDC$和$Rt\triangleCEB$中$\begin{cases}BC=CB（公共斜边）\\\(BD=CE\)（已知）\end{cases}$$\thereforeRt\triangleBDC\congRt\triangleCEB(\text{HL})$2.证明：$\becauseAD$是高\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)$\therefore\angleADB=\angleADC=90^\circ$$\therefore\triangleADB\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)\triangleADC$是直角三角形在$Rt\triangleADB$和$Rt\triangleADC$中$\begin{cases}AD=AD（公共直角边）\\\(BD=CD\)（已知）\end{cases}$$\thereforeRt\triangleADB\congRt\triangleADC(\text{SAS})$$\therefore\(AB=AC\)$##易错警示1.写HL证明\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)**必须先说明是直角三角形**\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)否则直接扣分；2.HL条件：斜边+一条直角边\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)不是两条直角边；两条直角边相等用SAS；3.只有角相等\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)没有边相等\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)不能证全等；4.直角三角形五种判定方法：SSS\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)SAS\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)ASA\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)AAS\(Rt\triangleABC\(C\(D\(\therefore\triangleACB\(\therefore\triangleABC\(\therefore\triangleABC\(\therefore\triangleBDC\(\therefore\triangleADB、\triangleADC\)\triangleCEB\)\triangleDEF\)\triangleBAD\)\triangleADB\)E\)D\)Rt\triangleDEF\)HL\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)HL是直角三角形专属。本章第五章直角三角形到此结束！下一章：**6.1函数**\(AC\perpBC\(BC=EF\(CD\perpAB\(\becauseAC\perpBC\(\becauseCD\perpAB，BE\perpAC\)AD\perpBD\)BE\perpAC\)AB=DE\)AD\perpBD\)继续同模板出题吗？湘教版（新教材）数学八年级上册5.3直角三角形全等的判定授课教师：.

班

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8年级（

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2026/10/10

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5章直角三角形123熟练掌握“斜边、直角边定理”（重点）熟练利用HL和判定一般三角形全等的方法判定两个直角三角形全等．（难点）掌握尺规作图：已知斜边和直角边会作直角三角形（重点）导入新课具有下列条件的Rt△ABC与Rt△A′B′C′(其中∠C=∠C′=90°)是否全等？如果全等在括号里填写理由，如果不全等在括号里打“×”。(1)AC=A′C′，∠A=∠A′()(2)AC=A′C′，BC=B′C′()(3)∠A=∠A′，∠B=∠B′()(4)AB=A′B′，∠B=∠B′()(5)AB=A′B′，AC=A′C′()ASASAS×？\\\\\\\\\\\\AAS我们学了哪些方法判定两个三角形全等？03新知探究思考问题1：在Rt△ABC和Rt△A

B

C

中，∠C=∠C=90°，若有一锐角和一边分别相等，这两个直角三角形全等吗？AAS或ASA03新知探究思考问题2：在Rt△ABC和Rt△A

B

C

中，∠C=∠C=90°，若有两直角边分别相等，这两个直角三角形全等吗？SAS03新知探究思考问题3：在Rt△ABC和Rt△A

B

C

中，∠C=∠C=90°，若有一条直角边和斜边分别相等，这两个直角三角形全等吗？03新知探究已知：在Rt△ABC和Rt△A′B′C′中，∠C=∠C′