14.2.5用“HL”判定三角形全等-课件-2026-2027学年人教版数学八年级上册

#人教版八年级上册14.2.5用“HL”判定直角三角形全等专项练习（全文约1500字\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)可直接打印）##知识要点速记1.**HL判定定理（斜边\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)直角边）**：斜边和一条直角边分别相等的两个**直角三角形**全等。>注意：HL**只适用于直角三角形**\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)普通三角形不能用HL！符号语言：在$Rt\triangleABC$和$Rt\triangleDEF$中$\begin{cases}AB=DE（斜边）\\AC=DF（直角边）\end{cases}$$\thereforeRt\triangleABC\congRt\triangleDEF(\text{HL})$2.直角三角形全等判定汇总通用4种：SSS\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)SAS\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)ASA\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)AAS；专属：HL（斜边+一条直角边）。3.常见隐含条件-垂直\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)$\boldsymbol{\perp}$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)直接得到直角相等；-公共斜边\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)公共直角边；-同角（等角）的余角相等；⚠️易错提醒：①HL只能给直角三角形用；②“一条斜边+一条直角边”\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)不是两条直角边（两条直角边相等用SAS）；③SSA在普通三角形不行\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)但在直角三角形中HL成立；4.解题思路：先标注$\angleC=\angleF=90^\circ$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)说明是直角三角形\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)再写HL条件。##一\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)基础过关（判断+填空+选择）###判断题（对√\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)错×）1.HL可以用来判定任意两个三角形全等。（）2.斜边和一条直角边对应相等的两个直角三角形全等。（）3.两个直角三角形\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)两条直角边对应相等\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)可用HL证明全等。（）4.直角三角形全等\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)除HL外\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)也能用SAS\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)ASA\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)AAS\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)SSS。（）5.斜边相等的两个直角三角形一定全等。（）###填空题6.HL是专门判定________三角形全等的方法\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)全称是________。7.在$Rt\triangleABC$和$Rt\triangleA'B'C'$中\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$\angleC=\angleC'=90^\circ$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)斜边$AB=A'B'$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)直角边$AC=A'C'$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)则依据________全等。8.两个直角三角形\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)斜边相等\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)一条直角边相等\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)那么它们的另一条直角边一定________。9.使用HL证明全等\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)第一步要先说明两个三角形是________。10.若$Rt\triangleABC$中\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)直角边$AC\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)BC$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)斜边是________。###选择题11.下列条件中\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)能用HL判定全等的是（）A.一个锐角和斜边相等B.斜边和一条直角边相等C.两条直角边相等D.两个锐角相等12.$Rt\triangleABC$与$Rt\triangleDEF$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$\angleC=\angleF=90^\circ$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)下列不能判定全等的是（）A.$AB=DE\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)AC=DF$B.$AB=DE\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)BC=EF$C.$\angleA=\angleD\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)BC=EF$D.$\angleA=\angleD\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)\angleB=\angleE$13.如图\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$AC\perpBC\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)AD\perpBD$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$\(AC=BD\)$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)判定$Rt\triangleABC\congRt\triangleBAD$的依据是（）A.HLB.SASC.AASD.ASA14.对于直角三角形全等\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)说法正确的是（）A.HL属于SSA\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)所以不能判定全等B.HL仅直角三角形可用C.三个角相等的直角三角形全等D.斜边相等就一定全等##二\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)中档解答题15.已知：$AC\perpBC\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)AD\perpBD$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)垂足分别为$C\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)D$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$\(AC=BD\)$。求证：$Rt\triangleABC\congRt\triangleBAD$。16.已知：$AB\perpCD$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$\angleACB=\angleADB=90^\circ$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$\(AC=AD\)$。求证：$\(BC=BD\)$。17.已知：如图\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$DE\perpAB\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)CF\perpAB$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)垂足$E\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)F$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$DE=CF\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)AE=BF$。求证：$Rt\triangleADE\congRt\triangleBCF$。18.已知：$AD$是$\triangleABC$的高\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$\(AB=AC\)$。求证：$\(BD=CD\)$。##三\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)拓展提升题19.已知：$AE\perpBC\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)DF\perpBC$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)垂足$E\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\triangleACE、\triangleBCD\)\triangleDFC\)\triangleADC\)\triangleBCF\)\triangleADB\)\triangleBAD\)F\)D\)BC\)F$\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$AB=DC\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)BE=CF$。求证：$AB\parallelCD$。20.已知：$C$是$AB$中点\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)$AC\perpCE\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)BC\perpCD\(AB=DE\(AB=DE\(\angleA=\angleD\(\angleA=\angleD\(AC\perpBC\(DE\perpAB\(DE=CF\(AE\perpBC\(AB=DC\(\becauseAC\perpBC\(\becauseDE\perpAB\(\thereforeAD\perpBC\(\becauseAE\perpBC\(\becauseAC\perpCE，BC\perpCD\)DF\perpBC\)\angleADB=\angleADC=90^\circ\)CF\perpAB\)AD\perpBD\)BE=CF\)DF\perpBC\)AE=BF\)CF\perpAB\)AD\perpBD\)\angleB=\angleE\)BC=EF\)BC=EF\)AC=DF\)AE=BD$。求证：$\angleA=\angleB$。##参考答案1.×2.√3.×（两条直角边用SAS）4.√5.×（还需要一条直角边相等）6.直角；斜边\(AC\(C\(E\(\therefore\triangleABC\(\therefore\triangleACB\(\therefore\triangleADE\(\therefore\triangleADB\(\therefore\triangleAEB\(\therefore\tri