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\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)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$Rt\triangle ABC$和$Rt\triangle A'B'C'$中\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\) $\begin{cases}AB=A'B'（斜边）\\AC=A'C'（一条直角边）\end{cases}$ $\therefore Rt\triangle ABC\cong Rt\triangle A'B'C'(\text{HL})$ ✅补充：直角三角形也可以用普通全等判定：$\boldsymbol{SSS\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)SAS\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)ASA\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)AAS}$ > 注意：HL**只适用于直角三角形**\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)普通三角形不能用HL。 ✅解题要点 1. 使用HL前\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)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\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)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\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)再列出斜边\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)直角边两组相等条件。 满分：100分 时间：40分钟 ## 一\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)选择题（每题4分\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)共20分） 1. 判定两个直角三角形全等特有的方法是（） A.SSS B.SAS C.AAS D.HL 2. 在$Rt\triangle ABC$与$Rt\triangle DEF$中\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\angle C=\angle F=90^\circ$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)斜边$\(AB=DE\)$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)直角边$\(AC=DF\)$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)则判定全等依据是（） A.HL B.ASA C.SSS D.SAS 3. 下列条件\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)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\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)两条直角边对应相等\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)判定全等依据是（） A.HL B.SAS C.AAS D.ASA 5. 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\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)斜边和一条直角边对应相等 D.两个直角三角形\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)斜边相等就一定全等 ## 二\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)填空题（每题4分\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)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\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)除HL外\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)还可以用________\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)SAS\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)ASA\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)AAS。 4. 在$Rt\triangle ABC\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)Rt\triangle DEF$中\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\angle C=\angle F=90^\circ$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)若$\(AB=DE\)$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\(BC=EF\)$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)则依据________全等。 5. 两个直角三角形只有锐角相等\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)只能证明它们________\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)不能证明全等。 ## 三\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)基础解答题（每题8分\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)共32分） 1. 已知：$\angle C=\angle D=90^\circ$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\(AC=AD\)$。求证：$Rt\triangle ACB\cong Rt\triangle ADB$。 2. 判断对错\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)并说明理由 （1）两个直角三角形\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)斜边相等\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)一条直角边相等\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)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\perp BC\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)AD\perp BD$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)垂足分别是$C\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)D$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\(AC=BD\)$。求证：$Rt\triangle ABC\cong Rt\triangle BAD$。 4. 已知：$Rt\triangle ABC$和$Rt\triangle DEF$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\angle A=\angle D=90^\circ$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\(BC=EF\)\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)\(AB=DE\)$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)求证两三角形全等。 ## 四\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)拓展提升题（每题14分\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)共28分） 1. 已知：$CD\perp AB\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)BE\perp AC$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)垂足分别为$D\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)E$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\(BD=CE\)$。求证：$Rt\triangle BDC\cong Rt\triangle CEB$。 2. 已知：$AD$是$\triangle ABC$的高\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\(BD=CD\)$。求证：$\(AB=AC\)$。 --- # 参考答案与解析 ## 一\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)选择题 1.D 2.A 3.C 4.B 5.C ## 二\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)填空题 1. **HL（斜边\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)直角边定理）** 2. **斜边** 3. **SSS** 4. **HL** 5. **相似** ## 三\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)基础解答题 1. 证明： $\because \angle C=\angle D=90^\circ$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\therefore \triangle ACB\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)\triangle ADB$是直角三角形 在$Rt\triangle ACB$和$Rt\triangle ADB$中 $\begin{cases}AB=AB（公共斜边）\\\(AC=AD\)（已知）\end{cases}$ $\therefore Rt\triangle ACB\cong Rt\triangle ADB(\text{HL})$ 2.（1）√；HL定理内容\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)斜边+一条直角边对应相等\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)两直角三角形全等。 （2）×；HL**只适用于直角三角形**\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)普通三角形不能使用。 3. 证明： $\because AC\perp BC\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)AD\perp BD$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\therefore \angle C=\angle D=90^\circ$ $\therefore \triangle ABC\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)\triangle BAD$是直角三角形 在$Rt\triangle ABC$和$Rt\triangle BAD$中 $\begin{cases}AB=BA（公共斜边）\\\(AC=BD\)（已知）\end{cases}$ $\therefore Rt\triangle ABC\cong Rt\triangle BAD(\text{HL})$ 4. 证明： $\because \angle A=\angle D=90^\circ$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\therefore \triangle ABC\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)\triangle DEF$是直角三角形 在$Rt\triangle ABC$和$Rt\triangle DEF$中 $\begin{cases}\(BC=EF\)（斜边）\\\(AB=DE\)（直角边）\end{cases}$ $\therefore Rt\triangle ABC\cong Rt\triangle DEF(\text{HL})$ ## 四\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)拓展提升题 1. 证明： $\because CD\perp AB\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)BE\perp AC$\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\therefore \angle CDB=\angle BEC=90^\circ$ $\therefore\triangle BDC\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)\triangle CEB$是直角三角形 在$Rt\triangle BDC$和$Rt\triangle CEB$中 $\begin{cases}BC=CB（公共斜边）\\\(BD=CE\)（已知）\end{cases}$ $\therefore Rt\triangle BDC\cong Rt\triangle CEB(\text{HL})$ 2. 证明： $\because AD$是高\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)$\therefore \angle ADB=\angle ADC=90^\circ$ $\therefore \triangle ADB\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)\triangle ADC$是直角三角形 在$Rt\triangle ADB$和$Rt\triangle ADC$中 $\begin{cases}AD=AD（公共直角边）\\\(BD=CD\)（已知）\end{cases}$ $\therefore Rt\triangle ADB\cong Rt\triangle ADC(\text{SAS})$ $\therefore \(AB=AC\)$ ## 易错警示 1. 写HL证明\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)**必须先说明是直角三角形**\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)否则直接扣分； 2. HL条件：斜边+一条直角边\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)不是两条直角边；两条直角边相等用SAS； 3. 只有角相等\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)没有边相等\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)不能证全等； 4. 直角三角形五种判定方法：SSS\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)SAS\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)ASA\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)AAS\(Rt\triangle ABC\(C\(D\(\therefore \triangle ACB\(\therefore \triangle ABC\(\therefore \triangle ABC\(\therefore\triangle BDC\(\therefore \triangle ADB、\triangle ADC\)\triangle CEB\)\triangle DEF\)\triangle BAD\)\triangle ADB\)E\)D\)Rt\triangle DEF\)HL\(AC\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)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\perp BC\(BC=EF\(CD\perp AB\(\because AC\perp BC\(\because CD\perp AB，BE\perp AC\)AD\perp BD\)BE\perp AC\)AB=DE\)AD\perp BD\)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 级： 8年级（ ）班 . 时 间： . 2026/10/8 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 1 2 3 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） 学习目标 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 ′( ) ASA SAS × ？ \ \ \\ \\ \\\ \\\ 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ASA 4 03 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？ SAS 5 03 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？ 6 03 新知探究 已知：在Rt△ABC和Rt△A′B′C′中，∠C=∠C′=90°，AB=A′B′，AC=A′C′. 求证： Rt△ABC≌Rt△A′B′C′ 思考：AAS证明核心思维是将AAS转化为ASA，用三角形内角和定理解决角相等，再用已学的ASA判定全等。你能利用转化思想进行证明吗？ 将它转化为什么定理进行证明？ 可以利用什么得到该条件？ 7 03 新知探究 证明：在Rt△ABC中，由勾股定理得，， 同理，在Rt△A′B′C′中， . 由于AB=A′B′，AC=A′C′， 因此=，从而BC=B′C′. 在 △ABC与△A′B′C′中， 因此△ABC≌△A′B′C′（SSS）. 8 03 新知探究 “斜边、直角边” 定理：斜边和一条直角边分别相等的两个直角三角形全等. 几何语言 在Rt△ABC和Rt△中， ∴ Rt△ABC≌Rt△(HL). 03 新知探究 如图，BD，CE是△ABC的高，且BE=CD. 求证：Rt△BEC≌Rt△CDB. 例1 证明:因为BD，CE是△ABC的高， 所以∠BEC=∠CDB=90°. 在Rt△BEC和Rt△CDB中， 所以Rt△BEC ≌ Rt△CDB（HL）. 1. 两个同样大小的直角三角板按如图所 示的方式摆放，其中两条一样长的直角 边交于点，另一直角边， 分别落 在的边和上，且 ， 作射线，则在说明为 的平分 线的过程中，证全等的依据是( ) C A. 边角边 B. 角边角 C. 斜边、直角边 D. 边边边 举一反三 11 返回 举一反三 2．如图所示，已知在△ABC中，∠C＝90°，AD＝AC，DE⊥AB交BC于点E，若∠B＝28°，则∠AEC＝(　　) A．28° B．59° C．60° D．62° B 举一反三 13 返回 举一反三 03 新知探究 已知一直角边和斜边作直角三角形. 例2 如图，已知线段. 求作 Rt△ABC，使得斜边AB=，一条直角边BC=. 问题1：要先确定哪条边？ 问题2：定好了BC=，接下来要满足“直角”这个条件，该怎么做？ 问题3：作完垂线后，怎么确定点A的位置？​ 问题4：最后一步需要做什么，才能形成完整的Rt△ABC？ 03 新知探究 作法 ：（1）作一条直线l，在直线l 上截取BC=； （2）过点C作直线l 的垂线CD； （3）以点B为圆心，以为半径画圆弧，交CD于点A，连接AB，于是△ABC为所求作的直角三角形. 04 课堂练习 【知识技能类作业】必做题： 1.如图，已知AB⊥BD，CD⊥BD，若用“HL”判定Rt△ABD和Rt△CDB全等，则需要添加的条件是（　　） A．AD=CB B．∠A=∠C C．∠ADB=∠CBD D．AB=CD A 17 04 课堂练习 2.如图，在△ABC中，∠C＝90°，ED⊥AB于点D，BD＝BC，若AC＝6cm，则AE+DE等于（　　） A．4cm B．5cm C．6cm D．7cm C 18 新知应用 基础巩固题 1.下面说法是否正确?为什么? （1）两个锐角对应相等的两个直角三角形全等; （2）两条直角边对应相等的两个直角三角形全等. × √ 因为要判断两个三角形全等至少要有一组边对应相等. SAS AAA 新知应用 基础巩固题 2．如图，OD⊥AB于点D，OP⊥AC于点P，且OD＝OP，则△AOD与△AOP全等的理由是(　　) A．SSS  B．ASA  C．SSA  D．HL 3.如图，BE＝CF，AE⊥BC，DF⊥BC，要根据“HL”证明Rt△ABE≌Rt△DCF，则还需要添加的一个条件是(　　) A．AE＝DF    B．∠A＝∠D C．∠B＝∠C  D．AB＝DC D D 新知应用 基础巩固题 4.如图，CD⊥AB，BE⊥AC，垂足分别为点D，E，BE与CD相交于点O，且∠1＝∠2，则下列结论正确的个数为(　　) ①∠B＝∠C； ②△ADO≌△AEO； ③△BOD≌△COE； ④图中有4对三角形全等． A．4个  B．3个  C．2个  D．1个 A 3. 如图，在和 中， ，添加一个条件，可使用“斜边、直角边”判 定 ，则添加的条件是_________________ _______________. （答案不唯一） 举一反三 22 4．如图所示的网格是正方形网格(小正方形的边长均为1)，图形的各个顶点均在格点上，则∠1＋∠2的度数是________． 45° 返回 举一反三 23 5．[长沙市雅礼教育集团期末]如图，AD是△ABC的高，E是AD上一点，连接BE，AD＝BD，BE＝AC.若AD＝4，S△ABC＝14，则线段AE的长为________． 1 举一反三 24 返回 举一反三 6．如图，在四边形ABCD中，∠ABC＝90°，连接对角线AC，且AC＝AD，点E在边BC上，连接DE，过点A作AF⊥DE，垂足为F，若AB＝AF.求证： (1)∠DAC＝∠FAB； 【证明】因为AF⊥DE，所以∠AFD＝90°. 在Rt△AFD和Rt△ABC中，AD＝AC，AF＝AB，所以Rt△AFD≌Rt△ABC(斜边、直角边)．所以∠DAF＝∠CAB. 所以∠DAF＋∠CAF＝∠CAB＋∠CAF，即∠DAC＝∠FAB. 举一反三 26 返回 (2)DF＝CE＋EF. 【证明】如图，连接AE， 易知∠AFE＝90°. 在Rt△AEF和Rt△AEB中，AE＝AE，AF＝AB， 所以Rt△AEF≌Rt△AEB(斜边、直角边)，所以EF＝BE. 因为Rt△AFD≌Rt△ABC，所以DF＝BC. 因为BC＝CE＋BE＝CE＋EF，所以DF＝CE＋EF. 举一反三 27 课堂小结 SAS ASA AAS SSS HL 全等直角三角形的判定 谢谢观看 THANKS 【点拨】由题意得∠ABM＝∠ACM＝90°.在Rt△ABM和Rt△ACM中，所以Rt△ABM≌Rt△ACM(斜边、直角边)，所以∠BAM＝∠CAM，所以AM是∠PAQ的平分线． 【点拨】在△ABC中，∠C＝90°，AD＝AC，DE⊥AB交BC于点E，且AE＝AE，所以Rt△CAE≌Rt△DAE(斜边、直角边)，所以∠CAE＝∠DAE＝∠CAB. 因为∠B＋∠CAB＝90°，∠B＝28°，所以∠CAB＝90°－28°＝62°，所以∠AEC＝90°－∠CAB＝90°－31°＝59°. 【点拨】因为AD是△ABC的高，所以∠ADB＝∠ADC＝90°.在Rt△BDE和Rt△ADC中，所以Rt△BDE≌Rt△ADC(斜边、直角边)．所以DE＝CD.因为S△ABC＝14，所以BC·AD＝14.又因为AD＝BD＝4，所以(4＋CD)·4＝14，所以CD＝3＝DE，所以AE＝AD－DE＝1. $