22.3.3 相似三角形的判定定理2、3 课件 -2026-2027学年华东师大版数学九年级上册

#华东师大版九年级上册第3课时相似三角形判定定理2、3##一、知识点梳理>判定定理1：AA（两角分别相等）→相似（上一节）###✅判定定理2（SAS相似）**两边对应成比例\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)且夹角相等的两个三角形相似。**条件：①两组对应边的比相等；②**这两组边的夹角相等**（重点！必须是夹角\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)不\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)BC=6\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)AC=8$；$\triangleDEF$三边$DE=2\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)EF=3\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)DF=4$。求证：$\triangleABC\backsim\triangleDEF$。3.如图\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)点$D$在$AB$上\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)$AD=2\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)AB=6\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)AC=4$\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)$AE=\dfrac43$\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)$\angleA$公共角。求证：$\triangleADE\backsim\triangleACB$。4.拓展：$\triangleABC$中\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)$AB=10\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)AC=8$；$\triangleADE$中$AD=5\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)AE=4$\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)$\angleA$公共角。求证：$\triangleADE\backsim\triangleABC$。##五、参考答案与解析###选择题1.B解析：夹角$\angleA$与$\angleD$相等\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)满足SAS相似2.B解析：$\dfrac24=\dfrac36=\dfrac48=\dfrac12$\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)三边成比例\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)SSS相似3.C解析：两边成比例\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)夹角相等\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)SAS4.A解析：$\dfrac36=\dfrac48=\dfrac5{10}=\dfrac12$\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)对应边SSS相似5.B解析：A是SSA错误；C错；D是全等条件不是相似###填空题1.夹角2.三边3.$\boldsymbol{\dfrac12}$（$\triangleABC:\triangleA'B'C'$）4.$\backsim$（相似）5.SSS###解答题1.证明：$\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2$$\therefore\dfrac{AB}{A'B'}=\dfrac{AC}{A'C'}$又$\because\angleA=\angleA'$$\therefore\triangleABC\backsim\triangleA'B'C'$（两边对应成比例且夹角相等\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)SAS）2.证明：$\dfrac{AB}{DE}=\dfrac42=2\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)\dfrac{BC}{EF}=\dfrac63=2\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)\dfrac{AC}{DF}=\dfrac84=2$$\therefore\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}$$\therefore\triangleABC\backsim\triangleDEF$（三边对应成比例\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)SSS）3.证明：$AD=2\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)AC=4\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)AE=\dfrac43\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)AB=6$$\dfrac{AD}{AC}=\dfrac24=\dfrac12\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29$❌修正题目数据\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)更换标准版本：标准题：$AD=2\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)AC=4\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)AE=3\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)AB=6\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)\angleA$公共角$\dfrac{AD}{AC}=\dfrac24=\dfrac12\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)\dfrac{AE}{AB}=\dfrac36=\dfrac12$$\dfrac{AD}{AC}=\dfrac{AE}{AB}$\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)$\angleA=\angleA$$\therefore\triangleADE\backsim\triangleACB$（SAS）4.证明：$\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)\dfrac{AE}{AC}=\dfrac48=\dfrac12$$\therefore\dfrac{AD}{AB}=\dfrac{AE}{AC}$又$\angleA=\angleA$（公共角）$\therefore\triangleADE\backsim\triangleABC$（SAS）---下一节：**22.4相似三角形的性质**\(AB=4\(DE=2\(AB=10\(AD=5\(\because\dfrac{AB}{A'B'}=\dfrac{6}{3}=2\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AC}=\dfrac24=\dfrac12\(\dfrac{AD}{AB}=\dfrac{5}{10}=\dfrac12，\dfrac{AE}{AC}=\dfrac48=\dfrac12\)\dfrac{AE}{AB}=\dfrac36=\dfrac12\)\dfrac{AE}{AB}=\dfrac{\frac43}{6}=\dfrac29\)\dfrac{AC}{A'C'}=\dfrac{8}{4}=2\)AE=4\)AC=8\)DF=3\)AC=6\)继续吗？华东师大版（新教材）数学九年级上册22.3.3相似三角形的判定定理2、3授课教师：.

班

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9年级（

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2026/10/11

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22章图形的相似复习回顾相似三角形的判定方法有哪些？①