31.2.1.2 用平行线判定三角形相似 课件 2026-2027学年人教版数学九年级下册

# 人教版九年级下册 27.2.1.2 用平行线判定三角形相似专项练习题（约1500字） ## 核心知识点梳理 **平行线判定三角形相似定理（A型\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)X型模型）** 平行于三角形一边的直线和其他两边相交\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)所构成的三角形与原三角形相似。 在$\triangle ABC$中\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)若$DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$DE$交$AB\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)AC$于点$D\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)E$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)则$\boldsymbol{\triangle ADE \backsim \triangle ABC}$。 - A型：平行线在三角形内部\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)两个三角形共顶角； - X型（8字型）：平行线在三角形外\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)两边延长线相交\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)依然满足相似。 > 书写注意：相似三角形顶点字母必须一一对应； > 思路：看到平行线 → 得到同位角相等 → 证两角对应相等 → 三角形相似； > 结合上节课平行线分线段成比例：平行既能得到线段成比例\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)又能直接判定三角形相似。 易错点： 1. 必须是**平行于三角形一边**\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)不是任意平行线都能证相似； 2. 相似后\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)对应边成比例\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)对应角相等\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)比例式要匹配对应顶点； 3. 区分：平行线分线段成比例是线段关系；本节是直接判定三角形**相似**。 ## 一\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)选择题（每题6分\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)共24分） 1. 在$\triangle ABC$中\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)点$D\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)E$分别在$AB\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)AC$上\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)若$DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)则下列结论正确的是（） A.$\triangle ADE \backsim \triangle ACB$ B.$\triangle ADE \backsim \triangle ABC$ C.$\triangle AED \backsim \triangle ABC$ D.$\triangle BDE \backsim \triangle ABC$ 2. 如图\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\(AD:DB=\(1:2\)\)$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)则$\triangle ADE$与$\triangle ABC$的相似比为（） A.$\(1:2\)$　B.$\(1:3\)$　C.$\(2:3\)$　D.$\(1:4\)$ 3. 已知$AB\parallel CD$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$AD\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)BC$交于点$O$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)则下列相似正确的是（） A.$\triangle AOB \backsim \triangle DOC$ B.$\triangle AOB \backsim \triangle COD$ C.$\triangle AOD \backsim \triangle BOC$ D.$\triangle ABD \backsim \triangle DCA$ 4. 在$\triangle ABC$中\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$AD=4\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)AB=6$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\(DE=3\)$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)则$BC$长为（） A.$\(4.5\)$ B.$5$ C.$6$ D.$8$ ## 二\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)填空题（每题6分\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)共24分） 5. $DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\triangle ADE\backsim\triangle ABC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\(AD:AB=\(\boldsymbol{\(2:5\)}\)\)$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)若$\(DE=4\)$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)则$\(BC=\)$________。 6. 如图\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$AB\parallel CD$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$AC\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)BD$交于$O$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\(AO:OC=\(\boldsymbol{3:2}\)\)$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)则$\triangle AOB$与$\triangle COD$相似比为________。 7. 在$\triangle ABC$中\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$AD=2\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)DB=3$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)则$\triangle ADE$与$\triangle ABC$周长比为________。 8. $DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\triangle ADE \backsim\triangle ABC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$AD=3\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)AB=7$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)若$\triangle ADE$面积为$18$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)则$\triangle ABC$面积为________。 ## 三\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)解答题（共52分） 9.（12分）如图\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)在$\triangle ABC$中\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)点$D$在$AB$上\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)点$E$在$AC$上。已知$AD=5\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)DB=\(\boldsymbol{10}\)\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)\(DE=4\)$。 （1）求证：$\triangle ADE \backsim \triangle ABC$； （2）求$BC$的长度。 \(\boldsymbol{10}\).（14分）已知：$AB\parallel CD$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)线段$AD\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)BC$相交于点$O$。$OA=4\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)OD=6\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)OB=3$。 （1）证明：$\triangle AOB \backsim \triangle DOC$； （2）求线段$OC$的长。 11.（12分）在$\triangle ABC$中\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$EF\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$AE:AB=\(2:3\)$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\(AF=6\)$。 （1）求证：$\triangle AEF \backsim \triangle ABC$； （2）求$FC$的长。 12.（14分）如图\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)在$\triangle ABC$中\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)点$D$在$AB$延长线上\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)点$E$在$AC$延长线上\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$DE\parallel BC$。$AD=12\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)AB=8\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)\(BC=\)6$。 （1）求证：$\triangle ABC \backsim \triangle ADE$； （2）求$DE$的长。 --- # 参考答案与详细解析 ## 一\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)选择题 1. **答案：B** 解析：$DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\angle ADE=\angle ABC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\angle AED=\angle ACB$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)公共角$\angle A$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\triangle ADE \backsim\triangle ABC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)顶点顺序A对应A\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)D对应B\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)E对应C。 2. **答案：B** 解析：$\(AD:DB=\(1:2\)\)$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)则$AD:AB=1:(1+2)=\(1:3\)$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)相似比等于对应边之比\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)相似比$\(1:3\)$。 3. **答案：B** 解析：$AB\parallel CD$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\angle A=\angle D$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\angle B=\angle C$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)对顶角$\angle AOB=\angle DOC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\triangle AOB \backsim\triangle COD$（A对应C\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)B对应D\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)O对应O）。 4. **答案：A** 解析：$\because DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\therefore\triangle ADE\backsim\triangle ABC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\dfrac{AD}{AB}=\dfrac{DE}{BC}$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\dfrac{4}{6}=\dfrac{3}{BC}$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)解得$\(BC=\)\(4.5\)$。 ## 二\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)填空题 5.$\boldsymbol{\(\boldsymbol{10}\)}$ 解析：相似比$\dfrac{AD}{AB}=\dfrac{DE}{BC}$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\dfrac{2}{5}=\dfrac{4}{BC}$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)得$\(BC=\)\(\boldsymbol{10}\)$。 6.$\boldsymbol{\(\boldsymbol{3:2}\)}$ 解析：$AB\parallel CD$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\triangle AOB \backsim\triangle COD$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)相似比$\(AO:OC=\(\boldsymbol{3:2}\)\)$。 7.$\boldsymbol{\(\boldsymbol{\(2:5\)}\)}$ 解析：$AD=2\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)DB=3\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)AB=5$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)相似比$\(AD:AB=\(\boldsymbol{\(2:5\)}\)\)$；相似三角形周长比等于相似比\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)周长比$\(\boldsymbol{\(2:5\)}\)$。 8.$\boldsymbol{\(\boldsymbol{98}\)}$ 解析：相似比$\(3:7\)$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)面积比等于相似比平方$\(9:49\)$；$\dfrac{18}{S_{\triangle ABC}}=\dfrac{9}{49}$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)解得$S_{\triangle ABC}=\(\boldsymbol{98}\)$。 ## 三\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)解答题 9.解： （1）证明：$\because DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\therefore \angle ADE=\angle ABC\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)\angle AED=\angle ACB$。 又$\because \angle A=\angle A$（公共角）\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\therefore\triangle ADE \backsim\triangle ABC$（平行于三角形一边的直线和其他两边相交\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)构成三角形与原三角形相似）。 （2）$\because\triangle ADE \backsim\triangle ABC$ $\therefore\dfrac{AD}{AB}=\dfrac{DE}{BC}$。 $AD=5\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)DB=\(\boldsymbol{10}\)\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)AB=AD+DB=15$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\dfrac{5}{15}=\dfrac{4}{BC}$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)解得$\(BC=\)12$。 答：$\(BC=\)12$。 \(\boldsymbol{10}\).解： （1）证明：$\because AB\parallel CD$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\therefore \angle A=\angle D\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)\angle B=\angle C$。 又$\because \angle AOB=\angle DOC$（对顶角相等）\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\therefore\triangle AOB \backsim\triangle DOC$。 （2）$\because\triangle AOB \backsim\triangle DOC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\therefore\dfrac{OA}{OD}=\dfrac{OB}{OC}$。 代入数值：$\dfrac{4}{6}=\dfrac{3}{OC}$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)解得$OC=\dfrac{18}{4}=\(4.5\)$。 答：$OC=\(4.5\)$。 11.解： （1）证明：$\because EF\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\therefore \angle AEF=\angle ACB\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)\angle AFE=\angle ABC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\angle A=\angle A$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\therefore\triangle AEF \backsim\triangle ABC$。 （2）$\because\triangle AEF \backsim\triangle ABC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)$\dfrac{AE}{AB}=\dfrac{AF}{AC}$ $\dfrac{2}{3}=\dfrac{6}{AC}$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)解得$\(AC=9\)$。 $FC=AC-AF=9-6=3$。 答：$\(FC=3\)$。 12.解： （1）证明：$\because DE\parallel BC$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\therefore\angle A\(BC=\)\angle ADE\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)\angle ACB=\angle AED$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\because \angle A=\angle A$（公共角）\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\therefore\triangle ABC \backsim\triangle ADE$。 （2）$\because\triangle ABC \backsim\triangle ADE$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\) $\therefore\dfrac{AB}{AD}=\dfrac{BC}{DE}$。 $\dfrac{8}{12}=\dfrac{6}{DE}$\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)解得$\(DE=9\)$。 答：$\(DE=9\)$。 ## 本节解题总结 本节课核心两个模型：A型图\(AB\(D\(AB\(AD\(AC\(AD、BC\)BD\)BC\)AC\)E\)AC\)8字型（X型）。解题固定步骤： ① 找到平行线\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)写出平行条件； ② 直接使用定理判定三角形相似； ③ 根据相似\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)写出对应边比例式； ④ 代入已知线段长度\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)列方程求解。 做题重中之重：**相似三角形字母必须对应**\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)字母对应错\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)比例式就会写错。相似三角形周长比等于相似比\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)面积比是相似比的平方\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)经常混合考查。 平行线分线段成比例只得到线段成比例；而本节定理可以直接得到**三角形相似**\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)是证明三角形相似第一种判定方法\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)是后续相似证明的基础。 全文合计约1490字\(AD=4\(AD=2\(AD=3\(\therefore \angle ADE=\angle ABC\(\therefore \angle A=\angle D\(\therefore \angle AEF=\angle ACB\(\therefore\angle ABC=\angle ADE，\angle ACB=\angle AED\)\angle AFE=\angle ABC\)\angle B=\angle C\)\angle AED=\angle ACB\)AB=7\)DB=3\)AB=6\)满足1500字要求。 需要我给你**纯题目无答案版本**方便打印练习吗？ 人教版（新教材）数学九年级下册 31.2.1.2 用平行线判定三角形相似 授课教师： . 班 级： 9年级（ ）班 . 时 间： . 2026/10/19 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。 第三十一章 相似 复习引入 平行线分线段成比例 基本事实 两条直线被 所 截，所得的对应线段 . 基本事实的推论 平行于三角形一边的直线截 ，所得的对应线段成比例. 一组平行线 成比例 其他两边(或两边的延长线) 复习引入 相似 相似多边形 从一般到特殊 相似三角形 相似图形 三角形相似的判定 平行线分线段成比例 合作探究 思考 由前面所学的结论，在一个三角形中，你能作出一个与这个三角形相 似的三角形吗? 如图，在△ABC中 ，作DE//BC，且与AB，AC分别相交于点D，E. 求证：△ADE∽△ABC. 对应边成比例 对应角相等 相似三角形 合作探究 分析： 在△ABC中，由 ， 可知△ADE 与△ABC的角分别相等. 由前面基本事实的推论， 可知 ，若能再证明 ， 就能证明这两个三角形的边成比例， 从而证明△ADE∽△ABC. DE//BC = = 合作探究 分析： 从要证的 = 可以看出， 除DE外 ，AE，AC，BC 都在△ABC 的边上， 因此只需先将DE平移到BC边上， 如图，使BF=DE， 再证明 就可以了. = 证明： 过点E作EF//AB，交 BC 于点F. ∴ = . ∵DE//BC，∴∠ADE=∠B，∠AED=∠C， = . 根据平行四边形的定义得：四边形BFED是平行四边形， ∴ BF=DE， ∴ = =， ∴△ADE与△ABC的边成比例， 又∵△ADE与△ABC的角分别相等， ∴△ADE∽△ ABC. 合作探究 平行于三角形一边的直线，与其他两边相交，所构成的三角形与原三角形相似. 合作探究 思考 当平行于三角形一边的直线与其他两边的延长线相交时，所构成的三角形与原三角形相似吗？ 如图，DE∥BC，点D，E分别在BA，CA的延长线上. 求证：△ADE∽△ABC. 如何转化DE呢？ 证法1： 过点D作DF//AC，交BC的延长线于点F. ∴ = . ∵DE//BC，∴∠ADE=∠B，∠E=∠ACB， = . 根据平行四边形的定义得：四边形CFDE是平行四边形， ∴ CF=DE， ∴ = = ， ∴△ADE与△ABC的边成比例， 又∵△ADE与△ABC的角分别相等， ∴△ADE∽△ ABC. 合作探究 证法2： 在AB上截取AD'=AD，过点D'作D'E'//BC，交AC于点E'. ∵DE//BC，∴D'E'//DE， ∴∠D=∠AD'E'，∠E=∠AE'D'， ∴△ADE≌△AD'E'， ∵D'E'//BC， ∴△AD'E'∽△ABC. ∴△ADE∽△ABC. 合作探究 平行于三角形一边的直线，与其他两边的延长线相交，所构成的三角形与原三角形相似. 符号语言 ∵DE//BC， ∴△ADE∽△ ABC. 合作探究 三角形相似的判定定理 的预备定理 平行于三角形一边的直线，与其他两边(或两边的延长线)相交，所构成的三角形与原三角形相似. 返回 C 1. 举一反三 12 返回 2. C 举一反三 返回 3. 3 如图，AB∥DF，DE∥BC，则图中的相似三角形有________对. 举一反三 典例分析 例1 如图，在5×3的正方形网格中，每个小格的顶点叫做格点，B，C，D都在格点上，线段AB与CD相交于点E，AE:BE=(　　) A．2：1 B．3： C．3：2 D．4：1 A 典例分析 例2 如图，在▱ABCD中，点E在边BC的延长线上，连接AE交CD于点F．图中的相似三角形有（    ）． A．1对 B．2对 C．3对 D．4对 C 解：∵ 四边形ABCD是平行四边形， ∴AB∥CD，AB=CD. ∴△ABF∽△CEF. ∴ = ， 又E是CD的中点， ∴ AB=CD=2CE. ∴=2. 典例分析 例3 如图，在▱ABCD中 ，E是CD的中点，AC，BE相交于点F.求的值. 典例分析 例4 如图，在Rt△ABC中，∠BAC=90°，中线AD，BE相交于点F、EG∥BC，交AD于点G．GF=2．求BC的长． 解：如图，连接DE，∵中线AD，BE相交于点F， ∴DE∥AB，DE=AB，∴△ABF∽△DEF， ∴BF:EF=AF:DF=AB:DE=2:1，∴AD:DF=3:1， ∵EG∥BC，∴△BDF∽△EGF， ∴DF:GF=BF:EF=2:1, ∴DF=2GF=4，∴AD=3DF=12， ∵∠BAC=90°，AD是中线，∴BC=2AD=24． 返回 4. 如图，在△ABC中，∠ABC＝90°，∠A＝60°，直尺的一边与BC重合，另一边分别交AB，AC于点D，E，点B，C，D，E处的读数分别为15，12，0，1，则直尺宽BD＝______cm. 举一反三 返回 5. (4分)如图，F是▱ABCD的边CD上一点，连接AF并延长，交BC的延长线于点E.若AB＝6，AF＝2EF，求CF的长. 举一反三 返回 6. B [2025河南中考]如图所示的网格中，每个小正方形的边长均为1，△ABC的三个顶点均在网格线的交点上，点D，E分别是边BA，CA与网格线的交点，连接DE，则DE的长为(　　) 举一反三 返回 7. B 如图，在▱ABCD中，AC，BD相交于点O，E为OC的中点，EF∥AB交BC于点F.若AB＝4，则EF的长为(　　) 举一反三 返回 8. 6 如图，在平面直角坐标系中，C为△AOB的边OA上一点，AC∶OC＝1∶2，过点C作CD∥OB交AB于点D，C，D两点的纵坐标分别为1，3，则点B的纵坐标为________. 举一反三 9. 6 (1)CG的长为________； 举一反三 24 (2)若CD＝2，求EF的长. 返回 举一反三 25 归纳总结 三角形相似的判定定理的预备定理 文字语言 平行于三角形一边的直线，与 相交，所构成的三角形与原三角形相似. 符号语言 图形语言 其他两边(或两边的延长线) ∵DE//BC， ∴△ADE∽△ ABC. 谢谢观看 THANKS 如图，在△ABC中，DE∥AB，若＝，AB＝10，则DE的长为(　　) A．3 B．5 C．6 D．8 如图，已知AB∥CD，AD与BC交于点E，若AB＝3，CD＝2，则的值为(　　) A. B. C. D. 解：∵四边形ABCD是平行四边形， ∴AB∥CD，∴△EBA∽△ECF，∴＝. ∵AF＝2EF，∴＝，∴＝，∴CF＝2. A.　 B．1　 C.　 D. A. B．1 C. D．2 (8分)如图，D是△ABC的边BC上一点，连接AD，过AD上的点E作EF∥BD，交AB于点F，过点F作FG∥AC，交BC于点G，已知＝，BG＝4. 解：∵CD＝2，CG＝6，∴DG＝CG－CD＝4. ∵BG＝4，∴BD＝BG＋DG＝8.∵EF∥BD， ∴＝＝，∴＝. ∵EF∥BD，∴△AFE∽△ABD， ∴＝，即＝，∴EF＝. $