中小学教育资源及组卷应用平台 2025年中考二轮专题复习-几何压轴题高效拆分特训 专题一 全等模型高效拆分特训 INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET 特训1 旋转全等(一) 一线三等角模型INCLUDEPICTURE"标2.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标2.tif" \* MERGEFORMATINET INCLUDEPICTURE"条单.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\条单.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\条单.tif" \* MERGEFORMATINET 模型解读 条件 A,P,B三点共线,CP=PD,∠1=∠2=∠3. A,P,B三点共线,CP=PD,∠1=∠2=∠DPC. 图示 INCLUDEPICTURE"SK1-1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-1.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-1.tif" \* MERGEFORMATINET 同侧型 异侧型 结论 △ACP≌△BPD. 背景 常以等腰三角形、等边三角形、等腰直角三角形、矩形、正方形为背景. INCLUDEPICTURE"条单.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\条单.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\条单.tif" \* MERGEFORMATINET 典题训练 1.如图,AC=AB=BD,∠ABD=90°,BC=6,则△BCD的面积为_____. INCLUDEPICTURE"SK1-3.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-3.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-3.tif" \* MERGEFORMATINET 2.如图,把两个腰长相等的等腰三角形拼接在一起,腰AD=AF=AC,∠DAC=90°,作CE⊥AF于E,连接DE.若CE=12,DE=DF,求AC的长. INCLUDEPICTURE"SK1-4.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-4.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-4.tif" \* MERGEFORMATINET INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET 特训2 旋转全等(二)手拉手模型INCLUDEPICTURE"标2.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标2.tif" \* MERGEFORMATINET INCLUDEPICTURE"条单.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\条单.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\条单.tif" \* MERGEFORMATINET 模型解读 条件:如图,AB=AC,AD=AE,∠BAC=∠DAE.INCLUDEPICTURE"SK1-5.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-5.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-5.tif" \* MERGEFORMATINET 结论:△ACD≌△ABE. 顶角相等且顶点重合的两个等腰三角形→全等三角形. INCLUDEPICTURE"条单.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\条单.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\条单.tif" \* MERGEFORMATINET 典题训练 【熟悉模型】 如图①,已知△ABC与△ADE都是等腰三角形,AB=AC,AD=AE,且∠BAC=∠DAE,求证:BD=CE; 【运用模型】 如图②,P为等边三角形ABC内一点,且PA∶PB∶PC=3∶4∶5,求∠APB的度数.小明在解决此问题时,根据前面的“手拉手模型”,以BP为边构造等边三角形BPM,这样就有两个等边三角形共顶点B,然后连接CM,通过转化的思想求出了∠APB的度数,则∠APB的度数为_____; 【深化模型】 如图③,在四边形ABCD中,AD=4,CD=3,∠ABC=∠ACB=∠ADC=45°,求BD的长. INCLUDEPICTURE"SK1-6.tif" INCLUDE ... ...
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