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" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-6.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-6.tif" \* MERGEFORMATINET
INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\标1.tif" \* MERGEFORMATINET 特训3 旋转全等(三)半角模型INCLUDEPICTURE"标2.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标2.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\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
INCLUDEPICTURE"SK1-7.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-7.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-7.tif" \* MERGEFORMATINET 等腰直角三角形中的“半角模型”(如图)：条件：在△ABC中，AB＝AC，∠BAC＝90°，∠DAE＝45°.方法：将△ABD绕点A逆时针旋转90°，得到△ACF，连接EF.结论：△ADE≌△AFE，DE2＝BD2＋CE2.
正方形中的“半角模型”(如图)：INCLUDEPICTURE"SK1-8.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-8.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-8.tif" \* MERGEFORMATINET 条件：在正方形 ABCD 中，∠EAF＝45°.方法：将△ADF绕点A顺时针旋转90°，得到△ABG.结论：△AEF≌△AEG，EF＝BE＋DF，EF2＝CE2＋CF2，FA平分∠DFE，EA平分∠BEF.
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
如图，在正方形ABCD中，E，F分别是边BC，CD上的点(不与端点重合)，且∠EAF＝45°.
(1)求证：EF＝BE＋DF；
(2)连接BD，分别交AE，AF于点M，N，试探究BM，MN，DN之间的数量关系，并说明理由．
INCLUDEPICTURE"专项Z-35.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\专项Z-35.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\专项Z-35.tif" \* MERGEFORMATINET
INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\标1.tif" \* MERGEFORMATINET 特训4 中点模型(一)倍长中线(或作平行线)INCLUDEPICTURE"标2.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标2.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\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
题眼：中线、中点INCLUDEPICTURE"SK1-9.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-9.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-9.tif" \* MERGEFORMATINET 条件：如图，AO为△ABC的中线．方法1：如图①，延长中线AO至D, 使得DO＝AO，连接BD.方法2：如图②，过点B作BD∥AC交AO的延长线于点D.结论：△AOC≌△DOB.
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，AD是△ABC的中线，BE交AC于E，交AD于F，且AC＝BF.求证：AE＝FE.
INCLUDEPICTURE"SK1-10.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-10.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-10.tif" \* MERGEFORMATINET
2．如图，在△ABC中，∠A＝90°，D为BC的中点，DE⊥DF，DE交AB于点E，DF交AC于点F，连接EF.试猜想线段BE，CF，EF三者之间的数量关系，并证明你的结论．
INCLUDEPICTURE"SK1-11.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-11.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-11.tif" \* MERGEFORMATINET
INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\标1.tif" \* MERGEFORMATINET 特训5 中点模型(二)构造中位线INCLUDEPICTURE"标2.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标2.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\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
题眼：中线、中点INCLUDEPICTURE"SK1-12.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-12.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-12.tif" \* MERGEFORMATINET 方法1：再构造另一个中点，变中线为中位线．示例：如图①，AO为△ABC的中线，延长BA至D，使得AD＝AB，连接CD, 可证OA＝CD，OA∥CD.方法2：构造双中位线示例：如图②， 在四边形ABCD中，E，F分别是AD，BC的中点，取对角线BD的中点M，连接ME，MF.
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，在正方形ABCD中，AB＝2，E，F分别是边AB，BC的中点，连接EC，FD，G，H分别是EC，FD的中点，连接GH，则GH的长度为________．
INCLUDEPICTURE"专项Z-15.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\专项Z-15.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\专项Z-15.tif" \* MERGEFORMATINET
2．如图，四边形ABCD中，AB＝CD，E，F分别是边BC，AD的中点，延长BA，CD，分别交EF的延长线于P，Q.求证：∠BPE＝∠Q.
INCLUDEPICTURE"专项Z-16.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\专项Z-16.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\专项Z-16.tif" \* MERGEFORMATINET
INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\标1.tif" \* MERGEFORMATINET 特训6 中点模型(三)平行线证中点INCLUDEPICTURE"标2.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标2.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\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
方法：作平行或作垂直，证中点．在未知中点的问题中，不能采取倍长中线法，可作平行或垂直，通过三角形的全等证中点.
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
在△ABC中，AB＝AC，点D在射线BA上，点E在AC的延长线上，且BD＝CE.连接DE，DE与BC边所在的直线交于点F.
(1)当点D在线段BA上时，如图所示，求证：DF＝EF；
(2)过点D作DH⊥BC交直线BC于点H.若BC＝4，CF＝1，求BH的长．
INCLUDEPICTURE"SK1-13.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-13.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-13.tif" \* MERGEFORMATINET
INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\标1.tif" \* MERGEFORMATINET 特训7 中点模型(四)斜边中线INCLUDEPICTURE"标2.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标2.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\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，在Rt△ABC和Rt△ABD中，∠ACB＝∠ADB＝90°.INCLUDEPICTURE"SK1-14.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-14.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-14.tif" \* MERGEFORMATINET 方法：如图，取AB的中点O，连接OC，OD，得到OC＝OD＝AB，∠ODB＝∠OBD＝∠AOD，∠ODA＝∠OAD＝∠BOD，∠OCB＝∠OBC＝∠AOC，∠OAC＝∠OCA＝∠BOC.
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，BN，CM分别是△ABC的两条高，D是BC的中点，DE⊥MN于点E.
INCLUDEPICTURE"SK1-15.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-15.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-15.tif" \* MERGEFORMATINET
(1)求证：E是MN的中点；
(2)若BC＝12，MN＝8，则DE＝________．
2．如图，在四边形ABCD中，∠BAD＝∠BCD＝90°，O为BD的中点，连接OA，AC.若∠ADC＝135°，求证：AC＝OA.
INCLUDEPICTURE"SK1-16.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-16.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-16.tif" \* MERGEFORMATINET
INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\标1.tif" \* MERGEFORMATINET 特训8 角平分线模型INCLUDEPICTURE"标2.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标2.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\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
INCLUDEPICTURE"SK1-17.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-17.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-17.tif" \* MERGEFORMATINET 条件：OP平分∠AOB.方法一：(垂两边)如图①，过点P分别作PD⊥OA于D，PE⊥OB于E；方法二：(垂中间)如图②，过点P作DE⊥OP，交OA于D，OB于E；方法三：(截等线段)如图③，截取OD＝OE(点D，E分别在OA，OB上)，连接PD，PE.
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，△ABC的面积为10，P为△ABC内一点，BP平分∠ABC，AP⊥BP，则△PBC的面积为________．
INCLUDEPICTURE"专项Z-21.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\专项Z-21.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\专项Z-21.tif" \* MERGEFORMATINET
2．如图，在四边形ABCD中，∠BAD＝120°，∠C＝60°，BD平分∠ABC.求证：AD＝CD.
INCLUDEPICTURE"专项Z-19.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\专项Z-19.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\专项Z-19.tif" \* MERGEFORMATINET
3．如图，在△ABC中，∠A＝2∠B，CD平分∠ACB交AB于点D.求证：BC＝AD＋AC.
INCLUDEPICTURE"专项Z-26.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\专项Z-26.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\专项Z-26.tif" \* MERGEFORMATINET
INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\标1.tif" \* MERGEFORMATINET 特训9 十字架模型INCLUDEPICTURE"标2.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标2.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\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
图示 INCLUDEPICTURE"SK1-18.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-18.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-18.tif" \* MERGEFORMATINET INCLUDEPICTURE"SK1-19.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-19.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-19.tif" \* MERGEFORMATINET INCLUDEPICTURE"SK1-20.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-20.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-20.tif" \* MERGEFORMATINET
结论 正方形ABCD中，若AM⊥BN，则△ADM≌△BAN，∴AM＝BN，即＝1. 将AM，BN如上图所示进行平移，易得HK＝BN＝AM＝EF，∴＝1. 正方形ABCD中，若EF＝HK，则过点E，K分别作EN⊥CD于N，KM⊥BC于M，易证△ENF≌△KMH，从而得到EF⊥HK.
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)感知：如图①，在正方形ABCD中，E，F分别是AB，BC上的点，连接DE，AF，若BE＝CF，求证：DE＝AF.
(2)应用：在(1)的条件下，求证：AF⊥DE.
(3)探究：如图②，在正方形ABCD中，E，F分别为边AB，CD上的点(点E，F不与正方形的顶点重合)，连接EF，作EF的垂线分别交边AD，BC于点G，H，垂足为O，若E为AB的中点，DF＝1，AB＝4，则GH的长为________．
INCLUDEPICTURE"SK1-21.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-21.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-21.tif" \* MERGEFORMATINET
INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\标1.tif" \* MERGEFORMATINET 特训10 对角互补模型INCLUDEPICTURE"标2.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标2.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\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
类型 90°的对角互补模型 60°、120°的对角互补模型
图示 INCLUDEPICTURE"SK1-22.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-22.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-22.tif" \* MERGEFORMATINET INCLUDEPICTURE"SK1-23.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-23.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-23.tif" \* MERGEFORMATINET
条件 ∠ABC＝∠ADC＝90°，BD平分∠ABC. ∠ABC＝120°，∠ADC＝60°，BD平分∠ABC.
作法 过点D分别作DE⊥BC于点E，DF⊥BA交BA的延长线于点F.
结论 1.AD＝CD；2．AB＋BC＝BD；3．S四边形ABCD＝BD2. 1.AD＝CD；2．AB＋BC＝BD；3．S四边形ABCD＝BD2.
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，正方形ABCD的对角线相交于点O，E为AB上一动点．连接OE，作OF⊥OE交BC于点F，已知AB＝2，则四边形EBFO的面积为________．
INCLUDEPICTURE"SK1-24.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-24.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-24.tif" \* MERGEFORMATINET
2．如图，在四边形ABCD中，∠BAD＋∠BCD＝180°，DB平分∠ADC.若∠ADB＝60°，求证：△ABC是等边三角形．
INCLUDEPICTURE"SK1-25.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-25.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-25.tif" \* MERGEFORMATINET
INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\标1.tif" \* MERGEFORMATINET 特训11 设参导等角INCLUDEPICTURE"标2.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标2.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\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 方法技巧
含有等腰三角形的图形中，要证明两个角相等，可以考虑设α，β导角证等角.
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，在△ABC中，AB＝AC，点D在BC的延长线上，∠ADB＝45°， 过点C作CE⊥AB于点E，延长EC，交 AD的延长线于点F，求证：AC＝FC.
INCLUDEPICTURE"SK1-26.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-26.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-26.tif" \* MERGEFORMATINET
2．如图，在△ABC中，D，E分别在AB，BC边上，连接AE，CD相交于点F，若∠AFD＝2∠ABC，AE＝AC, 求证：CD＝AC.
INCLUDEPICTURE"SK1-27.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\SK1-27.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SK1-27.tif" \* MERGEFORMATINET
专题一 全等模型高效拆分特训 答案
INCLUDEPICTURE"标1.tif" INCLUDEPICTURE "D:\\课件\\中考福建数学\\标1.tif" \* MERGEFORMATINET INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\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 "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\标2.tif" \* MERGEFORMATINET
1．9
2．解：如图，过D作DH⊥EF于H.
INCLUDEPICTURE"SKD-1.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SKD-1.tif" \* MERGEFORMATINET
∵∠DAC＝90°，
∴∠DAH＋∠CAE＝90°.
∵CE⊥AF于E，
∴∠ACE＋∠CAE＝90°，
∴∠DAH＝∠ACE.
∵∠AHD＝∠AEC＝90°，AD＝AC，
∴△ADH≌△CAE，∴AH＝CE＝12.
设AC＝x，则AF＝AD＝x，∴FH＝AF－AH＝x－12.
∵DE＝DF，DH⊥FE，∴EH＝FH＝x－12，
∴AE＝AH－EH＝12－(x－12)＝24－x.
∵AC2＝AE2＋CE2，∴x2＝(24－x)2＋122，
∴x＝15.∴AC的长为15.
特训2 旋转全等(二)手拉手模型
【熟悉模型】证明：∵∠BAC＝∠DAE，∴∠BAD＝∠CAE.
在△ABD和△ACE中，
∵AB＝AC，∠BAD＝∠CAE，AD＝AE，
∴△ABD≌△ACE，∴BD＝CE.
【运用模型】150°
【深化模型】解：∵∠ACB＝∠ABC＝45°，
INCLUDEPICTURE"SKD-2.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SKD-2.tif" \* MERGEFORMATINET
∴∠BAC＝90°，且AC＝AB.
将△ADB绕点A顺时针旋转90°，得到△AEC，连接DE，如图．
∴AD＝AE，∠DAE＝90°，BD＝CE.
∴∠EDA＝45°，DE＝AD＝4.
∵∠ADC＝45°，∴∠EDC＝45°＋45°＝90°.
在Rt△DCE中，CE＝＝＝，
∴BD＝CE＝.
特训3 旋转全等(三)半角模型
(1)证明：如图，将△ADF绕点A旋转至△ABG的位置，
∴BG＝DF，AG＝AF，∠GAB＝∠FAD，
∴∠GAF＝∠BAD＝90°，
∴∠EAG＝∠GAF－∠EAF＝90°－45°＝45°＝∠EAF.
INCLUDEPICTURE"专项D-15.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\专项D-15.tif" \* MERGEFORMATINET
又∵AE＝AE，∴△EAG≌△EAF，
∴GE＝EF.
又∵GE＝BE＋BG＝BE＋DF，
∴EF＝BE＋DF.
(2)解：MN2＝BM2＋DN2.
理由：如图，将△ABM 绕点A旋转至△ADH的位置，连接HN，
∴AH＝AM, DH＝BM，∠ADH＝∠ABM，∠HAD＝∠MAB，
∴∠HAM＝∠BAD＝90°，
∴∠HAN＝∠HAM－∠EAF＝90°－45°＝45°＝∠MAN.
又∵AN＝AN，
∴△HAN≌△MAN，∴HN＝MN.
∵四边形ABCD是正方形，∴∠ABM＝∠ADN＝45°，
∴∠ADH＝∠ABM＝45°，
∴∠HDN＝∠ADH＋∠ADN＝90°，
∴HN2＝DH2＋DN2，∴MN2＝BM2＋DN2.
特训4 中点模型(一)倍长中线(或作平行线)
1．证明：延长AD到M，使AD＝DM，连接BM，如图．
INCLUDEPICTURE"SKD-3.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SKD-3.tif" \* MERGEFORMATINET
∵AD是△ABC的中线，∴CD＝BD，
在△ADC和△MDB中，
∴△ADC≌△MDB，∴BM＝AC，∠CAD＝∠M.
∵AC＝BF，∴BM＝BF，∴∠M＝∠BFM.
∵∠AFE＝∠BFM，∴∠CAD＝∠AFE，
∴AE＝FE.
2．解：猜想：BE2＋CF2＝EF2.
证明：延长ED到点G，使DG＝ED，连接GF，GC，如图．
INCLUDEPICTURE"SKD-4.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SKD-4.tif" \* MERGEFORMATINET
∵ED⊥DF，DG＝ED，∴EF＝GF.
∵D是BC的中点，∴BD＝CD.
在△BDE和△CDG中，
∴△BDE≌△CDG，∴BE＝CG，∠B＝∠GCD.
∵∠A＝90°，∴∠B＋∠ACB＝90°，
∴∠GCD＋∠ACB＝90°，即∠GCF＝90°，
∵在Rt△CFG中，GC2＋CF2＝GF2，
∴BE2＋CF2＝EF2.
特训5 中点模型(二)构造中位线
1．1
2．证明：连接AC，取AC的中点M, 连接ME, MF，则ME, MF 分别为△ABC 和△ACD的中位线，
∴ME＝AB，MF＝CD，ME∥AB，MF∥CD.
∵AB＝CD，∴ME＝MF，∴∠MEF ＝∠MFE.
∵ME∥AB，MF∥CD，
∴∠BPE ＝∠MEF，∠Q＝∠MFE，
∴∠BPE＝∠Q.
特训6 中点模型(三)平行线证中点
(1)证明：如图①，过点D作DG∥AC，交BC于点G.
INCLUDEPICTURE"SKD-5.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SKD-5.tif" \* MERGEFORMATINET
∴∠DGB＝∠ACB.
∵AB＝AC，∴∠B＝∠ACB，
∴∠DGB＝∠B，∴BD＝GD.
∵BD＝CE，∴GD＝CE.
∵DG∥AC，
∴∠GDF＝∠CEF，∠DGF＝∠ECF，
∴△DGF≌△ECF，∴DF＝EF.
(2)解：当点D在线段BA上时，过点E作EO⊥BC，交BC的延长线于O，如图②所示，
INCLUDEPICTURE"SKD-7.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SKD-7.tif" \* MERGEFORMATINET
∵AB＝AC，∴∠B＝∠ACB＝∠OCE.
又∵∠DHB＝∠EOC＝90°，
BD＝CE，∴△DHB≌△EOC，
∴BH＝CO，
∴HO＝HC＋CO＝HC＋HB＝BC＝4.
∵∠DHF＝∠EOF＝90°，∠DFH＝∠EFO，
由(1)得DF＝EF，
∴△DHF≌△EOF，∴HF＝OF＝HO＝2.
∵CF＝1，∴BH＝CO＝OF－CF＝2－1＝1；
当点D在BA的延长线上时，过点E作EO⊥BC交BC的延长线于点O，如图③，
INCLUDEPICTURE"SKD-8.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SKD-8.tif" \* MERGEFORMATINET
同理可证△DHB≌△EOC，△DHF≌△EOF，
∴BH＝OC，HF＝OF.
∴HO＝HC＋CO＝HC＋HB＝BC＝4，∴HF＝OF＝HO＝2.
∵CF＝1，∴BH＝CO＝OF＋CF＝2＋1＝3.
综上所述，BH的长为1或3.
特训7 中点模型(四)斜边中线
1．(1)证明：如图，连接DM，DN.
INCLUDEPICTURE"SKD-9.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SKD-9.tif" \* MERGEFORMATINET
∵BN，CM分别是△ABC的两条高，
∴∠BMC＝∠CNB＝90°.
∵D是BC的中点，
∴DM＝BC，DN＝BC.
∴DM＝DN.
∵DE⊥MN，∴E是MN的中点．
(2)2
2．证明：连接OC，如图．∵∠BAD＝∠BCD＝90°，O为BD的中点，
INCLUDEPICTURE"SKD-10.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SKD-10.tif" \* MERGEFORMATINET
∴∠ABC＋∠ADC＝180°，OA＝BD＝OB，OC＝BD＝OB，
∴OA＝OC.
∵∠ADC＝135°，∴∠ABC＝45°.
∵OB＝OC，∴∠OBC＝∠OCB，
∴∠COD＝2∠CBO，同理可得∠AOD＝2∠ABO.
∴∠AOC＝∠AOD＋∠COD＝2∠ABO＋2∠CBO＝2(∠ABO＋∠CBO)＝2∠ABC＝90°，
∴△AOC为等腰直角三角形，
∴AC＝＝OA.
特训8 角平分线模型
1．5
2．证明：如图，过点D分别作DE⊥BC于E，DF⊥BA，交BA的延长线于点F.
INCLUDEPICTURE"专项D-5.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\专项D-5.tif" \* MERGEFORMATINET
∵点D在∠ABC 的平分线上，
∴DE＝DF.
∵∠BAD＝120°，∴∠DAF ＝60°，
∴∠DAF ＝∠C.
∵DE⊥BC, DF⊥AF，
∴∠F＝∠DEC＝90°，
∴△ADF≌△CDE，∴AD＝CD.
3．证明：在BC边上截取EC＝AC，连接DE，则 BC＝BE＋EC.
∵CD是∠ACB的平分线，∴∠DCE＝∠DCA.
在△CDE和△CDA中，
∴△CDE≌△CDA，∴∠CED＝∠A，ED＝AD.
∵∠CED＝∠B＋∠BDE，∠A＝2∠B，
∴∠B＝∠BDE，∴BE＝ED，∴BE＝AD，
∴BE＋EC＝AD＋AC，即BC＝AD＋AC.
特训9 十字架模型
(1)证明：∵四边形ABCD是正方形，
∴AD＝AB＝BC，∠DAB＝∠B＝90°.
∵BE＝CF，∴AE＝BF，
∴△DAE≌△ABF，∴DE＝AF.
(2)证明：∵△DAE≌△ABF，∴∠ADE＝∠BAF.
∵∠BAF＋∠DAF＝∠DAB＝90°，
∴∠ADE＋∠DAF＝90°，
∴∠DOA＝90°，∴AF⊥DE.
(3)
特训10 对角互补模型
1．1
2．证明：如图，过点B分别作BF⊥DC于点F，BE⊥DA交DA的延长线于点E，
INCLUDEPICTURE"SKD-11.tif" INCLUDEPICTURE "C:\\Users\\庞建宇\\Desktop\\中考数学福建\\SKD-11.tif" \* MERGEFORMATINET
则∠BEA＝∠BFC＝90°.
∵DB平分∠ADC，∴BE＝BF.
∵∠BAD＋∠BCD＝180°，∠BAD＋∠BAE＝180°，
∴∠BCF＝∠BAE.
在△BEA和△BFC中，
∴△BEA≌△BFC，∴AB＝CB.
∵∠ADB＝60°，DB平分∠ADC，
∴∠ADC＝2∠ADB＝120°.
∵∠BAD＋∠BCD＝180°，∴∠ADC＋∠ABC＝180°，
∴∠ABC＝180°－∠ADC＝60°，
∴△ABC是等边三角形．
特训11 设参导等角
1．证明：设∠B＝α.∵AB＝AC，∴∠B＝∠ACB＝α.
∵∠ACB是 △ACD的外角，且∠ADB＝45°，
∴∠CAD＝∠ACB－∠ADB＝α－45°.
∵CE⊥AB，∴∠CEB＝90°.∴∠BCE＝90°－∠B＝90°－α.
∴∠DCF＝∠BCE＝90°－α.
∵∠ADC是△CDF的外角，∴∠F＝∠ADC－∠DCF＝α－45°，
∴∠CAD＝∠F，∴AC＝FC.
2．证明：设∠CAE＝2α，∠ABC＝β，则∠AFD＝2β.
∵AE＝AC，∴∠AEC＝∠ACE＝90°－α.
∵∠AFD是△ACF的外角，
∴∠ACD＝∠AFD－∠CAE＝2β－2α.∴∠BCD＝∠ACE－∠ACD＝90°－α－(2β－2α)＝90°＋α－2β.
∵∠ADC是△BCD 的外角，
∴∠ADC＝∠ABC＋∠BCD＝β＋90°＋α－2β＝90°＋α－β，
∴∠CAD＝180°－∠ACD－∠ADC＝90°＋α－β.
∴∠ADC＝∠CAD，∴CD＝AC.
精品试卷·第 2 页 （共 2 页）
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