【精品】几何必会模型：手拉手模型带答案

1、【精品文档，百度专属】 本文为word版资料，可以任意编辑修 本文为word版资料，可以任意编辑修手拉手模型模型手拉手EBC图BC图BC图如图， ABC是等腰三角形、 ADE是等腰三角形，AB = AC, AD = AE，/ BAC = Z DAE =:.结论：连接 BD、CE，则有 BADCAE .模型分析如图，/ BAD = Z BAC-Z DAC，/ CAE =Z DAE-Z DAC ./ BAC =Z DAE =:-，Z BAD = Z CAE .在厶BAD和厶CAE中，AB = AC ,:_BAD = CAE ,IAD =AE ,图、图同理可证.(1) 这个图形是由两个共顶点且顶角相

2、等的等腰三角形构成.在相对位置变化的同时， 始终存在一对全等三角形.(2) 如果把小等腰三角形的腰长看作小手，大等腰三角形的腰长看作大手，两个等腰 三角形有公共顶点，类似大手拉着小手，所以把这个模型称为手拉手模型.(3) 手拉手模型常和旋转结合，在考试中作为几何综合题目出现.模型实例例1如图， ADC与厶EDG都为等腰直角三角形， 连接AG、CE,相交于点H，问:E/ GDE + Z CDG , / ADC = Z EDG = 90°,(1) AG与CE是否相等？(2) AG与CE之间的夹角为多少度?解答:(1) AG = CE.理由如下：/ ADG = Z ADC + Z CDG

3、, / CDE/ ADG = Z CDE .在厶ADG和厶CDE中,AD =CD ,；._ADG = CDE ,DG 二DE , ADE CDE . AG = CE .(2)v ADG CDE , / DAG = Z DCE ./ COH = Z AOD ,/ CHA = Z ADC = 90°. AG与CE之间的夹角是 90°.ABD和厶BCE,A ABD和厶BCE都是等边三角例2如图，在直线AB的同一侧作 形，连接AE、CD，二者交点为 H .求证：(1 ) ABE DBC ;(2) AE = DQ ;(3) / DHA = 60°(4) A AGB DFB

4、;(5) A EGB CFB ;(6) 连接 GF , GF / AC ;(7)连接 HB, HB 平分/ AHC .证明：(1 )Z ABE = 120° / CBD = 120° 在厶ABE和厶DBC中，BA 二 BD ,/ABE ZDBC ,BE = BC , ABE DBC .(2) tA ABE DBC , AE= DC.(3) A ABE DBC,/ 1 = Z 2./ DGH =Z AGB ./ DHA =Z 4= 60°(4) vZ 5= 180° / 4-Z CBE= 60° 4=/ 5./ ABE也厶 DBC , / 1 =

5、 / 2.又 AB = DB, AGB DFB (ASA).(5) 同(4)可证 EGB也厶 CFB ( ASA ).图(6) 如图所示，连接 GF .由(4)得， AGB DFB . BG= BF .又/ 5 = 60° BGF是等边三角形. / 3= 60° / 3=/ 4 . GF / AC .(7) 如图所示，过点 B作BM丄DC于M，过点B作BN丄AE于点NA图C/ ABE DBC , SABE = Sa DBC .11丄 X AE X BN=丄 X CD X BM .22/ AE= CD, BM = BN .点B在/ AHC的平分线上. HB 平分/ AHC .

6、练习:1. 在厶ABC中，AB= CB,/ ABC = 90 ° F为AB延长线上一点，点 E在BC上，且 AE=CF .(1) 求证：BE= BF ;(2) 若/ CAE = 30° 求/ ACF 度数.答案：(1) 证明：/ ABC = 90° .在 Rt ABE 和 Rt CBF 中，CF = AE ,AB =CB , Rt ABE 也 Rt CBF (HL ). BE= BF.(2) v AB = CB,/ ABC = 90°/ BAC =/ BCA = 45° / CAE = 30° / BAE = 45° 30&

7、#176; = 15° ./ Rt ABE也 Rt CBF , / BCF = / BAE = 15° . / ACF = / BCF + / BCA = 15° + 45° = 60° .2. 如图， ABD与厶BCE都为等边三角形，连接求证：(1) AE= DC ;(2) / AHD = 60°(3) 连接 HB, HB 平分/ AHC .答案：AE与CD，延长AE交CD于点H .(1)v/ ABE =/ ABD / EBD，/ DBC =/ EBC/ EBD，/ ABD = / EBC = 60° / ABE =/ D

8、BC .在厶ABE和厶DBC中,AB 二 DB , 上ABE ZDBC ,BE =BC , ABE DBC . AE= DC.(2)v ABE DBC ,/ EAB =Z CDB .又/ OAB + Z OBA = Z ODH +Z OHD ,/ AHD =Z ABD = 60°(3)过B作AH、DC的垂线，垂足分别为点 M、N .ABE DBC , SaaBE = Sa dbc .11即AE BM = - CD BN.22又 AE = CD, BM = BN . HB 平分/ AHC .3. 在线段AE同侧作等边 ABC和等边 CDE (/ ACEv 120 ° ,点P与

9、点M分别是线D段BE和AD的中点.求证： CPM是等边三角形.答案：证明： ABC和厶CDE都是等边三角形, AC = BC, CD = CE . / ACB =Z ECD = 60°. / BCE =Z ACD . BCE ACD . / CBE =Z CAD , BE = AD .又点P与点M分别是线段 BE和AD的中点, BP= AM .在厶BCP和厶ACM中,BC =AC ,；_CBE = CAD ,BP 二 AM , BCP ACM . PC = MC,Z BCP =Z ACM . / PCM =Z ACB = 60° . CPM是等边三角形.4. 将等腰Rt A

10、BC和等腰Rt ADE按图方式放置，/ A= 90° AD边与AB边重合，AB= 2AD = 4.将厶ADE绕A点逆时针方向旋转一个角度 « ( 0 °v « V 180 °, BD的延长线交CE于P.(1) 如图，求明： BD = CE, BD丄CE;(2) 如图，在旋转的过程中，当AD丄BD时，求CP长.图E图答案：(1)v 等腰 Rt ABC 和等腰 Rt ADE , AB= AC, AD = AE,Z BAC =Z DAE = 90° / DAB = 90° / CAD，/ CAE= 90° / CAD ,

11、 / DAB = / CAE . ABD 也厶 ACE . BD = CE. / DBA = / ECA. / CPB =/ CAB . ( 8 字模型) BD丄CE.(2)由(1)得 BP 丄 CE .又 AD 丄 BD, / DAE = 90° AD = AE,四边形ADPE为正方形.AD = PE= 2. / ADB = 90° AD = 2, AB= 4, BD = CE= 2、3 . CP= CE PE= 2 3 -2 .Baidubaidu badiubaidubaidubaidu baidubaidubadiu baidubaidubadiu baidubai

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