中考必会几何模型：8字模型

1、8字模型【结论】如图，AC与BD相交于点O,则/A+/B=ZC+/D.【证明】在ABDK/A+/B+/AOB=180.在ACDOK/C+/D+/COD=180./AOB=COD,/A+/B=/C+/D.拓展模型1i若BP,DP分别是/ABC/ADC勺平分线，则/P1(/A+/C).拓展模型2若/CB*/ABC，/CD*/ADC则/P=yA+；/C.名师小技巧看到相交线，就联想到8字模型，利用结论进行解题,8字模型是经典的2换2模型，已知两角之和，可知另两角之和.经典例题典例1如图，/C=ZD=90,/A=20,则/COA=/B=典例2如图，AABCAADE且/CAD=1O,/B=/D=25,/

2、EAB=120,求/DFBR!/DGB勺度数.AB典例3(1)如图，五角星的顶点为A,B,C,D,E,则/A+/B+/C+/D+/E的度数为().A.90AD.360(2)如图，是由图中点A向下移动到BE上得到的，/CAD廿B+ZE+ZC+ZD五角之和有无变化？证明你的结论;图(3)如图，是由中点C移动到BD上得到的，/CAD廿B+/ACE+/D+ZE五角之和有无变化？证明你的结论.D典例4如图，/A+/B+/C+/D+/E+/F=初露锋芒ZA.40B.502.如图是由线段AB,CDDF,E则/A+/B+/C+/F的度数为(C.60D.703F,CA组成的平面图形，/D=28,)1.如图，已知

3、AE?BD,A（XCD./A=40,则/D的度数为（）CA.62B.1523.如图所示，/1=60,贝U/二BA.240B.280一714C.208D.236A+/B+/C+/D+/E+/F=()CC.360D.54044 .如图，求：/5 .已知：如图，/A+/B+/C+/ADB1.（2020浙江绍）A.10A+ZB+ZC+/D+/E+/F+ZG+/H=Ik.f/ZrGCl/lIJJ、/Tj/Stajr露riBE与CF相交于点G.求证：D+/E+/F=360.直击中考工中考模拟）如图所示，/a的度数是（）.B.20C.30D.40答案典例1:【答案】700;20【解析】./C=90,/A=2

4、0,/COA=180-/C-/A=180-90-20=70.由8字模型的结论，知/A+/C=/B+/D.又/C=/D,.B=/A=20.典例2:【答案】900;65i.i【解析】ABC2AADIEDAEhBAC=-(ZEAB-ZCAD)=-X(120-10)=55./DFB=/FAB吆B=/FAC吆CAB廿B=10+55+25=90,./DGB=ZDFB-/D=90-25=65.典例3:(1)【答案】B/【解析】如图，连接CD.;,由8字模型的结论，知/B+/E=/1+/2.岳透心/A+/ACD+ADC=180,/ACDN1+/3,/ADCN2+/4./A+/B+/ACE+ADB+E=180.

5、故选B.(2)【答案】无变化.【解析】根据平角的定义，得出/BACVCAD廿DAE=180./BAChC+/E,/EAD=B+/D,./CAD+B+/C+/D+/E=/BAC+CAD+DAE=180;(3)【答案】无变化.【解析】ACB=CAD+D,/ECD=B+ZE,/CAD+B+ZACE+D+ZE=ZACB+ACE+ECD=180.典例4:【答案】360【解析】/A+/B=/1+/2.（角的8字模型）/A+/B+/C+/ADC+FEB吆F=/1+Z2+ZC+/ADC+FEB吆F=360.（四边形内角和360）初露锋芒1 .【答案】A【解析】由8字模型的结论，知/A+/B=/C+/D.ABL

6、BD.ACLCD.：/B=/C=90,./D=/A=40.故选A.2 .【答案】C【解析】由8字模型的结论知，/A+/C=/D+/DEG/B+/F=/D吆DGE/A+/B+C吆F=/D+/DEG+DGE+D./D=28,/D+/DEG+DGE=180./A+/B+/C+/F=180+28=208.故选C.3 .【答案】A【解析】如图，连接EF.由8字模型的结论知，/A+/D=Z2+/3,/B+/C=/CFE吆BEF,./A+ZD+ZBEA吆CFD=2+/3+/BEA+CFD=CFE吆BEF.vZ1=60,/CFE吆BEF=180-Z1=180-60=120./A+ZB+/C+/D+ZBEA吆CFD=22CFE吆BEF）=2X120=2404 .【答案】360.【解析】如图，连接GHCD./E+/B=/1+/2.（8字模型）/A+/F=/3+/4.（8字模型）/A+/B+/FCH+ADG+E+/F+/DGB+/EHC=1+/2+/3+/4+/GDA+FCH+DGB+EHC=360.（四边形内角和360）5 .【答案】360.【解析】如图，连接BC./1=/E+/F,/1=/GBC+GCB,8 /E+/F=/GBC+GCB, /A+ZB+ZC+ZD+ZE+ZF=/A+ZABC+B