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1、-1 1-2.5相交弦定理精 品 数 学 课 件2019 届 北 师 大 版 -* *-2 2.5 5相交弦定理ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1.能结合具体图形,准确表述相交弦定理.2.掌握相交弦定理,并能利用相交弦定理进行有关的证明和计算.ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航ZHISHI SHULI知识梳理ZHONGN
2、AN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航与圆有关的比例线段问题剖析:与圆有关的比例线段问题,主要是圆与相似形的综合,其解法大致可分以下几种:(1)直接由相似形得到,即先
3、由已知条件证得两个三角形相似,从而直接得到有关对应线段成比例.这是简单型的比例线段问题.(2)利用“等线段”代换得到,在证明形如a2=bc的“等积式”时,如果其中有三条线段共线,那么一般往往把平方项线段用“等线段”进行代换.(3)利用“中间积”代换得到,在证明形如a2=bc的“等积式”时,如果其中有三条线段共线,不妨可以把平方项线段利用中间积进行代换试试.(4)利用“中间比”代换得到,在证明比例线段时(不论共线与否),如果不能直接运用有关定理,不妨寻找“中间比”进行代换试试.ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI
4、TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知识梳理知识梳理知识梳理ZHONGNAN JVJIAOZHONG
5、NAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN随堂演练随堂演练随堂演练随堂演练DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航题型一题型二解析:OA=2,OA=OM,OM=2.AM=OA-OM=2-2=2(
6、-1),CM=OC+OM=2+2=2(+1).BNA=45,BOA=2BNA=90.BM=4.ACBN=M,MNBM=CMAM.4MN=2(+1)2(-1).MN=2.答案:2ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知识梳理知识梳理知识梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JV
7、JIAO重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN随堂演练随堂演练随堂演练随堂演练DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航题型一题型二证明:AD平分BAC,BAE=DAC.又ACB=AEB,ABEADC,ABAC=ADAE.又AE=AD+DE,ADAE=AD(AD+DE)=
8、AD2+ADDE,ABAC=AD2+ADDE.又AE和BC是O的相交弦,ADDE=BDDC,ABAC=AD2+BDDC,AD2=ABAC-BDDC.ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANG
9、YANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二证明证明:AB与与CD是是O1的相交弦的相交弦,AGBG=CGDG.同理在同理在O2中中,AGBG=EGFG.CGDG=EGFG,CGFG=EGDG,CGFG-CG=EGDG-EG.CGCF=EGED,CGED=EGCF.ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI S
10、HULI知识梳理知识梳理知识梳理知识梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN随堂演练随堂演练随堂演练随堂演练DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航
11、1 2 3 4 5解析:AM和BM是3x2+2mx+18=0的两根,AMBM=6.又AB和CD相交于点M,CMMD=AMBM=6.CDCD=6.CD=3.答案:CZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知识梳理知识梳理知识梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO
12、重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN随堂演练随堂演练随堂演练随堂演练DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航1 2 3 4 5解析:在O中,OF=2,CF=4,OC=OD=6.AFFB=CFFD=4(2+6)=32.在O中,EFOF=AFFB=32.EF=16,即EF
13、=EC+CF=EC+4=16.EC=12.答案:AZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知识梳理知识梳理知识梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLI
14、ANSUITANGYANLIAN随堂演练随堂演练随堂演练随堂演练DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航1 2 3 4 5解析:由相交弦定理得AFFB=EFFC,FC=2.由AFCABD,可知,BD=.由切割线定理得DB2=DCDA.又由得DA=4CD,4DC2=DB2=,DC=(负值舍去).答案:ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUI
15、TANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知识梳理知识梳理知识梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN随堂演练随堂演练随堂演练随堂演练DIANLI TOUXIDIANLI TOUXIDIANLI
16、TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航1 2 3 4 5解析:因为AB和CD是O的两条相交弦,所以AEEB=CEED.即2EB=34.所以EB=6.所以AB=AE+EB=2+6=8.所以OB=AB=4.因为MD为O的切线,D为切点,所以MD2=MBMA=MB(MB+AB),所以42=MB(MB+8),解得MB=-44.又MB0,所以MB=4-4.答案:44-4ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5
