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1、-1 1-2.4切割线定理精 品 数 学 课 件北 师 大 版-* *-2 2.4 4切割线定理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.理解并掌握切割线定理及推论.2.理解并掌握切割线定理的逆定理.3.能够熟练地应用切割线定理及推论解决相关问题.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123ZHISHI SHULI知识梳理ZHON
2、GNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航123ZHISHI 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123ZHISHI 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123ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练
3、DIANLI TOUXI典例透析MUBIAODAOHANG目标导航123ZHISHI 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123ZHISHI 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目标导航题型一
4、题型二题型三题型四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 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MUBIAODAOHANGMU
5、BIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航题型一题型二题型三题型四证明:如图所示,连接BC,BD.E为的中点,DBE=CBE.又AB是O的切线,ABC=CDB.ABC+CBE=DBE+CDB.又ABF=ABC+CBE,AFB=DBE+CDB,ABF=AFB.AB=AF.又AB是O的切线,ACD为割线,由切割线定理可知AB2=ACAD,AF2=ACAD.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
6、二题型三题型四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 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MUBIAODAOHANGMUBI
7、AODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航题型一题型二题型三题型四证明:PA与圆相切于A,MA2=MBMC.M为PA的中点,PM=MA,PM2=MBMC,.BMP=PMC,BMPPMC,MCP=MPB.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题型三题型四ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MU
8、BIAODAOHANG目标导航题型一题型二题型三题型四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题型三题型四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 JVJIAO重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANG
9、YANLIANSUITANGYANLIAN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题型三题型四解:(1)设PC=x.CD=4cm,PD=PC+CD=(x+4)(cm).AB=3cm,PA=2cm,PB=AB+PA=5(cm).由切割线定理,得PE2=PAPB.PE2=25=10.PE=(cm).由切割线定理的推论,得PCPD=PA
10、PB.x(x+4)=25,化简整理得x2+4x-10=0,解得x=-2+或x=-2-(舍去).x=(-2)(cm),即PC=(-2)cm.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 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
11、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题型三题型四(2)由弦切角定理,得CEP=D,CPE=EPD,CPEEPD.PD=PC+CD=-2+4=(2+)(cm),.DE=)a(cm).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题型三题型四ZHISHI SHUL
12、IZHISHI 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 TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAO
13、HANGMUBIAODAOHANG目标导航目标导航目标导航目标导航题型一题型二题型三题型四解:PB=PA+AB=3+3=6,PAPB=36=18.又PC2=(3)2=18,PC2=PAPB,PC与O切于点C,PCA=ABC.又ABC=35,PCA=35.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题型三题型四ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAO
14、DAOHANG目标导航题型一题型二题型三题型四解析:PM是O1的切线,PM2=PBPA.又PM=PN,PN2=PBPA,PN与O2相切,PN与O2仅有1个公共点.答案:1ZHISHI 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题型三题型四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
15、 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题型三题型四错解:由题意得PAAB=PMMN,22=MN,MN=.故填.错因分析:错解混淆了割线定理中成比例的线段,应该是PAPB=P
16、MPN.正解:由题意得PAPB=PMPN,即PA(PA+AB)=PM(PM+MN),故2(2+2)=(+MN),解得MN=.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 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SUITANGYANLIA
17、N随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5解析:AB,AC分别与O相切于点B,C,ADE是O的割线,由切割线定理,得AB2=ADAE,故A不正确,D不正确;由ACDAEC,得CDAE=ACCE,故B不正确;由ACDAEC,得ADCE=ACCD,由ABDAEB,得ADBE=ABBD.又AB=AC,故BECD=BDCE.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 SHULIZHIS
18、HI 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:由于BE是O的切线,则CBE=BAC=70.由切割线定理知,EB
19、2=EDEC.又BE=2,CE=4,则ED=1,故CD=CE-ED=4-1=3.答案:703ZHISHI 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DIANLI TOUXIDIANLI
20、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:由题意得DC2=DBDA.DA=DB+BA=DB+3,(2)2=DB(DB+3),解得DB=4(负值舍去).D=D,BCD=BAC,BCDCAD.AC=.答案: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 5
21、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 JVJIAO重难聚焦重难聚焦重难聚焦重难聚焦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:如图所示,延长PO交O于点E,则PAPE=PBPC.设PC=x(x0),PB=BC,PB=x.又PE=PA+AE=PA+2AO=16,216=xx,解得x=8.又x0,x=8.PC=8.
