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1、-* *-2 2.4 4切割線定理ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航1.理解并掌握切割線定理及推論.2.理解并掌握切割線定理的逆定理.3.能夠熟練地應(yīng)用切割線定理及推論解決相關(guān)問題.ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航123ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂
2、演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航123ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航123ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航123ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)
3、導(dǎo)航123ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航123ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三題型四ZHISHI SHULIZHISHI SHU
4、LIZHISHI SHULIZHISHI SHULI知識(shí)梳理知識(shí)梳理知識(shí)梳理知識(shí)梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重難聚焦重難聚焦重難聚焦重難聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN隨堂演練隨堂演練隨堂演練隨堂演練DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAOD
5、AOHANG目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航題型一題型二題型三題型四證明:如圖所示,連接BC,BD.E為的中點(diǎn),DBE=CBE.又AB是O的切線,ABC=CDB.ABC+CBE=DBE+CDB.又ABF=ABC+CBE,AFB=DBE+CDB,ABF=AFB.AB=AF.又AB是O的切線,ACD為割線,由切割線定理可知AB2=ACAD,AF2=ACAD. ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三題型四ZHISHI SHULIZHISHI SHUL
6、IZHISHI SHULIZHISHI SHULI知識(shí)梳理知識(shí)梳理知識(shí)梳理知識(shí)梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重難聚焦重難聚焦重難聚焦重難聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN隨堂演練隨堂演練隨堂演練隨堂演練DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODA
7、OHANG目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航題型一題型二題型三題型四證明:PA與圓相切于A,MA2=MBMC.M為PA的中點(diǎn),PM=MA,PM2=MBMC,.BMP=PMC,BMPPMC,MCP=MPB. ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三題型四ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三題型四ZH
8、ISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三題型四ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知識(shí)梳理知識(shí)梳理知識(shí)梳理知識(shí)梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重難聚焦重難聚焦重難聚焦重難聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN隨堂演練隨堂演練
9、隨堂演練隨堂演練DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航題型一題型二題型三題型四解:(1)設(shè)PC=x.CD=4cm,PD=PC+CD=(x+4)(cm).AB=3cm,PA=2cm,PB=AB+PA=5(cm).由切割線定理,得PE2=PAPB.PE2=25=10.PE=(cm).由切割線定理的推論,得PCPD=PAPB.x(x+4)=25,化簡(jiǎn)整理得x2+4x-10=0,
10、解得x=-2+或x=-2-(舍去).x=(-2)(cm),即PC=(-2)cm. ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知識(shí)梳理知識(shí)梳理知識(shí)梳理知識(shí)梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重難聚焦重難聚焦重難聚焦重難聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN隨堂演練隨堂演練隨堂演練隨堂演練DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUX
11、I典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航題型一題型二題型三題型四(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知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三題型四ZHISHI SHULIZHISHI SHULIZHISHI SHULIZH
12、ISHI SHULI知識(shí)梳理知識(shí)梳理知識(shí)梳理知識(shí)梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重難聚焦重難聚焦重難聚焦重難聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN隨堂演練隨堂演練隨堂演練隨堂演練DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)
13、導(dǎo)航目標(biāo)導(dǎo)航題型一題型二題型三題型四解:PB=PA+AB=3+3=6,PAPB=36=18.又PC2=(3)2=18,PC2=PAPB,PC與O切于點(diǎn)C,PCA=ABC.又ABC=35,PCA=35. ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三題型四ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三題型四解析:
14、PM是O1的切線,PM2=PBPA.又PM=PN,PN2=PBPA,PN與O2相切,PN與O2僅有1個(gè)公共點(diǎn).答案:1ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三題型四ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知識(shí)梳理知識(shí)梳理知識(shí)梳理知識(shí)梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重難聚焦重難聚焦重難聚焦重難聚焦SUI
15、TANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN隨堂演練隨堂演練隨堂演練隨堂演練DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航題型一題型二題型三題型四錯(cuò)解:由題意得PAAB=PMMN,22=MN,MN=.故填.錯(cuò)因分析:錯(cuò)解混淆了割線定理中成比例的線段,應(yīng)該是PAPB=PMPN.正解:由題意得PAPB=PMPN,即PA(P
16、A+AB)=PM(PM+MN),故2(2+2)=(+MN),解得MN=. ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航1 2 3 4 5ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航1 2 3 4 5ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBI
17、AODAOHANG目標(biāo)導(dǎo)航1 2 3 4 5解析:AB,AC分別與O相切于點(diǎn)B,C,ADE是O的割線,由切割線定理,得AB2=ADAE,故A不正確,D不正確;由ACDAEC,得CDAE=ACCE,故B不正確;由ACDAEC,得ADCE=ACCD,由ABDAEB,得ADBE=ABBD.又AB=AC,故BECD=BDCE.答案:CZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航1 2 3 4 5ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISH
18、I SHULI知識(shí)梳理知識(shí)梳理知識(shí)梳理知識(shí)梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重難聚焦重難聚焦重難聚焦重難聚焦DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航1 2 3 4 5解析:由于BE是O的切線,則CBE=BAC=70.由切割線定理知,EB2=EDEC.又BE=2,CE=4,則ED=1,故
19、CD=CE-ED=4-1=3.答案:703 ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航1 2 3 4 5ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知識(shí)梳理知識(shí)梳理知識(shí)梳理知識(shí)梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重難聚焦重難聚焦重難聚焦重難聚焦DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典
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