新編高中數(shù)學(xué) 1.3.2托勒密定理課件 北師大版選修41

1、-1 1-*3.2托勒密定理北 師 大 版 數(shù) 學(xué) 課 件精 品 資 料 整 理 -* *-*3 3.2 2托勒密定理zhishi shuli知識梳理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.理解并掌握托勒密定理.2.知道托勒密定理的推廣.zhishi shuli知識梳理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知識梳理zhongnan jvjiao重難聚焦suitangy

2、anlian隨堂演練dianli touxi典例透析mubiaodaohang目標(biāo)導(dǎo)航zhishi shulizhishi shulizhishi shulizhishi shuli知識梳理知識梳理知識梳理知識梳理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【做一做】 在圓內(nèi)接四邊形abcd中,ab=2,bc=2,cd=2,da=2,則acbd=().a.8+4 b.8-4c.32d.不確定解析:acbd=abcd+adbc=22+22=8+4.答案:azhishi shulizhishi shuliz

3、hishi shulizhishi shuli知識梳理知識梳理知識梳理知識梳理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mubiaodaohangmubiaodaohangmubiaodaohangmubiaodaoh

4、ang目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航托勒密定理的逆命題剖析:逆命題:如果一個四邊形的兩對邊乘積之和等于兩條對角線的乘積,那么這個四邊形的四個頂點(diǎn)共圓.這個逆命題成立,其原因如下:如圖所示,在任意四邊形abcd中,連接ac,在四邊形abcd的內(nèi)部取點(diǎn)e,使得1=2,3=4,則abeacd,所以.則有beac=abcd,.zhishi shulizhishi shulizhishi shulizhishi shuli知識梳理知識梳理知識梳理知識梳理zhongnan jvjiaozhongnan jvjiaozhongnan jvjiaozhongnan jvjiao重難聚焦重難聚焦重難聚焦重難聚

5、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)導(dǎo)航目標(biāo)導(dǎo)航又bac=1+eac,dae=2+eac,所以bac=dae,所以abcaed,所以,即edac=bcad,且5=6,所以beac+edac=abcd+bcad,即ac(be+ed)=abc

6、d+adbc,很明顯be+edbd, 所以abcd+adbcacbd,當(dāng)be+ed=bd時,即點(diǎn)b,e,d共線,此時bd是對角線,則有abcd+adbc=acbd,3=4,5=6,則在abd中,1+2+eac+3+6=180,所以1+2+eac+4+5=180.又bad=1+2+eac,bcd=4+5,所以bad+bcd=180,所以a,b,c,d四點(diǎn)共圓.即當(dāng)abcd+adbc=acbd時,a,b,c,d四點(diǎn)共圓.zhishi shuli知識梳理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

7、型二zhishi shulizhishi shulizhishi shulizhishi shuli知識梳理知識梳理知識梳理知識梳理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mubiaodaohangmubiaodao

8、hangmubiaodaohangmubiaodaohang目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航題型一題型二證明:如圖所示,連接ef,df.aef=cdf,eaf=dcf,aefcdf,afdc=aecf,af(bc-bd)=cf(be-ba),afbc+abcf=becf+bdaf,在圓內(nèi)接四邊形abcf中,有afbc+abcf=bfac,bfac=becf+bdaf.又ac=be=bd,bf=af+cf.zhishi shuli知識梳理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zhis

9、hi shulizhishi shulizhishi shulizhishi shuli知識梳理知識梳理知識梳理知識梳理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mubiaodaohangmubiaodaohangmu

10、biaodaohangmubiaodaohang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【變式訓(xùn)練1】 在例1中,已知條件不變,利用圓內(nèi)接四邊形bdfe中的托勒密定理重新證明.證明:acf=abf,abf=edf,acf=edf,同理可證def=caf,afcefd,.令=m,ef=maf,de=mac,df=mcf.在圓內(nèi)接四邊形bdfe中,有bfde=bdef+bedf,bfmac=bdmaf+bemcf,bfac=bdaf+becf.又ac=bd=be,bf=af+cf.zhishi shuli知識梳理zhongnan jvjiao重難聚焦suitangyanlian隨堂演練dian

11、li touxi典例透析mubiaodaohang目標(biāo)導(dǎo)航題型一題型二【例2】 若a,b,x,y是正實數(shù),且a2+b2=1,x2+y2=1.求證:ax+by1.分析:構(gòu)造圓內(nèi)接四邊形,借助托勒密定理來證明.證明:如圖所示,rtacb和rtadb位于ab的兩側(cè),且公共斜邊ab=1.根據(jù)勾股定理,則可以設(shè)ac=b,bc=a,ad=x,bd=y.因為acb+adb=90+90=180,所以四邊形acbd內(nèi)接于圓,且ab為直徑,所以acbd+bcad=abcd,且cdab,acbd+bcad=abcdab2=12=1,所以acbd+bcad1,即ax+by1.zhishi shuli知識梳理zhong

12、nan 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zhishi shulizhishi shulizhishi shulizhishi shuli知識梳理知識梳理知識梳理知識梳理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 touxidianl

13、i 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【變式訓(xùn)練2】 解方程:3+2x.解:顯然x6,如圖所示.設(shè)o的直徑ab=x,c,d是ab異側(cè)圓周上的兩點(diǎn),且ac=4,ad=6,則bc=,bd=.連接cd,由托勒密定理,得6+4=xcd.與已知方程比較,得cd=2,所以coscad=,故cad=60,連接do,co,則od=oc,doc=120,因此原方程的解為x=ab=2od=2.zhishi shu

14、li知識梳理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 3zhishi shulizhishi shulizhishi shulizhishi shuli知識梳理知識梳理知識梳理知識梳理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

15、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證明:連接pr,qr,如圖所示.在圓內(nèi)接四邊形aprq中,由托勒密定理,得apqr+aqpr=arpq.由圓周角定理的推論,得1=2,3=4.又adbc,所以4=5,所以3=5,所以pqrcab,所以.令=k,所以qr=kab,pr=kbc,pq=kac,所以apkab+aqkbc=arkac,所以apab+aqbc=arac.又四邊形a

16、bcd為平行四邊形,所以bc=ad,所以apab+aqad=arac.zhishi shuli知識梳理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 3zhishi shulizhishi shulizhishi shulizhishi shuli知識梳理知識梳理知識梳理知識梳理zhongnan jvjiaozhongnan jvjiaozhongnan jvjiaozhongnan jvjiao重難聚焦重難聚焦重難聚焦重難聚焦suitangyanliansuitangyanliansuitan

17、gyanliansuitangyanlian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1 2 3解:如圖所示,取的中點(diǎn)c(c與p在線段ab的兩側(cè)),連接pc,bc,ac,則ac=bc.在圓內(nèi)接四邊形acbp中,由托勒密定理知acpb+bcpa=abpc.因為ac=bc,所以ac(pb+pa)=pcab,即pb+pa=pc,顯然ab,ac均為定值,則為定值,所以要使pa+pb取最大值,只需pc取最大值.因為a,b是定點(diǎn),所以c也為定點(diǎn),則當(dāng)pc為直徑時,pc取最大值,所以當(dāng)p為的中點(diǎn)時,