高中数学 1.3.2托勒密定理课件 北师大版选修41

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1、-1 1-*3.2托勒密定理精 品 数 学 课 件北 师 大 版-* *-*3 3.2 2托勒密定理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.理解并掌握托勒密定理.2.知道托勒密定理的推广.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

2、练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知识梳理知识梳理知识梳理MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航【做一做】 在圆内接四边形ABCD中,AB=2,BC=2,CD=2,DA=2,则ACBD=().A.8+4 B.8-4C.32D.不确定解析:ACBD=ABCD+ADBC=22+22=8+4.答案:AZHISHI SHULIZHISHI SHULIZHISHI SHU

3、LIZHISHI 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MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标

4、导航目标导航目标导航托勒密定理的逆命题剖析:逆命题:如果一个四边形的两对边乘积之和等于两条对角线的乘积,那么这个四边形的四个顶点共圆.这个逆命题成立,其原因如下:如图所示,在任意四边形ABCD中,连接AC,在四边形ABCD的内部取点E,使得1=2,3=4,则ABEACD,所以.则有BEAC=ABCD,.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SUITANGY

5、ANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN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)=ABCD+ADBC,很明

6、显BE+EDBD, 所以ABCD+ADBCACBD,当BE+ED=BD时,即点B,E,D共线,此时BD是对角线,则有ABCD+ADBC=ACBD,3=4,5=6,则在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四点共圆.即当ABCD+ADBC=ACBD时,A,B,C,D四点共圆.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

7、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MUBIAODAOHANGMUBIAODAOHANGMUBIA

8、ODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航题型一题型二证明:如图所示,连接EF,DF.AEF=CDF,EAF=DCF,AEFCDF,AFDC=AECF,AF(BC-BD)=CF(BE-BA),AFBC+ABCF=BECF+BDAF,在圆内接四边形ABCF中,有AFBC+ABCF=BFAC,BFAC=BECF+BDAF.又AC=BE=BD,BF=AF+CF.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 SHULIZ

9、HISHI 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MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHA

10、NGMUBIAODAOHANG目标导航目标导航目标导航目标导航题型一题型二【变式训练1】 在例1中,已知条件不变,利用圆内接四边形BDFE中的托勒密定理重新证明.证明:ACF=ABF,ABF=EDF,ACF=EDF,同理可证DEF=CAF,AFCEFD,.令=m,EF=mAF,DE=mAC,DF=mCF.在圆内接四边形BDFE中,有BFDE=BDEF+BEDF,BFmAC=BDmAF+BEmCF,BFAC=BDAF+BECF.又AC=BD=BE,BF=AF+CF.ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典

11、例透析MUBIAODAOHANG目标导航题型一题型二【例2】 若a,b,x,y是正实数,且a2+b2=1,x2+y2=1.求证:ax+by1.分析:构造圆内接四边形,借助托勒密定理来证明.证明:如图所示,RtACB和RtADB位于AB的两侧,且公共斜边AB=1.根据勾股定理,则可以设AC=b,BC=a,AD=x,BD=y.因为ACB+ADB=90+90=180,所以四边形ACBD内接于圆,且AB为直径,所以ACBD+BCAD=ABCD,且CDAB,ACBD+BCAD=ABCDAB2=12=1,所以ACBD+BCAD1,即ax+by1.ZHISHI SHULI知识梳理ZHONGNAN JVJIA

12、O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 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 TOUXIDI

13、ANLI 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【变式训练2】 解方程:3+2x.解:显然x6,如图所示.设O的直径AB=x,C,D是AB异侧圆周上的两点,且AC=4,AD=6,则BC=,BD=.连接CD,由托勒密定理,得6+4=xCD.与已知方程比较,得CD=2,所以cosCAD=,故CAD=60,连接DO,CO,则OD=OC,DOC=120,因此原方程的解为x=AB=2OD=2.ZHISHI SHULI知识梳理ZHO

14、NGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3ZHISHI 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 TOUXIDIAN

15、LI 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证明:连接PR,QR,如图所示.在圆内接四边形APRQ中,由托勒密定理,得APQR+AQPR=ARPQ.由圆周角定理的推论,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.又四边形ABCD为平行四边形

16、,所以BC=AD,所以APAB+AQAD=ARAC.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 3ZHISHI 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SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANS

17、UITANGYANLIAN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解:如图所示,取的中点C(C与P在线段AB的两侧),连接PC,BC,AC,则AC=BC.在圆内接四边形ACBP中,由托勒密定理知ACPB+BCPA=ABPC.因为AC=BC,所以AC(PB+PA)=PCAB,即PB+PA=PC,显然AB,AC均为定值,则为定值,所以要

18、使PA+PB取最大值,只需PC取最大值.因为A,B是定点,所以C也为定点,则当PC为直径时,PC取最大值,所以当P为的中点时,PA+PB取最大值.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 33利用托勒密定理证明“勾股定理”.证明:如图所示,在RtABC中,ABC=90,作以RtABC的斜边AC为一条对角线的矩形ABCD,显然ABCD是圆内接四边形.由托勒密定理,得ACBD=ABCD+ADBC,又四边形ABCD是矩形,则AC=BD,AB=CD,AD=BC,所以AC2=AB2+BC2.即直角三角形的两直角边的平方和等于斜边的平方.