6平面几何托勒迷定理二解答

6平面几何托勒迷定理二

例1:圆O是?ABC的外接圆,I是?ABC的内心,射线AI交圆O于D,求证AB,BC,CA成等差数列的 充要条件是S?IBC?S?DBC

A13证明:由?BID??5??1??2??3??2??4??2??DBI知DI?BD?DC必要性:若AB,BC,CA成等差数列,即AB?AC?2BC而?IBA,?ICA,?IBC有相等的高,则S?IAB?S?IAC?2S?IBC又由托勒迷定理,有AB?DC?AC?BD?AD?BCADAB?AC即(AB?AC)?DI?AD?BC,??2,即I是AD的中点,于是DIBCS?AIB?S?IBD,S?IAC?S?ICD,2S?IBC?S?IAB?S?IAC?SBDCI?S?IBC?S?BDC故S?IBC?S?DBC1111充分性:若S?IBC?S?DBC即IB?BC?sin?B?DB?BC?sin?A有2222?A?BIB:DB?sin:sin比较上述得IA?BD,但DI?DB即知AD?2DI,22AB?ACAD由托勒迷定理知??2即AB?AC?2BC故AB,BC,CA为等差数列BCDI2OB

24I5CDA13O2I5

B24CDIA2IB2IC2例2.如图设I为?ABC的内心,角A,B,C所对的边长分别为a,b,c求证???1

bcacab

证明:设I在三边上的射影分别为D,E,F设?ABC的外接圆半径及内切圆半径分别为R,r,则ID?IE?IF?r由B,D,I,F四点共圆,且IB为其圆的直径,应用托勒迷定理,有DF?IB?ID?BF?IF?BD?r(BD?BF)b?IB即有b?IB2?2Rr(BD?BF)2R同理有a?IB2?2Rr(AF?AE),c?IC2?2Rr(CD?CE),从而a?IA2?b?IB2?c?IC2?2Rr(a?b?c)由正弦定理,有DF?IB?sin?B?又由S?ABC?1abcr(a?b?c)?有2Rr(a?b?c)?abc故24R2IA2IB2IC2222a?IA?b?IB?c?IC?abc即???1bcacab

AIBC

AFIBD1

C6平面几何托勒迷定理二

例3.如图若?ABC与?A'B'C'的边长分别为a,b,c与a',b'c'且?B??B',?A??A'?1800则aa'?bb'?cc'

BAcbCA'aBCB'C' b D

证明:作?ABC的外接圆,过C作CD//AB交圆于D,连AD，BD因?A??A'?1800??A??D,?BCD??B??B'则?A'??D,?B'??BCD从而?A'B'C'??DCB,有A'B'DC?B'C'CB?A'C'DB即c'a'b'ac'ab' DC?a?DB故DC?a',DB?a'又AB//CD知BD?AC?b,AD?BC?a,由托勒迷定理得AD?BC?AB?DC?AC?BD即a2?c?ac'ab'a'?b?a'故aa'?bb'?cc'

例4.如图在?ABC中,B1,C1分别是AB,AC延长线上的点,D1为B1C1的中点,连AD1交ABC外接圆于D,求证AB?AB1?AC?AC1?2AD?AD

1

证明:连BD,CD设?ABD??,?CAD??,?ABC外接圆的半径为R,因为DBS11为1C1的中点,知?ABD1?S?AC1D1?2S?AB1C1在?BCD中,由正弦定理有BD?2Rsin?,CD?2Rsin?,BC?2Rsin(???)在圆内接四边形ABCD中,由托勒密定理得,AB?CD?AC?BD?AD?BC即AB?2R?sin??AC?2R?sin??AD?2Rsin(???)

两边同乘以14RAB1?AC1?AD1得AB?AB1?S?AC1D1?AC?AC1?S?AB1D1?AD?AD1?S?AB1C1即AB?AB1?AC?AC1?2AD?AD1

ABDCB1D1C1ABDCB1D1C12

6平面几何托勒迷定理二

例5:如图,设M,N是?ABC内部的两个点,且满足?MAB??NAC,?MBA??NBC AM?ANBM?BNCM?CN 证明???1AB?ACBA?BCCA?CB

证明设.K是射线BN上的点,且满足?BCK??BMA.因?BMA??ACB,则K在?ABC的外部,ABBMAM又?MBA??CBK,则?ABM??KBC,即有??BKBCCKABBMAM由?ABK??KBC即有??BKBCCKABBMABBKAK由?ABK??MBC,?知?ABK??MBC于是??KBBCBMBCCMA由?CKN??MAB??NAC知A,N,C,K四点共圆,应用托勒密定理 有AC?NK?AN?KC?CN?AK或AC?(BK?BN)?AN?KC?CN?AK将AM?BCBK?CNAB?BCKC?,AK?,BK?代入得BMBMBMAB?BCAN?AM?BCCN?BK?CMAC?(?BN)??BMBMBMAM?ANBM?BNCM?CN即???1AB?ACBA?BCCA?CB

BNMCBNAMKC例6.在?ABC中,AB?AC,线段AB上有一点D,线段AC延长线上有一点E,使得DE?AC.线段DE与?ABC的外接圆交于T,P是线段AT延长线上的一点,若P在?ADE的外接 圆上,求证PD?PE?AT证明:连BT,CT,由A,B,T,C;A,D,P,E 分别四点共圆,知?CBT??CAT??EDP,?BCT??BAT??DEP,于是DPPEDE?BTC∽?DPE,可设???kBTCTBC

BDA

CTPE对四边形ABTC应用托勒密定理,有AC?BT?AB?CT?BC?AT注意到AB?AC?DE,即得PD?PE?AT

ADCBTPE3

6平面几何托勒迷定理二

7.已知AB为O1的直径,圆周上的点C,D分别在AB的两侧,过CD中点M分别作AC,AD的垂线,垂足为P,Q,求证BC?MP?BD?MQ?2MC2

解.对四边形ABCD应用托勒密理.BC?AD?BD?AC?AB?CD

即BC?AD?BD?AC?CD,又?ABD∽?MCP及?ABC∽?MDQABAB

有AD?MP,AC?MQ,于是BC?MP?BD?MQ?CDABMCABMDMCMD

注意到CD?2MC?2MD

PAMCBQD8.已知平行四边形ABCD中,过B的圆分别交AB,BC,BD于E,F,G,求证BE?AB?BF?BC?BG?BD

连EG,FG和EF对四边形BFGE应用托勒密定理,有BE?FG?BF?EG?BG?EF又?FEG??FBG??ADB,?EFG??EBG则?EFG∽?ABD,有 FGEGEF??,令其比值为t,ABADBD则t?BE?AB?t?BF?AD?t?BG?BDDCGFAEB消去t,消去t,注意AD?BC即证

9.设AF为O1与O2的公共弦,点B,C分别在O1,O2上,且AB?AC,?BAF,?CAF的平分线交O1,O2于点D,E,求证DE?AF解:作DG//AF交O1与G，则AG?FD,GF?AD

对四边形AGDF应用托勒密定理AD?FG?AG?FD?AF?GD由AD平分?BAF,知FD?BD,即AG?BD,由此知GB//DA,有GD?AB故AD2?FD2?AF?GD?FD2?AF?AB同理,有AE2?FE2?AF?AC此两式相减有DA2?EA2?DF2?EF2故DE?AF

DFEB AC4

6平面几何托勒迷定理二

5

本文来源：https://www.bwwdw.com/article/zmdp.html