如图4,等边三角形ABC中,P,Q各为AB,AC的中点,D为PQ上一点,直线CD交AB于F,直线BD交AC于E.求证:1/CE+1/BF=3/BC相似的题目.用相似做.
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![如图4,等边三角形ABC中,P,Q各为AB,AC的中点,D为PQ上一点,直线CD交AB于F,直线BD交AC于E.求证:1/CE+1/BF=3/BC相似的题目.用相似做.](/uploads/image/z/7202443-67-3.jpg?t=%E5%A6%82%E5%9B%BE4%2C%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CP%2CQ%E5%90%84%E4%B8%BAAB%2CAC%E7%9A%84%E4%B8%AD%E7%82%B9%2CD%E4%B8%BAPQ%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E7%9B%B4%E7%BA%BFCD%E4%BA%A4AB%E4%BA%8EF%2C%E7%9B%B4%E7%BA%BFBD%E4%BA%A4AC%E4%BA%8EE.%E6%B1%82%E8%AF%81%EF%BC%9A1%2FCE%2B1%2FBF%3D3%2FBC%E7%9B%B8%E4%BC%BC%E7%9A%84%E9%A2%98%E7%9B%AE.%E7%94%A8%E7%9B%B8%E4%BC%BC%E5%81%9A.)
如图4,等边三角形ABC中,P,Q各为AB,AC的中点,D为PQ上一点,直线CD交AB于F,直线BD交AC于E.求证:1/CE+1/BF=3/BC相似的题目.用相似做.
如图4,等边三角形ABC中,P,Q各为AB,AC的中点,D为PQ上一点,直线CD交AB于F,直线BD交AC于E.求证:1/CE+1/BF=
3/BC
相似的题目.用相似做.
如图4,等边三角形ABC中,P,Q各为AB,AC的中点,D为PQ上一点,直线CD交AB于F,直线BD交AC于E.求证:1/CE+1/BF=3/BC相似的题目.用相似做.
三角形EDQ相似于三角形EBC,得
EC/QC=BC/(BC-DQ)
BC/CE=(BC-DQ)/QC
同理
BF/BP=BC/(BC-PD)
BC/BF=(BC-PD)/BP
BC/CE+BC/BF=(BC-DQ)/QC+(BC-PD)/BP (QC=AC/2=BC/2,BP=AB/2=BC/2)
=2(BC-DQ+BC-PD)/BC
=2(2BC-PQ)/BC (PQ=BC/2)
=3
1/CE+1/BF=3/BC
证明:连接AD并延长,交BC于G; 过点A作BC的平行线,与BE,CF的延长线分别交于M,N; 则BC∥PQ∥MN. 故:MN/BC=ND/DC=AD/DG=AP/PB=1,即:(AN+MA)/BC=1,AN/BC+MA/BC=1; 又AN/BC=AF/BF; MA/BC=AE/CE. ∴AF/BF+AE/CE=1. 则(AF/BF+1)+(AE/CE+1)=3. 即:(AF+BF)/BF+(AE+CE)/CE=3; 即:AB/BF+AC/CE=3; 故BC/BF+BC/CE=3;(两边同除以BC) 1/BF+1/CE=3/BC.
(1)先证明三角形EDQ相似于三角形EBC,得
EC/QC=BC/(BC-DQ)
BC/CE=(BC-DQ)/QC
同理
BF/BP=BC/(BC-PD)
BC/BF=(BC-PD)/BP
BC/CE+BC/BF=(BC-DQ)/QC+(BC-PD)/BP (QC=AC/2=BC/2,BP=AB/2=BC/2)
=2(BC...
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(1)先证明三角形EDQ相似于三角形EBC,得
EC/QC=BC/(BC-DQ)
BC/CE=(BC-DQ)/QC
同理
BF/BP=BC/(BC-PD)
BC/BF=(BC-PD)/BP
BC/CE+BC/BF=(BC-DQ)/QC+(BC-PD)/BP (QC=AC/2=BC/2,BP=AB/2=BC/2)
=2(BC-DQ+BC-PD)/BC
=2(2BC-PQ)/BC (PQ=BC/2)
=3
1/CE+1/BF=3/BC
(2)证明:连接AD并延长,交BC于G;
过点A作BC的平行线,与BE,CF的延长线分别交于M,N;
则BC∥PQ∥MN.
故:MN/BC=ND/DC=AD/DG=AP/PB=1,即:(AN+MA)/BC=1,AN/BC+MA/BC=1;
又AN/BC=AF/BF; MA/BC=AE/CE.
∴AF/BF+AE/CE=1.
则(AF/BF+1)+(AE/CE+1)=3.
即:(AF+BF)/BF+(AE+CE)/CE=3;
即:AB/BF+AC/CE=3;
故BC/BF+BC/CE=3;(两边同除以BC)
1/BF+1/CE=3/BC.
收起
=(BC-DQ)/QC
同理
BF/BP=BC/(BC-PD)
BC/BF=(BC-PD)/BP
BC/CE+BC/BF=(BC-DQ)/QC+(BC-PD)/BP (QC=AC/2=BC/2,BP=AB/2=BC/2)
=2(BC-DQ+BC-PD)/BC
=