Math CE paper2 2007 Q19 how to do it?
please help me
Math CE 2007 Q19 how to do it?
2009-01-21 5:32 am
回答 (2)
2009-01-22 4:29 am
✔ 最佳答案
角DFC=角BFE (對頂角)角FCD=角FEB (錯角,EB平行DC)
角FDC=角FBE (錯角,EB平行DC)
所以三角形FDC形相似三角形FBE (AAA)
因此得知
DF/FB=CF/FE
DF*FE=CF*FB
角DFE=角CFB (對頂角)
面積DEF=DF*FE*sin角DFE/2
面積CBF=CF*FB*sin角CFB/2
面積比 1:1
參考: ME
2009-01-22 4:19 am
Area of EDC=Area of BDC
=>Area of EDF +Area of DFC = Area of BCF +Area of DFC
=>Area of EDF= are BCF
so Area of EDF : Area of BCF = 1 :1 (A)//
=>Area of EDF +Area of DFC = Area of BCF +Area of DFC
=>Area of EDF= are BCF
so Area of EDF : Area of BCF = 1 :1 (A)//
收錄日期: 2021-04-13 16:22:54
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090120000051KK01653