1.圖中，ABD和ACE是直線，角DBC=角ECB。求證三角形ABC是等腰三角形。
圖:http://pic.qnpic.com:83/r.jsp?fn=//qiannao/share/2011/1/6/-8b49-660e1.jpg
2.圖中，ABC和DEF均是空線,DF//AC,DB和FB分別是角ABE和角CBE的角平分線。求證BE是三角形BDF的一條中線。
圖:http://pic.qnpic.com:83/r.jsp?fn=//qiannao/share/2011/1/6/-8b49-660e81294262277421.jpg
三角形證明(兩條數)急!
2011-01-06 1:19 pm
回答 (2)
2011-01-06 2:56 pm
✔ 最佳答案
1. ∠ ABC + ∠DBC = 180° 和 ∠ ACB + ∠ECB = 180° (鄰角互補)把兩等式相減, ∠ ABC + ∠DBC – ACB – ∠ECB = 180° – 180° = 0°∠ ABC = ∠ ACB (∠DBC = ∠ECB)=> Δ ABC 是等腰三角形 (底角相等) 2. ∠ EBD = ∠ ABD = ∠ EDB (內錯角, AC // DF)=> BE = DE (底角相等)∠ EBF = ∠ CBF = ∠ EFB (內錯角, AC // DF)=> BE = FE (底角相等)i.e. BE = DE = FEBE為Δ BDF的中線2011-01-06 06:59:52 補充：
And actually I couldn't open the links you posted here... but I drew the diagrams myself following you detailed description... hope that could help you~
參考: myself, sorry that I'm not quite good at presenting it in chinese as I learned that in english...
2011-01-07 4:18 am
收錄日期: 2021-04-12 00:59:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110106000051KK00160