在 △ABC 中 , ㄥA = 60° , 求證 :
1/(a+b) + 1/(a+c) = 3/(a+b+c)
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2010-12-29 2:08 am
回答 (2)
2010-12-29 2:17 am
✔ 最佳答案
1/(a+b) + 1/(a+c) = 3/(a+b+c)<=> 1/(a+b) - 1/(a+b+c) + 1/(a+c) - 1/(a+b+c) = 1/(a+b+c)
<=> c/(a+b)(a+b+c) + b/(a+c)(a+b+c) = 1/(a+b+c)
<=> c/(a+b) + b/(a+c) = 1
<=> c(a+c) + b(a+b) = (a+b)(a+c)
<=> c^2 + b^2 = a^2 + bc
<=> (c^2 + b^2 - a^2)/(2bc) =1/2
<=> cos ㄥA =1/2
<=> ㄥA = 60°
2010-12-29 2:21 am
cosine law: a^2=b^2+c^2-bc,
a^2+bc=b^2+c^2, then (a+b)(a+c)=b(a+b)+c(a+c)
hence, 1= c/(a+b) + b/(a+c)
3= 1+ c/(a+b) + 1+ b/(a+c)
3= (a+b+c)/(a+b) + (a+b+c)/(a+c)
3/(a+b+c) = 1/(a+b) + 1/(a+c)
a^2+bc=b^2+c^2, then (a+b)(a+c)=b(a+b)+c(a+c)
hence, 1= c/(a+b) + b/(a+c)
3= 1+ c/(a+b) + 1+ b/(a+c)
3= (a+b+c)/(a+b) + (a+b+c)/(a+c)
3/(a+b+c) = 1/(a+b) + 1/(a+c)
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101228000051KK01169