Question

在△ABC中,A+B=120°角A,B,C所对的边分别为a,b,c,求证:a/b+c+b/a+c=1

Answer 1

要证明的a/(b＋c)＋b/(a＋c)=1就是a(a＋c)＋b(b＋c)=(b＋c)(a＋c)，就是a²＋ac＋b²＋bc=ab＋bc＋ac＋c²，就是a²＋b²=ab＋c²，就是a²＋b²－c²=ab。考虑到C=60°，则cosC=1/2=[a²＋b²－c²]/(2ac)，化简就得a²＋b²－c²=ab，这样就证明了。

Answer 2

a/sinA=b/sinB=c/sinC=k
a/(b+c)+b/(a+c)=sinA/(sinB+sinC)+sinB/(sinA+sinC)
=[(sinA)^2+(sinB)^2+sinC(sinA+sinB)]/[(sinA+sinB)sinC+(sinC)^2+sinAsinB]
c^2=a^2+b^2-2abcos60
(sinC)^2=(sinA)^2+(sinB)^2-(sinAsinB)
(sinC)^2+sinAsinB=(sinA)^2+(sinB)
a/(b+c)+b/(a+c)=1

Answer 3

A+B=120°，C=60°
余弦定理cosC=1/2=(a^2+b^2-c^2)/2ab a^2+b^2-c^2=ab a^2+b^2=c^2+ab
a/b+c+b/a+c=(a^2+ac+b^2+bc)/(ab+ac+bc+c^2)
=(a^2+ac+b^2+bc)/(ac+bc+a^2+b^2)=1