已知三角形ABC三内角A,B,C成等差数列,求证:对应三边abc满足1/(a+b)+1/(b+c)=3/(a+b+c)

Question

已知三角形ABC三内角A,B,C成等差数列,求证:对应三边abc满足1/(a+b)+1/(b+c)=3/(a+b+c)

Answer 1

设 ∠A=a°,公差是x°
a+a+x+a+x+x=180°
3a+3x=180°
a+x=60°
∠B=60°
BD=c/2
CD=a-c/2
AD²=AB²-BD²=c²-(c/2)²=3c²/4
AC²=AD²+CD²=3c²/4+(a-c/2)²=3c²/4+a²-ac+c²/4=a²-ac+c²
即：b²=a²-ac+c²


再来看
1/(a+b)+1/(b+c)-3/(a+b+c)

=[(b+c)(a+b+c)+(a+b)(a+b+c)-3(a+b)(b+c)]/[(a+b)(b+c)(a+b+c)]
把分母放在一边不管,只看分子：
(b+c)(a+b+c)+(a+b)(a+b+c)-3(a+b)(b+c)
=(ab+b²+bc+ac+bc+c²)+(a²+ab+ac+ab+b²+bc)-3(ab+ac+b²+bc)
=3ab+2b²+3bc+2ac+c²+a²-3ab-3ac-3b²-3bc
=a²-ac+c²-b²
=0
∴1/(a+b)+1/(b+c)-3/(a+b+c)=0
∴1/(a+b)+1/(b+c)=3/(a+b+c)