已知,(y+z)/2x = (z+x)/2y = (x+y)/2z = k ,
可得：y+z = 2kx ,z+x = 2ky ,x+y = 2kz ,
三式相加可得：2(x+y+z) = 2k(x+y+z) ,
整理得：2(k-1)(x+y+z) = 0 ,
该等式对任意 x、y、z 都成立,
则有：k-1 = 0 ,
解得：k = 1 .