因为x/(a-b)=y/(b-c)=z/(c-a)
可得x=y(a-b)/(b-c)
z=y(c-a)/(b-c)
x+y+z=y(a-b)/(b-c)+y+y(c-a)/(b-c)
=y(a-b+c-a)/(b-c)+y
=y(c-b)/(b-c)+y
=-y+y=0
x/(a-b)=y/(b-c)=z/(c-a) 求x+y+z
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