`a(y+z)=b(z+x)=c(x+y)`

`=> (a(y+z))/(abc) = (b(z+x))/(abc) = (c(x+y))/(abc)`

`=> (y+z)/(bc) = (z+x)/(ac) = (x+y)/(ab)`

Ta có:

`(y+z)/(bc) = (z+x)/(ac) = ((z+x)-(y+z))/(ac-bc) = (z+x-y-z)/(c(a-b)) = (x-y)/(c(a-b)) (1)`

`(z+x)/(ac) = (x+y)/(ab) = ((x+y)-(z+x))/(ab-ac) = (x+y-z-x)/(a(b-c)) = (y-z)/(a(b-c)) (2)`

`(y+z)/(bc) = (x+y)/(ab) = ((y+z)-(x+y))/(bc-ab) = (y+z-x-y)/(b(c-a)) = (z-x)/(b(c-a)) (3)`

`(1)(2)(3)=> (x-y)/(c(a-b)) = (y-z)/(a(b-c)) = (z-x)/(b(c-a))`

`=>đpcm`.