Question 914044
b+c-a/a, c+a-b/b, a+b-c/c in AP


{{{ (c+a-b)/b - (b+c-a)/a = (a+b-c)/c - (c+a-b)/b}}}


{{{ (a(c+a-b) - b(b+c-a))/ab = (b(a+b-c)-c(c+a-b))/bc}}}


{{{(ac+a^2-ab-b^2-bc+ab)/ab = (ab+b^2-bc-c^2-ac+bc)/bc}}}



{{{(a^2-b^2+ac-bc)/ab = ( b^2-c^2+ab-ac)/bc}}}


{{{((a+b)(a-b)+c(a-b))/ab=((b+c)(b-c)+a(b-c))/bc}}}



{{{((a-b)(a+b+c))/ab= ((b-c)(a+b+c))/bc}}}


{{{(a-b)/a= (b-c)/c}}}


ac-bc = ab-ac

/abc

1/b -1/a = 1/c-1/b

Hence they are in AP