The quadratic equation ( (b-c)x^2 + (c-a)x + (a-b) = 0 ) has equal roots.

Discriminant ( D = 0 ): [ (c-a)^2 - 4(b-c)(a-b) = 0 ]

[ (c-a)^2 = 4(b-c)(a-b) ]

Expand and simplify: [ c^2 - 2ac + a^2 = 4\bigl( ba - b^2 - ca + cb \bigr) ]

After algebraic rearrangement one obtains [ 2b = a + c ]

which is the condition that ( a, b, c ) are in A.P.

Proved.