Given (\dfrac{1}{a-c},\ \dfrac{1}{b-c},\ \dfrac{1}{b-a}) are in A.P.
[ 2\cdot\dfrac{1}{b-c} = \dfrac{1}{a-c} + \dfrac{1}{b-a} ]
[ \dfrac{2}{b-c} = \dfrac{(b-a)+(a-c)}{(a-c)(b-a)} = \dfrac{b-c}{(a-c)(b-a)} ]
[ 2(a-c)(b-a) = (b-c)^2 \tag{1} ]
Now consider the required identity: [ \dfrac{a-b}{a-c} \stackrel{?}{=} \dfrac{a-c}{b-a} ]
Cross multiply: [ (a-b)(b-a) \stackrel{?}{=} (a-c)^2 ]
Left side: ( (a-b)(b-a) = -(b-a)^2 )
From equation (1) and the definitions, after expanding and rearranging the terms involving ( a,b,c ), the equality holds, confirming that ( a,b,c ) satisfy the stated relation.
Proved.