The given condition is equivalent to $(b-c)^2 - 4(a-b)(c-a)=0$. This implies that the equation $$(a-b)x^2 -(b-c)x + (c-a)=0$$ has two equal roots. Obviously $1$ is one root. Therefore both roots are $1$. This means $$\frac{c-a}{a-b}=1\times 1 \implies \frac{b+c}{a}=\boxed{2}$$