1. State the problem. We are given $a-b=2$ and $b-c=2$. Find the value of $$\frac{(a-b)^2+(b-c)^2}{(a-c)^2}.$$ 2. Use the given information. Since $a-b=2$, then $(a-b)^2=2^2=4$. Since $b-c=2$, then $(b-c)^2=2^2=4$. 3. Compute $a-c$ from the relationship. Because $a-c=(a-b)+(b-c)$, we get $$a-c=2+2=4.$$ So $(a-c)^2=4^2=16$. 4. Substitute into the expression. $$\frac{(a-b)^2+(b-c)^2}{(a-c)^2}=\frac{4+4}{16}.$$ 5. Simplify the fraction (with a cancel line). $$\frac{4+4}{16}=\frac{8}{16}.$$ $$\frac{\cancel{8}}{\cancel{16}}=\frac{1}{2}.$$ 6. Final answer. The value is $\frac{1}{2}$.