1. **Problem:** Evaluate $$\frac{(a-b)^2+(b-c)^2}{(a-c)^2}$$ given that $a-b=b-c=2$. 2. **Formula and rule:** Use the given values directly, and also note that $a-c=(a-b)+(b-c)$. 3. Substitute the given information: $a-b=2$ and $b-c=2$. 4. Compute the numerator: $$\frac{(2)^2+(2)^2}{(a-c)^2}$$ $$\frac{4+4}{(a-c)^2}$$ $$\frac{8}{(a-c)^2}$$ 5. Find $a-c$ using the add rule: $$a-c=(a-b)+(b-c)=2+2=4$$ 6. Square the denominator: $$ (a-c)^2=4^2=16 $$ 7. Substitute and simplify: $$\frac{8}{16}=\frac{1}{2}$$ 8. **Final answer:** $$\frac{1}{2}$$