1. **Problem:** Find the value of $$\frac{(a-b)^2+(b-c)^2}{(a-c)^2}$$ given that $a-b=b-c=2$. 2. **Formula and rule:** Since $a-c=(a-b)+(b-c)$, we first use the relation between the differences. 3. **Substitute the given values:** $$a-b=2$$ $$b-c=2$$ So, $$a-c=(a-b)+(b-c)=2+2=4$$ 4. **Compute the numerator:** $$ (a-b)^2+(b-c)^2=2^2+2^2=4+4=8 $$ 5. **Compute the denominator:** $$ (a-c)^2=4^2=16 $$ 6. **Form the fraction and simplify:** $$\frac{(a-b)^2+(b-c)^2}{(a-c)^2}=\frac{8}{16}$$ $$\frac{\cancel{8}}{\cancel{16}}=\frac{1}{2}$$ 7. **Final answer:** $$\frac{1}{2}$$