In triangle $ABC$ we have $\angle C = 90^o$ and $AC = BC$. Furthermore $M$ is an interior pont in the triangle so that $MC = 1 , MA = 2$ and $MB =\sqrt2$. Determine $AB$
**
   - If $s^2 = 1$, then $s = 1$, which leads to an impossible configuration given the distances.
   - Therefore, $s^2 = 5$, so $s = \sqrt{5}$.

7. **Finding $AB$:**
   - Since $AB$ is the hypotenuse of the right triangle $ABC$ with legs $AC = BC = \sqrt{5}$:
     \[
     AB = \sqrt{AC^2 + BC^2} = \sqrt{(\sqrt{5})^2 + (\sqrt{5})^2} = \sqrt{5 + 5} = \sqrt{10}
     \]

The final answer is $\boxed{\sqrt{10}}$