jmc

Let $F$ be the point at which $\overline{CE}$ is tangent to the semicircle, and let $G$ be the midpoint of $\overline{AB}$. Because $\overline{CF}$ and $\overline{CB}$ are both tangents to the semicircle, $CF=CB=2$. Similarly, $EA=EF$. Let $x=AE$. The Pythagorean Theorem applied to $△CDE$ gives $$(2-x{)}^2+2^2=(2+x{)}^2.$$It follows that $x=1/2$ and $CE=2+x=\boxed{\frac{5}{2}}$.