Rectangle $ABCD$ has $AB=6$ and $BC=8$. A semicircle with diameter $\overline{AB}$ is constructed inside the rectangle, and the tangent to the semicircle from $C$ (other than $\overline{CB}$) intersects side $\overline{AD}$ at $E$. What is the length of $\overline{CE}$?

1. Let the tangent from $C$ touch the semicircle at $F$. The side $CB$ is perpendicular to the diameter $AB$ at $B$, so $CB$ is also tangent to the circle, and the two tangent segments from $C$ are equal: $CF=CB=8$.
2. Likewise $AD\perp AB$ at $A$, so $AD$ is tangent at $A$, and the two tangent segments from $E$ are equal: $EF=EA=x$. Then $CE=CF+FE=8+x$ and $ED=8-x$.
3. In the right triangle $CDE$ (right angle at $D$, with $CD=AB=6$), $(8-x)^2+6^2=(8+x)^2$. This simplifies to $100-16x=64+16x$, so $32x=36$ and $x=\dfrac98$.
4. Therefore $CE=8+\dfrac98=\boxed{\dfrac{73}8}$.