jmc

Notice that after folding the paper, $\overline{CF}$ becomes $\overline{EF}$ (we are basically reflecting the segment across the crease line). If $FD=x$, then $CF=EF=6-x$. Angle $FDE$ is a right angle since $ABCD$ is a square, so $△FDE$ is a right triangle. We also know the length of $\overline{ED}$ is $3$ since $E$ is the midpoint of $\overline{AD}$. By the Pythagorean Theorem, $(6-x{)}^2=x^2+3^2$ and we can solve for $x$. $$\begin{array}{rl}(36-12x+x^2)& =x^2+9\Rightarrow \\ 36-12x& =9\Rightarrow \\ 27& =12x\Rightarrow \\ \frac{9}{4}& =x\end{array}$$ The length of $\overline{FD}$ is $\boxed{\frac{9}{4}}$ cm.