Make the opposite vertices $A, C$ coincide, and fold the paper. Find the length of the crease.



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Given that the crease $EF$ is the perpendicular bisector of the diagonal $AC$, $O$ is the midpoint of $AC$ and also the midpoint of $EF$. Let $OE = OF = x \, \text{cm}$. Then
\[
\begin{aligned}
AF & = \sqrt{AO^2 + OF^2} = \sqrt{\left(\frac{5}{2}\right)^2 + x^2} \\
& = \frac{1}{2} \sqrt{25 + 4x^2} \, (\text{cm}), \\
FD & = BE = \sqrt{AE^2 - AB^2} \\
& = \frac{1}{2} \sqrt{4x^2 - 11} \, (\text{cm}).
\end{aligned}
\]

From $AF + FD = AD$, we have
\[
\frac{1}{2} \sqrt{4x^2 + 25} + \frac{1}{2} \sqrt{4x^2 - 11} = 4.
\]

Solving for $x$ gives $x = \pm \frac{18}{5}$. (Discard the negative value)
\[
\therefore EF = 2x = \frac{15}{4} = 3.75 \, (\text{cm}).
\]