In a square $A B C D$ with side length 2, moving points $E$ and $F$ are both on side $A D$, satisfying $A E=D F$. Connecting $C F$ intersects the diagonal $B D$ at point $Q$, and connecting $A Q$ and $B E$ intersect at point $P$. Find the minimum value of $D P$.

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Solve as shown in Figure 3.
From the given conditions,
$$
\begin{array}{l}
A B=D C=D A, \\
\angle B A E \\
=\angle C D F \\
=90^{\circ},
\end{array}
$$
$\angle C D Q=\angle A D Q$.
Since $A E=D F, D Q=D Q$, therefore,
$\triangle A B E \cong \triangle D C F, \triangle C D Q \cong \triangle A D Q$.
Thus, $\angle A E B=\angle D F C, \angle D A Q=\angle D C Q$.
Therefore, $\angle A E B+\angle D A Q$
$$
=\angle D F C+\angle D C Q=90^{\circ} \text {. }
$$

Hence, $\angle A P B=90^{\circ}$.
Therefore, point $P$ lies on the circle $\odot O$ with $A B$ as its diameter.
Since point $E$ moves along side $A D$, the trajectory of point $P$ is the arc of $\odot O$ subtended by $\angle A B D$.
Connect $O P, O D$, and let $O D$ intersect $\odot O$ at point $H$.
It is easy to see that $D H=O D-O H=O D-O P \leqslant D P$.
Thus, the minimum value of $D P$ is the length of $D H$.
In the right triangle $\triangle O A D$,
$$
A D=2, O A=\frac{1}{2} A B=1 \text {. }
$$

By the Pythagorean theorem,
$$
O D=\sqrt{O A^{2}+A D^{2}}=\sqrt{5} \text {. }
$$

Therefore, $D P_{\text {min }}=D H=O D-O H=\sqrt{5}-1$.