Example 4 Let the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ be $1, P$ is a moving point on the face diagonal $B C_{1}$, and $Q$ is a moving point on the base $A B C D$. Try to find the minimum value of $D_{1} P+P Q$.
(1998 "Hope Cup" Competition Question)
Fig. 19-6
Solution From the given conditions, we know that when $D_{1} P+P Q$ is at its minimum, point $Q$ must be the projection of point $P$ on the base, as shown in Fig. 19-6 (a). $D_{1} P$ and $P Q$ are in the two faces of the dihedral angle $D_{1}-B C_{1}-C$. For this purpose, rotate $\triangle B C_{1} C$ around $B C_{1}$ by $90^{\circ}$, so that $\triangle B C_{1} C$ and the diagonal plane $A B C_{1} D_{1}$ are in the same plane, as shown in Fig. 19-6(b).

Since $P Q \perp B C$, when $D_{1}, P, Q$ are collinear and perpendicular to $B C_{1}$, $D_{1} P+P Q$ is minimized. It can be found that $D_{1} Q_{1}=$ $D_{1} P_{1}+P_{1} Q_{1}=\sqrt{2}+\frac{\sqrt{2}}{2}(\sqrt{2}-1)=1+\frac{\sqrt{2}}{2}$. Therefore, the required minimum value is $1+\frac{\sqrt{2}}{2}$.