Consider a cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. Point $X$ moves uniformly along the square $A B C D$ in the direction $A B C D A$, and point $Y$ moves uniformly along the square $B^{\prime} C^{\prime} C B$ in the direction $B^{\prime} C^{\prime} C B B^{\prime}$ at the same speed, and points $X$ and $Y$ start from points $A$ and $B^{\prime}$ respectively at the same time. Find the trajectory of the midpoint of segment $X Y$, and draw its graph.
Analysis: Divide the positions of points $X, Y$ into the following four cases:
(1) When $X$ is on $AB$, $Y$ is on $B^{\prime}C^{\prime}$, and $AX = B^{\prime}Y$. Draw a plane $\alpha$ parallel to $ABCD$ through the midpoint $A_{1}$ of $AA^{\prime}$, intersecting $BB^{\prime}$, $CC^{\prime}$, and $DD^{\prime}$ at $B_{1}$, $C_{1}$, and $D_{1}$ respectively. Draw $XX_{1} \perp \alpha$ and $YY_{1} \perp \alpha$, with feet of the perpendiculars at $X_{1}$ and $Y_{1}$, which lie on $\Lambda_{1}B_{1}$ and $B_{1}C_{1}$ respectively, and $A_{1}X_{1} = AX$, $B_{1}Y_{1} = B^{\prime}Y$, hence $A_{1}X_{1} = B_{1}Y_{1}$.

Let $\alpha$ intersect $XY$, $AB^{\prime}$, and $BC^{\prime}$ at $Z$, $E$, and $F$ respectively. It is easy to see that $Z$, $E$, and $F$ are the midpoints of $XY$, $AB^{\prime}$, and $BC^{\prime}$ respectively. Therefore, the trajectory of the midpoint of $XY$ is the trajectory of the midpoint of segment $X_{1}Y_{1}$ within the half-plane $\alpha$, and this trajectory is the segment $EF$.
(2) When $X$ is on $BC$, $Y$ is on $C^{\prime}C$, and $BX = C^{\prime}Y$. It is easy to see that the trajectory of the midpoint $Z$ of $XY$ is the height $CF$ of the isosceles right $\triangle BC^{\prime}C$.
(3) When $X$ is on $CD$, $Y$ is on $CB$. Let the intersection of $AC$ and $BD$ be $G$. By analogy with (2), the trajectory of the midpoint $Z$ of $XY$ is the segment $CG$.
(4) When $X$ is on $DA$, $Y$ is on $BB^{\prime}$. By analogy with (1), the trajectory of the midpoint $Z$ of $XY$ is the segment $GE$.
Since $EF \perp \frac{1}{2}AC$ and $GE \perp \frac{1}{2}B^{\prime}C$, the trajectory of the midpoint $Z$ of $XY$ is the rhombus $EFCG$.
Comment: Since the moving points $X$ and $Y$ are both on the edges of the cube, using a spatial coordinate system to solve this problem would also be a good choice.