Proof: As shown in Figure 5, let the lines $B D$ and $A C$ intersect the sides of the square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ at points $T$, $S$, $X$, and $Y$. Since $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ are both squares, and $A C \perp B D$, it follows that $T X \perp S Y$, and thus $T X = S Y$. But $B D = A C$, so,
$$\begin{array}{l} B T + D X \\ = C S + A Y. \end{array}$$

Since
$\angle E A^{\prime} M + \angle E B M = 90^{\circ} + 90^{\circ} = 180^{\circ}$,
then $A^{\prime}, E, B, M$ are concyclic,
hence $\angle B E F = \angle A M N$.
Since $\angle N A^{\prime} F + \angle N A F = 90^{\circ} + 90^{\circ} = 180^{\circ}$,
then $A^{\prime}, F, A, N$ are concyclic,
hence $\angle B F E = \angle A N M$.
Therefore, $\triangle B E F \sim \triangle A M N$.
Similarly, we can obtain
$\triangle B E F \sim \triangle C G H, \triangle C G H \sim \triangle D P Q$.
Thus, $\triangle B E F \sim \triangle C G H \sim \triangle D P Q \sim \triangle A M N$.
Since $B T, C S, D X, A Y$ are the angle bisectors of the corresponding angles of the four similar triangles, it must be that
$$\frac{B T}{C S} = \frac{E F}{G H}, \quad \frac{C S}{D X} = \frac{G H}{P Q}, \quad \frac{D X}{A Y} = \frac{P Q}{M N}, \quad \frac{A Y}{B T} = \frac{M N}{E F},$$

or $\frac{B T}{E F} = \frac{C S}{G H} = \frac{D X}{P Q} = \frac{A Y}{M N} (= \lambda)$.
Thus, $B T = E F \cdot \lambda, C S = G H \cdot \lambda$,
$$D X = P Q \cdot \lambda, A Y = M N \cdot \lambda.$$

Substituting the above four equations into (1) and eliminating the parameter $\lambda$, we get
$$E F + P Q = G H + M N.$$