Given that $ABCD$ is a square, point $M$ (different from points $B, C$) is on side $BC$, and the perpendicular bisector $l$ of line segment $AM$ intersects $AB$ and $CD$ at points $E$ and $F$ respectively.
(1) Which is longer, $BE$ or $DF$? Please explain your reasoning.
(2) If $AB=1$, find the range of $|BE - DF|$ as point $M$ moves through every point on the segment $\dot{B}C$ internally.
(1) As shown in Figure 3, let $A C$ and $B D$ intersect at point $O$, where $O$ is the center of the square $A B C D$. Line $l$ intersects $A M$ at the midpoint $N$ of $A M$. Connect $NO$, then $NO // B C$. Therefore, $\angle O N M = \angle N M B = \angle A M B B E' = D F' > D F$, which means $B E$ is longer than $D F$.
(2) Let $B M = x$, then
$$
\begin{array}{l}
M C = 1 - x (0 < x < 1), A M = \sqrt{1 + x^{2}}, \\
A N = N M = \frac{\sqrt{1 + x^{2}}}{2} .
\end{array}
$$

Extend $A M$ and $D C$ to intersect at point $P$.
From the similarity of right triangles $\triangle M C P \sim \triangle M B A$, we get $\frac{C P}{A B} = \frac{C M}{M B}$.
Thus, $C P = \frac{A B \cdot C M}{M B} = \frac{1 - x}{x} = \frac{1}{x} - 1$,
$$
\begin{aligned}
M P & = \sqrt{(1 - x)^{2} + \left(\frac{1}{x} - 1\right)^{2}} \\
& = \sqrt{(1 - x)^{2} + \frac{(1 - x)^{2}}{x^{2}}} = \frac{1 - x}{x} \cdot \sqrt{x^{2} + 1} .
\end{aligned}
$$

Clearly, points $B, E, N, M$ and $C, F, N, M$ are concyclic.
By the secant theorem, we have
$$
A E \cdot A B = A N \cdot A M, P C \cdot P F = P M \cdot P N .
$$

Thus, $A E = \frac{1}{2} A M^{2} = \frac{1}{2}(1 + x^{2})$,
$$
\begin{array}{l}
B E = 1 - A E = 1 - \frac{1 + x^{2}}{2} = \frac{1 - x^{2}}{2}, \\
P F = \frac{P M \cdot P N}{P C} \\
= \frac{\frac{(1 - x) \sqrt{x^{2} + 1}}{x} \left[\frac{\sqrt{1 + x^{2}}}{2} + \frac{(1 - x) \sqrt{x^{2} + 1}}{x}\right]}{\frac{1 - x}{x}} \\
= \frac{x^{2} + 1}{2} + (x^{2} + 1) \frac{1 - x}{x} \\
= x + \frac{1}{x} - \frac{x^{2} + 1}{2} .
\end{array}
$$

Therefore, $C F = P F - C P$
$$
\begin{array}{l}
= x + \frac{1}{x} - \frac{x^{2} + 1}{2} - \left(\frac{1}{x} - 1\right) = x + \frac{1 - x^{2}}{2}, \\
D F = 1 - C F = -x + \frac{x^{2} + 1}{2} .
\end{array}
$$

Thus, $|B E - D F| = B E - D F$
$$
= \frac{1 - x^{2}}{2} + x - \frac{x^{2} + 1}{2} = x(1 - x) .
$$

The range of the function $y = x(1 - x) (0 < x < 1)$ is $\left(0, \frac{1}{4}\right]$, therefore, the range of $|B E - D F|$ is $\left(0, \frac{1}{4}\right]$.