一、(50 points) As shown in Figure 3, in the isosceles right triangle $\triangle A B C$, $D$ is the midpoint of the hypotenuse $A B$. A line $C C^{\prime} / / A B$ is drawn through the right-angle vertex $C$. Points $E$, $F$, and $M$ are on sides $A C$, $B C$, and $E F$ respectively, such that $\frac{C E}{A E}=$ $\frac{B F}{C F}=\frac{F M}{E M}$. A perpendicular line $M H \perp C C^{\prime}$ is drawn from point $M$ to $H$. Compare the lengths of $M D$ and $M H$.
From $A C = B C$ and $\frac{C E}{A E} = \frac{B F}{C F}$, we know $C E = B F$.
As shown in Figure 9, connect
$D F$, rotate $\triangle B D F$ counterclockwise by $90^{\circ}$ around point
$D$, then point $B$ moves to point $C$
and point $F$ moves to point $E$,
thus $\triangle C D E \cong$
$\triangle B D F$. Therefore, $\triangle D E F$ is an isosceles right triangle.
Let $C D$ intersect $E F$ at point $N$, draw $E P \perp M H$ at point $P$, $F Q \perp C D$ at point $Q$. In Rt $\triangle E C F$, since $C N$ is the angle bisector of $\angle E C F$ and by the given conditions, we have
$$
\frac{E N}{N F} = \frac{C E}{C F} = \frac{B F}{C F} = \frac{F M}{E M} \text{. }
$$

By the property of the ratio, we get
$$
\frac{E N}{E N + N F} = \frac{F M}{F M + E M} \Rightarrow \frac{E N}{E F} = \frac{F M}{E F} \text{. }
$$

Thus, $E N = F M$.
Also, since $D E = D F, \angle D E N = \angle D F M = 45^{\circ}$, we have
$\triangle D E N \cong \triangle D F M$,
so $D N = D M$.
In Rt $\triangle E P M$ and Rt $\triangle F Q N$, we have
$$
E M = E F - M F = E F - E N = F N \text{. }
$$

Furthermore, since $M H \perp C C^{\prime}, C D \perp A B, A B \parallel C C^{\prime}$, we have
$E P \parallel C C^{\prime} \parallel A B \parallel F Q$,
thus $\angle P E M = \angle Q F N$.
Therefore, Rt $\triangle E P M \cong$ Rt $\triangle F Q N$.
Hence, $P M = Q N$.
From equations (1) and (2), we get
$$
\begin{array}{l}
M D = D N = D Q + Q N = B F \sin 45^{\circ} + Q N \\
= C E \sin 45^{\circ} + P M = P H + P M = M H .
\end{array}
$$

Note: A Cartesian coordinate system can be established, and it can be shown that the locus of point $M$ is a parabola, with $D$ being the focus of the parabola and $C C^{\prime}$ being the directrix. By the definition of a parabola, we have $M D = M H$.