One, (50 points) As shown in the figure, in the isosceles right $\triangle ABC$, $D$ is the midpoint of the hypotenuse $AB$. A line $l \parallel AB$ is drawn through the right-angle vertex $C$. Points $K$, $M$, and $N$ are on sides $AC$, $BC$, and $KM$ respectively, such that $\frac{CK}{AK} = \frac{BM}{MC} = \frac{FN}{NK}$. A perpendicular $NH \perp l$ is drawn from point $N$ to $H$. Compare the lengths of $ND$ and $NH$.
One, Solution: Connect $D K, D M, D C$, let $C D$ intersect $K M$ at $E$, draw $K P \perp N H$ at $P$, and draw $M Q \perp C D$ at $Q$.
$$
\because \frac{C K}{K A}=\frac{B M}{M C}, C A=C B \quad \therefore C K=B M
$$
$\therefore \triangle C D K \cong \triangle B M D$,
$$
\therefore D K=D M, \angle C D K=\angle M D B \text {, }
$$
$\therefore \triangle K D M$ is an isosceles right triangle.
Furthermore, since $C E$ bisects $\angle A C B$, we have $\frac{C K}{C M}=\frac{K E}{E M}$, and $C K=$ $B M$.
$$
\therefore \frac{K E}{E M}=\frac{M N}{N K} \quad \therefore K E=M N \text {. }
$$

Thus, $\triangle D K E \cong \triangle D M N \quad \therefore D E=D N$
It is also easy to prove that $\triangle K P N \cong \triangle M Q E$
$$
\begin{array}{l} 
\therefore N H=P N+P H=P N+C K \sin 45^{\circ}=E Q+ \\
B M \sin 45^{\circ}=E Q+Q D=E D=D N .
\end{array}
$$

Proof completed.