一、(20 points) As shown in Figure 3, the diameter $C D$ of $\odot O$ is perpendicular to the chord $A B$ at point $M$. A chord $E F$ is drawn through $M$, with point $E$ inside arc $\overparen{B C}$. Connect $C F$ to intersect $A B$ at point $N$. When $A B$ is translated within $\odot O$, compare the lengths of $E M$ and $C N$.
Given that diameter $C D \perp A B$ at point $M$, $M$ is the midpoint of $A B$. When $A B$ is also a diameter, $B M, C M, E M$ are all radii of the circle. By the property "in a right triangle, the hypotenuse is greater than the legs," we have $C N > C M = E M$.
Next, we prove that when $A B$ is not a diameter, we also have
$$
C N > E M.
$$

As shown in Figures 8 and 9, connect $C E$, and we get $\triangle C M E$ as an acute triangle. Draw $M H \perp C E$ at point $H$, and it is easy to see that
$$
\begin{array}{l}
C M > M H. \\
\text{By } \angle N C M = \frac{\pi}{2} - \angle F E C = \angle E M H \\
\Rightarrow \mathrm{Rt} \triangle C N M \backsim \mathrm{Rt} \triangle M E H \\
\Rightarrow \frac{C N}{E M} = \frac{C M}{M H} > 1 \Rightarrow C N > E M.
\end{array}
$$

In summary, when $A B$ moves within $\odot O$, we always have $C N > E M$.