One, (20 points) As shown in the figure, given that $AB, CD$ are perpendicular chords in a circle $\odot O$ with radius 5, intersecting at point $P$. $E$ is the midpoint of $AB$, $PD=AB$, and $OE=3$. Try to find the value of $CP + CE$.

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By the intersecting chords theorem, we have
$$
C P \cdot P D=A P \cdot P B \text {. }
$$

Also, $A E=E B=\frac{1}{2} A B$,
$$
\begin{aligned}
\because C P \cdot P D & =\left(\frac{1}{2} A B+E P\right)\left(\frac{1}{2} A B-E P\right) \\
& =\frac{1}{4} A B^{2}-E P^{2} .
\end{aligned}
$$

And $P D=A B$,
$$
\therefore \quad C P=\frac{1}{4} A B-\frac{E P^{2}}{A B} \text {. }
$$

In the right triangle $\triangle C E P$,
$$
\begin{array}{l}
C E^{2}=C P^{2}+E P^{2}=\left(\frac{1}{4} A B-\frac{E P^{2}}{A B}\right)^{2}+E P^{2} \\
=\left(\frac{1}{4} A B+\frac{E P^{2}}{A B}\right)^{2}, \\
\therefore \quad C E=\frac{1}{4} A B+\frac{E P^{2}}{A B} .
\end{array}
$$

Thus, $C P+C E=\frac{1}{2} A B=A E=\sqrt{O A^{2}-O E^{2}}=4$.