One. (20 points) As shown in Figure 5, $P$ is a point outside circle $\odot O$, $PA$ is tangent to $\odot O$ at point $A$, and $PBC$ is a secant of $\odot O$. $AD \perp PO$ at point $D$. If $PB=4$, $CD=5$, $BC=6$, find the length of $BD$.
Connect $O A, O B, O C$. Then $O B=O C$. It is easy to see that
$$
\begin{array}{l}
P A \perp O A, \\
P A^{2}=P B \cdot P C=P D \cdot P O \\
\Rightarrow \frac{P O}{P C}=\frac{P B}{P D} .
\end{array}
$$

Since $\angle B P D$ is a common angle, we have
$$
\begin{array}{l}
\triangle P O B \sim \triangle P C D . \\
\text { Then } \frac{P C}{C D}=\frac{P O}{O B}=\frac{P O}{O C} .
\end{array}
$$

At the same time, $\triangle P O C \sim \triangle P B D \Rightarrow \frac{P O}{O C}=\frac{P B}{B D}$.
$$
\text { Hence } \frac{P C}{C D}=\frac{P B}{B D} \Rightarrow B D=\frac{P B \cdot C D}{P C}=\frac{4 \times 5}{4+6}=2 \text {. }
$$