II. (25 points) As shown in Figure 2, given that $O$ is the circumcenter of $\triangle ABC$, $AB=AC$, and $D$ is a point on the circumcircle of $\triangle OBC$. A line perpendicular to $OD$ is drawn through point $A$, and the foot of the perpendicular is $H$. If $BD=7, DC=3$, find $AH$.
II. As shown in Figure 5, extend $BD$ and $OD$, intersecting $\odot O$ at points $N$ and $E$ respectively.
From the problem, we have
$$
\begin{array}{l}
\angle NDE \\
=\angle ODB \\
=\angle OCB \\
=\angle OBC \\
=\angle CDE.
\end{array}
$$

Thus, $DE$ is the external angle bisector of $\angle BDC$.
Since point $D$ lies on the radius $OE$ of $\odot O$, and points $C$ and $N$ lie on $\odot O$, points $C$ and $N$ are symmetric with respect to line $OE$, so $DN = DC$.
Extend $AH$ to intersect $\odot O$ at point $M$.
Since $O$ is the center and $AM \perp OD$, points $A$ and $M$ are symmetric with respect to line $OD$, so $AH = MH$.
Therefore, $MN = AC = AB$.
Also, $\angle FNM = \angle FAB$, $\angle FBA = \angle FMN$, so $\triangle ABF \cong \triangle NMF \Rightarrow MF = BF, FN = AF$.
Thus, $AM = AF + FM = FN + BF = BN$
$$
\begin{array}{l}
= BD + DN = BD + DC = 10 \\
\Rightarrow 2AH = 10 \Rightarrow AH = 5.
\end{array}
$$