Example 4 As shown in Figure 4, given that $AB$ is the diameter of $\odot O$, $CE \perp AB$ at point $H$, intersects $\odot O$ at points $C$ and $D$, and $AB = 10, CD = 8, DE = 4, EF$ is tangent to $\odot O$ at point $F, BF$ intersects $HD$ at point $G$.
(1) Find $GH$;
(2) Connect $FD$, determine whether $FD$ is parallel to $AB$, and provide a proof.
(2012, Joint Autonomous Admission Examination of Excellent Talent Cultivation Cooperative Universities)
Solve (1) Connect $A F, O E, O F$. Then $A, F, G, H$ are concyclic.
Since $E F$ is a tangent, $O F \perp E F$.
Thus, $\angle F G E=\angle B A F=\angle E F G$.
Therefore, $E F=E G$.
Also, $O E^{2}=O H^{2}+H E^{2}=O F^{2}+E F^{2}$, so
$E F^{2}=O H^{2}+H E^{2}-O F^{2}=3^{2}+8^{2}-5^{2}=48$.
Hence, $E F=E G=4 \sqrt{3}$.
Thus, $G H=E H-E G=8-4 \sqrt{3}$.
(2) $F D$ is not parallel to $A B$.
Proof by contradiction.
Assume $F D / / A B$. Establish a Cartesian coordinate system with $O$ as the origin and the line containing $A B$ as the $y$-axis.
Then $F(4,3), E(8,-3)$.
So, $k_{E F}=\frac{-3-3}{8-4}=-\frac{3}{2}$.
And $k_{O F}=\frac{3}{4}, k_{E F} k_{O F}=-\frac{3}{2} \times \frac{3}{4}=-\frac{9}{8} \neq-1$,

This contradicts the fact that $E F$ is a tangent to the circle.
Therefore, the conclusion holds.