II. (25 points) As shown in Figure 4, $AB$ is the diameter of $\odot O$, $C$ is a point outside $\odot O$, $CD \perp AB$ at point $D$, intersecting $\odot O$ at point $E$. $EF$ is a non-diameter chord, and
$$
CF = CE. FG \perp AB
$$

at point $G$, $CH \perp EF$ at point $H$, points $D$ and $G$ are on opposite sides of point $O$, $\angle DHG = 45^{\circ}$. Find the value of $EF^2 : AB^2$.
II. As shown in Figure 10, connect $O H, O E, O F$.
Given $C E=C F, C H \perp E F$, we have $E H=F H$.
By the perpendicular chord theorem, $O H \perp E F$.
Also, $F G \perp A B \Rightarrow \angle F H O+\angle F G O=180^{\circ}$
$\Rightarrow O, G, F, H$ are concyclic $\Rightarrow \angle G F O=\angle G H O$.
Similarly, $\angle D E O=\angle D H O$.
Thus, $\angle G F O+\angle D E O=\angle G H O+\angle D H O$ $=\angle D H G=45^{\circ}$.
Since $E D / / F G \Rightarrow \angle G F E+\angle D E F=180^{\circ}$
$\Rightarrow \angle O F E+\angle O E F=135^{\circ}$
$\Rightarrow \angle E O F=180^{\circ}-(\angle O F E+\angle O E F)=45^{\circ}$.
Let the radius of $\odot O$ be $R$. Then $A B=2 R$.
Draw $F I \perp O E$ at point $I$. Then
$$
O I=F I=F O \sin \angle E O F=R \sin 45^{\circ}=\frac{\sqrt{2}}{2} R \text {. }
$$

Thus, $E I=O E-O I=\left(1-\frac{\sqrt{2}}{2}\right) R$.
By the Pythagorean theorem,
$$
\begin{array}{l}
E F^{2}=F I^{2}+E I^{2}=\left(\frac{\sqrt{2}}{2} R\right)^{2}+\left[\left(1-\frac{\sqrt{2}}{2}\right) R\right]^{2} \\
=(2-\sqrt{2}) R^{2} .
\end{array}
$$

Therefore, $E F^{2}: A B^{2}=(2-\sqrt{2}) R^{2}:(2 R)^{2}$
$$
=(2-\sqrt{2}): 4 \text {. }
$$