II. (25 points) As shown in Figure 3, in $\triangle ABC$, $AB > AC$, $\angle BAC = 45^\circ$, and $E$ is the intersection of the external angle bisector of $\angle BAC$ with the circumcircle of $\triangle ABC$. Point $F$ is on $AB$ such that $EF \perp AB$. Given that $AF = 1$ and $BF = 5$. Find the area of $\triangle ABC$.
As shown in Figure 6, take point $D$ on $FB$ such that $FD = AF$. Connect $ED$ and extend it to intersect the circumcircle of $\triangle ABC$ at point $G$.
Since $EF \perp AD$ and $AF = FD$,
$$
\begin{aligned}
\Rightarrow & \angle AED \\
& = 180^{\circ} - 2 \angle EAD \\
& = \angle BAC \\
\Rightarrow & \overparen{AG} = \overparen{BC} \\
\Rightarrow & \overparen{AC} = \overparen{BG} \\
\Rightarrow & AC = BG. \\
& \text{Also, } \angle BGE = \angle BAE = \angle ADE = \angle BDG, \text{ then} \\
& BG = BD \Rightarrow AC = BD = 4.
\end{aligned}
$$

It is easy to see that the altitude from vertex $A$ to side $AB$ of $\triangle ABC$ is
$$
h = AC \sin 45^{\circ} = 2 \sqrt{2}.
$$

Thus, $S_{\triangle ABC} = \frac{1}{2} AB \cdot h = \frac{1}{2} \times 6 \times 2 \sqrt{2} = 6 \sqrt{2}$.