II. (25 points) As shown in Figure $1 . D$ is the midpoint of the base $BC$ of isosceles $\triangle ABC$, and $E, F$ are points on $AC$ and its extension, respectively. Given that $\angle EDF=90^{\circ}, ED=DF=1, AD=5$. Find the length of segment $BC$.
II. As shown in Figure 7, draw $EG \perp AD$ at point $G$. Draw $FH \perp AD$ at point $H$. Then $\angle EDG = \angle DFH$. Therefore,
$\mathrm{Rt} \triangle EDG \cong \mathrm{Rt} \triangle DFH$.
Let $EG = x, DG = y$. Then
$$
DH = x, FH = y \text{. }
$$
II. $x^{2} + y^{2} = 1$.
Also, $\mathrm{Rt} \triangle AEG \sim \mathrm{Rt} \triangle AFH$, so
$$
\frac{EG}{FH} = \frac{AG}{AH} \text{. }
$$

That is, $\frac{x}{y} = \frac{5 - y}{5 + x}$.
Simplifying, we get $x^{2} + y^{2} = 5(y - x)$.
From equations (1) and (2), we know
$$
y - x = \frac{1}{5} \text{. }
$$

From $(x + y)^{2} + (y - x)^{2} = 2(x^{2} + y^{2})$, we know
$$
(x + y)^{2} = 2 - \frac{1}{25} = \frac{49}{25} \text{. }
$$

Thus, $x + y = \frac{7}{5}$.
From equations (3) and (1), we know $x = \frac{3}{5}, y = \frac{4}{5}$.
Since $\mathrm{Rt} \triangle AEG \sim \mathrm{Rt} \triangle ACD$, then $\frac{CD}{AD} = \frac{EG}{AG}$.
Therefore, $CD = AD \cdot \frac{EG}{AG} = 5 \times \frac{\frac{3}{5}}{5 - \frac{4}{5}} = \frac{5}{7}$.
Thus, $BC = 2CD = \frac{10}{7}$.