II. (15 points) As shown in Figure 4, in the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, point $D$ is on side $CA$ such that
$$
\begin{array}{l}
CD=1, DA=3, \text { and } \\
\angle BDC=3 \angle BAC .
\end{array}
$$

Find the length of $BC$.
As shown in Figure 7, construct the angle bisector $AE$ of $\angle ABD$, and let $AE = x$.
Then $AE = x$,
$$
\begin{array}{l}
BC^2 = x^2 - (4 - x)^2 = 8x - 16, \\
BD^2 = BC^2 + 1^2 = 8x - 15, \\
AB^2 = BC^2 + 4^2 = 8x.
\end{array}
$$

From $\frac{BD}{BA} = \frac{DE}{EA}$, we get
$$
\frac{8x - 15}{8x} = \frac{(3 - x)^2}{x^2}.
$$

Solving this, we find $x = \frac{24}{11}$.
Thus, $BC = \sqrt{8x - 16} = \frac{4\sqrt{11}}{11}$.

Solution 2: Let $\angle BAC = \alpha$. As shown in Figure 7, construct $\angle BDF = \alpha$, and let the ray $DF$ intersect the extension of $AB$ at point $F$.

Since $\angle FDC = 3\alpha - \alpha = 2\alpha$, by the exterior angle theorem of a triangle, we have $\angle F = \alpha$.
Thus, $BF = BD$, and $FD = DA$.
Let $BC = y$. Then
$$
BF = BD = \sqrt{1 + y^2}, \quad BA = \sqrt{16 + y^2}.
$$

Since $\triangle BFD \sim \triangle DAF$, we have
$$
\frac{\sqrt{1 + y^2}}{3} = \frac{3}{\sqrt{1 + y^2} + \sqrt{16 + y^2}}.
$$

Solving this, we find $y = \frac{4}{\sqrt{11}}$, i.e., $BC = \frac{4}{\sqrt{11}}$.
Solution 1: Let $\angle BAC = \alpha$. Then $\angle ABD = 3\alpha - \alpha = 2\alpha$.