Example 1 As shown in Figure 1, in the right triangle $\triangle ABC$, $\angle ACB=$ $90^{\circ}$, point $D$ is on side $CA$, such that $CD=1, DA=3$, and $\angle BDC=3 \angle BAC$. Find the length of $BC$. ${ }^{[1]}$
(2009, Xin Zhi Cup Shanghai Junior High School Mathematics Competition)
From the problem, we know
$$
\angle A B D=2 \angle B A C \text {. }
$$

Draw the angle bisector of $\angle A B D$ through point $B$, intersecting $D A$ at point $E$. Then $\triangle A E B$ is an isosceles triangle.
Let $A E=x$. Then $B E=x, D E=3-x$. Applying properties to $\triangle E B C$ and $\triangle A B C$ respectively, we have
$$
\begin{array}{l}
x^{2}=B E^{2}=B D^{2}+D E^{2}+2 D E \cdot D C \\
=B D^{2}+(3-x)^{2}+2(3-x) \\
\Rightarrow B D^{2}=8 x-15, \\
A B^{2}=B D^{2}+D A^{2}+2 D A \cdot D C \\
=8 x-15+9+2 \times 3=8 x .
\end{array}
$$

By the angle bisector property, we have
$$
\frac{B D}{B A}=\frac{D E}{E A} \Rightarrow \frac{8 x-15}{8 x}=\frac{(3-x)^{2}}{x^{2}} \Rightarrow x=\frac{24}{11} \text {. }
$$

Thus, $B C=\sqrt{B D^{2}-C D^{2}}=\sqrt{8 x-16}=\frac{4 \sqrt{11}}{11}$.