Example 2 In $\triangle A B C$, it is known that the coordinates of vertex $A$ are $A(1,4), \angle A B C$ and $\angle A C B$'s bisector line equations are $x-2 y=0$ and $x+y-1=0$, respectively. Find the equation of line $B C$.

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As shown in Figure 7-1, let $A(1,4)$ have a symmetric point $A^{\prime}$ $\left(x_{11}, y_{n 1}\right)$ with respect to the line $x-2 y=0$. By the property of the angle bisector, it is easy to know that $A^{\prime}$ lies on the line $B C$. From the system of equations
$$
\left\{\begin{array}{l}
\frac{x_{0}+1}{2}-2 \cdot \frac{y_{0}+4}{2}=0, \\
\frac{y_{0}-4}{x_{n}-1}=-2
\end{array}\right. \text {. }
$$

we solve to get $A^{\prime}\left(\frac{19}{5},-\frac{8}{5}\right)$.
Similarly, the symmetric point of $A(1,4)$ with respect to the line $x+y-1=0$ is $A^{\prime \prime}(-3,0)$, which also lies on the line $B C$.
Thus, the equation of the line passing through $A^{\prime}$ and $A^{\prime \prime}$ is the equation of line $B C$, which is $4 x+17 y+12=0$.