100 Given $\triangle A B C$, the coordinates of vertex $A$ are $(2,-4) . B D, C E$ are the angle bisectors of $\angle B, \angle C$ respectively, and their line equations are $x+y-2=0, x-3 y-6=0$, then the equation of the line on which side $B C$ lies is $\qquad$ .
$100 x+7 y-6=0$. It is easy to find that the symmetric points of point $A(2,-4)$ with respect to the lines
$$
\begin{array}{l}
x+y-2=0, \\
x-3 y-6=0,
\end{array}
$$

are $A_{1}(6,0), A_{2}\left(\frac{2}{5}, \frac{4}{5}\right)$, respectively, and the line passing through $A_{1}, A_{2}$ is the line $B C$.