Four, in isosceles $\triangle A B C$, the base $B C$ lies on the line $x+y=0$, the vertex $\mathrm{A}(2,3)$, and it is known that one of its equal sides $\mathrm{AB}$ is parallel to the line $\mathbf{x}-4 y-6=0$. Find the equation of the other equal side $\mathrm{AC}$.
Let the equation of the line $\mathrm{AB}$ be
$$
\mathbf{x}-4 \mathrm{y}+\mathrm{m}=0 \text{. }
$$
Since point $\mathrm{A}(2,3)$ lies on this line, $\therefore \mathrm{m}=10$.
Solve the system of equations
$$
\left\{\begin{array}{l}
x-4 y+10=0, \\
\mathbf{x}+\mathbf{y}=0 .
\end{array}\right.
$$

to get the coordinates of point $B$ as $(-2,2)$.
The base $\mathrm{BC}$. The equation of the altitude $\mathrm{AD}$ is $x-y+1=0$.
Solve the system of equations $\left\{\begin{array}{l}x-y+1=0, \\ x+y=0 .\end{array}\right.$

to get the coordinates of point $D$ as $\left(-\frac{1}{2}, \frac{1}{2}\right)$.
By the midpoint formula, the coordinates of point $C$ are $(1,-1)$, hence the equation of the line $\mathrm{AC}$ is
$$
\frac{x-1}{2-1}=\frac{y+1}{3+1}
$$

which simplifies to $4 x-y-5=0$.