Step 1: Understand the role of a perpendicular bisector.
The perpendicular bisector of $AB$ is the mirror line for $A$ and $B$, so $B$ is the reflection of $A$ across $x-y+5=0$. Similarly $C$ is the reflection of $A$ across $x+2y=0$. Here $A=(1,-2)$.
Step 2: Reflect $A$ across $x-y+5=0$ to get $B$.
With $a=1,\,b=-1,\,c=5$, evaluate $ax+by+c = 1+2+5 = 8$, and $a^2+b^2 = 2$. The reflection formula gives \[ B = \left(1 - \frac{2(1)(8)}{2},\; -2 - \frac{2(-1)(8)}{2}\right) = (1-8,\,-2+8) = (-7,6). \]
Step 3: Reflect $A$ across $x+2y=0$ to get $C$.
With $a=1,\,b=2,\,c=0$, evaluate $ax+by+c = 1-4 = -3$, and $a^2+b^2 = 5$. Then \[ C = \left(1 - \frac{2(1)(-3)}{5},\; -2 - \frac{2(2)(-3)}{5}\right) = \left(1+\frac{6}{5},\,-2+\frac{12}{5}\right) = \left(\frac{11}{5},\,\frac{2}{5}\right). \]
Step 4: Find the slope of $BC$.
Using $B=(-7,6)$ and $C=\left(\frac{11}{5},\frac{2}{5}\right)$, the slope is \[ m = \frac{\frac{2}{5}-6}{\frac{11}{5}+7} = \frac{\frac{2-30}{5}}{\frac{11+35}{5}} = \frac{-28}{46} = -\frac{14}{23}. \]
Step 5: Write the line through $B$.
Using point $B=(-7,6)$: \[ y - 6 = -\frac{14}{23}(x+7). \] Multiplying through by $23$ gives $23y - 138 = -14x - 98$.
Step 6: Tidy into standard form.
Bringing all terms to one side: $14x + 23y - 40 = 0$, which is the equation of line $BC$.
\[ \boxed{14x+23y-40=0} \]