$ABC$ be a triangle with $A(1,3)$ and $y=x,$ $y=-2x$ are the equation of either both the internal angle bisectors or both the external angle bisectors of angles $B$ and $C$ respectively. Then find the equation of $BC$ and the length of the inradius of the triangle $ABC.$
In my opinion, the question firsts wants the reader to determine whether the equations are of internal angle bisectors or external angle bisectors. And then he will solve accordingly. I assumed that the equations are of internal angle bisectors and By God's Grace, I think that I assumed the truth and solved the question. But I'm not able to prove that those equations can't be of external angle bisectors. For internal angle bisectors, my work is like this
Reflect $A$ on $y=x$ as $A'.$ Now $A'$ must lie on $BC.$ Now reflect $A'$ on $y=-2x$ as $A''.$ Now $A''$ must lie on $AC.$ Now we find equation of $AC$ and we'll solve it simultaneously with $y=-2x$ to get the coordinates of $C$ then we'll solve for the equation of $BC$ then we'll solve equation of $BC$ simultaneously with $y=x$ to get coordinates of $B.$ Now for inradius use the formula $r=\frac{\Delta}{s}$
I got equation of $BC$ as $y=1$ and inradius as $0.5$
Now I want someone to help me proving that those equations can't be of external angle bisectors (or if they can be then what are the consequent answers) and also please check my answers for the internal case.