In $\Delta ABC, A\equiv (1,2)$. If the angle bisector of $\angle B$ is $y=2x+10$ and perpendicular bisector of $AC$ is $y-x=0$, find the equation of $BC$
The solution given says that:
The image of point $A$ about the angle bisector of $C$ lies on $BC$.
The problem becomes easy after this since we now know $2$ points on the line $BC$(Another being $C$ itself)
Is the statement given in the answer a well known theorem/property? Im not exactly sure how I can prove its validity.
Hint: As I said in comment the equation given for angle bisector of angle $B$ can be that of exterior, as can be seen in figure. We must use this fact that the intersection of perpendicular bisector of $AC$ and bisector of angle $B$ is on circumscribed circle of triangle. So we extend these two to intersect at $D (-10, -10)$ and draw a circle center at $O(-3, -3)$ and radius ($R=6.4$) passing $A,C$ and D. Vertex $B$ is on this circle. Vertex $C$ is the mirror of $A$ and has coordinates $C( 2, 1)$, the coordinates of $B$ is $B(-3.2, -3.4)$ so equation of $BC$ is:
$$y+3.4=\frac{1+3.4}{2+3.2}(x+3.2)$$
$$y=(\frac{4.4}{5.2}=\frac{11}{13})(x+3.2)-3.4$$
Maybe the solution should say '...about the angle bisector of $∠B$ lies on $BC'$. This is true, if we reflect $A$ in the angle bisector it will land on $BC$ (not necessarily at point $C$, maybe $BC$ produced), it will form an isosceles triangle with $A,B$ and the image of $A$, ($A'$), so will have a line of symmetry that is the angle bisector, try a quick sketch of any triangle.
The point $M(-3,4)$ is where $AA'$ crosses and is perpendicular to $y=2x+10$ so the other point on $BC$ is $(-7,6)$, (AM is 4 left 2 up, then repeat).
However the original question doesn't seem realistic, as can be seen by drawing diagrams, i.e is it really possible to have an angle bisector at $B$ of the equation given? There seems to be a mistake somewhere in the solution or the question you were set, but hope this helps.
