Prove that, reflection of vertex of a triangle about angle bisector through other vertex lie on opposite side of triangle. 
$A(1,3)$ and $C({-2\over5},{-2\over5})$ are the vertices of a triangle $ABC$ and equation of angle bisector of $\angle ABC$ is $x+y=2$. Find equation of side $BC$

I tried a lot to solve this question, but was unable to. So I decided to look at the solution.
The solution included a line

Reflection of a vertex about angle bisector lie on the opposite side of triangle.

This line is enough to solve this question. 
But why is that true. Can anyone provide me a proof for it.

By drawing the reflection of A, we get A'.

P  is a point on x + y = 2 and is on the opposite side of AC other than B.
$\angle ABP = \angle A'BP$ because B lies on the perpendicular bisector of AC.
Also, $\angle ABP = \angle CBP$ because BP is given as the angle bisector of $\angle ABC$.
Therefore, B is on the extension of CA'. 
