I am having difficulty realizing what would be sufficient to prove problem #82 asked in Kiselev's Geometry Book I.
82.* On one side of an angle $A$, the segments $AB$ and $AC$ are marked, and on the other side the segments $AB' = AB$ and $AC' = AC$. Prove that the lines $BC'$ and $B'C$ meet on the bisector of the angle $A$.
Very well, I have made this drawing and using $SAS$ congruence test (learned in the chapter) that the triangles marked $\alpha $ and $\alpha'$ are congruent, same for $\beta$ and $\beta'$. Now, is proving that sufficient? Or what should I be specifically proving to say that the point $P$ lies on the bisector and that the lines $BC'$ and $B'C$ meet at $P$?
Edit: I have thought of this solution: through $SAS$ I proved that the triangles $\alpha = \alpha'$. That means that the internal angles are also congruent between them. So the angle formed between segment $BP$ and the bisector is equal to the angle between segment $B'P$ and the bisector. So the bisector of angle $A$ is also the bisector of the new angle $BPB'$, with $P$ as the vertex. From that, $P$ must lie on the bisector. Is that sufficient?