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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 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?

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No.

Clearly $ABC'$ and $AB'C$ are congruent (SAS). Let $P$ be the intersection of $BC'$ and $B'C$. Then triangles $BCP$ and $B'C'P$ are congruent (ASA) and so you get $BP=B'P$. Hence triangles $ABP$ and $AB'P$ are congruent and the result follows.

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  • $\begingroup$ This was really confusing to me, because it seems I did the congruence proofs you talked about, but I don't follow how proving $BCP$ and $B'C'P$ are congruent and $BP=B'P$, so the congruence of $ABP$ and $AB'P$ is what proves that P lies on the bisector. $\endgroup$ – Júlio Cezar Jan 20 '19 at 21:44
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Alright, so I have improved on what the post by user10354138 laid down, so that the proof is crystal clear (to me).

drawing

Let $P$ be the intersection of segments $BC'$ and $B'C$. It is evident from $SAS$ that triangles $ABC'$ and $AB'C$ are congruent. So are segments $BC' = B'C$, and the supplementary angles for $B = B'$ (in red) and the angles $C = C'$ (in blue).

Using $ASA$ with those angles and the sides $BC = B'C'$ (from $AB' = AB$ and $AC' = AC$), we find that the triangles $BPC$ and $B'PC'$ are congruent, so are the segments $BP = B'P$.

We let line $L$ be a line that cuts the angle $A$, passing through the intersection $P$. We let segment $AP$ of the line $L$ be a common side to triangles $ABP$ and $AB'P$.

Using segment $AB = AB'$, angle $B = B'$ and segment $BP = B'P$ from earlier proofs, we meet criteria for $SAS$. Thus triangles $ABP$ and $AB'P$ are congruent and angles $a$ and $b$ are equal.

Thus line $L$ is the bisector to angle $A$, where $a = b$. Since line $L$ passes through intersection $P$ forming segment $AP$, point $P$ lies on the bisector, so lines $BC'$ and $B'C$ meet on the bisector, as stated in the problem.

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