# What is sufficient to prove Kiselev's Geometry #82?

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?

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.

• 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. – Júlio Cezar Jan 20 '19 at 21:44

Alright, so I have improved on what the post by user10354138 laid down, so that the proof is crystal clear (to me).

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.