Given a triangle $ABC$, whose (one of the) longest side is $AC$, consider the two circles with centers in $A$ and $C$ passing by $B$.
(The part in italic is edited after clever observations pointed out buy some users: see below for details).
EDIT: You may be interested also in this other question Another conjecture about a circle intrinsically bound to any triangle.
The two circles determine two points $D$ end $E$, where they intersect the side $AC$.
We draw two additional circles: one with center in $A$ and passing by $D$, and the other one with center in $C$ and passing by $E$.
The new circles determines two points $F$ and $G$ where they intersect the sides $AB$ and $BC$, respectively.
My conjecture is that the points $BGEDF$ always determine a circle, whose center coincides with the incenter of the triangle.
Is there an elementary proof for such conjecture?
Since I am not an expert in the field, this can be a very well known theorem. I apologize in that case. Thanks for your help.