A conjecture related to a circle intrinsically bound to any triangle

Question

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.

Answer 1

We have $AF=AD$ and $AB=AE$, so the triangles $AFD$ and $ABE$ are isosceles, so $FD\|EB$ and $BEDF$ is isosceles, thus inscriptible.

This shows $F$ is on the circle through $B,D,E$.

By analogy/symmetry, $G$ is also on it.

Answer 2

There can be no proof of the conjecture since it is false, for if the $\triangle ABC$ is obtuse, then one cannot guarantee that the circles will intersect the third side $AC$ at $D$ and $E.$

Unfortunately, one cannot bystep this by considering the line through $AC$ instead.

Edit: OP has found a way around this; he need only state as hypothesis that $A$ and $C$ be the acute angles of the triangle.

Answer 3

I am writing this to add to both the conjecture and the proof given.

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If the angle at $B$ is a greatest angle of the triangle $ABC,$ then the conjecture is true (provided we allow the points of the cyclic "pentagon" to coincide). Moreover, this pentapunctual circle is unique. This is clear when $ABC$ is scalene, for if one used $A$ or $C$ instead of $B,$ the largest angle, then necessarily the points $D$ and $E$ cannot exist since $AC$ is the longest side, thus it exceeds both $AB$ and $BC.$ If the triangle is isosceles (or even equilateral), then since $\hat B$ is equal to at least one other angle, the symmetry imposes that only one such circle exists.

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I notice that you have added something about the identity of this "$5$-point" circle. Again, it is indeed the case that its centre coincides with the incentre $I$ of the $\triangle ABC,$ for since $DF$ and $EG$ are chords of this circle, their perpendicular bisectors must intersect at its centre $I'.$ But we also know that the triangles $ADF$ and $CEG$ are isosceles with $AD=AF$ and $CE=CG$ respectively. Therefore the perpendicular bisectors of the sides $DF$ and $EG$ must also be angle bisectors of $\hat A$ and $\hat C$ respectively. This shows that their intersection $I'$ is not different from $I.$

PS. This is not directly related, but let me point out an interesting relationship between the incircle and the circumcircle of any triangle $ABC$ which I had not learnt of before. Let the intersection of the incircle with $AB,BC,CA$ be $C',A',B'$ respectively. Then the lines $AA',BB',CC'$ intersect in a centre of the triangle $Q$ which I have called the quasicentroid. Of course, there are thousands of known triangle centres (cf. The Encyclopedia of Triangle Centres), but I have been unable to determine whether $Q$ is part of the categorised centres, and if so under what name, hence the tentative term "quasicentroid."

Edit: The point I called $Q,$ I subsequently found out, is more well known as the gergonne point of the triangle.

Answer 4

By construction $ADF$, $CEG$ and $BFG$ are isoceles so that the bisectors of $DF$, $EG$ and $FG$ are also bisectors of the angles of the triangle and meet at the incenter.

By symmetries, $IB=ID=IE=IF=IG$.