Consider an equilateral triangle $ADB$, and draw any triangle $ACB$, where $C$ is contained in the equilateral triangle $ADB$.
We draw two circles: one with center in $A$ and passing by $C$ (obtaining the point $K$ on the side $AB$ and the point $G$ on the side $AD$); the other one with center in $B$ and passing by $C$ (obtaining the point $J$ on the side $AB$ and the point $H$ on the side $DB$).
The side $AB$ results subdivided in three segments $AJ$ (blue), $JK$ (red), and $KB$ (green).
Now, we draw two other circles: one with center in $A$ and passing by $J$ (obtaining the point $F$ on the side $AD$ and the point $M$ on the side $AC$), and the other one with center in $B$ and passing by $K$ (obtaining the point $I$ on the side $DB$ and the point $N$ on the side $CB$).
This way, also the other sides $AD$ and $DB$ of the equilateral triangles result subdivided in three segments.
Let us now draw the angle bisectors of $F,B,G$ (green) and of $H,A,I$ (blue), intersecting in the point $L$.
My conjecture is that, no matter the exact position of $C$, the angle bisectors of $F,B,G$ and of $H,A,I$ form always and angle equal to $\frac{2\pi}{3}$.
Moreover, the point $L$ always lies on an arc of circle (whose center is located on the specular point of $D$ with respect to the side $AB$).
It is probably a very obvious result, and I apologize for the naivety in this case. However,
Is there an elementary proof for such conjecture?
Thanks for your help and for your suggestions!
As it seems you have noted by the coloring of your segments,
$$AF=DH,FG=HI,GD=IB.$$
We then see that, as (degenerate) pentagons,
$$BAFGD\sim ADHIB.$$
Let $L$ be the intersection of the bisectors angles $HAI$ and $FBG$. We then have
$$\angle LAB=\angle IAB + \frac{\angle HAI}{2}=\angle GBD+\frac{\angle FBG}{2}=\angle LBD$$
(where the second equality is by our similarity), so
$$\angle LAB=60^{\circ}-\angle LBA \implies \angle BLA=120^{\circ}.$$