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Consider an equilateral triangle $ADB$, and draw any triangle $ACB$, where $C$ is contained in the equilateral triangle $ADB$. enter image description here

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 $DB$).

enter image description here

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 $CB$).

enter image description here

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 $L$. enter image description here

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!

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    $\begingroup$ Rotating $\triangle ABD$ by $120^\circ$ (clockwise) about its center carries the bisector through $B$ onto the (original) bisector through $A$. $\endgroup$ – Blue Jul 18 '18 at 23:44
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    $\begingroup$ BTW, here's an equivalent construction that avoids the point $C$: Let $\overline{JK}$ be any subsegment of $\overline{AB}$ in equilateral $\triangle ABD$. Transfer the endpoints of that segment onto $\overline{AD}$ and $\overline{BD}$ via circles about $A$ and $B$; this gives your segments $\overline{FG}$ and $\overline{HI}$. Then, the bisectors of $\angle FBG$ and $\angle HAI$ make an angle of $120^\circ$. $\endgroup$ – Blue Jul 18 '18 at 23:55
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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

$$\angle LAB=60^{\circ}-\angle LBA \implies \angle BLA=120^{\circ}.$$

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