We draw the circles with centers in $A$ and in $B$ passing by $D$, determining the new points $E$ and $F$ on the side $AB$.
If now we draw the two circles with center in $A$ and in $B$ and passing by $E$ and by $F$, respectively, we determine two new points $G$ and $H$ on the sides $AD$ and $DB$.
This post A conjecture related to a circle intrinsically bound to any triangle shows that the points $EGDHF$ determines always a circle.
Now we focus on the segments $DG$ and $CD$, and we draw their perpendicular bisectors. They intersect in the point $I$.
The circle with center in $I$ and passing by $C$, pass also through $G$ and $D$, for any $D$. Moreover, it always determines a point $J$ on the side $AC$ of the equilateral triangle.
A similar construction can be done starting from the perpendicular bisectors of $CD$ and $DH$, obtaining the center $K$ and the point $L$ on the side $CB$ of the equilateral triangle.
My conjecture is that the points $CJEFL$ always determine a circle.
Please, can you help me to find an elementary proof of such conjecture? Thanks for your suggestions!