Given an equilateral triangle $ABC$, we choose a point $D$ inside it, determining a new triangle $ADB$.
We draw the circles with centers in $A$ and in $B$ passing by $D$, determining the new points $F$ and $G$ on the side $AB$.
These two points define the segment $AF$ (blue), $FG$ (green), and $GB$ (red). We denote $\alpha=\overline{AF}$, $\gamma=\overline{FG}$, and $\beta=\overline{GB}$.
So far, I was able to prove that if $\frac{\gamma^2}{2\alpha\beta}=1$, then $D$ lies on the arc of the circle with center in the midpoint $M$ of the side $AB$, and with diameter equal to $\overline{AB}$ (in these conditions, $ABD$ is a right triangle).
On which curves lies the point $D$ if $$ \frac{\gamma^2}{2\alpha\beta}=n, $$ where $n\in\{2,3,4\ldots\}$?
Thanks for your suggestions!