A conjecture related to a circle intrinsically bound to any triangle

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.

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.