Four concyclic triangle centers

Question

Can you prove the claim given below? Inspired by Lester's theorem I have formulated the following claim:

Claim. Given any scalene triangle $\triangle ABC$ . Let $D$ be the reflection of incenter in sideline $AB$, and define $E$ and $F$ cyclically. The lines $CD$, $BF$, $AE$ concur in X(79) . Then, the two Fermat points , incenter and $X(79)$ lie on the same circle.

GeoGebra applet that demonstrates this claim can be found here.

Answer 1

There is a standard method to solve this problem (using the $p,q$ method) which can be used to attack the general one of when four given triangle centres are cyclic for any triangle. We suppose that the centres have functions $f_1,\dots,f_4$ where we can assume wlog that each $f$ is homogeneous and has cyclic sum $1$ (we are using the concepts and terminology of the Encyclopedia of Triangle Centers). Then we can regard the determinant of the $4$ by $4$ matrix with rows $$x_i^2+y_i^2\quad x_i\quad y_i\quad 1,$$ as a function of $p,q$ where $$(x_i,y_i)=(f_i(b,1,a)+pf_i(1,a,b),\ qf_i(1,a,b)),$$ $a^2=(p-1)^2+q^2$ and $b^2=p^2+q^2$. The required condition is that this be the zero function. For simple centre functions it can be computed by hand and it is easy to write a programme, say in Mathematica, to attack the general case.