# Four concyclic triangle centers

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.

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.