Let $ABC$ be a triangle, $P$ be a point in the plane, $A'B'C'$ be the cevian triangle of $P$. Let point $A_b$ chosen on $CA$, point $A_c$ chosen on $AB$ such that $A'A_bA_c$ be an equilateral triangle and $A'A_bA_c$ with the same orientation than $ABC$. Let point $B_c$ chosen on $AB$, point $B_a$ chosen on $BC$ such that $B'B_cB_a$ be an equilateral triangle and $B'B_cB_a$ with the same orientation than $ABC$. Let point $C_a$ chosen on $BC$, point $C_b$ chosen on $AC$ such that $C'C_aC_b$ be an equilateral triangle and $C'C_aC_b$ with the same orientation than $ABC$. How can show that centroids of three equilateral triangles $A'A_bA_c$, $B'B_cB_a$, $B'B_cB_a$ are collinear?
Given some $V_C\in AB$, there is a unique equilateral triangle inscribed in $ABC$ with a vertex at $V$: its other vertices $V_A,V_B$ are given by the intersection of $BC$ with the line $AC$ rotated by $60^\circ$ clockwise around $V_C$, the intersection of $AC$ with the line $BC$ rotated by $60^\circ$ counter-clockwise around $V_C$. In particular $V_A=f(V_C)$ and $V_B=g(V_C)$ with $f,g$ being affine maps. This implies that the centroid of $V_A V_B V_C$, i.e. $\frac{V_A+V_B+V_C}{3}$, lies on a line.