An equilateral triangle constructed from a reference triangle is a topic which is intersested by plane geometry lovers. See Napoleon equilateral triangle, Morley equilateral triangle....In this topic I propose more than 3000 equilateral triangles.
Consider $ABC$ be a triangle, $B_a$, $C_a$ be two points lie on $BC$, $C_b, A_b$ be two point lie on $CA$; $A_c, B_c$ be two points on the $AB$ such that $AB_aC_a$, $BC_bA_b$, $CA_cB_c$ be three equilateral triangle. Let $n$ be a positive integer number.
Let $A'$=triangle center $n$-th of $\triangle AA_bA_c$; $B'$=triangle center $n$-th of $\triangle BC_bA_b$; $C'$=triangle center $n$-th of $\triangle CC_aB_a$.
Let $A''$=triangle center $n$-th of $\triangle AB_cC_b$; $B''$=triangle center $n$-th of $\triangle BC_aA_c$; $C''$=triangle center $n$-th of $\triangle CA_cC_a$.
You can see define triangle center $n$-th in here and here
Problem:
For any positive integer $n$, any triangle $ABC$ then $A'B'C'$ be an equilateral triangle
For any positive integer $n$, any triangle $ABC$ then $A''B''C''$ be an equilateral triangle.
Triangles $ABC$ and $A'B'C'$ are perspective; $ABC$ and $A''B''C''$ are perspective
My question: I am looking for a proof of the problem above. You can check the conjecture above hold true for $n=1,\cdots ,3000$ in here the geogebra applet
Some new equilateral triangles I discovered recently in here: