First, I would like to show where I want to go...
... because I find this picture beautiful.
Now, consider any triangle $ABD$, and build an equilateral triangle $ABC$ on one of its longest sides.
Let draw two circles, the first passing by $B$, $C$, $D$ and the second with center in $A$ and passing by $D$ (i.e. with center on the opposite vertex with respect to the side $BC$).
These two circles determine a point $P$, in correspondence of their (other) intersection.
Let us do the same for the other two sides, obtaining other two points $N$ and $O$.
My conjecture is that the triangle $ONP$ is always equilateral.
Although it is likely and obvious/known result, my question is
Is there an elementary proof for such conjecture?
Thank you for your suggestions!
This is true. It has little to do with the original triangle. What you're doing is to reflect the point $D$ in each of the three angle bisectors of an equilateral triangle. The resulting points form an equilateral triangle because the product of any two of these reflections is a rotation through $\frac{2\pi}3$ around the centre of the triangle.