# A conjecture about an equilateral triangle bound to any triangle

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?

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.

• I see. But this remains true also when the point $D$ is outside the equilateral triangle used for the construction. But, of course, it may be obvious.
– user559615
Jul 6, 2018 at 15:30
• @andrea.prunotto: I don't see where I assumed that $D$ is inside the triangle. Jul 6, 2018 at 15:36
• But I disagree with the fact that this has little to do with the original triangle, because the area of the obtained equilateral triangle strongly depends on the position of $D$ with respect to $AB$, i.e. on the shape of the original triangle, don't you think?
• @andrea.prunotto: I'm not sure I understand what you mean by that -- I just didn't use the original triangle in the proof; I'm not sure what it would mean for something to be related to that. $D$ does lie on that circle, because the reflections in the angle bisectors preserve the distance from the centre. Jul 6, 2018 at 16:05