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First, I would like to show where I want to go...

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... because I find this picture beautiful.

Now, consider any triangle $ABD$, and build an equilateral triangle $ABC$ on one of its longest sides.enter image description here

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 $BC$).

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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 $O$.

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My conjecture is that the triangle $ONP$ is always equilateral.

enter image description here

Although it is likely and obvious/known result, my question is

Is there an elementary proof for such conjecture?

Thank you for your suggestions!

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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.

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  • $\begingroup$ 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. $\endgroup$
    – user559615
    Jul 6, 2018 at 15:30
  • $\begingroup$ @andrea.prunotto: I don't see where I assumed that $D$ is inside the triangle. $\endgroup$
    – joriki
    Jul 6, 2018 at 15:36
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    $\begingroup$ Yep, got it! Ignore my comment. Thanks for your answer! $\endgroup$
    – user559615
    Jul 6, 2018 at 15:39
  • $\begingroup$ 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? $\endgroup$
    – user559615
    Jul 6, 2018 at 15:47
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    $\begingroup$ @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. $\endgroup$
    – joriki
    Jul 6, 2018 at 16:05

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