In more than 2300 years since [Euclid's Elements](https://en.wikipedia.org/wiki/Euclid%27s_Elements) appear, there were only two equilateral triangles become famous: [The Morely equilateral triangle](http://mathworld.wolfram.com/FirstMorleyTriangle.html) and [the Napoleon equilateral triangle](https://en.wikipedia.org/wiki/Napoleon%27s_theorem). In more than 20000 points of [Encyclopedia of Triangle Centers](http://faculty.evansville.edu/ck6/encyclopedia/ETCPart10.html) don't has any triangle perspective triplet to $ABC$. 

But 4 years ago, Professor Paul Yiu had sent to me a very new nice equilateral triangle associated with two [Fermat points](http://mathworld.wolfram.com/FermatPoints.html) and [Kiepert hyperbola](http://mathworld.wolfram.com/KiepertHyperbola.html) of a [reference triangle](http://mathworld.wolfram.com/ReferenceTriangle.html). But I did not have a 
 proof. I posed at here and hope that have a solution.

1. Let $ABC$ be a triangle with two Fermat points $F_1$, $F_2$. Circles with center $F_1$ radius $F_1F_2$ meets the Kiepert hyperbola again at three points $M$, $N$, $P$.

Then triangle $MNP$ is equilateral and perspective to $ABC$ at triplet points. This mean: 

- $MA$, $NB$, $PC$ are concurrent
- $MB$, $NC$, $PA$ are concurrent
- $MC$, $NA$, $PB$ are concurrent

Three points above collinear.

[![enter image description here][1]][1]

2. Let $ABC$ be a triangle with two Fermat points $F_1$, $F_2$. Circles with center $F_2$ radius $F_2F_1$ meets the Kiepert hyperbola again at three points $M$, $N$, $P$.

Then triangle $MNP$ is equilateral and perspective to $ABC$ at triplet points. This mean: 

- $MA$, $NB$, $PC$ are concurrent
- $MB$, $NC$, $PA$ are concurrent
- $MC$, $NA$, $PB$ are concurrent

Three points above collinear

**See also:**

* [Morley's trisector theorem](https://en.wikipedia.org/wiki/Morley%27s_trisector_theorem)


  [1]: https://i.sstatic.net/EZ9FU.png