There's a nice feature I came up with using GeoGebra about four years ago. I have some kind of argument for it but it's not entirely rigorous, I hope to formalize it in the answers.
I would also be happy if anyone could provide a proof of pure Euclidean geometry
Given: $∆ABC$ is a triangle, the equilateral triangles $∆BCX,∆CAY,∆ABZ$ are constructed towards the outside of the triangle $∆ABC$, point $L$ is the center of $∆BCX$, point $M$ is the center of $∆CAY$, point $N$ is the center of $∆ABZ$, and $L',M',N'$ are three points such that $L'M'∥LM$, $M'N'∥MN$ and $N'L'∥NL$ such that $M',A,N'$ lie on one line and $N',B,L'$ lie on one line and $L',C,M'$ lie on one line.
Required: Prove that the area of the triangle $∆LMN$ is equal to a quarter of the area of the triangle $∆L'M'N'$.
An idea I have is to use the tiling property of Napoleon's triangle...
Adding this network of triangles we get a network like this:
Now it is clear that the large network is a withdrawal of a double network from the network of infinite Napoleon's triangles, and therefore each triangle has an area equal to four times the area of Napoleon's triangle.
I think the argument needs some modification to become rigorous. What should we do?
Is the property I discovered already known? If so, please provide a reference that mentions it
Edit: Regarding the tiling, it is copied from this video: https://youtu.be/KQ8cSuoopyc?si=rzq4soIOjSbBdasu
The circumcircles of the outer equilateral triangles, centred at $L$, $M$, $N$, meet at $F$, the Fermat point of triangle $ABC$. Note that $LM$ is the perpendicular bisector of common chord $CF$, $MN$ is the perpendicular bisector of $AF$, $LN$ is the perpendicular bisector of $BF$.
If we construct then diameters $FL'$, $FM'$, $FN'$ (see figure below) we get that $L'M'\parallel LM$, $L'M'=2LM$, $C\in L'M'$, and analogous relations for the other sides of $L'M'N'$. It follows that $L'M'N'$ are the same points defined in the question and we immediately get the requested property.