Given the midpoint (or centroid) $D$ of any triangle $\triangle ABC$, we build three squares on the three segments connecting $D$ with the three vertices. Then, we consider the centers $K,L,M$ of the three squares.
My conjecture is that
The area of the triangle $\triangle KLM$ is equal to half of the area of the triangle $\triangle ABC$.
This is for sure a well known result (well, if true!). In this case, sorry for the trivial problem!
However, It would be great to have suggestions for developing a proof without words of such simple claim (again, if true), i.e. avoiding trigonometry, etc. Thanks for your help!
EDIT: The conjecture can be easily extended to any regular polygon built on the described segments (e.g. equilateral triangles yield to $1/3$ of the $\triangle ABC$ area, etc.).
EDIT (2): The (extended) conjecture appears to be true also by building the segments starting from the orthocenter (red, left), instead of the centroid (grey, right). The area of the final triangle $\triangle KLM$ is however the same!
Best Answer
Observe a spiral similarity with center at $D$ which takes $A$ to $K$. Then it takes $B$ to $M$ and $C$ to $L$. So this map takes triangle $ABC$ to triangle $KML$ which means that they are similary with dilatation coefficient $k={\sqrt{2}\over 2}$ So the ratio of the areas is $k^2 =1/2$.