Proof without words of a simple conjecture about any triangle

Question

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!

Answer 1

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 $k^2 =1/2$.

Answer 2

The two triangles share a centroid. The small triangle vertices are one half of each square's diagonal while the larger triangle has the length of a side. The triangles are similar since lengths are proportional by a factor of sqrt(2) which we square for an area comparison of 2:1 in favor of the larger one.

Answer 3

This could be an idea for a "proof without words":

Answer 4

In the special case where $\triangle ABC$ is right and isosceles, it seems true almost visually that, in triangles $KLM$ and $ABC$, base$$KM=BC$$and altitude$$BA=2BE$$making$$\triangle ABC=2\triangle KLM$$

On the other hand, if we build the squares on the sides of $\triangle ABC$, as in the figure below, the reverse appears to happen:$$\triangle KLM=2\triangle ABC$$since base$$KL=AC$$and altitude$$BM=2BE$$

OP's conjecture seems, as some are suggesting, a particular result of a larger general theory?

The above figures suggest perhaps the further question, whether a "proof without words" is ever really possible; it seems words are implicitly present in any train of thought, and a proposition that was truly self-evident would not be a proof strictly speaking but rather a basis or element of a proof.