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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.

enter image description here

My conjecture is that

The area of the triangle $\triangle KLM$ is equal to half of the area of the triangle $\triangle ABC$.

enter image description here

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!

enter image description here

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  • $\begingroup$ Would you know a reference for the proof of this please? Thanks. $\endgroup$ – NoChance Nov 11 '18 at 17:36
  • $\begingroup$ @NoChance Well, I am not fully sure that this claim is true. Therefore I have no clue of a proof. Maybe should I remove the tag "alternative-proof"? Sorry for confusion. $\endgroup$ – user559615 Nov 11 '18 at 17:38
  • $\begingroup$ Thank you for your response, somehow I though there is a proof because of the wording "This is for sure a well known result". Very interesting idea. $\endgroup$ – NoChance Nov 11 '18 at 17:40
  • $\begingroup$ A proof without words generally takes a known proof, and finds a geometric representation of what's happening. If you don't yet know that it's true, then finding a proof would be a better place to start than looking for a specific type of proof. $\endgroup$ – Teepeemm Nov 11 '18 at 20:44
  • $\begingroup$ @Teepeemm Sure, I definitely agree. However, now greedoid provided a very nice and simple proof. So we can maybe focus on the proof without words! ; ) $\endgroup$ – user559615 Nov 11 '18 at 21:25
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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 $k^2 =1/2$.

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  • $\begingroup$ Thanks, greedoid for your always neat answers. I thought there could be some even easier solution, tangram/mosaic-like. But I am not sure. Thanks again however! $\endgroup$ – user559615 Nov 11 '18 at 17:45
  • $\begingroup$ Oh, sorry, I'm a bad reader (and writter). Anyway you should develop whole theory before you can play with ,,proofs without words''. You can not do anything just by the picture. $\endgroup$ – Aqua Nov 11 '18 at 17:50
  • $\begingroup$ True. I see your point. Say, I thought it should have been not too difficult to prove this (e.g. with trigonometry), but I was looking for a more visual approach. But, true: first one has to have a proof! Therefore, thanks again for yours! $\endgroup$ – user559615 Nov 11 '18 at 18:26
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    $\begingroup$ I've just realized that the conjecture can be extended by building any regular polygon on these segments (with equilateral triangles, the centers define a triangle of area $1/3$ of the initial triangle, with hexagons, the areas coincide and so on). I think your proof can be nicely generalized! $\endgroup$ – user559615 Nov 11 '18 at 21:51
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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.

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    $\begingroup$ Perhaps as a visual you can show the construction of the squares, show a copy of each side as it contracts in length and rotates into position along the diagonal to the shared centroid. Then connect the vertices to show the small triangle. I believe this would have the desired effect except for deriving the factor of sqrt(2) which would require a visual proof of this case of the pythagorean theorem. I imagine there are several of those available for such a popular theorem. $\endgroup$ – Nathan Nov 11 '18 at 23:16
  • $\begingroup$ Thanks Nathan. And welcome on MSE! $\endgroup$ – user559615 Nov 12 '18 at 4:31
  • $\begingroup$ Thanks for the interesting challenge and welcome. It just popped up on my Google feed! $\endgroup$ – Nathan Nov 13 '18 at 18:46
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This could be an idea for a "proof without words": enter image description here

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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$$\triangle ABC=2\triangle KLM$$ABC=2KLM

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$$BM=2BE$$KLM=2ABC

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.

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