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Given any triangle $\triangle ABC$, and given one of its side, we can draw two lines perpendicular to that side passing through its two vertices. If we do this construction for each side, we obtain the points $D,E,F$ where two of these perpendicular lines meet at the minimum distance to each side.

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These three points can be used to build three triangles on each side of the starting triangle.

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The conjecture is that

The sum of the areas of the triangles $\triangle AFB$, $\triangle BDC$, and $\triangle CEA$ is equal to the area of $\triangle ABC$.

This is likely an obvious and very well known result. But I cannot find an easy proof of this. Therefore I apologize for possible triviality, and I thank you for any suggestion.

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    $\begingroup$ I am not too sure but did you consider that there might be a point inside the triangle, $G$ such that $AFBG$, $BDCG$ and $AECG$ are parallelograms? $\endgroup$ Nov 9, 2018 at 7:58
  • $\begingroup$ I agree with Raptor, and would guess it is the center of the inner circle. i.e. intersection of angle symmetrals. however, have you also tried it for triangles with an obtuse angle? there should be weird stuff happening there, since one of you orthogonals actually will go into the triangle! $\endgroup$
    – Felix
    Nov 9, 2018 at 8:14
  • $\begingroup$ The heights are meeting at that point $\endgroup$
    – Moti
    Nov 9, 2018 at 8:14
  • $\begingroup$ Thanks all for your comments! $\endgroup$
    – user559615
    Nov 9, 2018 at 9:38
  • $\begingroup$ @Raptor@Enkidu. Working on it! $\endgroup$
    – user559615
    Nov 9, 2018 at 9:52

3 Answers 3

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Draw the orthocenter. You get three parallelograms which immediately provide the answer.

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  • $\begingroup$ Nice and easy solution! $\endgroup$
    – YiFan Tey
    Nov 9, 2018 at 8:31
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The answer of Moti is perfect. Note, though, that this also means that this property holds only if the orthocenter is inside the triangle, otherwise the external triangles overlap and the required property does not hold any more.

Internal orthocenter.

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External orthocenter.

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    $\begingroup$ No, I do not think so. Maybe that we must consider the oriented area of the triangles build with respect to each side. I guess that the sum of the oriented areas of these triangles (some positive some negative) still equals the area of the original triangle. $\endgroup$ Nov 9, 2018 at 11:51
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    $\begingroup$ Signed areas do, indeed, save the result. $\endgroup$
    – Blue
    Nov 9, 2018 at 11:54
  • $\begingroup$ @FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter. $\endgroup$
    – Vaelus
    Nov 9, 2018 at 12:57
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Here is a diagram with brief explanation (using signed areas and color code) to show that the result holds true for obtuse triangles also, that is the case when the orthocenter $H$ lies outside $\triangle ABC$.

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$$[ABC] = \color{red}{[HBC]} - \color{green}{[HBA]} - \color{blue}{[HAC]}$$ $$[ABC] = \color{red}{[BDC]} - \color{green}{[ABF]} - \color{blue}{[CAE]}$$ $$[ABC] = \color{red}{[BDC]} + \color{green}{[AFB]} + \color{blue}{[CEA]}$$

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