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Suppose you have a scalene triangle as shown in the picture below. Let's call it triangle $ABC$.

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Inside that triangle you do the following things: Draw a point $F$, that lies on $\overline {AB}$, and so that $\overline {AF} =1$. Draw a point $G$, that lies on $\overline {BC}$, so that $\overline {CG}= 1$. Unite the points $F$ and $G$. Then draw the midpoints of $\overline {AC}$ and $\overline {FG}$. Let's call them $I$ and $H$ respectively. And then draw the line segment $\overline {IH}$. You will have the following figure:

enter image description here

But what I am really interested in is the quadrilateral, so here is a picture of it a little bit zoomed

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This was a part of a problem I was solving, but I added the fact that the triangle was scalene. While I was solving the problem I thought that the quadrilaterals $AFHI$ and $HICG$ were congruent, and that there might exist some kind of relationship between their angles. I first thought $ \angle {FAI} \cong \angle {ICG}$, but because the triangle is scalene, that can´t be true. Then I thought that probably one of the angles of $AFHI$ could be congruent with one of the angles of $HICG$. So I drew in Geogebra but found that none of them were congruent. I thought that if two figures were congruent their angles would be congruent too. Or are this figures still congruent? And what relationship exists between their angles?.

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  • $\begingroup$ "I thought that if two figures were congruent their angles would be congruent too." Yes, that's how it works. By definition, figures are congruent if and only if all their "corresponding parts" (sides and angles) are congruent. Since simple GeoGebra measurement tells you the angles of these quadrilaterals don't match, those quadrilaterals simply are not congruent. Now, one could derive some relation between the angles using trigonometry, but they aren't particularly nice. $\endgroup$ – Blue Apr 2 '18 at 7:54
  • $\begingroup$ Mmm. So how many non-congruent quadrilaterals but with the same sides can there exist? $\endgroup$ – Vmimi Apr 2 '18 at 17:52
  • $\begingroup$ I also thought that if you draw lines $ \overline {AG} $ and $ \overline {FC} $, and if you apply the cosine rule you may derive a relationship, but I don't know. $\endgroup$ – Vmimi Apr 2 '18 at 17:57
  • $\begingroup$ There's a whole continuum of non-congruent quadrilaterals with the same sides. Take the simplest example: all sides equal. There are infinitely-many rhombi with a given side length ... square, "flat", and everything in between. Generally, imagine building a quadrilateral with fixed-length rods and flexible hinges. Unlike a triangle, the figure isn't "rigid"; you can push and pull it into various shapes. $\endgroup$ – Blue Apr 2 '18 at 18:18
  • $\begingroup$ As I mentioned, you can use trig to get relations among angles. For example, with $b := |\overline{AC}|$ and $d := |\overline{AF}| = |\overline{CG}|$, one can derive $$\frac{\cos\angle AFG}{\cos\angle CGF} = \frac{b \cos A - d(1-\cos B)}{b \cos C - d (1-\cos B)}$$ (barring typos and/or errors made in haste). I'll leave you to relate the angles at $H$ and $I$. $\endgroup$ – Blue Apr 2 '18 at 19:13

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