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Given any triangle $\triangle ABC$, let denote with $D$, $E$ and $F$ the midpoints of the three sides, and draw the three circles with centers in $D,E,F$ and passing by $A,B,C$, respectively.

enter image description here

These three circles determine other three points $G,H,I$ in correspondence of their intersections (other than $A,B,C$).

My conjecture (likely pretty obvious, like this one) is that

The points $D,E,F,G,H,I$ are always concyclic.

enter image description here

In order to prove this, I think one must show that, e.g. the points $D,F,E,H$ lie on a regular trapezoid, but my attempts so far yield to a really muddled reasoning.

Any suggestion how to sketch a simple proof of such conjecture?

Thanks for your help!

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    $\begingroup$ $G$, $H$, $I$ are the feet of the altitudes from $A$, $B$, $C$ to the respective opposite side-lines, so that $\bigcirc DEFGHI$ is the nine-point circle of $\triangle ABC$. $\endgroup$ – Blue Aug 20 '18 at 8:53
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enter image description here By Kawhi. A and G are symmetry by DF. This is because AD=DG, DF=DF, AF=FG, ∆ADF≅∆GDF (SSS). In this way, ∠DGF=∠DAF. Since DE//AF, and AD//EF, we have ∠DAF=∠DEF. So, ∠DGF=∠DEF  G,D,F,E are on the same circle.

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    $\begingroup$ Please, do not provide only a link, type an explanation. $\endgroup$ – Taroccoesbrocco Aug 20 '18 at 20:11
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enter image description here

Triangle FHE and triangle FCE are congruent. (observe circles centered at E and F)

Triangle FCE and triangle DEF are congruent. ( $\because$ D, E, F are midpoints)

Therefore, Triangle FHE and triangle DEF are congruent.

Let S be the circle that D, E, F lie on. Then S is symmetric about the perpendicular bisector of line segment EF. ( $\because$ the center of S lies on the bisector)

The union of triangle FHE and triangle DEF is also symmetric about the perpendicular bisector of line segment EF. ( $\because$ the two triangles are congruent)

Since D is on the circle S, H is also on S.

Hence the points D, F, E, H are concyclic.


Note:

The name of the circle S is nine-point circle.

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  • $\begingroup$ Thanks a lot! Any idea if the center of the red circle is a known one? I checked en.wikipedia.org/wiki/Encyclopedia_of_Triangle_Centers, but I did not find anything. Thanks again! $\endgroup$ – user559615 Aug 20 '18 at 5:05
  • $\begingroup$ True. I was however referring to a center related to the initial triangle $\triangle ABC$, but I got your idea. Thanks! $\endgroup$ – user559615 Aug 20 '18 at 5:11
  • $\begingroup$ @AndreaPrunotto I have completely rewritten my answer, hoping that it is now rigorous. Please have a look at it, thank you! $\endgroup$ – user529760 Aug 21 '18 at 10:08
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First of all, the red circle in the following picture is the circumscribed circle to the triangle $\Delta DFA$ and its center $K$ is the common intersection to the red axes $d_1, e_1$ and $f_1$ of its sides. Circumscribed circle

Now consider point H. Point H and J

It is the intersection of the circle having the side $\overline{BC}$ as diameter and the circle having $\overline{AC}$ as diameter.

For this reason the triangle $\Delta{BCH}$ is right in H. Also the triangle $\Delta{AHC}$ is a right triangle being $\overline{AC}$ the diameter of the other circle. For this reason we demonstrate that $H$ is the intersection of the prolongation of side $\overline{AB}$ with the red circle.

Consider now the axis $d_1$ of side $\overline{FE}$: it is orthogonal both to $\overline{FE}$ and to $\overline{DH}$ and it is moreover an axis of symmetry. So segments $\overline{HF}$ and $\overline{DE}$ intersect in point $N$ which is on the axis $d_1$.

Now as $\overline{DB}$ and $\overline{FE}$ are parallel then $F\hat{E}D = E\hat{D}B$ and being symmetric as considered before, $H\hat{D}E = D\hat{H}F$ so the inscribed angle of $H$ and that of $E$ are equal, therefore they belong to the same red circle.

Using the same discussion, we can conclude that also the inscribed angle of point $G$ is the same of point $E$.

Point G

The last step is to demonstrate that point I lies on the same circle.

In order to do so, $I$ is the intersection of both circles centered in the midpoints $E$ and $D$, therefore the line $\overline{DE}$ contains the diameters of both circles. So point $I$ is symmetrical to $B$ with respect to this axis and moreover triangle $\Delta BDE$ is similar to $\Delta ABC$ with a similarity ratio of $\frac{1}{2}$ so the distance of $B$ and $I$ foromt $\overline{DE}$ is the same and therefore $I$ lies on side $\overline{AC}$.

Now $K$, the center of the circumscribed red circle, lies on the axis $f_1$ of symmetry of side $\overline{DE}$ and it is orthogonal to side $\overline{AC}$

Point S

For the same reasons of symmetry as before, the intersection $S$ of segment $\overline{ID}$ with segment $\overline{FE}$ lies on the axis $f_1$ therefore angle $D\hat{E}F = D\hat{I}F$ so even point I has the same inscribed angle of the other points.

For this reason all the points $D,F,I,E,H$ and $G$ lie on the same circle.

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  • $\begingroup$ Thanks for your nice work, Francesco! $\endgroup$ – user559615 Nov 4 '18 at 20:32

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