A little conjecture about a circle related to any triangle 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.

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.


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!
 A: 
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. 
A: 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.
A: 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.

Now consider point H.

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 $E$.

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 $\overline{AC}$

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.
