# 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?

• $G$, $H$, $I$ 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 $\triangle ABC$. – Blue Aug 20 '18 at 8:53

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.

• Please, do not provide only a link, type an explanation. – Taroccoesbrocco Aug 20 '18 at 20:11

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.

• 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! – user559615 Aug 20 '18 at 5:05
• True. I was however referring to a center related to the initial triangle $\triangle ABC$, but I got your idea. Thanks! – user559615 Aug 20 '18 at 5:11
• @AndreaPrunotto I have completely rewritten my answer, hoping that it is now rigorous. Please have a look at it, thank you! – user529760 Aug 21 '18 at 10:08

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.

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