Six points on an ellipse Can you prove the following proposition:

Proposition. Let $\triangle ABC$ be an arbitrary triangle with centroid $G$. Let $D,E,F$ be the points on the sides $AC$,$AB$ and $BC$ respectively , such that $GD \perp AC$ , $GE \perp AB$ and $GF \perp BC$ . Now let $H$ be a point on the extension of the segment $AB$ beyond $A$ such that $AH=GF$ . Similarly, define the points $I,J,K,L,M$ so that the point $I$ lies on the extended segment $AC$ and $AI=GF$ , the point $J$ lies on the extended segment $BC$ and $BJ=GD$ , the point $K$ lies on the extended segment $AB$ and $BK=GD$ , the point $L$ lies on the extended segment $AC$ and $CL=GE$ and the point $M$ lies on the extended segment $BC$ and $CM=GE$ . I claim that the points $H,I,J,K,L,M$ lie on an ellipse.


GeoGebra applet that demonstrates this proposition can be found here.
 A: It is easy to see that $IJ$ is parallel to $AB$, etc. The result follows from the converse of the Pascal's theorem: Consider the hexagon $MHKLIJ$, then the intersection points of the three pairs of opposite sides of this hexagon all lie on the infinite line.
A: We'll prove in two ways (via Pascal's Theorem and Carnot's Theorem) that the six points lie on a conic.  We'll switch to barycentric coordinates to show that the conic is an ellipse.
Let $a=\lvert BC \rvert, b=\lvert CA \rvert,c=\lvert AB \rvert.$
The medians through $G$ divide $\triangle ABC$ into six equal triangles.  These triangles make up, in pairs, the three triangles whose base  and apex are respectively a side of $\triangle ABC$ and $G$.  So these latter triangles have equal area and thus
$$
\begin{aligned}
& \lvert \triangle ABG \rvert=\lvert \triangle BCG \rvert=\lvert \triangle CAG \rvert \\
&\implies b\lvert GD \rvert=c\lvert GE \rvert=a\lvert GF \rvert \\
&\implies \lvert GD \rvert=k/b,\lvert GE \rvert=k/c,\lvert GF \rvert=k/a.
\end{aligned}
$$
for $k$, where $\lvert \triangle ABC \rvert=3k/2.$
Proof via the converse of Pascal's Theorem: (this is a slightly more detailed version of @Saginomiya's answer.).
Note that
$$
\frac{\lvert AI\rvert}{\lvert BJ\rvert}=\frac{\lvert GF\rvert}{\lvert GD\rvert}=\frac{b}{a}=\frac{\lvert AC\rvert}{\lvert BC\rvert}.
$$
Thus $HK\parallel IJ.$ Similarly, $LI\parallel MH$ and $LK\parallel MJ$.  These pairs are opposite sides of the  hexagon $MHKLIJ$.  By the converse of Pascal's Theorem, since opposite sides meet on a line (the projective line at infinity), the points $M,H,K,L,I,J$ lie on a common conic.
Proof by Carnot's Theorem:
Carnot's Theorem is like Menalaus' Theorem, except that we intersect a triangle with a conic instead of a line.  That theorem says that the six intersection points lie on a conic if and only if
$$
\frac{\lvert AH \rvert}{\lvert BH \rvert} \cdot
\frac{\lvert AK \rvert}{\lvert BK \rvert} \cdot
\frac{\lvert BJ \rvert}{\lvert CJ \rvert} \cdot
\frac{\lvert BM \rvert}{\lvert CM \rvert} \cdot
\frac{\lvert CL \rvert}{\lvert AL \rvert} \cdot
\frac{\lvert CI \rvert}{\lvert AI \rvert} =1.
$$
Filling in the lengths, we have
$$
\begin{aligned}
&  \frac{\lvert AH \rvert}{\lvert BH \rvert} \cdot
\frac{\lvert AK \rvert}{\lvert BK \rvert} \cdot
\frac{\lvert BJ \rvert}{\lvert CJ \rvert} \cdot
\frac{\lvert BM \rvert}{\lvert CM \rvert} \cdot
\frac{\lvert CL \rvert}{\lvert AL \rvert} \cdot
\frac{\lvert CI \rvert}{\lvert AI \rvert} \\
&= \frac{\lvert k/a \rvert}{\lvert BH \rvert} \cdot
\frac{\lvert AK \rvert}{\lvert k/b \rvert} \cdot
\frac{\lvert k/b \rvert}{\lvert CJ \rvert} \cdot
\frac{\lvert BM \rvert}{\lvert k/c \rvert} \cdot
\frac{\lvert k/c \rvert}{\lvert AL \rvert} \cdot
\frac{\lvert CI \rvert}{\lvert k/a \rvert} \\ 
&= \frac{\lvert AK \rvert}{\lvert BH \rvert} \cdot
\frac{\lvert BM \rvert}{\lvert CJ \rvert} \cdot
\frac{\lvert CI \rvert}{\lvert AL \rvert} \\ 
&= \frac{\lvert c+k/b \rvert}{\lvert c+k/a \rvert} \cdot
\frac{\lvert a+k/c \rvert}{\lvert a+k/b \rvert} \cdot
\frac{\lvert b+k/a \rvert}{\lvert b+k/c \rvert} \\ 
&= \frac{\lvert bc+k \rvert}{\lvert ac+k \rvert} \cdot
\frac{\lvert ca+k \rvert}{\lvert ba+k \rvert} \cdot
\frac{\lvert ab+k \rvert}{\lvert cb+k \rvert} \\ 
&= 1 
\end{aligned}
$$
To show that the conic is an ellipse:
Using Mathematica and the baricentricas.nb package I computed the conic's discriminant
$$
\frac{1}{4} a^2 b^4 c^4 k^4 \left(a^2-2 a b-2 a c+b^2-2 b c+c^2\right) (a b+k)^2 (b  
c+k)^2 (a b c+a k+b k)^2 (a b c+a k+c k)^2
$$
which has the same sign as $$d=a^2-2 a b-2 a c+b^2-2 b c+c^2.\tag{1}$$
The Hadwiger–Finsler inequality states that for a triangle with side lengths $a,b,c$ and area $T$
$$
a^2+b^2+c^2\ge  (a-b)^2+(b-c)^2+(c-a)^2+4\sqrt 3 T\tag{HF}
$$
But $(HF)$ implies
$$
-4\sqrt 3 T\ge a^2-2 a b-2 a c+b^2-2 b c+c^2,
$$
so the conic's discriminant is negative and therefore the conic is an ellipse.
A further observation:
None of the  proofs here depend on $k$ being a specific value. Therefore the six points lie on an ellipse as long as the lengthenings or shortenings are in proportion to $1/a,1/b,1/c$. So by varying $k$ (including values $k\lt 0$) we get a family of ellipses. Their centers are on the line $X(1)X(6)$, i.e the line through the incenter and symmedian point.  There's some more discussion of a related general case at
Bradley, Hexagons with Opposite Sides Parallel.
A: Here is a simple, if rather tedious, way to do this from scratch, using the $p,q$ method.  One can assume that the vertices are $(0,0)$, $(1,0)$ and $(p,q)$.  $G$ is then $\frac 13(1+p,q)$.  Using the unit normals to the sides one can easily calculate the lengths of $GH$, etc. and so the coordinates of the new points and thus verify that they lie on an ellipse. One advantage of this method is that it can potentially be used to generalise and deepen the result.
A: The OP has already an accepted answer, but ok... Here is a two parameters thematic generalization (and one parameter goes in a non-trivial direction).

In a triangle $\Delta ABC$ let $a,b,c$ and $h_a,h_b,h_c$ be the lengths of the sides and of the heights corresponding to the vertices $A,B,C$. Fix now $r_a,r_b,r_c>0$ so that we have the equal proportions:
$$ 
r_a:r_b:r_c=h_a:h_b:h_c=\frac 1a:\frac 1b:\frac 1c\ .
$$
Write $K=ar_a=br_b=cr_C>0$, a first parameter.
We draw the circles $(A)$, $(B)$, $(C)$ centered in $A,B,C$, having as radius respectively $r_a,r_b,r_c$.
Let $I$ be the incenter of $\Delta ABC$, and consider the parameter $k\in\Bbb R$.
The circle $(A)$ intersects the line $BA$ in two points, $A_B^+\in[AB$, and $A_B^+$, and $A_B^-$. Simiarly consider the other points $A_C^\pm$, $B_A^\pm$, $B_C^\pm$, $C_A^\pm$, $C_B^\pm$ as in the figure. The points with the upper minus are the ones from the OP. Then:
$$
\begin{aligned}
&A_C^+B_C^+\ \|\ 
AB\ \|\ 
A_C^-B_C^-\ \|\ 
A_\gamma B_\gamma
\ ,\ 
\\
&A_B^+C_B^+\ \|\ 
AC\ \|\ 
A_B^-C_B^-\ \|\ 
A_\beta C_\beta
\ ,\ 
\\
&B_A^+C_A^+\ \|\ 
BC\ \|\ 
B_A^-C_A^-\ \|\ 
B_\alpha C_\alpha
\ .
\end{aligned}
$$
Let $A_\beta$ be the point between $A_B^+$ and $A_B^-$ so that $AA_\beta =|mr_a|$, and so that the sign of $m$ corresponds. (A plus sign places $A_\beta$ between $A$ and $A_B^+$.) Similarly consider the points $A_\gamma$; $B_\alpha$, $B_\gamma$; $C_\alpha$, $C_\beta$.
The line $A_\beta A_\gamma$ is perpendicular in $S$ on $IA$ and intersects the circle $A$ in two points $A_B$ in the half-plane containing $B$ w.r.t. $IA$, and $A_C$ in the other one. Consider analogously the line $B_\gamma B_\alpha$, intersecting $IB$ in $T$, and the circle $(B)$ in $B_A,B_C$, and the line $C_\alpha C_\beta$, intersecting $IC$ in $U$, and the circle $(C)$ in $C_A,C_B$.

Then the six points $A_B,A_C$; $B_A$, $B_C$; $C_A$, $C_B$ are on a conic.


Note: The OP is obtained for the special constellation $r_a=\frac13h_a$, and $m=-1$. The generalization covers generically cases with sides not parallel to any diagonal.

Proof: For the parallelisms relations, it is enough to show $AB\|A_C^+B_C^+$. This is because of:
$$
\frac{AA_C^+}{BB_C^+}=\frac {r_a}{r_b}=\frac{1/a}{1/b}=\frac ba=\frac{AC}{BC}\ .
$$
(Thales in $\Delta ABC$.)
For the main part, i have a proof using baricentric coordinates. Hard to type in detail here. After the post of brainjam, using the same idea of involving Carnot's theorem i was also searching for a solution along such lines, using the triangle $A'B'C'$ from the picture, but the trigonometric relations involved are also hard to typeset. So...
Barycentric coordinates. Notations are following (bary-short.pdf by Evan Chen + Max Schindler) .
The displacement between $A(1,0,0)$ and a point $P(x,y,z)$ on the circle $A$ is
$(1-x,-y,-z)=(y+z,-y,-z)$, so the (homogeneous) equation of the circle $(A)$ is:
$$
(A)\ :\qquad
-a^2yz+b^2(y+z)z+c^2(y+z)y=\frac{K^2}{a^2}(x+y+z)^2\ .
$$
The point $A_B^⁺=\left(1-\frac K{ac}\right)A+\frac K{ac}B
=\left(1-\frac K{ac},\ \frac K{ac},\ 0\right)$ is verifying for instance this equation. We consider now the points $A_\beta,A_\gamma$,
they have correspondingly the coordinates
$$
\begin{aligned}
\left(1-\frac {mK}{ac},\ \frac {mK}{ac},\ 0\right)
&=[ac-mK:mK:0]\ ,
\\
\left(1-\frac {mK}{ab},\ 0,\ \frac {mK}{ab}\right)
&=[ab-mK:0:mK]\ ,
\end{aligned}
$$
and the line $A_\beta A_\gamma$ has the equation
$$
\begin{vmatrix}
x & y & z\\ac-mK & mK & 0\\ ab-mK & 0 & mK
\end{vmatrix}
=0\ .
$$
Its intersections with the circle $(A)$ are the points $A_B$, $A_C$ with coordinates $(x(A),y(A),z(A))$ given by the formulas:
$$
\begin{aligned}
x(A) &= 
1-\frac K{2abc}\Big(\ m(b+c) \pm (b-c)\sqrt D\ \Big)\ ,
\\
y(A) &=
\frac {K}{2abc}\Big(\ mb \pm  b\sqrt D\ \Big)\ ,
\\
z(A) &=
\frac {K}{2abc}\Big(\ mc \mp  c\sqrt D\ \Big)\ ,
\qquad\text{ where }\\[2mm]
D&=\frac{4bc - m^2((b+c)^2-a^2)}{(a + b - c)(a - b + c)}>0\ .
\end{aligned}
$$
For the other four points we have similar expressions. Using computer support, it turns out that there exist $P,Q,R;U,V,W$ (algebraic expressions in $a,b,c;K,m$) so that these points satisfy:
$$
g(x,y,z):=
Px^2 + Qy^2 + Rz^2 + 2Uyz + 2Vzx + 2Wxy = 0\ .
$$
(The expressions are rather complicated.) To obtain two linear equations corresponding to the above two points, we isolate the parts in $\sqrt D$ and "not in $\sqrt D$" obtained after expanding $g(x(A),y(A),z(A))$. The obtained six linear equation have a solution, sage code can be postponed.
The conic is an ellipse, for this we can proceed computationally, or give a deformation argument supported by convexity. For $K\to 0$ the limiting conic is an ellipse, being bounded inside the limit of the triangle $A'B'C'$ from the picture, where $B'C'\perp IA$, etc. and a continuous deformation changes the type only going through a parabola. But there is no such constellation of three chords $A_\beta A_\gamma$, ... of a parabola.
$\square$

Note: There may be a way using Carnot's reciprocal for the triangle $A'B'C'$ with orthocenter $I$, so we have to compute the powers like $C'A_B\cdot C'A_C=C'S^2-r_a^2$. In $\Delta C'IS$ the angle in $C'$ is $B/2$, so $C'S$ is $\cot \frac B2$ times $IS=IA+AS=4R\cos\frac B2\cos \frac C2$. We obtain for this power of $C'$ w.r.t. $(A)$ a certain expression involving the constants $K,m$, and the trigonometric functions $\sin$, $\cos$ computed in $\frac A2$, $\frac B2$, $\frac C2$. It should be brought in a form offering the symmetry for simplifications.

Bonus: A final picture showing some individual ellipses of the family, together with two other "thematic ellipses". (Involving $A_{CB}^+=B_A^+C_A^+\cap C$, and similar points.)

