Six points on an ellipse

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.

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.

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.