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.)