An interesting concurrence problem First, construct a triangle $\triangle ABC$. After that, erect three equilateral triangles $\triangle ABF$, $\triangle BCD$, $\triangle CAE$ externally. Suppose the midpoints of segment  $AF$, $BF$, $BD$, $CD$, $CE$, and $AE$ are $M_1$, $M_2$, $M_3$, $M_4$, $M_5$, and $M_6$. Connect the lines $M_1M_2$, $M_3M_4$, and $M_5M_6$. Let $M_1M_2\cap M_5M_6=P$, $M_1M_2\cap M_3M_4=Q$, $M_3M_4\cap M_5M_6=R$. Last, connect lines $AP$, $BQ$, and $CR$.
It seems that the three lines $AP$, $BQ$, and $CR$ concur at a point $X$. However, I'm not able to prove that. If my hypothesis is true, is there a special name for that point? (e.g. incenter, orthocenter...)
I suspect that the point $X$ is the symmedian point (Kimberling center $X_6$) of $\triangle ABC$. But still, it's only a guess. I'm thinking of a rigorous proof.

 A: Here is a slight generalization of the given proposition.
To show it, it is simplest to use barycentric coordinates. A quick reference for this is:
Barycentric Coordinates for the Impatient, Max Schindler and Evan Chen
The solution will (easily) compute the point of intersection $PA\cap QB\cap RC$
in its barycentric coordinates. The computation can be reduced to only a few lines, i postpone this simplified version, the expert can switch directly to that point.


Proposition: Let $ABC$ be a given triangle, denote the lengths of the sides by $a,b,c$ as usual. Consider some "(signed) parameter" $s\ne 0$, and draw parallels $M_a$, $M_b$, $M_c$ to the sides of the triangle at (signed) distance $ta$, $tb$, $tc$ from the sides with the lengths $a,b,c$ respectively. (A positive sign means going towards the exterior of the triangle.) The three parallels determine a triangle $\Delta PQR$, where
$P=M_b\cap M_c$,
$Q=M_a\cap M_c$,
$R=M_a\cap M_b$.


Then the lines $PA$, $QB$, $RC$ are the concurrent in $X(6)$, the symmedian (alias Lemoine, alias Grebe) point of $\Delta ABC$.

(Alternatively, construct similar rectangles on the sides $AB$, $BC$, $CA$, all of them being either in the opposite half plane, or in the same half-plane w.r.t. the given triangle, and take $M_a$, $M_b$, $M_c$ to be the lines defined by the parallel sides to $AB$, $BC$, $CA$ respectively in these rectangles.)

For $X(6)$ the barycentric coordinates and further properties are described in the ETC.
Before we start the proof, here is just an observation related to the many comments to the OP. For two general triangles $\Delta ABC$, $\Delta A'B'C'$ with three pairs of
parallel sides,
$AB\|A'B'$,
$BC\|B'C'$,
$CA\|C'A'$, the lines $AA'$, $BB'$, $CC'$ are concurrent. This is a corollary to theorems in projective geometry (Desargues). Or it can be shown by taking the intersection $X=AA'\cap BB'$, and using the similarity of the two triangles and from
$$
\frac{XA}{XA'} =
\frac{AB}{A'B'} =
\frac{AC}{A'C'} \ ,
$$
so $\Delta XAC\sim \Delta XA'C'$, they have the same angle in $X$, so $X,C,C'$ are on a line.

Proof:
Let $A_H$ be the projection of $A$ on $BC$. We do not need this - but to have an explicit situation the barycentric coordinates of $A_H$ are
$$
A_H = [0:\tan B:\tan C]=\left(\ 0\ ,\ \frac{\tan B}{\tan B+\tan C}\ ,\ \frac{\tan C}{\tan B+\tan C}\ \right)\ .
$$
Let $T$ be the point of intersection of $AA_H$ with $M_a$. Then we have
$\displaystyle AA_H=h_a=\frac{2S}a$, the (signed) length $A_HT$ is $ta$, as given between $BC$ and $M_a$, so we can write the equality:
$$
A_H = \frac{h_a}{h_a+ta}T + \frac{ta}{h_a+ta}A\ .
$$
Here, instead of the points $A_H$, $T$, $A$ we plug in the barycentric coordinates.
Looking at the first component only, we get from
$$
\begin{aligned}
(0,*,*) &= \frac{h_a}{h_a+ta}T + \frac{ta}{h_a+ta}(1,0,0)\qquad\qquad\text{ the shape for the point $T$:}\\
T &= \frac{h_a+ta}{h_a}(0,*,*) - \frac{ta}{h_a}(1,0,0) 
= \left(\ -\frac {ta}{h_a}\ , \ *\ ,\ *\ \right)
= \left(\ -\frac {ta^2}{2S}\ , \ *\ ,\ *\ \right)
\ .
\end{aligned}
$$
A line parallel to $BC$ has an equation of the shape $x=$constant, so the equation of $M_a$ is in barycentric coordinates $(x,y,z)$:
$$
\begin{aligned}
&(M_a)\qquad &x &=  -\frac {ta^2}{2S}\ ,
&&\text{ and similarly:}\\
&(M_b)\qquad &y &= -\frac {tb^2}{2S}\ , \\
&(M_c)\qquad &z &= -\frac {tc^2}{2S}\ .
\end{aligned}
$$
The intersection $P=M_b\cap M_c$ has thus the coordinates:
$$
P=\left(\ *\ ,\ -\frac {tb^2}{2S}\ ,\ -\frac {tc^2}{2S}\ \right)
=[\ *\ :\ b^2\ :\ c^2\ ]
\ .
$$
The euqation of the line $PA$ is given by the vanishing of the following $3\times 3$-matrix determinant, where the second and third row in it are related to barycentric coordinates for $A$ and $P$:
$$
(PA)\qquad
0 
=
\begin{vmatrix}
x & y & z\\
1 & 0 & 0\\
* & b^2 & c^2
\end{vmatrix}
= -
\begin{vmatrix}
y & z\\
b^2 & c^2
\end{vmatrix}
\qquad\text{ i.e. }\qquad
\frac{y}{b^2} = \frac{z}{c^2} 
\ .
$$
Similar equation hold for the lines $QB$ and $RC$. The point $[a^2:b^2:c^2]$
satisfy all these equation, so it is the needed intersection.

Short version of the proof: The point $T$ on both $M_a$ and the height $AA_H$ from $A$ has the $x$-coordinate equal to the signed proportion $-ta/h_a$, so the $M_a$ has the equation $x=-ta/ha=-ta^2/(2S)$. For $M_b$ and $M_c$ we extrapolate the equations $y=-tb^2/(2S)$ and $z=-tc^2/(2S)$. The point $P$ is thus $[*:b^2:c^2]$, a point lying as $A$ on the line $y/b^2=z/c^2$, the symmedian line through $A$. So the wanted intersection is the symmedian point $X(6)=[a^2:b^2:c^2]$.
A: Referencing the comments where it is observed that the triangles PQR and ABC are homothetic, here is another proof requiring only homothetic triangles, that does not use Desargues' theorem, in the image below. The vertices of the outer triangle are not labeled, but they are A, B, and C, with A at the top and going counterclockwise. Also note that as a proof by contradiction the vertex A' of the inner triangle is shown out of position for the contradiction.

