Problem with several Simson lines in need of complete geometry solution The problem:

$A,B,C,A’,B’,C’$ are concyclic. The three Simson lines of $A,B,C$ about $\triangle A'B'C'$ intersect at $D,E,F$. The three Simson lines of $A’,B’,C’$ about $\triangle ABC$ intersect at $D’,E’,F’$.
Show that $D,E,F,D’,E’,F’$ are concyclic.
Moreover, if $H,H’$ are the orthocenters of $\triangle ABC,\triangle A’B’C’$. Show that the center of $\odot(DEFD’E’F’)$ is the midpoint of $HH’$.

It might be no good drawing the graph. When orthocenters and Simson lines appear at the same time, I think of the Steiner theorem, which states that

$A,B,C,D$ concyclic, $H$ is the orthocenter of $\triangle ABC$, then the midpoint of $DH$ is on the Simson line of $D$.

There’s also a property that might help:

Let $A,B\in\odot O$. The intersection angle of Simson lines of $A,B$ equals to the inscribed angle of $\overset{\LARGE\frown}{AB}$ in $\odot O$.


I’ll show my complex number method. (Barely needs figure)
First let $\odot{ABC}$ be the unit circle.
There’s a formula of Simson line of $P$ and $\triangle ABC$ that is
$$Pz-ABC\overline z=\frac1{2P}(P^3+\sigma_1P^2-\sigma_2P-\sigma_3).$$
So we can get the Simson line of $A$ and $B4$ then put them together to find intersection.
\begin{cases}Az-A’B’C’\overline z=\frac1{2A}(A^3+\sigma_1A^2-\sigma_2A-\sigma_3),\\
Bz-A’B’C’\overline z=\frac1{2B}(B^3+\sigma_1B^2-\sigma_2B-\sigma_3).\end{cases}
Subtract, and it is not hard to solve $\displaystyle z=D=\frac12(A+B+\sigma_1+\frac{\sigma_3}{AB}).$
The midpoint of $HH’$ is $M=\dfrac12(A+B+C+A’+B’+C’).$
Then $$|D-M|=\frac12|-C+\frac{A’B’C’}{AB}|=\frac12|ABC-A’B’C’|.$$
Thus all of $DM,EM,FM,D’M,E’M,F’M$ equal to this. $\#$
 A: The question was asked (in 1906) and answered by Absolonne (in 1907) in a French language publication Mathesis Question 1575, pg 23.  The answer here is organized into

*

*a diagram

*a screen shot of the original

*my translation of the original French, along with commentary

*some other reference materials

A caveat: this solution is not mine, I'm merely translating Absolonne's proof with small corrections, readability enhancements, and annotations.
Figure

Screen shot of 1907 solution

Translation and Commentary
[my comments will be in bold. Note that we start with two triangles $ABC,DEF$ instead of OP's triangles $ABC,A'B'C'$ ]
We consider two triangles $ABC, DEF$ inscribed in the same circle. Prove 1) that the Simson lines from the points $D,E,F$ with respect to the triangle $ABC$ form a triangle $D'E'F'$ similar to $DEF$ and
that the Simson lines of the points $A,B,C$ with respect to the triangle $DEF$ form a triangle $A'B'C'$ similar to $ABC$; 2) that the triangles $D'E'F' ,
A'B'C'$ are inscribed in the same circle whose center is the midpoint of the orthocenters of the triangles $ABC, DEF$. ([proposed by] S. KANTOR)
1) Solution by M. ABSOLONNE.  The Simson line of point $D$
with respect to triangle $ABC$ is perpendicular to the isogonal $AD_i$ of $AD$ with respect to angle $A$ and $DD_i$ is parallel to $BC$; similarly, the Simson line of $A$ with respect to the triangle $DEF$ is perpendicular to
the isogonal $DA_i$ of $DA$ relative to the angle $D$. The angles $A_iDD_i$ and
$A_iAD_i$ being equal, we conclude that the Simson line $E'F'$ of $D$
makes with $EF$ the same angle as the Simson line $B'C'$ of $A$ makes with $BC$. The sides of the triangle $D'E'F'$ being perpendicular to the isogonals
of $AD, AE, AF$, this triangle is directly similar to $DEF$; $A'B'C'$
will also be directly similar to ABC.
[The main takeaway here is that angle between $EF,E'F'$ is the same as that between $BC,B'C'$, etc.  This angle will be referred to as $\alpha$ in the following.]
2) $H$ being the orthocenter of $ABC$, the sides of the triangle $D'E'F'$ pass
through the midpoints $d,e,f$ of $HD,HE,HF$, and the triangle $def$ being homothetic to $DEF$, its orthocenter $O$ is the midpoint of $H,H'$, the orthocenters of $ABC, DEF$. [As will be shown below, this] point $O$ is also the center of the circumcircle of the triangle $D'E'F'$.
Let $R$ be the radius of the circle $ABC$, $\alpha$ the angle of the corresponding sides of $D'E'F'$ and $DEF$ which is also that of the sides of $A'B'C'$ and $ABC$; we will have in the triangle $OF'e$, for example:
$$\dfrac{OF'}{Oe}=\dfrac{\cos(\alpha)}{\cos(e)}.$$
[The latter follows from the law of sines on the triangle $OF'e$, and the fact that the relevant angles are complements of $\alpha$ and $e=\angle{def}$]
Now, if $M$ is the midpoint of $df$,
$$Oe = 2OM = R \cos(e) \\
\text{whence}\quad OF'=R \cos(\alpha).
$$
[He's saying that the distance of $D',E',F'$ from $O$ is $R \cos(\alpha)$, therefore $O$ is their circumcenter.  But there appears to be a typo in the line $Oe = 2OM = R \cos(e)$.  If O' is the circumcenter of $def$ then the line should read $Oe = 2O'M = R \cos(e)$. $2O'M$ is a well known formula for the distance $Oe$ between a vertex and the orthocenter $O$.  But $Oe = 2r \cos(e)$, where $r=R/2$ is the circumradius of $def$, is also well known as a formula for that distance.]
The circumcircle of the triangle A'B'C' will obviously have the same center and, according to the preceding equality, the same radius.
Other references
I was lead to Absolonne's proof by this note by Goormaghtigh, who has written related papers.
Johnson, Modern Geometry, pg 211, Section 338 has a special case of the theorem where the Simson lines are concurrent rather than forming a triangle.
Zaharinov, The Simson Triangle and Its Properties discusses Simson triangles, and this theorem in particular, using complex numbers.
