Point of concurrency I am looking for the proof of the following claim:

Claim: Let $\triangle ABC$ be an arbitrary triangle, $D$ its nine-point center and $E,F,G$ are the nine-point centers of the triangles $\triangle ABD$ , $\triangle BCD$ and $\triangle ACD$ respectively. Then, the three straight lines $AF$ , $BG$ and $CE$ are concurrent.


GeoGebra applet that demonstrates this claim can be found here.
 A: There is a systematic way to solve such concurrency questions. It may seem rather tedious and inelegant but has several advantages, amongst them being, as an example, that
it can be used to consider other related results without additional effort. For example,  one can ask the corresponding question for other triangle centres.
Step 1.  Given four points in general position in the plane one defines a function $f$ which associates to $(A,D,B,E)$ the intersection of $AD$ and $BE$.  This can easily be computed by hand or set up, say,  in Mathematica.
One then composes  the map which associates a sixtuple $$(A,D ,B,E, C,F)$$ the triple $$(X,Y, Z)$$ where $X$ is the intersection of $BE$ and $CF$ etc. with
$$(X,Y,Z)\mapsto Y\wedge Z+Z\wedge X+X\wedge Y.$$
The vanishing of this function establishes that $AD$, etc., are concurrent.
Added remark: this multi-purpose function can be used to mechanise proofs of many of the classical concurrency results. One can try Fermat-Torricelli for starters.
Step 2.  One now uses the $p,q$ method to simplify the computations, i.e., assumes that the vertices of $  ABC$
are $(0,0)$, $(0,1)$ and $(p,q)$.  Using the expression for the barycentric coordinates of the nine point centre of a triangle (in terms of the side lengths) which can easily be found on the net, one can compute the coordinates of $  D$, $E$ and $F$ in terms of $p$ and $q$.  One can then plug these into the function of step 1 to finish the proof.
Confession.  The last part becomes too complicated to perform by hand but can be done with Mathematica.  For other centres with less intricate barycentric coordinates, it is more manageable.  Due to corona, I don't have access to my office or Mathematica and so haven't been able to actually compute these functions explicitly.
