I saw the main problem, which I have described as Theorem 5 below, on C. Kimberling's website. I made some attempts to prove the problem with geometric methods. I wrote notes. I will present them below. I don't know only the proof of Theorem 5 (Main Problem).
This question doesn't includes multiple questions in one. It focus on Theorem 5 only. So the main problem is Theorem 5. I have given other theorems with their proofs. Because they contain my efforts that can be useful in solving the main problem.
My Efforts:
Theorem 1. Let $ABC$ be any triangle. $O$ is the circumcenter, $DEF$ is the orthic triangle, $KLM$ is the tangential triangle of the triangle $ABC$. (In other words, the triangle $ABC$ is the contact triangle of the triangle $KLM$.) Then $DE \parallel KL$, $DF \parallel KM$, $EF \parallel LM$.
Theorem 2. $OA \perp EF$, $OB \perp DF$, $OC \perp DE$.
Theorem 3. The lines $KD$, $LE$, $MF$ are concurrent at the point denoted by $X_{25}$.
Theorem 4. If the midpoints of the triangle $ABC$ are $P$, $R$, $S$ then the lines $KP$, $LR$, $MS$ intersect at the point $O$.
Proof of Th 1. Let $\angle BAC =\alpha $. By properties of the orthic triangle $ \angle EDC = \angle FDB = \alpha $. $\angle BOC = 2 \angle BAC = 2\alpha$. $OB \perp KM$ and $OC \perp KL$, so $ \angle BKC = 180^\circ -2\alpha$. Since $|KB|=|KC|$, we get $ \angle KBC = \angle KCB = \alpha $. By the equality of alternate interior angles, we conclude that $DF \parallel KM$, $DE \parallel KL$. Similarly, we understand that $EF \parallel LM$.
Proof of Th 2. The circumcircle of $ABC$ is the incircle of $KLM$. Hence $OA \perp LM$. By Theorem 1, $EF \parallel LM$. Similarly, we get $OB \perp DF$, $OC \perp DE$.
Proof of Th 3. By theorem 1, $DF \parallel KM$, $DE \parallel KL$, $EF \parallel LM$. So $DEF \sim KLM$ are similar triangles and they are homothetic. Therefore, the lines $KD$, $LE$, $MF$ are concur at a point (in the homothetic center $X_{25}$).
Proof of Th 4. Since $BOCK$ is a kite, we find $OK \perp BC$ and $OK$ bisect $[BC]$. So $OK$ passes through $P$ that the midpoint of $[BC]$. Similarly, $OL$ passes through $R$ and $OM$ passes through $S$. That is, all the lines $KP$, $LR$, $MS$ pass through the point $O$.
$\color{red}{\textbf{Main Problem (Theorem 5.)}}$ The point $X_{25}$ is on the Euler line of triangle $ABC$.
I will be glad if you share your ideas and solutions for the proof of Theorem 5.
Other Useful Notes and My Comments:
1. Orthic Triangle is here and there
2. Tangential Triangle is here and Contact Triangle is there
3. Euler Line is here
4. Theorems that are equivalent to the statements of Theorems 1-2-4 or that will enable us to obtain these theorems as a result are given in Advanced Euclidean Geometry by Roger Johnson (1929), page 172.
5. R. Johnson did not specify the concurrency in Theorem 3. I understand that he saw the homothetic between $DEF$ and $KLM$. Therefore, I don't think that he coluld't see concurrency of the lines at the homotethic center. (My opininon: R. Johnson knew but didn't write it, probably because it was easy.)
6. Theorem 5 does not exist in R. Johnson either. C. Kimbeling's website is given as a feature without proof.
Let $A_1,B_1,C_1$ be the mid points of the segments $HA$, $HB$, $HC$, so that the Euler circle $\odot(9)$ of $\Delta ABC$ passes through the nine points $D,E,F;P,Q,R;A_1,B_1,C_1$. Let $9$ be the center of $\odot(9)$.
The lines $AB$ and $DE$ are "antiparallel" w.r.t. the lines $CEA$ and $CDB$, so $ABCE$ is cyclic, so the point $H$ has the same power $\Pi_1$ w.r.t. the circle $\odot(ABCE)$ when computed as $HA\cdot HD$, respectively as $HB\cdot HE$. After a cyclic permutation of $A,B,C$ we see that $HC\cdot HF$ is equal to the same power $\Pi_1$.
It is known that $AH=2AA_1=2A_1H=2OP$. To see this, one can argument as follows. The triangle $\Delta A_1DP$ has the vertices on $(9)$, and a right angle in $D$, so $A_1P$ is a diameter of $\odot(9)$, $9$ is the mid point, so it is on the perpendicular bisector for the segment $DP$, and with the same argument also for $QE$ and $RF$. The mid point of $OH$ is the only point with this property, so it is $9$. Then the diagonals of $OA_1HP$ intersect in their mid points, so it is a parallelogram, so $A_1H=OP$. The similar relations $B_1H=OQ$, $C_1H=OR$ are then valid with the same argument.
The direction $QP\|AB$ is again antiparallel to the direction $KL\|DE$, so $QPKL$ is cyclic, so the point $O$ has the same power $\Pi_2$ w.r.t. the circle $\odot(QPKL)$ when computed as $OP\cdot OK$, respectively as $OQ\cdot OL$. After a cyclic permutation of $A,B,C$ we see that $OR\cdot OM$ is equal to the same power $\Pi_2$. Putting all together (and ignoring signs): $$ \begin{aligned} \Pi_1 &=HA\cdot HD=HB\cdot HE=HC\cdot HF\ ,\\ 2\Pi_2 &=2OP\cdot OK=2OQ\cdot OL=2OR\cdot OM\ ,\\[3mm] \frac{\Pi_1}{2\Pi_2} & =\frac{HA\cdot HD}{2OP\cdot OK} =\frac{HB\cdot HE}{2OQ\cdot OL} =\frac{HC\cdot HF}{2OR\cdot OM} \\ & =\color{blue}{ \frac{HD}{OK} =\frac{HE}{OL} =\frac{HF}{OM}}\ . \end{aligned} $$ Let $Y$ be the point on the Euler line $OH$ of $\Delta ABC$ with the property that $$ \color{blue}{ \frac {YH}{YO} } $$ has the same common value of the previous three fractions. Then we have the similarities of triangles $$ \begin{aligned} \Delta YHD&\sim\Delta YOK\ ,\\ \Delta YHE&\sim\Delta YOL\ ,\\ \Delta YHF&\sim\Delta YOM\ , \end{aligned} $$ so $Y$ is also on the lines $DK$, $EL$, $FM$, i.e. $Y=X_{25}$. (It is the center of homothey that maps $\Delta DEF$ to $\Delta KLM$.)
$\square$