Strange point lies on common tangent of 9-point circle and incircle
The question has been asked and answered at Art of Problem Solving. User amar_04 gives a synthetic proof. Reproduced here, in case the URL breaks.
Just use Menelaus repeatedly
$\textbf{LEMMA 1:-}$ The Tangent to the Incircle at the Feuerbach Point and the Nagel Line of $\Delta ABC$ are Isotomic Lines. Proof:- See Theorem 10 here
$\textbf{LEMMA 2:-}$ Let $\tau_1,\tau_2$ be two Isotomic Lines in $\angle A$ cutting $BC$ at $X,Y$. Then $BX=CY$
Proof:- Let $\tau_1$ cut $\overline{AB},\overline{AC}$ at $\{P,M\}$ and let $\tau_2$ cut $\overline{AB},\overline{AC}$ at $Q,N$. Apply Menelaus Theorem on $\Delta ABC$ cut by the transversals $\tau_1,\tau_2$ $$\begin{align*}\frac{CX}{XB}\cdot\frac{BP}{PA}\cdot\frac{AM}{MC}=-1=\frac{BY}{YC}\cdot\frac{CN}{NA}\cdot\frac{AQ}{QB}&\implies\frac{CX}{XB}=\frac{BY}{YC}\\ &\implies\frac{CB}{XB}=\frac{CB}{YC}\\ &\implies BX=CY\end{align*}$$
$\textbf{LEMMA 3:-}$ If the Tangent $(\ell)$ to the Incircle at the Feuerbach point of $\Delta ABC$ cuts $\overline{BC}$ at $X$. Then $BX=\frac{a(b-a)}{b+c-2a}$
Proof:- Let $G,N_a$ be the Centroid and Nagel Point of $\Delta ABC$ and let $\overline{AG}\cap\overline{BC}=\{M\}$ and $\overline{AN_a}\cap\overline{BC}=\{P\}$. Let the Nagel Line of $\Delta ABC$ cut $\overline{BC}$ at $Y$. Now applying Menelaus Theorem on $\Delta AMP$ cut by the Nagel Line gives
\begin{align*}\frac{PY}{YM}\cdot\frac{MG}{GA}\cdot\frac{AN_a}{N_aP}=-1 &\implies\frac{PY}{YM}=\frac{2(s-a)}{a}\\ &\implies\frac{PM}{MY}=\frac{2s-3a}{a}\\ &\implies\frac{\frac{a}{2}-(s-b)}{MY}=\frac{2s-3a}{a}\\ &\implies MY=\frac{ab-ac}{2(b+c-2a)}\\ &\implies CY=MY+CM\\ &=\frac{ab-ac}{2(b+c-2a)}+\frac{a}{2}\\ &=\frac{a(b-a)}{b+c-2a}\end{align*} Now by $\textbf{LEMMA 1}$ we get that $\ell$ and the Nagel Line are Isotomic Lines and if $\ell\cap\overline{BC}=X$. Then from $\textbf{LEMMA 2}$ we get $CY=BX=\frac{a(b-a)}{b+c-2a}$.
Now come back to the Problem at hand. Applying Menelaus Theorem on $\Delta ADP$ cut by the transversal $\overline{BI}$ we get $$\frac{DB}{BP}\cdot\frac{PK}{AK}\cdot\frac{AJ}{JD}=-1\implies\frac{AK}{PK}=\frac{a}{2(s-b)}=\frac{AU}{XU}$$ Again by Menelaus Theorem we get $$\frac{XT}{TB}\cdot\frac{BR}{RA}\cdot\frac{AU}{XU}=-1\implies > \frac{BX}{TB}=\frac{2s-3a}{a}\implies TB=\frac{a(b-a)}{b+c-2a}$$ So by $\textbf{LEMMA 3}$ get that $T\in$ on the Tangent to the Incircle at the Feuerbach Point. $\blacksquare$