Strange point lies on common tangent of 9-point circle and incircle 
Let $ABC$ be a triangle, with medial triangle $DEF$ and intouch triangle $PQR$. Let $J$ be the midpoint of $\overline{AD}$, and let $BJ$ meet $AP$ at $K$. Let $X$ be the point on ray $\overrightarrow{CB}$ such that $CX=CA$. Let the line through $K$ parallel to $BC$ meet $AX$ at $U$. Let $RU$ meet $BC$ at $T$. Prove that $T$ lies on the common tangent of the nine-point circle and the incircle of $\triangle ABC$.


I have a proof with barycentric coordinates. We can compute $K=(2(s-b):s-c:s-b)$, $X=(0:b:a-b)$, $U=(a-b+c:b:a-b)$ and $T=(0:a-c:a-b)$. The common tangent of the nine-point circle and incircle has equation $$(a-b)(a-c)x+(b-c)(b-a)y+(b-c)(a-c)z=0,$$ and the result follows.
However, as you might guess, I'm interested in a synthetic solution to this problem.
 A: 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. $\blacksquare$

