Claim. Quadrilateral $WXYZ$ is cyclic, and its circumcentre is $J$.
Proof. As $N$ is the midpoint of $\widehat{BAC}$, $W$ and $Z$ lie on $BC$, and $X$ and $Y$ are the second intersections of $NW$ and $NZ$ with $\Gamma$, we have that $WXYZ$ is cyclic.
Let the parallel to $BC$ through $N$ intersect $TU$ and $TV$ at $U'$ and $V'$, respectively. Then $U'$ is the intersection of the tangents to $\Gamma$ at $N$ and $X$, so $U'N = U'X$. As $NU' \parallel BC$, $U'NX$ is similar to $UWX$, so $UW = UX$ as well. Hence, the perpendicular bisector of $WX$ is the internal bisector of $\angle XUW$, which is the external bisector of $\angle VUT$. Analogously, the perpendicular bisector of $YZ$ is the external bisector of $\angle TVU$. This means that the circumcentre of $WXYZ$ is the intersection of the external bisectors of $\angle VUT$ and $\angle TVU$, which is $J$. $\square$

Let $AN$ intersect $BC$ at $L$, so $XY$ passes through $L$ as well. By power of a point from $L$ to $\Gamma$ and circle $WXYZ$, we have that $LA \cdot LN = LX \cdot LY = LW \cdot LZ$, so $WANZ$ is also cyclic. Thus, $A$ is the Miquel point of quadrilateral $WXYZ$. As $WXYZ$ is cyclic with circumcentre $J$ and its opposite sides $WX$ and $YZ$ intersect at $N$, we have that $AN \perp AJ$. Since $AN$ is the external bisector of $\angle BAC$, this implies that $AJ$ is the internal bisector of $\angle BAC$.

Solution 2:

Let the internal and external angle bisectors of $\angle BAC$ intersect $BC$ at $K$ and $L$, respectively. Let $AK$ intersect circle $ABC$ again at $M$, and let $D$ be the intersection of the tangents to $\Gamma$ at $B$ and $C$. Let $\Omega$ be the $T$-excircle of $TUV$, and let $\omega$ be the incircle of $DBC$.

Claim. The points $T$, $K$, and $D$ are collinear.
Proof. Note that $BC$ and $XY$ are the polars of $T$ and $D$ with respect to $\Gamma$. By La Hire's Theorem, $TD$ is the polar of $L$ with respect to $\Gamma$. As $(B, C ; K, L) = -1$, $K$ also lies on the polar of $L$, thus proving the collinearity. $\square$

Claim. The incentre of $DBC$ is $M$.
Proof. We have that $\angle MBC = \angle MAC = \frac{1}{2} \angle BAC = \frac{1}{2} \angle DBC$, so $BM$ bisects $\angle DBC$. Similarly, $CM$ bisects $\angle BCD$, so $M$ is the incentre of $DBC$. $\square$

Claim. The intersection of the common external tangents of $\Omega$ and $\omega$ is $K$.
Proof. Let $K'$ be the intersection of the common external tangents of $\Omega$ and $\omega$. As $\Omega$ and $\omega$ are both tangent to $BC$ and lie on the same side of $BC$ opposite to $A$, $K'$ lies on $BC$. As $T$ is the intersection of the common external tangents of $\Gamma$ and $\Omega$ and $D$ is the intersection of the common external tangents of $\Gamma$ and $\omega$, by Monge's theorem $K'$ lies on $TD$. As $K'$ lies on both $BC$ and $TD$, it is the same point as $K$. $\square$

Hence, $K$ is collinear with the centres of $\Omega$ and $\omega$, which are $M$ and $J$, respectively. As $K$ and $M$ both lie on the bisector of $\angle BAC$, so does $J$.