Wow, this is very genius problem, it kinda hides incenter config very well, took me a while to realice. Let $I_A$ center of $\omega$ let $KK'$ diameter of $\omega$ and let the parallel from $K'$ to $BC$ hit $AB,AC$ at $B', C'$ respectivily, let $I$ the incenter of $\triangle ABC$ and $N$ the midpoint of the big arc $BAC$ in $\Omega$, and let the tangents from $P,Q$ to $\omega$ hit at $R$. Claim 1: $E,F,I$ are colinear Proof: Let the tangent from $A$ to $\Omega$ hit $EF$ at $S$ now by radax $ST$ is tangent to $\Omega$ and now in general its known that $S$ is the center of the A-apollonian circle in $\triangle AEF$ (also T-apollonian circle of $\triangle TEF$) so let the angle bisectors of $\angle EAF, \angle ETF$ meet $EF$ at $I'$, now as by $\sqrt{bc}$ inversion we have $T,I,N$ colinear and notice $A,I,I'$ colinear but also $T,I,I',N$ colinear so $I=I'$ meaning that $E,F,I$ are colinear as desired. Finishing: Let $YZ \cap KI_A=M$ now $AM$ is median of $\triangle ABC$ due to a well known lemma and also by homothety at the end we can get that $I_A$ is the midpoint of minor arc $BC$ in $\Omega$, also notice that $YPZQ$ is harmonic and that from here u can identify the whole config, indeed by some IMO 2019/6 config u can see that $K'P \perp YZ$ so $PK \parallel YZ$ so $V$ is the midpoint of $YZ$ and notice that $AV \cdot VI_A=YV \cdot VZ$ so $V$ lies on radax of $\Omega, \omega$, now by Reim's $ANQK'$ is cyclic so by reim's again $TI_A \parallel QK'$ so as we have $PK \parallel AI_A$ we get that $\angle TAI_A=\angle QPK'$ so $RQ \parallel UT$ so $UP=UT$ so $U$ lies on the radax of $\Omega, \omega$ as well, now notice that this radax is clearly parallel to $BC$ so $UV$ parallel to $BC$ as desired.

Let $I$ be the incenter of triangle $ABC$ and $M,N$ be the midpoints of the minor and major arcs $BC$, respectively. Lastly, let $S$ be the midpoint of $BC$. We start with some Claims. Claim 1: Points $F,I,E$ are collinear. Proof: It's clear that there exists exactly one line parallel to $BC$ satisfying that $(TEF)$ is tangent to $\Omega$. Define $F,E$ as the points at which the line through $I$ parallel to $BC$ intersects $AB,AC$ respectively. It is enough to prove that $(TEF)$ is tangent to $\Omega$. Let the tangent to $A$ at $\Omega$ intersect $EF$ at point $X$. Then, $\angle XAI=\angle XAF+\angle FAI=\angle AEF+\angle FAI=\angle AEF+\angle IAE=\angle AIX,$ and so $XA=XI$. Moreover, $\angle AXI=\angle AFE-\angle XAF=\angle B-\angle AEF=\angle B-\angle C=2\angle AMN=2\angle ATN,$ with the last equality being true as it is well known that points $N,I,T,$ are collinear. Therefore, $\angle AXI=2\angle ATN$ and $XA=XI$, hence $X$ is the circumcenter of $AIT$, which implies that $XA=XT,$ and so $XT$ is tangent to $\Omega$. To finish, note that $XT^2=XA^2=XF \cdot XE,$ and so $XT$ is tangent to $(TEF),$ too, as desired $\blacksquare$ Claim 2: $M$ is the $A-$ excenter of triangle $AEF$. Proof: Note that $M$ belongs in the internal angle bisector of angle $\angle FAE$. It is enough to prove that $MK=MZ$ (with $K,Z$ defined as the projections of $M$ on $FE,AB$ respectively). Indeed, $\dfrac{MK}{MA}=\dfrac{MI}{MN}=\dfrac{MC}{MN}=\sin \dfrac{\angle A}{2}=\dfrac{MZ}{MA},$ and so $MK=MZ$, as desired $\blacksquare$ Claim 3: Points $I,P,U$ are collinear. Proof: It is enough to prove that points $P,K$ are symmetric across the internal angle bisector of angle $\angle A$. To that end, we may note that $AT,AK$ are isogonal. Indeed, if $AK$ intersects $BC$ at $D$, then by homothety $D$ is the touchpoint of the $A-$ excircle with $BC,$ and it is well known that $AT,AD$ are isogonal, as desired $\blacksquare$ Claim 4: $V$ is the midpoint of $YZ$. Proof: It follows by the Ratio Lemma in quadrilateral $QYKZ$ and the fact that $ZPKY$ is a trapezoid (plus two sine laws in triangles $APZ$ and $APY$ $\blacksquare$ To the problem, note that $IU$ and $AX$ are parallel, as $\angle PIM=\angle KIM=\angle ANM= \angle AIX=\angle XAI,$ as desired. We have to prove that $UV$ is parallel to $XI$, which is equivalent to $\dfrac{WA}{WI}=\dfrac{IA}{IV}$. Note that $\dfrac{WA}{WI}=\dfrac{WA/XA}{WI/XI}=\dfrac{\sin \angle AXT}{\sin \angle IXT}=\dfrac{\sin 2(\angle AIN)}{\sin 2(\angle IAT)}=\dfrac{\sin \angle IAN \cos \angle IAN}{\sin \angle IAT \cos \angle IAT}=\dfrac{AN \cdot AI}{IK \cdot KN},$ and so it remains to prove that $IV=\dfrac{IK \cdot KN}{AN}$. We compute $IV$ now: $IV=MI-MV=MI-MA\sin^2\dfrac{\angle A}{2}=MB-MA\sin^2\dfrac{\angle A}{2}$ Moreover, $\dfrac{IK \cdot KN}{AN}=\dfrac{MI}{MN} \cdot KN=\dfrac{MI}{MN} \cdot (MN-MK)=\dfrac{MI}{MN} \cdot (MN-\dfrac{MA \cdot MI}{MN})=$ $=MI-\dfrac{MA \cdot MI^2}{MN^2}=MB-\dfrac{MA \cdot MB^2}{MN^2}=MB-MA\sin^2 \dfrac{\angle A}{2},$ which is the same as the expression obtained for $IV$, hence we may conclude.

Let $S_a$ be the midpoint of arc $BC$ not containing $A$, and define $S_b$, $S_c$ similarly. Redefine $E = AC \cap S_aS_b$, $F = AB \cap S_aS_c$. We claim that this is equivalent to the original definition. Claim 1. $EF \parallel BC$. Proof. Note that $\frac{AF}{FB} = \frac{AS_c \cdot AS_a}{BS_c \cdot BS_a} = \frac{AS_a}{BS_a}$ which is symmetric w.r.t. swapping $B$ and $C$, so $\frac{AF}{FB} = \frac{AE}{EC}$, hence proved. Claim 2. Let $D = S_bS_c \cap EF$. Then $AD$ is tangent to $(ABC)$. Proof. Note that \[ \frac{DS_c}{DS_b} = \frac{FS_c}{FS_a} \cdot \frac{ES_a}{ES_b} = \frac{AS_c \cdot BS_c}{AS_a \cdot BS_a} \cdot \frac{AS_a \cdot CS_a}{AS_b \cdot CS_b} = \frac{AS_c^2}{AS_b^2} \]which implies the result. Now it is well-known that $AS_bTS_c$ is harmonic, so actually $DT$ is also tangent to $(ABC)$. By radical center on $(AEF), (ABC), (TEF)$, we get that DT is tangent to $(TEF)$ too, as desired. Since $AK$ passes through the $A$-extouch point by homothety, $AP$ and $AK$ are isogonal in $\angle BAC$ by a well-known fact. Hence $PK \parallel YZ$, so combining this with $PYQZ$ harmonic we get that $V$ is the midpoint of $YZ$, and hence lies on $AS_a$. Finally we show that $UV$ is the radical axis of $(ABC)$ and $(KYZ)$, which would finish as the line of centers is the perpendicular bisector of $BC$. By symmetry we know that $PU$ and $EF$ are antiparallel w.r.t. $AB$ and $AC$, so $EF \parallel BC$ implies $PU \parallel AD$. So from $DA = DT$ we get $UP = UT$, so $U$ lies on the radical axis. On the other hand, from $VA \cdot VS_a = VY \cdot VZ$ we get that $V$ lies on the radical axis as well, so we are done.