Let $S = \overline{AT} \cap \overline{BC}$. By ratio lemma, if $f(X) = \frac{XB}{XC}$ for any $X$, the desired result is equivalent to $$f(P) = f(Q)f(R) = \frac{f(Q)}{f(A)} = \frac{f(Q)f(T)}{f(S)} \iff f(P)f(S) = f(Q)f(T),$$or that $TPQS$ is cyclic. Since $\angle AIV = \angle IUV = 90^\circ - \tfrac 12 \angle B$, $\overline{AI}$ is tangent to $(IVU)$. It is well known that $\angle AIS = 90^\circ$, so $\overline{AI}$ is also tangent to $(ITS)$. Thus $$AT \cdot AS = AI^2 = AV \cdot AU,$$meaning $STVU$ is cyclic, as desired.

First of all observe that $AR||BC \iff BC\cap AT,T,U,V$ are concyclic. This means that it's enough to prove that $AT\cap BC=E$ lies on $(TUV)$. Radical center of $(AI),(BIC),(ABC)$ is $E$, becuse $ET\cdot EA=EB\cdot EC$. It's well-known that $(AI)$ touches $(BIC)$ $\implies$ $EI$ touches $(AI)$ Easy angle chasing shows that $\angle VUI=\angle VIA \implies AI^2=AV\cdot AU$ $T$ is foot of perrpendicular in right triangle $AIE \implies AI^2=AT\cdot AE$ Finally, $AT\cdot AE=AI^2=AV\cdot AU \implies TVUE$ is cyclic $\blacksquare$

[asy][asy] import olympiad; size(12cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,I,K,U,V,M,T,D,L,P,Q,R; O=(0,0);A=dir(120);B=dir(200);C=dir(340);path w=circumcircle(A,B,C); I=incenter(A,B,C);K=foot(A,B,C);U=extension(B,I,A,K);V=extension(C,I,A,K);M=extension(O,midpoint(B--C),A,I);D=foot(I,B,C); T=intersectionpoints(w,M--100D-99M)[0];L=extension(A,T,B,C);path q=circumcircle(T,U,V); Q=intersectionpoints(q,w)[1];P=intersectionpoints(q,C--100B-99C)[1]; R=intersectionpoints(w,Q--100P-99Q)[0]; draw(A--B--C--cycle,deep);draw(w,deep); draw(K--A,deep);draw(A--L,deep); draw(M--T,deep);draw(M--A,deep);draw(q,heavyblue);draw(L--B,deep); draw(R--Q,heavymagenta);draw(A--R,heavymagenta);draw(C--V,deep);draw(B--I,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$I$",I,dir(I)); dot("$U$",U,dir(U)); dot("$V$",V,dir(V)); dot("$M$",M,dir(M)); dot("$D$",D,dir(D)); dot("$T$",T,dir(T)); dot("$L$",L,dir(L)); dot("$Q$",Q,dir(Q)); dot("$P$",P,dir(P)); dot("$R$",R,dir(R)); [/asy][/asy] Let $M$ be the midpoint of arc $BC$ not containing $A$. Let $D=TM\cap BC$, let $L=AT\cap BC$. Claim. $AI$ is tangent $(UVI)$. Proof. Obvious due to angle chase. $$\measuredangle VIA=\measuredangle CIA=90^\circ+\measuredangle CBI= \measuredangle VUB=\measuredangle VUI.\quad \square$$ Claim. $AI$ is tangent $(TID)$. Proof. Obvious due to inversion at $M$ with radius $MB=MI$. $\square$ Note that $L$ lies on $(TID)$, hence by PoP, $LTUV$ is cyclic. Now, $$\measuredangle PLT=\measuredangle PQT=\measuredangle RQT=\measuredangle RAT,$$thus $AR\parallel BC$.

$\overline{AR}||\overline{BC}$ is equivalent to $T,U,V,\overline{AT}\cap \overline{BC}$ concyclic by angle chase (it's a general version of shooting lemma). Let $T_A$ be the point on $\overline{BC}$ such that $\angle AIT=90$. Then radical axis on $(AI),(BIC),(ABC)$ means that $A-T-T_A$. Let $D$ be the foot from $I$ to $\overline{BC}$. $\measuredangle T_ATI=\measuredangle ITA=90=\measuredangle IDT_A$, so $T_A,T,I,D$ cyclic. Let $(T_ATID)\cap \overline{AV}=U',V'$. Let $T_A'$ be a point such that $\overline{IT_A'}||\overline{BC}$. Then $\overline{IU'},\overline{IV'}$ are isogonal in $\angle T_AIT_A'$. $\overline{IT_A},\overline{IT_A'}$, as well as $\overline{IA},\overline{ID}$, are isogonal. Projecting this involution onto $\overline{AV}$, we have $(A,P_{\infty});(U,V);(U',V')$ are pairs under an involution. Thus $AU\cdot AV=AU'\cdot AV'=AT\cdot AT_A$ as desired.

$X = \overline{AT} \cap \overline{(TVU)}$ $X,B,C$ Collinear.