Let $V,W,U,X$ be the points of tangency with $AB,BC,CD,DA$, respectively, $O$ the center of $w$, and $S$ the foot of the altitude from $O$ to $PQ$. Then it's fairly well known that $R$ lies on $XU,WV$. (Apply Pascal to $XXUWWV,UUWVVX$). Let $L$ be the intersection of $PQ$ and $BD$. Extend $IT$ to intersect $w$ again at $Z$. Then $OT,OW$ are perpendicular to $QP,QC$ respectively, so $\angle PQC =\angle TOW$, which means that $Q,I,W,Z$ are concyclic. So $\angle IZQ =\angle IWC$. Similarly $\angle IZP = \angle IUC$. But $I$ lies on $OC$, the perpendicular bisector of $UW$, so $ZI$ bisects $\angle QZP$. Let $QI$ intersect $PQ$ at $R'$, and let $L'$ be the intersection of $PQ$ and the external bisector of $\angle QZP$. Then $L',Q,R',P$ are harmonic. Let $X',W'$ be the second intersections of $L'V, L'U$ with $w$. Since $O,S,U,P,V$ are concyclic, $\angle L'SU =\angle UVP = \angle UX'V$, so $L',S,U,X'$ are concyclic. $R'$ is clearly the radical center of $w,(L',S,T,Z), (L',S,U,X')$, so $R'$ lies on $X'U$, and similarly on $VW'$. Then the tangents to $w$ at $X',W'$ meet at $Q'$, which is the point on $PQ$, such that $L',Q',R',P$ are harmonic. Thus $Q=Q'$, which in turn means that $R=R'$.

Nice problem! Brazil 2017/3 wrote: A quadrilateral $ABCD$ has the incircle $\omega$ and is such that the semi-lines $AB$ and $DC$ intersect at point $P$ and the semi-lines $AD$ and $BC$ intersect at point $Q$. The lines $AC$ and $PQ$ intersect at point $R$. Let $T$ be the point of $\omega$ closest from line $PQ$. Prove that the line $RT$ passes through the incenter of triangle $PQC$. Observe that $\overline{OT} \perp \overline{PQ}$. Let $S=\overline{OT} \cap \overline{PQ}$, $X$ be the inverse of $S$ in $\omega$. Let $Z=\overline{CI}\cap \overline{PQ}$. Claim. $\overline{TI}, \overline{CX}, \overline{PQ}$ concur. (Proof) Let $A', B'$ lie on $\overline{CP}, \overline{CQ}$ respectively. Suppose $\overline{A'B'}$ is tangent to $\odot(O)$ at point $X$. Consider $\triangle CA'B'$ as reference. Let $\lambda=\tfrac{PQ}{A'B'}$. Let $a,b,c$ denote the sides, $r$ the inradius, $r_C$ the $C$-exradius, $\ell_C$ the length of $C$-angle bisector, $k_C$ the length $CO$, $h_C$ the length of $C$-altitude, $\phi$ the angle $COT$, in $\triangle A'B'C$ Observe that $$(OT; XS)=\frac{OT}{OS}=\frac{r_C}{k_C\cos \phi+\lambda h_C}=\frac{\ell_Cr_C}{h_C(k_C+\lambda \ell_C)}.$$Also, we have \begin{align*}(OI, CZ)=\frac{k_C}{\left(\frac{a+b}{a+b+c}\right)\lambda \ell_C} \div \frac{k_C+\lambda \ell_C}{\left(\frac{c}{a+b+c}\right)\lambda \ell_C}=\frac{ck_C}{(a+b)(k_C+\lambda \ell_C)}. \end{align*} Now we see $$\frac{\ell_Cr_C}{h_C}=\frac{2abc \cos \tfrac{C}{2}}{(a+b)(a+b-c)}=\frac{ck_C}{(a+b)}$$hence these cross-ratios are equal. Clearly, we obtain $\overline{TI}, \overline{XC}, \overline{SD}$ concur. $\blacksquare$ By Pascal's Theorem, polars of $A$ and $C$ in $\omega$ meet on $\overline{PQ}$. Consequently, pole of $\overline{AC}$ lies on $\overline{PQ}$. Hence $\overline{AC}$ is the polar of $S$, so $A,C,X$ are collinear; proving the conclusion. $\blacksquare$

Let $(J)$ and $(K)$ be the incircles of $\triangle PCQ$ and $\triangle PAQ,$ resp tangent to $PQ$ at $X$ and $X'.$ Since $ABCD$ is tangential $\Longrightarrow$ $AP-AQ=CP-CQ,$ but $QX=\tfrac{1}{2}(CQ+PQ-CP)$ and $QX'=\tfrac{1}{2}(AQ+PQ-AP)$ $\Longrightarrow$ $QX=QX'$ $\Longrightarrow$ $X \equiv X'.$ Since $X$ is the exsimilicenter of $(J) \sim (K)$ and $A$ is the exsimilicenter of $\omega \equiv (I) \sim (K),$ then by Monge & d'Alembert theorem it follows that the exsimilicenter of $(I) \sim (J)$ is on $AX.$ But as their radii $\overline{IT}$ and $\overline{JX}$ are parallel with the same direction, then $A,T,X$ are collinear. Let $E \equiv AC \cap BD$ and let $Y$ be the antipode of $X$ on $(J).$ Since $PQ$ is the polar of $E$ WRT $(I),$ then $IE \perp PQ,$ i.e $I,E,T.$ Since $C$ is the insimilicenter of $(I) \sim (J),$ then $C,T,Y$ are collinear. From $TI \parallel XY,$ we get $T(J,X,Y,I)=(J,A,C,E)=-1$ and from the complete $ABCD,$ we get $T(R,A,C,E)=-1$ $\Longrightarrow$ $T,J,R$ are collinear.

Proof : Let $\omega’$ be the incircle of $\triangle PQC$. Let $I, I’$ be the centers of $\omega, \omega’$, respectively. Let $PQ\cap \omega’=S$ and $K$ be the exsimilicenter of $\omega$ and $\omega’$. Let $X=\omega\cap AB$, $Y=\omega\cap BC$, $Z=\omega\cap CD$, $W=\omega\cap DA$. Let $E=AC\cap BD$. Clearly, $ST$ passes through $K$. Hence $$A(I, I’; R, T)=(I, I’; C, K)=-1.$$And $$I(A, I’; R, T)=(\perp XW, \perp YZ; \perp BD, \perp PQ)$$$$=(XW, YZ; BD, PQ)=(A, C; R, E)=-1.$$Hence $A(I, I’; R, T)=I(A, I’; R, T)$ $\Longrightarrow$ $I’, R, T$ are collinear.

From $ABCD$ being tangential, we have that if the incircles of $PAQ$, $PCQ$ meet $PQ$ at $M,M'$, then, $PM = \frac{AP + PQ - AQ}{2} = \frac{CP + PQ - CQ}{2} = PM'$, so the incircles mentioned above are tangent to $PQ$ at the same point. The homothety at $A$ sending the incircle of $ABCD$ to that of $APQ$ sends $T$ to $M$ so $A,T,M$ are collinear. Let $N$ be the diametrically opposite point of $M$ in the incircle of $PCQ$. From homothety at $C$, we have $C, N, T$ are collinear. Suppose the incenters of $PCQ, ABCD$ are $I_1, I_2$. Then we have that since $PQ$ is the polar of $AC\cap BD \equiv X$, then $I_2X \perp PQ$, which means that $T$ is on $I_2X$. So we get $(I_1, A; C, X) = -1 = (R, A; C, X)$ which completes the proof.

Here, I present a lengthy and non-projective approach. First, we make some crucial observations. Let $E,\ F$ be the contact points of $\omega$ and $BC, \ AD $, respectively. Let $K=\overline{IT}\cap\overline{PQ}$, $S=\overline{EF}\cap \overline{IT}$, $I$ the incenter of $ABCD$ and $L $ the incenter of $\bigtriangleup PCQ$. It is known that $AC$ pass through $S$. Clearly, $\angle QKI=90^\circ$, thus $QKEIF$ is cyclic. Since $\angle EKI=\angle IKF$ and $IE=IT=IF$, $T$ is the incenter of $\bigtriangleup EKF$. We infer the following results: \begin{eqnarray*} \angle SET &=&\dfrac{\angle FEK}{2}=90^\circ-\dfrac{KQF}{2}=90^\circ-\angle LQI\\ \angle IET& =&90^\circ-\dfrac{\angle EIK}{2}=90^\circ-\dfrac{\angle KQE}{2}=90^\circ-\angle CQL \label{e1}\\ \dfrac{SE}{KS}&=&\dfrac{\sin \angle EKS}{\sin \angle KES}=\dfrac{IE}{KF} \label{e2} \end{eqnarray*} Let's turn to the solution. By Menelaus' theorem ($\bigtriangleup CIS,\ R-L-T$ ), it suffices to show that: $$\dfrac{RC}{RS}\cdot\dfrac{ST}{TI}\cdot \dfrac{IL}{CL}=1$$But, \begin{eqnarray*} \dfrac{RC}{RS}&=&\dfrac{CQ}{QS}\cdot\dfrac{\sin \angle RQC}{\sin \angle SQR}=\dfrac{CQ}{QS}\cdot \dfrac{2\sin \angle CQL\cdot \cos \angle CQL}{\sin \angle SQR}\\ \dfrac{ST}{TI}&=&\dfrac{SE}{IE}\cdot \dfrac{\sin \angle SET}{\sin \angle IET}=\dfrac{SE}{IE}\cdot \dfrac{\cos \angle LQI}{\cos \angle CQL} \\ \dfrac{IL}{CL}&=&\dfrac{QI}{QC}\cdot \dfrac{\sin \angle IQL}{\sin \angle CQL} \end{eqnarray*}Therefore, \begin{eqnarray*} \dfrac{RC}{RS}\cdot\dfrac{ST}{TI}\cdot \dfrac{IL}{CL} &=&\dfrac{CQ}{QS}\cdot \dfrac{SE}{IE}\cdot \dfrac{QI}{IC}\cdot \dfrac{2\sin \angle CQL\cdot \cos \angle CQL}{\sin \angle SQR} \cdot \dfrac{\cos \angle LQI}{\cos \angle CQL} \cdot \dfrac{\sin \angle IQL}{\sin \angle CQL}\\ &=& \dfrac{SE}{QS}\cdot \dfrac{QI}{IE}\cdot \dfrac{2\sin \angle IQL\cos\angle IQL}{\sin \angle SQR}\\ &=& \dfrac{SE}{IE} \cdot \dfrac{QI}{KS}\cdot \sin \angle KQF\\ &=& \dfrac{SE}{IE}\cdot \dfrac{QI}{KS}\cdot \dfrac{KF}{QI}\\ &=& 1 \end{eqnarray*}The proof is complete.

Notice that this problem is equivalent to IMO 2008 Problem 6

Solved with Gomes17. Let $U= AC \cap BD$, $X=AD \cap \omega, Y= AB \cap \omega, Z= BC \cap \omega, W= CD \cap \omega$, $I$ the center of $\omega$, $S=IT \cap PQ$ and $I_C$ the incenter of $PQC$. By Brianchon's Theorem $U$ lies on $XZ$ and $YW$, then $U \in \Pi_{\omega}(P), \Pi_{\omega}(Q) \implies PQ= \Pi_{\omega}(U) \implies UI \perp PQ$. Observe that since $T \in \omega$ such that $T$ is the closest one from $PQ$, we have that $IT \perp PQ$, so $I,T,U$ are collinear. By construction, $-1=(BD \cap PQ, R; P, Q) \stackrel{B}=(U,R;A,C)$. $\quad (\star)$ Now, the main claim, inspired on IMO 2008/6: Claim: Let $L$ the tangency point of the incircle $\omega_C$ of $PQC$ touches $PQ$. Then, the incircle $\omega_A$ of $APQ$ is tangent to $\omega_C$ at $L$. Proof: Let $L'$ the tangeny point of $\omega_A$ with $PQ$. Therefore, $2PL'=PA+PQ-AQ,2PL=PD+PQ-QD$. We want to prove that $PL=PL'$ (both lie inside the segment $PQ$), so we want to prove that $PA-AQ=PD-QD$. This is true if and only if $(PY+YA)-(AX+XQ)=(PW+WD)-(QX-XD)$, which is indeed true, since $AY=AX,DW=DX$ and $PY=PW$. $\square$ Now, observe that since $IT, I_CL \perp PQ \implies I_CL \parallel IT \implies TL \cap II_C=K$ is the exmilicenter of $\omega, \omega_C$. Also, oberve that $C$ is the insimilicenter of $\omega, \omega_C$, so $C \in II_C$, so $(I,I_C; K,C)=-1$. $\quad (\star \star)$ To finish, notice that by Monge's Theorem on $\omega, \omega_A, \omega_C$, their exmilicenters are collinear, so $A,K,L$ are collinear, then $A,K,L,T$ are collinear. From $(\star)$ and $(\star \star)$, $$(U,R;A,C) \stackrel{T}= (S,R; TI_C \cap PQ; L, TC \cap PQ)=-1=(I,I_C; K,C) \stackrel{T}= (S, TI_C \cap PQ; L, TC \cap PQ)$$so $TI_C \cap PQ= R$, so $R,T,I_C$ are collinear, as desired. $\blacksquare$

Solved with Olympikus. Let $X,Y,Z,W$ be the tangency points on sides $AB,BC,CD,AD$ respectively and let $S$ be the intersection of $XZ$ and $WY$. By brianchon's theorem, $S$ lies on $AC$ and $BD$. Now let $U$ be the foot of the perpendicular from $O$ to $PQ$, where $O$ is the center of $\omega$ and $G$ be the foot of the perpendicular from $I$ to $PQ$, where $I$ is the incenter of triangle $PQC$. Also let $V$ be the intersection of lines $XW$ and $YZ$. Pascal on hexagon $XXYZZW$ shows that $V,P,R'$ are collinear where $R' = WZ \cap XY$. In a similar way, $Q,R',V$ are collinear, hence $V$ lies on $PQ$. Now, Desargues theorem on triangles $CYZ$ and $AXW$ gives $XY,WZ,AC$ concurrent. Hence, $R = R'$ and by the Miquel's point theorem on quadrilateral $XYZW$, one gets that $S$ lies on $UT$ (which is the line joining the center and the miquel point of $XYZW$). Notice that $I,R,T$ are collinear $\iff$ triangles $RIH$ and $RTS$ are homothetic, which is iff $\frac{IH}{ST} = \frac{RH}{RS}.$ But one may notice that $\frac{RH}{RS} = \frac{HG}{SU}$ since triangles $RGH$ and $RUS$ are similar. Hence, our problem is iff $\frac{IH}{ST} = \frac{HG}{SU}$. Now, a straightforward calculation. Denote $r,r'$ the inradius of $ABCD$ and $PQC$, and let $OS = x$. First, note that $\frac{IH}{x} = \frac{CI}{IO} = \frac{r'}{r}$, hence $IH = \frac{r'.x}{r} (1)$. Next, $US = TU + (r-x)$, and $UT = \frac{r(r-x)}{x}$ applying inversion distance formula (recall that $S' = U$ when inverting w.r.t $\omega$). Hence $US = \frac{r(r-x)}{x} + (r-x)$ (2). Finally, $HG = IH + r' = \frac{r'.x}{r} + r'$ and $TS = r-x$. Therefore, $I,R,T$ collinear $$\iff \frac{IH}{ST} = \frac{HG}{SU} \iff \frac{\frac{r'x}{r}}{r-x} = \frac{\frac{r.r'}{r} + \frac{r'x}{r}}{r-x} + \frac{r(r-x)}{x}$$, which miraculously simplifies to $x + r = x + r$, which is true, as desired. $\blacksquare$

WizardMath wrote: From $ABCD$ being tangential, we have that if the incircles of $PAQ$, $PCQ$ meet $PQ$ at $M,M'$, then, $PM = \frac{AP + PQ - AQ}{2} = \frac{CP + PQ - CQ}{2} = PM'$, so the incircles mentioned above are tangent to $PQ$ at the same point. The homothety at $A$ sending the incircle of $ABCD$ to that of $APQ$ sends $T$ to $M$ so $A,T,M$ are collinear. Let $N$ be the diametrically opposite point of $M$ in the incircle of $PCQ$. From homothety at $C$, we have $C, N, T$ are collinear. Suppose the incenters of $PCQ, ABCD$ are $I_1, I_2$. Then we have that since $PQ$ is the polar of $AC\cap BD \equiv X$, then $I_2X \perp PQ$, which means that $T$ is on $I_2X$. So we get $(I_1, A; C, X) = -1 = (R, A; C, X)$ which completes the proof. How this line $(I_1, A; C, X) = -1 = (R, A; C, X)$ becomes? And the point $I_1$ is not collinear with $ACX$, so how you consider the cross ratio $(I_1, A; C, X)$?

Beautiful problem! The best Brazil National Olympiad I have seen up to now, maybe the best geo I have seen too! Solution using moving points (feat. Hyperbolas): Fix points $P,Q,R$ and and define $I$ as the incenter of $\triangle PQR.$ Then animate $O$ over line $CI$ and define $A$ as the intersection of the reflection of line $\overline{PC}$ with respect to line $PO$ with the reflection of line $\overline{QC}$ with respect to line $\overline{QO}.$ Then define the other points acordingly, i.e. $B:=\overline{PA}\cap \overline{CQ}, D:= \overline{QA}\cap \overline{PC}$ then points $R$ and $T$ are defined the same way. We then establish the following claims: Claim 1: Let $\Gamma$ be the incircle of $\triangle PQC$ and further let $W$ be the point of $\Gamma$ that is furthest from line $\overline{PQ}.$ Then point $R$ lies on the line $WC$ and, furthermore, the map $O\rightarrow R$ from line $\overline{IC}$ to line $\overline{DW}$ is linear. Proof: It is not hard to see that, from the definition of $R,$ the homothety centered at $C$ that sends $\Gamma$ to $\omega$ must send $W$ to $R.$ So that point $R$ lies on $DW.$ Then it follows that the map from $O\rightarrow D$ from line $\overline{IC}$ to $\overline{DW}$ is linear since now we can define $R$ as the projection of $O$ to line $DW$ centered at the point of infinity of the pencil of lines perpendicular to line $\overline{PQ}. \square$ Now let $\Gamma'$ be the incircle of $\triangle PAQ.$ Further, let $H$ be the touch point of $\Gamma$ with line $\overline{PQ}.$ Claim 2: Point $A$ satisfies $PA-AQ = PC -CQ.$ Equivalently, $\Gamma,\Gamma '$ and $\overline{AC}$ are tangent at $H$ and as $O$ moves at line $\overline{AC}$ the point $A$ moves at the Hyperbola with focus $P$ and $Q$ passing through $H$ and $C.$ Proof: Let $\omega$ touch lines $\overline{PA},\overline{QA},\overline {CB}$ and $\overline{CD}$ at $X,Y,U$ and $V,$ respectively. From the fact that the two tangents drawed from a point to a circle are equal it follows that $$PA=PX+XA=PV+XA=PC +DV + XA $$similarly we have $$AQ = CQ +UC + AY $$and since $DV=UC$ and $XA=AY$(again from the equal tangent fact) the Claim follows. The other parts of the Claim follows well known fact about the lenght of the tangent from a point of the triangle with respect to its incircle(i.e. equals to the semiperimeter minus the lenght of the oppositive side). $\square$ Now let $\mathcal{C}$ be the hyperbola with focus $P$ and $Q$ passing through $H$ (and hence to $C$). Further, let $K$ be center of the positive homothety that sends $\Gamma$ to $\omega$ (i.e. the intersection of the external common tangents of these two circles). Claim 3: The map $O\rightarrow A$ from $\overline{IC}$ to $\mathcal{C}$ is linear. Proof: By Claim 2 we have that $\Gamma$ and $\Gamma '$ are tangent at $H.$ So it follows from Monge's Theorem that $A,H$ and $K$ are colinear. Also, we have by definition that $K$ lies on line $\overline{CI}$ so that $K=\overline{IC} \cap \overline{AH}.$ Then we are done if we show that the map $O\rightarrow K$ from line $\overline{IC} \rightarrow \overline{IC}$ is linear, since the map $K\rightarrow A$ from $\overline{IC} \rightarrow \mathcal{C}$ by a projection at $H$ (which is a point of $\mathcal{C}$) is linear. From the last paragraph, it is sufficient to show that the map $O\rightarrow K$ from $\overline{IC}\rightarrow \overline{IC}$ is linear. Then recall that $K$ and $C$ are the centers of the positive (respectively negative) homothety that sends $\Gamma$ to $\omega$ so that it follows $(O,I;C,K)=-1.$ Then $K$ can be define as the harmonic conjugate of $C$ with respect to segment $\overline{OI}.$ Now one can assign an coordinate system to line $\overline{IC}$ and show that $K$ can be expressed as the image of $O$ under a composition of homothetys and inversions (i.e., if we let $O$ have a variable value $x$ in the assigned cordinate system, then $K$ can be expressed in the form $a+b{/}(x+c)$ for constants $a,b,c$ dependent on $I$ and $C$). It is well known that inversion (and of course homothetys) preserves cross ratio, so it follows that the map is indeed linear. $\square$ Now we are able to solve the problem. Define $T' = \overline{IR} \cap \overline{PQ}$ and as in the problem statement $T:=\overline{AC} \cap \overline{PQ}.$ Then by Claim 1 we have that the map $O\rightarrow T'$ from $\overline{IC}\rightarrow \overline{PQ}$ is linear. Also, since point $C$ lies on the conic $\mathcal{C}$ too, it follows by Claim 4 that the map $O \rightarrow A$ from $\overline{IC} \rightarrow \overline{PQ}$ is also linear. So to finish the problem it is sufficient to verify that $T=T'$ for only three choices of $O.$ So taking $O=1,$ the limit as $O$ approachs $C,$ and $O$ the intersection of $\overline{IC}$ and $\overline{PQ}$ is sufficient.

A quadrilateral $ABCD$ has the incircle $\omega$ and is such that the semi-lines $AB$ and $DC$ intersect at point $P$ and the semi-lines $AD$ and $BC$ intersect at point $Q$. The lines $AC$ and $PQ$ intersect at point $R$. Let $T$ be the point of $\omega$ closest from line $PQ$. Prove that the line $RT$ passes through the incenter of triangle $PQC$. Let $O$ be the center of $\omega$. Make a few (badly named) definitions: $W$, $X$, $Y$, and $Z$ are the tangency points of $\omega$ to $AB$, $BC$, $CD$, and $DA$. $U=XY\cap PQ$. $R'$ is the $C$-extouch point of $\triangle CPQ$. $I_C$ is the $C$-excenter of $\triangle CPQ$. $K$ is the $C$-intouch point of $\triangle CPQ$. $L$ is the antipode of $K$ in the incircle of $\triangle CPQ$. $N=OT\cap PQ$. The key now is to redefine $R$ to be the point such that $R=TI\cap PQ$. We will show that $U$ and $R$ are harmonic conjugates with respect to $PQ$, which solves. Let $O'$ be the harmonic conjugate of $O$ with respect to $II_C$. Then \[-1=(K,L;I,\infty_{KL})\stackrel{T}{=}(K,R';R,N)\stackrel{\infty_{OT}}{=}(I,I_C;R\infty_{OT}\cap II_C,O)\]thus $RO'\perp PQ$. It suffices to show the following (totally redefined points): Transformed Brazil 2017/3 wrote: Given $\triangle ABC$ with incenter $I$ and $A$-excenter $I_A$. Let $D$ be a point on the $A$-internal bisector and let $D'$ be its harmonic conjugate with respect to $II_A$. Let $E$ be the projection of $D'$ onto $BC$. Furthermore let $X$ and $Y$ be the projections of $D$ onto $AC$ and $AB$ and let $XY\cap BC=F$. Show that $E$ and $F$ are harmonic conjugates with respect to $BC$. As it turns out, I did some extra things I didn't need to do in the original problem. Re-project $D$ onto $BC$ at $G$. Let $M$ be the midpoint of $BC$. Animate $D$ linearly; then $y=\frac{MG}{MF}$ is constant. In particular let $U$ be the $A$-intouch point; then \[x=MU^2=MG\cdot ME=y\cdot MF\cdot ME\]is constant, thus $MF\cdot ME$ is constant. It suffices to prove the question for a single choice of $D$. Taking $D=A$ works fine, and we are done. $\blacksquare$ The last step is technically moving points, but it's degree $0$ so I figure it is fine.

Let $S$ be where the incircle of $PQC$ touches $PQ$. Claim: $S$ is also the intouch point in $APQ$. Proof. This is a length chase. First, By external pitot or whatever, $AP+CQ=AQ+CP$, which is checkable by using $\omega$ as the excircle of $PBC$ and $QCD$ and finding tangent lengths. Then, \[2PS=CP+PQ-CQ=PQ+(CP-CQ)=PQ+(AP-AQ)\]as desired. $\blacksquare$ Let $X=AC\cap BD$. The polar of $X$ with respect to $\omega$ is $PQ$, so $IX \perp PQ$ or $I,X,T$ collinear. Now let $I_C$ be the incenter of $PQC$ and $E$ be the exsimilicenter of $\omega$ and $(I_C)$. By Monge on $(I_C),$ $\omega$, and the incircle of $APQ$, we have $E,A,S$ are collinear. By homothety, $ETS$ are collinear. Thus, combined with above, $A,E,T$ are collinear. By the polar, $(AC;XR)=-1$, and by similicenters, $(II_C;CE)=-1$. Thus by prism lemma $IX, RI_C, AE$ concur. But $AE, IX$ pass through $T$ so $RI_C$ does as desired. Remark: This is 2008 IMO 6 but better

Let $\omega$ whose circumcenter is $I$ be tangent to $AB,BC,CD,DA$ at $K,L,M,N$ respectively. $IT$ intersects $PQ,\omega$ at $H,S$ respectively. Let $\gamma$ be the incircle of $PQC$ and $\gamma$ is tangent to $PQ,CP,CQ$ at $Y,E,F$. Let $AC\cap BD=X$. Note that by Brianchon theorem, $K,X,M$ and $L,X,N$ are collinear. Claim: $IH\perp PQ$. Proof: Let $l$ be the line tangent to $\omega$ at $T$. By the definition of $T$, we have that $l \parallel PQ$. Hence $90=\measuredangle (IT,l)=\measuredangle (IH,PQ)$.$\square$ Claim: $I,X,H$ are collinear. Proof: Let $\Psi(H)$ be the point $H$ swaps with under the inversion centered at $I$ with radius $IK$. Since $H\in (PMKI),(QNLI)$, this results in $\Psi(H)\in KM,LN$. Thus, $\Psi(H)=KM\cap LN=X$, subsequently $I,X,H$ are collinear.$\square$ Claim: $S,C,Y$ are collinear. Proof: $C$ is the center of the homothety sending $\omega$ to $\gamma$. Under this homothety, $M,E$ and $N,F$ swap with each other. Collinearity of $S,Y,C$ is equavilent to $YEF\overset{?}{\sim} SML$. \[\measuredangle YEF=\frac{\measuredangle YJF}{2}=\frac{\measuredangle YJC-\measuredangle FJC}{2}=\frac{\measuredangle SIC-\measuredangle LIC}{2}=\frac{\measuredangle SIL}{2}=\measuredangle SML\]\[\measuredangle EFY=\frac{180-\measuredangle ZJE}{2}=\frac{180-\measuredangle CJE+\measuredangle CJZ}{2}=\frac{180-\measuredangle CIM+\measuredangle CIT}{2}=\frac{180-\measuredangle TIM}{2}=\measuredangle MLS\]Which yields $YEF\sim SML$.$\square$ Claim: $R,J,T$ are collinear. Proof: Let $YJ\cap AC=Z$. We have $CZJY\sim CXIS$. $XI.XH=XK.XM=XT.XS$ hence $\frac{XI}{XS}=\frac{XT}{XH}$ which is folowed by $XIS\sim XTH$. \[\frac{JZ}{JY}=\frac{IX}{IS}=\frac{XT}{TH}\]Thus, by homothety centered at $R$ sends $YJZ$ to $HTX,$ we get that $R,J,T$ are collinear as desired.$\blacksquare$