Let the tangent at $P$ of $(ACD)$ intersects $CD$ at $X$ and the tangent at $R$ of $(ABD)$ intersects $AD$ at $Y$. Let $T_1=PQ\cap AB$ and $T_2=RS\cap AB$. By Pascal's Theorem with $C,Q,P,P,A,D$. we get that $T_1,I,X$ are collinear. Similarly, $T_2,I,Y$ are collinear. Now, if we can prove that $X=Y$, we will get $T_1=T_2$ which leads to $AB,PQ,RS$ concurrent. Let $Z$ be the intersection of the tangent at $P$ of $(ACD)$ and the tangent at $R$ of $(ABD)$. Note that $$X=Y\iff Z,C,D \text{ are collinear } \iff \mathrm{Pow} (Z,(ABD)) = \mathrm{Pow} (Z,(ACD)) \iff ZP=ZR.$$Since $\angle{CPI}+\angle{CRI}=\angle{CDA}+\angle{CDB}=180^{\circ}$, $C,P,R,I$ lie on a circle We get \begin{align*} \angle{ZPR}&=\angle{ZPC}+\angle{CPR}\\ &=\angle{CAP}+\angle{CIR}\\ &=\angle{CAI}+\angle{CBI}+\angle{ACI}\\ &=\angle{CAI}+\angle{CBI}+\angle{BCI}\\ & =\angle{CBR}+\angle{CIP}\\ & =\angle{ZRC}+\angle{CRP}\\ & =\angle{ZRP}. \quad \blacksquare \end{align*}

Excellent dynamic tutorial problem! Let $CL$ be the internal bisector of angle $C$ with $L \in \overline{AB}$. Fix $A, B, C, D$ and vary $I \in \overline{CL}$. Note that $Q, S$ remain fixed but $P, R$ vary and so $I \rightarrow P, I \rightarrow R$ are projectivities. Define $T_1=\overline{PQ} \cap \overline{AB}$ and $T_2=\overline{RS} \cap \overline{AB}$. Our aim is to show that $T_1=T_2$. Since $P \rightarrow T_1$ and $Q \rightarrow T_2$ are also projectivities, we need to show thus for exactly three positions of $I$. For $I \in \{C, L\}$ the result is obvious. Our third choice is $I=CL_{\infty}$. Fix $A, B, C$ now and vary $D$. Points $P, R$ move on the lines $\ell_A, \ell_B$ parallel to $CL$, passing through $A, B$ respectively. Note that $PQ \parallel CB$ and $RS \parallel CA$. It follows that $D \rightarrow P \rightarrow T_1$ are affine maps and same for $D \rightarrow T_2$. Evidently, it suffices to work it out for exactly two finite points $D$. The result obviously holds for $D \in \{D_1, D_2\}$ where $D_1, D_2$ lie on $\overline{AB}$ with $\angle AD_1C=\angle BD_2C=\angle ACL$. Note: It is remarkable how this *medium-ish*problem reduces to four "trivial" case checks, all of which are selectable as per our preferences.

Can someone check this solution ? I'm really new to this projective thingy Fix $D \in \overline{AB}$ and animate $I$ alongside the bisector. Also let $X_A = \overline{PQ} \cap \overline{AB}, X_B = \overline{RS} \cap \overline{AB}, J = \overline{CI} \cap \overline{BC}$. Observe $I \overset{A}{\mapsto} P \overset{Q}{\mapsto} X_A$ is a projective map, and simiarly $J \overset{B}{\mapsto} R \overset{S}{\mapsto} X_B$ is projective. To show that these points coincide, we need to check for three points, viz: $I = C$, then it's clear that $X_A = J = X_B$. $I = J$. Then it's clear $X_A = D = X_B$ $I = P^{\infty}$. Then $\Delta JQX_A \sim \Delta JCB$, and $JX_A = JQ \times (\frac{JB}{JC}) = (\frac{CJ \times JQ}{CJ})(\frac{JB}{JC}) = (\frac{AJ \times JC}{CJ})(\frac{JB}{JC}) = \frac{JB \times JA \times JD}{JC^2} = JX_B$, and we're done !

[asy][asy] unitsize(2inches); pair C=dir(100); pair A=dir(140); pair B=dir(40); pair X=extension(C,incenter(A,B,C),A,B); pair D=0.7*B+0.3*A; pair U=extension(C,X,A,(A-C)*(D-A)/(D-C)+A); pair V=extension(C,X,B,(B-C)*(D-B)/(D-C)+B); pair Q=2*foot(circumcenter(A,C,D),C,X)-C; pair S=2*foot(circumcenter(B,C,D),C,X)-C; pair I=0.4*C+0.6*X; pair P=2*foot(circumcenter(A,C,D),A,I)-A; pair R=2*foot(circumcenter(B,C,D),B,I)-B; draw(A--B--C--cycle); draw(S--Q); draw(A--U); draw(B--V); draw(A--P); draw(B--I); draw(circumcircle(A,C,D)); draw(circumcircle(B,C,D)); dot("$A$",A,dir(A)); dot("$B$",B,dir(55)); dot("$C$",C,dir(50)); dot("$X$",X,dir(230)); dot("$D$",D,dir(240)); dot("$U$",U,dir(0)); dot("$V$",V,dir(190)); dot("$Q$",Q,dir(270)); dot("$S$",S,dir(120)); dot("$P$",P,dir(260)); dot("$R$",R,dir(270)); dot("$I$",I,dir(130)); [/asy][/asy] We will be dealing extensively with non-harmonic cross ratios in this solution, so we'll start by clarifying notation. For three collinear points $P,A,B$, we define the ratio $(P;AB)$ to be the signed ratio $PA/PB$. For four collinear points $P,Q,A,B$, we define the cross ratio $(PQ;AB)=\frac{(P;AB)}{(Q;AB)}$. Let $X$ be the foot of the angle bisector from $C$ onto $AB$, and let $U=AA\cap CX$, $V=BB\cap CX$. By Pascal on $PPQCDA$, we have that $PP\cap CD$, $PQ\cap AD$, and $QC\cap AP$ are collinear, so $PP\cap CD$, $PQ\cap AB$, and $I$ are collinear. Similarly, we get that $RR\cap CD$, $RS\cap AB$, and $I$ are collinear, so it suffices to show that $PP$ and $RR$ concur on $CD$. Note that \[(PA;CD)\stackrel{P}{=}(PP\cap CD,AI\cap CD;C,D)\]and \[(RB;CD)\stackrel{Q}{=}(RR\cap CD,BI\cap CD;C,D).\]Thus, \[\frac{(PA;CD)}{(RB;CD)}=\frac{(PP\cap CD;C,D)}{(RR\cap CD;C,D)}\cdot(BI\cap CD,AI\cap CD;C,D)\stackrel{I}{=}\frac{(PP\cap CD;C,D)}{(RR\cap CD;C,D)}(BA;XD).\]However, we also have \[(PA;CD)\stackrel{A}{=}(I,AA\cap CX;C,X)\]and \[(RB;CD)\stackrel{B}{=}(I,BB\cap CX;C,X),\]so by similar logic, \[\frac{(PA;CD)}{(RB;CD)}=(VU;CX).\]Our goal is to show that $\frac{(PP\cap CD;C,D)}{(RR\cap CD;C,D)}=1$, so it suffices to show that $(BA;XD)=(VU;CX)$, which is what we'll show. Note that this is an $I$-independent statement that we've reduced the problem to. By the ratio lemma on $\triangle BCX$ and point $V$, we have that \[(V;CX)=\frac{CB}{XB}\cdot\frac{\sin\angle CBV}{\sin\angle XBV}=\frac{CB}{XB}\cdot\frac{\sin\angle CDB}{\sin\angle DCB},\]and similarly \[(U;CX)=\frac{CA}{XA}\cdot\frac{\sin\angle CDA}{\sin\angle DCA}.\]Since $CB/XB=CA/XA$ by the angle bisector theorem, we have that \[(VU;CX)=\frac{\sin\angle DCA}{\sin\angle DCB}=\frac{AD}{DB}\div\frac{AC}{BC}=\frac{AD}{DB}\div\frac{AX}{BX}=(BA;XD),\]as desired.

Fix $\triangle ABC,$ and animate $I$ on $\angle C$ bisector (which we denote by $\ell$), so $Q,S$ are fixed and $P,R$ move as a function of $I.$ Denote $E_1= \overline{RS} \cap \overline{AB},$ and $E_2 = \overline{PQ} \cap \overline{AB}.$ Note the map $I \xmapsto{B} R \xmapsto{S} E_1$ is projective, and $I \xmapsto{A} P \xmapsto{Q} E_2$ is also projective. It suffices to check these maps coincide for three positions of $I.$ $I=C$ is obvious. $I=\overline{AB} \cap \ell$ is obvious. $I=P_\infty$ in the direction of $\ell$ we will prove with animation again. Animate $D$ on $\overline{AB}.$ Note we have $\overline{AP} \parallel \overline{QC} \parallel \overline{BR},$ $\overline{PQ} \parallel \overline{BC},$ and $\overline{SR} \parallel \overline{AC}.$ In particular, we have $\angle BAP$ is fixed, so the map $D \mapsto \infty_\ell \xmapsto{A}P \mapsto{E_1},$ where the last map was induced by projection about $\infty_{BC}.$ Similarly, $\angle DBR$ is fixed, so we have $D \mapsto \infty_\ell \xmapsto{B} R \mapsto E_2,$ where the last map was induced by projection about $\infty_{AC}.$ It suffices to check these maps concur for three positions of $D.$ But this isn't hard to do. Take $D=A, D=B,$ and $D=\ell \cap \overline{AB}.$ wow that was funny, mmp in an mmp