It turns out that only angle chasing is enough (no similarities, no radical axis etc., but only cyclic quads), but I'll let somebody else post that. Let $AQ \cap CD = T$, with the aim to show that $B$, $R$ and $T$ are collinear. Observe that $\angle QRC = \angle QAP = \angle ATC = \angle QTC$ and so $CTRQ$ is cyclic, implying $\angle CRT = \angle CQT = \angle ADC = \angle ABC$. So if we show that $\angle BRC = \angle PBC$ we shall be done. The latter is equivalent to $\triangle PBC \sim \triangle BRC$ and hence to $BC^2 = CR \cdot CP$. But since $AC = BC$, we reduce to $AC^2 = CR \cdot CP$ and hence to $\triangle ACR \sim \triangle PCA$. Now it remains to observe that $\angle ACR = \angle ACP$ and $\angle CAR = \angle CAQ + \angle QAR = \angle CDQ + \angle QPR = \angle APD + \angle DPC = \angle APC$ and we are done.

Define $X = \overline{AQ} \cap \overline{CD}$ and let $R' = \overline{BX} \cap \overline{PC}$. We wish to show $APR'Q$ is cyclic. Let $M$ be the center of the parallelogram. Claim: $M$, $Q$, $R'$ are collinear. Proof. By Pappus theorem on $\overline{ABP}$ and $\overline{DCX}$. $\blacksquare$ [asy][asy] size(12cm); pair A = dir(90); pair D = dir(240); pair C = A*A/D; pair Q = dir(20); pair B = A+C-D; pair P = extension(D, Q, A, B); pair M = midpoint(A--C); pair X = extension(A, Q, C, D); pair Rp = extension(P, C, B, X); filldraw(unitcircle, invisible, blue); filldraw(A--B--C--D--cycle, invisible, red); draw(A--C, red); draw(B--D, red); draw(C--X, red); draw(B--P, red); draw(A--X, lightblue); draw(B--X, lightblue); draw(C--P, lightblue); draw(M--Rp, deepgreen); draw(D--P, lightblue); draw(Q--C, lightblue); pair Z = 2*A-P; pair W = -D+2*foot(origin, D, Z); draw(Q--W, deepgreen); draw(D--Z--A, orange); dot("$A$", A, dir(A)); dot("$D$", D, dir(D)); dot("$C$", C, dir(C)); dot("$Q$", Q, dir(72)); dot("$B$", B, dir(B)); dot("$P$", P, dir(P)); dot("$M$", M, dir(190)); dot("$X$", X, dir(X)); dot("$R'$", Rp, dir(Rp)); dot("$Z$", Z, dir(Z)); dot("$W$", W, dir(W)); /* TSQ Source: !size(12cm); A = dir 90 D = dir 240 C = A*A/D Q = dir 20 R72 B = A+C-D P = extension D Q A B M = midpoint A--C R190 X = extension A Q C D R' = extension P C B X unitcircle 0.1 lightcyan / blue A--B--C--D--cycle 0.1 lightred / red A--C red B--D red C--X red B--P red A--X lightblue B--X lightblue C--P lightblue M--Rp deepgreen D--P lightblue Q--C lightblue Z = 2*A-P W = -D+2*foot origin D Z Q--W deepgreen D--Z--A orange */ [/asy][/asy] Let $W$ be the point for which $AQCW$ is harmonic. Let $Z = \overline{DW} \cap \overline{AB}$. Claim: $\triangle ZAD \cong \triangle PAC$. Proof. Projecting from $D$ we get $-1 = (AC;QW) = (A\infty;PZ)$ so $A$ is the midpoint of $\overline{PZ}$. Since $AD=AC$ and $\measuredangle ZAD = \measuredangle CAP$, we're done. $\blacksquare$ To finish, \begin{align*} \measuredangle AQR' &= \measuredangle AQM = \measuredangle WQC = \measuredangle WDC = \measuredangle WDA + \measuredangle ADC \\ &= \measuredangle ACP + \measuredangle DCA = \measuredangle DCP = \measuredangle APR'. \end{align*} Remark: [Motivation] The Pappus step allows one to erase many points in the picture. After this step, rewrite the conclusion as $\measuredangle AQM = \measuredangle APC$; then the points $B$, $R'$, and $X$ may now be deleted.

Angle chase sol sketch: Let $AQ$ and $BR$ meet at $X$, I will prove that $CX||AB$. This is equivalent to $CQRX$ being cyclic. Note that $\angle ARC= \angle AQD= \angle ACD= \angle ABC$, so $ABRC$ is cyclic. Hence $\angle CRX= \angle BAC = \angle ADC = \angle CQX$ so we are done.

Let $T = AC\cap BD$. Note that $\angle QPA = \angle QDC = \angle QAC\to AC$ touches $(AQRP)$ As well as $$\angle ACQ = \angle ADQ = \angle ADC - \angle CDQ = \angle CBA - \angle CAQ = \angle CAB - \angle CAQ = \angle QAP = \angle CKQ$$ $\to AC$ touches $(CQRK).\hspace{10 cm}$ Now, we have $T\in QR$ since $QR$ bisects $AC$ and $R\in BK$ by Pappus' Theorem.

[asy][asy] unitsize(1cm); pair D, C, A, B, Q, P, R, X; D=(0,0); A=(2,5); C=(4,0); B=A+C-B; P=(10,5); Q=intersectionpoints(D--P,circumcircle(A,D,C))[1]; R=intersectionpoints(C--P,circumcircle(A,Q,P))[0]; X=extension(A,Q,B,R); draw(circumcircle(A,C,D)); draw(circumcircle(A,Q,P)); draw(A--D--X--C--A--P--C--Q--R--B--X--D--P--A--X); label("$D$", D, SW); label("$C$", C, SE); label("$A$", A, NE); label("$B$", B, N); label("$Q$", Q, NNE); label("$P$", P, SE); label("$R$", R, ESE); label("$X$", X, SE); [/asy][/asy] We claim that $ABRC$ is cyclic. We will first show that $AC$ is tangent to the circumcircle of $APQ$. We have \begin{align*} \angle QAC&=\angle QDC\\ &=\angle QPA, \end{align*}so $AC$ is tangent to the circumcircle of $APQ$. Now, we have \begin{align*} \angle ABC&=\angle CAB\\ &=180^{\circ}-\angle PRA\\ &=\angle ARC. \end{align*}This means that $ABRC$ is cyclic. If $AQ$ and $CD$ intersect at $X$, then we have \begin{align*} \angle RCX&=\angle RPA\\ &=180^{\circ}-\angle AQR\\ &=\angle RQX. \end{align*}Therefore, $RQCX$ is also cyclic. Since $AD=BC=AC$, we get $\angle CDA=\angle ACD$, so we get \begin{align*} \angle CRX&=\angle CQX\\ &=180^{\circ}-\angle AQC\\ &=\angle CDA\\ &=\angle ACD\\ &=\angle CAB\\ &=180^{\circ}-\angle BRC. \end{align*}Therefore, $B$, $R$, and $X$ are collinear, so $AQ$, $BR$, and $CD$ concur at $X$.

ISL 2021 G1. Let $ABCD$ be a parallelogram such that $AC=BC$. A point $P$ is chosen on the extension of segment $AB$ beyond $B$. Define $Q=(ACD)\cap PD$ and $R=(APQ)\cap PC$. Prove that $CD$, $AQ$, and $BR$ are concurrent. Solution. Define $E=AQ\cap CD$. Use directed angles modulo $360^\circ$. Claim 1. $C$, $Q$, $R$, $E$ are concyclic. Proof. We have \begin{align*}\measuredangle CEQ&=180^\circ-\measuredangle EAD-\measuredangle ADE\\&=180^\circ-(\measuredangle QAC+\measuredangle CAD)-(\measuredangle ADQ+\measuredangle QDC)\\&=180^\circ-\measuredangle QAC-\measuredangle QDC-\measuredangle CAD-\measuredangle BAQ\\&=180^\circ-2\measuredangle QAC-(180^\circ-\measuredangle ADC-\measuredangle DCA)-\measuredangle BAQ\\&=\measuredangle ADC+\measuredangle DCA-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADC-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADQ+2\measuredangle QDC-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADQ-\measuredangle BAQ\\&=\measuredangle BAQ\\&=\measuredangle CRQ.\end{align*} Claim 2. $B$, $R$, $E$ are collinear. Proof. We have \begin{align*}\measuredangle QRB&=180^\circ-\measuredangle CRQ-\measuredangle ERC\\&=180^\circ-\measuredangle PAQ-\measuredangle EQC\\&=180^\circ-\measuredangle PAQ-\measuredangle ADC\\&=180^\circ-\measuredangle ACQ-\measuredangle DCA\\&=\measuredangle QCE\\&=180^\circ-\measuredangle ERQ.\end{align*} Now, Claim 2 implies the result. [asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.195830202854993, xmax = 18.824455296769358, ymin = -6.88018031555221, ymax = 5.561592787377907; /* image dimensions */ /* draw figures */ draw((4.37,-1.48)--(9.429564356435643,-1.4836435643564359), linewidth(0.8)); draw((5.65,1.88)--(10.709564356435642,1.8763564356435638), linewidth(0.8)); draw((9.429564356435643,-1.4836435643564359)--(5.65,1.88), linewidth(0.8)); draw((5.65,1.88)--(3.733168316831683,-3.151683168316832), linewidth(0.8)); draw((3.733168316831683,-3.151683168316832)--(10.709564356435642,1.8763564356435638), linewidth(0.8)); draw(circle((8.179089667867803,0.9165372693836932), 2.7063915055664625), linewidth(0.8)); draw(circle((3.7352424196126757,-0.2715209217572098), 2.8801629933754995), linewidth(0.8)); draw((3.733168316831683,-3.151683168316832)--(9.429564356435643,-1.4836435643564359), linewidth(0.8)); draw((5.65,1.88)--(7.737054951313305,-5.926480752802572), linewidth(0.8)); draw((7.737054951313305,-5.926480752802572)--(10.709564356435642,1.8763564356435638), linewidth(0.8)); draw((7.737054951313305,-5.926480752802572)--(4.37,-1.48), linewidth(0.8) + linetype("2 2")); /* dots and labels */ dot((5.65,1.88),dotstyle); label("$A$", (5.706498873027804,2.0304132231404948), NE * labelscalefactor); dot((4.37,-1.48),dotstyle); label("$B$", (4.429263711495121,-1.335477084898571), NE * labelscalefactor); dot((9.429564356435643,-1.4836435643564359),dotstyle); label("$C$", (9.493125469571757,-1.335477084898571), NE * labelscalefactor); dot((10.709564356435642,1.8763564356435638),linewidth(4pt) + dotstyle); label("$D$", (10.77036063110444,2.0003606311044315), NE * labelscalefactor); dot((3.733168316831683,-3.151683168316832),dotstyle); label("$P$", (3.798159278737796,-3.0033959429000725), NE * labelscalefactor); dot((6.468215171114042,-1.1804757104902723),linewidth(4pt) + dotstyle); label("$Q$", (6.53294515401954,-1.0650037565740034), NE * labelscalefactor); dot((5.290538122451863,-2.6956484235895024),linewidth(4pt) + dotstyle); label("$R$", (5.345867768595046,-2.582659654395189), NE * labelscalefactor); dot((7.737054951313305,-5.926480752802572),linewidth(4pt) + dotstyle); label("$E$", (7.79515401953419,-5.813313298271971), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

The circumcircle $DCR$ is $w$ and $BR$ Intersects $w$ at $F$. Claim:$D,A,F$ are collinear. $\angle{PAF}=\angle{PRF}=\angle{CDF}$ and $CD||AB$ implies the result. Radical axis theorem on $(CDFR),(AQRP),(AQCD)$

Cute $$\measuredangle ARC = \measuredangle ARP = \measuredangle AQP = \measuredangle AQD = \measuredangle ACD = \measuredangle ABC$$meaning $ABRC$ is cyclic. Let $S= \overline{AD} \cap \overline{BR}$. $$\measuredangle SAP = \measuredangle ABC = \measuredangle CAB = \measuredangle CRB = \measuredangle SRP$$meaning $SAQRP$ is cyclic. $$\measuredangle SRC = \measuredangle BRC = \measuredangle BAC = \measuredangle ADC = \measuredangle SDC$$meaning $SRCD$ is cyclic. Finish by radical axis theorem on $(SAQR), (AQCD) , (SRCD)$.

Let $S=AD\cap BR.$ Note that $\measuredangle ABC=\measuredangle ACD=\measuredangle AQD=\measuredangle ARC\implies R\in \odot (ABC).$ Next $\measuredangle SDP=$ $=\measuredangle ADC=$ $\measuredangle SRP\implies S\in \odot (APQ)\cap \odot (CDR).$Radical axis on $\odot (APQRS),\odot (ACDQ),\odot (CDRS)$ finish.

Angles are directed. We have \[ \angle ARC = \angle ARP = \angle AQP = \angle AQD = \angle ACD = \angle ABC, \]so $BRCA$ is cyclic. Let ray $\overline{RB}$ intersect $(APQ)$ at $E$. We claim that $E$ lies on $\overline{AB}$. Indeed, \[ \angle EAP = \angle ERP = \angle BRC = \angle BAC = \angle DAB. \]Finally, as \[ \angle ERC = \angle BRC = \angle BAC = \angle EDC, \]quadrilateral $ERCD$ is cyclic. Considering the radical center of $(ERQA)$, $(AQCD)$, and $(ERCD)$ establishes the concurrency.

Let $\angle ABC=x$ Note that, $AB$ is tangent to $(ACD)$ since $\angle BAC=\angle ADC$. Similarly $AC$ is tangent to $(AQRP)$ since $\angle PQA=180-\angle AQD=180-\angle ACD=180-\angle CAP=180-x$ Note that $\angle ARP=180- \angle ARP=180- \angle AQP=x$, hence $R \in (ABC)$. Let $AQ \cap CD=E$ Then since $AB || CD \implies \angle RCE=\angle RPA=\angle RQE$ where the last equality holds because $Q \in (ARP)$ Hence $E \in (RCQ)$ Note that $\angle ERC=x$, since $\angle ERC=\angle CQE=180-\angle AQC=180-(180-x)$ But since $R \in (ABC)$ we have $\angle BRC=180-x$ Hence we get $B-R-E$

Yay I can solve something now Let $RB \cap \odot(APQ) = X$ $$\measuredangle ARC = \measuredangle ARP = \measuredangle AQP = \measuredangle AQD = \measuredangle ACD = \measuredangle ABC \implies \odot(ABRC)$$This implies that $D-A-X$, Now $$\measuredangle DXR = \measuredangle AXR = \measuredangle APR = \measuredangle DCR \implies \odot(CDXR)$$Now radical centre on $\odot(AQDC),\odot(AQRP),\odot(CDXR)$ finishes the problem $\blacksquare$ v_Enhance wrote: Define $X = \overline{AQ} \cap \overline{CD}$ and let $R' = \overline{BX} \cap \overline{PC}$. We wish to show $APR'Q$ is cyclic. Let $M$ be the center of the parallelogram. Claim: $M$, $Q$, $R'$ are collinear. Proof. By Pappus theorem on $\overline{ABP}$ and $\overline{DCX}$. $\blacksquare$ [asy][asy] size(12cm); pair A = dir(90); pair D = dir(240); pair C = A*A/D; pair Q = dir(20); pair B = A+C-D; pair P = extension(D, Q, A, B); pair M = midpoint(A--C); pair X = extension(A, Q, C, D); pair Rp = extension(P, C, B, X); filldraw(unitcircle, invisible, blue); filldraw(A--B--C--D--cycle, invisible, red); draw(A--C, red); draw(B--D, red); draw(C--X, red); draw(B--P, red); draw(A--X, lightblue); draw(B--X, lightblue); draw(C--P, lightblue); draw(M--Rp, deepgreen); draw(D--P, lightblue); draw(Q--C, lightblue); pair Z = 2*A-P; pair W = -D+2*foot(origin, D, Z); draw(Q--W, deepgreen); draw(D--Z--A, orange); dot("$A$", A, dir(A)); dot("$D$", D, dir(D)); dot("$C$", C, dir(C)); dot("$Q$", Q, dir(72)); dot("$B$", B, dir(B)); dot("$P$", P, dir(P)); dot("$M$", M, dir(190)); dot("$X$", X, dir(X)); dot("$R'$", Rp, dir(Rp)); dot("$Z$", Z, dir(Z)); dot("$W$", W, dir(W)); /* TSQ Source: !size(12cm); A = dir 90 D = dir 240 C = A*A/D Q = dir 20 R72 B = A+C-D P = extension D Q A B M = midpoint A--C R190 X = extension A Q C D R' = extension P C B X unitcircle 0.1 lightcyan / blue A--B--C--D--cycle 0.1 lightred / red A--C red B--D red C--X red B--P red A--X lightblue B--X lightblue C--P lightblue M--Rp deepgreen D--P lightblue Q--C lightblue Z = 2*A-P W = -D+2*foot origin D Z Q--W deepgreen D--Z--A orange */ [/asy][/asy] Let $W$ be the point for which $AQCW$ is harmonic. Let $Z = \overline{DW} \cap \overline{AB}$. Claim: $\triangle ZAD \cong \triangle PAC$. Proof. Projecting from $D$ we get $-1 = (AC;QW) = (A\infty;PZ)$ so $A$ is the midpoint of $\overline{PZ}$. Since $AD=AC$ and $\measuredangle ZAD = \measuredangle CAP$, we're done. $\blacksquare$ To finish, \begin{align*} \measuredangle AQR' &= \measuredangle AQM = \measuredangle WQC = \measuredangle WDC = \measuredangle WDA + \measuredangle ADC \\ &= \measuredangle ACP + \measuredangle DCA = \measuredangle DCP = \measuredangle APR'. \end{align*} Remark: [Motivation] The Pappus step allows one to erase many points in the picture. After this step, rewrite the conclusion as $\measuredangle AQM = \measuredangle APC$; then the points $B$, $R'$, and $X$ may now be deleted. It seems that vEnhance has forgotten simple thinking

Let $E\neq R$ be such that $E=BR\cap (APQ)$ and let $F=BC\cap PE.$ By parallelism we have $AD=AC=BC.$ Since $\angle BAC=\angle ADC$ and $\angle PAC=180^{\circ}-\angle AQP=\angle AQD=\angle ACD.$ we have that $AP$ is tangent with $(ADC)$ and $AC$ is tangent with $(APQ).$ Let $\angle BAC=\alpha.$ We have $\angle CDA=\angle CBA=\angle CAB=\angle ACD$, $\angle CAB=\angle CAP=\angle AEP,$ and $\angle CBA=\angle FBP.$ Since $\angle ABF=180^{\circ}- \alpha$ we have that $A,B,F,E$ are concyclic. Therefore, $\angle FAP=\angle FAB=\angle FEB=\angle PER=\angle PAR.$ We also have $\angle CPA=\angle CAR=\angle AER=\angle AFC$ hence $A,C,P,F$ are concyclic. So we get $\angle BAE=\angle CFP=\angle CAP=\alpha$ so $P, A,$ and $E$ are collinear. Since $\angle PRE=\angle PAE=\alpha$ we get that $\angle EDC+\angle CRE=180^{\circ}$ therefore $D,C,R,$ and $E$ are concyclic. Applying the radical axis theorem on $(AQCD), (AQRE)$ and $(DCRE)$ we get that $DC, AQ,$ and $RE$ concur. \qed

let $L$ is the intersection of $(CD)$ & $(BR)$ claim_1:$ACRB$ is cyclic proof :$\angle AQD=\angle ACD=\angle ADC=\angle ABC$ $\rightarrow \angle AQP=180^o -\angle ABC$ $angle ARP=\angle AQP$ $\rightarrow \angle ARC=\angle ABC$ so $ACRB$ is cyclic also $\angle LCB=\angle ABC=\angle CAB$ so $LC $ is tangent to the circle of $ACRB$ $\rightarrow LC^2=LR.LB$ let $T$ is the intersection of $BR$ and the circum circle of traingle $APQ$ claim_2: traingles $ABT$ & $LRC$ are similiarity proof: $\angle CLR=\angle BTA$ by paralle $\angle LRC=\angle PRT=\angle BAT$ so we proof our claim , this mean that $LR.BT=LC.AB=LC.CD$ $\rightarrow LC^2+LC.CD=LR.LB+LR.BT \rightarrow LC.LD=LR.LT$ this mean that $L\in(AQ)$, so we done

Let $E=BR\cap (APQ)$. Notice $\angle EAB=\angle EAP=\angle ERP=\angle BAC=\angle ABC$, which menas $AE\parallel BC$. So $E,A,D$ are collinear. Now $\angle EDC=\angle ABC=\angle BAC=\angle ERP$, which means points $E,R,C,D$ are cyclic. So by the radical center theorem on cyclic quadrilaterals $(AQCD), (AQEP), (ERCD)$ we get our desired result.

Finally a normal G1 . Let $BR$ meet $(APQ)$ again at $T$. Notice that $CA = CB = DA$, so $ACD$ is also isosceles. Now, observe $$\angle CAP = \angle ACD = \angle AQD = 180^{\circ} - AQP$$which implies $CA$ is tangent to $(APQ)$. Thus, $$\angle ABC = \angle CAB = \angle CAP = 180^{\circ} - \angle ARP = \angle ARC$$so $ABRC$ is cyclic. This yields $$\angle PAT = \angle PRT = \angle PRB = \angle PAC = \angle CBA = 180^{\circ} - \angle DAP$$which means $A, D, T$ are collinear. As a result, we have $$\angle CDT = \angle CDA = \angle DCA = \angle PAC = \angle PRT = 180^{\circ} - \angle CRT$$so $CDTR$ is cyclic. Now, Radical Axes on $(ACD), (APQ), (CDTR)$ finishes. $\blacksquare$ Remarks: Constructing $T$ and reverse engineering the fact that $CDTR$ is cyclic is very natural (due to Radical Axes). Then, blindly angle chasing to obtain the necessary angles ends up working out.

Let $BR\cap CD=X, Y=XB\cap (APRQ).$ Lemma 1 $AC$ tangent to $(APRQ)$ at $A$. Proof: As $AP\parallel CD$ and $AQCD$ concyclic therefore $$\angle APD=\angle CDP=\angle CAQ$$ Lemma 2 $BC$ is tangent to $(BPR)$ Proof: By Lemma 1 $$BC^2=AC^2=CR\cdot CP$$ Lemma 3 $D, A ,Y$ collinear Proof: As $A,Y,P,R$ concyclic $\angle ACPR=\angle AYR$, by Lemma 2 $\angle APR=\angle CBX \Rightarrow \angle AYR=\angle CBX\Rightarrow $ $BC\parallel AY$ but $BC\parallel AD$, therefore $D,A, Y$ are in the same line. Lemma 4 $DYRC$ cyclic Proof: By Lemma 2 and Lemma 3 $$\angle APR=\angle AYR$$as $AP\parallel DC \Rightarrow$ $$\angle APR=\angle XCP \Rightarrow \angle AYR=\angle XCR$$ Finally, the radical axes of $AQCD, AYPRQ, DYRC$ are concurrent, this is $CD, AQ,YR$ are concurrent $\blacksquare$

Define $X=AQ \cap BR$. We prove that $X$ lies on $CD$. Claim 1 : $\square ABRC$ cyclic. Proof : $\angle ARC = 180 -\angle ARP = 180- \angle AQP = \angle AQD = \angle ACD= \angle ADC =\angle ABC$. This gives $\square ABRC$ cyclic. $\blacksquare$ Claim 2 : $\square QRXC$ cyclic. Proof : $\angle XRC = 180 - \angle BRC = \angle BAC = \angle ABC = \angle ADC = 180- \angle AQC =\angle XQC$. This gives $\square QRXC$ cyclic. $\blacksquare$ Claim 3 : $X$ lies on $CD$. Proof : $\angle CXA=\angle CXQ =\angle CRQ = 180 -\angle PRQ =\angle PAQ=\angle PAX$. This gives $AP \parallel CX$. But we have $AP \parallel CD$. Hence $X$ lies on $CD$. $\blacksquare$ Hence we are done! Remark : $Q$ turns out to be the $A-$ Dumpty point of $\triangle ACP$!

Let $BR$ intersect $AD$ at $T$. By Radical Axis Theorem it is sufficient to show that $T$ lies on $(AQRP)$. Claim: $AC$ is tangent to $(AQP)$ $$\angle QAC=\angle QDC=\angle QPA$$Claim: $BC$ is tangent to $(RPB)$ $$CB^2=CA^2=CP\cdot CR$$Cliam: $T$ lies on $(AQRP)$ $$\angle RXA=\angle RXB=\angle RBC=\angle RTA$$

let $AQ$ intersect $CD$ at $S$ from angle chasing, we get that $CSQ=CRQ=QAP$ thus, $SCQR$ are cyclic more angle chasing yields $CDQ=CAQ=QRA=QPA$, and $BRQ=QAD=QCS$, so since $SCQR$ are cyclic $SRB$ are collinear thus, $CD$, $AQ$, and $BR$ are concurrent at $S$

Note that $$\measuredangle ADC =\measuredangle CBA= \measuredangle BAC.$$So $CD$ is tangent to $(ACD)$. Also note that $$ \measuredangle ACD = \measuredangle CAB = \measuredangle ABC.$$Thus $CD$ is tangent to $(ACB)$. Now note that $$ \measuredangle ABC = \measuredangle CAP = \measuredangle ARP =\measuredangle ARC.$$Therefore $ABRC$ is cyclic. Now define $T= AQ \cap CD$ then $$\measuredangle RCT = \measuredangle RPA = \measuredangle RQA = \measuredangle RQT.$$Thus $RQCT$ is cyclic. Now note that $$\measuredangle CRT = \measuredangle CQT = \measuredangle ADC = \measuredangle BCT.$$$$ \measuredangle RCT = \measuredangle RPB = \measuredangle RBC.$$Thus $\triangle CRT \sim \triangle BCT$, so we can conclude $B$, $R$, and $T$ are collinear. $\square$

Let $F = BR \cap CD$ and $E = BR \cap (AQP)$. Claim : $B,R,C,A$ are concyclic. Proof: $\angle ABC = \angle BAC = \angle ACD = \angle AQD = 180^{\circ} - \angle AQP = 180^{\circ} - \angle ARP = \angle ARC$ Claim : $E,A,D$ are collinear. Proof: $\angle EAP = \angle ERP = \angle BRP = \angle BAC = \angle ABC = 180^{\circ} - \angle BAD = 180^{\circ} - \angle PAD$ Claim : $R,F,C,Q$ are concylic. Proof: $\angle FCQ = \angle QAD = \angle ERQ = 180^{\circ} - \angle QRF$ Claim : $A,Q,F$ are collinear. Proof: $\angle FQC = \angle FRC = \angle PRE = \angle PAE = 180^{\circ} - \angle BAD = \angle ADC = 180^{\circ} - \angle AQC$. And we are done.

A tad bit harder than the solution suggests. Let $S = \overline{BR} \cap (AQP)$. By radical axis, it suffices to show that $SRCD$ is cyclic. Claim: $ABRC$ is cyclic. Proof. $\measuredangle ARC = \measuredangle ARP = \measuredangle AQD =\measuredangle ACD = \measuredangle ABC$. $\blacksquare$ So $\measuredangle SRC = \measuredangle BAC = \measuredangle SDC$, which finishes.

To prove the concurrence it is enough to prove that radical center of $(SDCR),(AQDC)$ and $(AQRP)$ exists or to prove $(SDCR)$ is conyclic. $$\measuredangle CRS=\measuredangle PRS =\measuredangle PAS=\measuredangle ADC \iff \text{$S,A,D$ are collinear}$$Now to prove S,A,D are collinear, this can be done by angle chasing: $$\measuredangle SAB= \measuredangle SRP= \measuredangle BRC=\measuredangle BAC \iff (ABCR)$$We do this because $\measuredangle DAC=180-\measuredangle CAS=180-2\measuredangle BAS=180-\measuredangle ADC-\measuredangle ACD \implies S,A,D$ are collinear. Now to proving $(ABCR)$:$$\measuredangle ABC= \measuredangle ACD= \measuredangle AQD= \measuredangle AQP=\measuredangle ARP=\measuredangle ARC.$$So $CD,AQ\; \text{and} \;BR$ concur at the radical center of of $(SDCR),(AQDC) \;\text{and} \;(AQRP)$.

Let $S = BR \cap DC$. Our goal is to prove that $\Delta ASC \sim \Delta APD$. First we claim that $\Delta APC \sim \Delta CBS$: $\angle SCB = \angle CBA = \angle BAC = \angle PAC$. Even more, since $\angle CAQ = \angle CDQ = \angle APQ$, $CA$ is tangent to $\Delta APQ$ and because of $AC = BC$ and PoP, $CB$ is tangent to $\Delta BPR$, in other words $\angle CBS = \angle APC$ and we have our claim. Now because we have $\angle SCA = 180^\circ - \angle ACD = \angle PBC = \angle PAD$ and from our previous claim $\frac{AC}{AP} = \frac{CS}{BC} = \frac{CS}{BD}$, our desired similarity is indeed true and $\angle CAQ = \angle CAS$, meaning $A, Q, S$ are collinear.

First AoPS Post Let $X=\overline{RB} \cap \overline{DA}$ I claim that $X$ lies on $(AQRP)$. Simple angle chasing gives that $\overline{CA}$ is tangent to $(APQR)$. Hence $CB^{2}=CA^{2}=CR \cdot CP$ by power of a point. Therefore $\measuredangle RPA= \measuredangle CPB= \measuredangle CBR= \measuredangle AXR$ as desired. Now note that \[\measuredangle XRC= \measuredangle XRP =\measuredangle XAB =\measuredangle XDC\]Hence $RXDC$ is cyclic. Hence by Radical Axis Theorem, $\overline{XBR}$, $\overline{AQ}$, and $\overline{BC}$ are concurrent.

Claim 1: $ABCR$ cyclic. Proof: Consider $\angle ABC = \angle ACD = \angle AQD = 180 - \angle AQP = 180 - \angle ARP = \angle ARC$. Claim 2: $AD \cap BR = K$ lies on $(AQR)$ Consider $\angle APR = \angle APC = 180 - \angle PAC - \angle PCA = 180 - \angle BCA - \angle PBR = 180 - \angle KAB - \angle KBA = \angle AKR$, as desired. Claim 3: $(KCDR)$ cyclic. Proof: Consider $\angle KDC + \angle CRK = \angle ADC + \angle BRC = \angle CAB + (180 - \angle CAB) = 180$. To finish, apply the radical axis theorem on $(AQCD), (AQKR), (KRCD)$, to get $CD, AQ, KR$ concur, and since $K,B,R$ are collinear, we are done.

So many concyclic quadrilaterals and paralellism implies that Reim's Theorem would be useful here. Let $BD\cap PC=H$ and $AQ\cap CD=J$. Claim: $C,H,Q,R$ and $J$ are concyclic. Proof: Since $\angle QRC=\angle PAQ=\angle QJC$ implies points $C,Q,R$ and $J$ are collinear. On the other hand $$\angle QHC=\pi-\angle BHQ=\pi-\angle PAQ=\angle PRQ$$Thus, $H\in (CQRJ)$. Claim: $ABQH$ and $ABRC$ are concyclic quads. Proof: $AC=BC$ is given. Notice that $\angle AQH=\angle ACD=\angle ABH$ implies points $A,B,Q$ and $H$ lie on a circle. On the other hand, PoP gives $$PB\cdot PA=PQ\cdot PH=PR\cdot PC\Longleftrightarrow R\in (ABC) $$Claim: Points $B$, $R$ and $J$ are collinear. Proof: Observe that $AB\parallel CJ$ and common chord of $(ABRC)$ and $(CHQRJ)$ is $CR$. By Converse of Reim's Theorem, $B$, $R$ and $J$ are collinear, as desired.

Let $AQ \cap BR = K$. It suffices to show $C$, $D$, $K$ collinear, or $KC \parallel AB$: Notice $ABCR$ is cyclic, as \[\measuredangle CBA = \measuredangle DCA = \measuredangle DQA = \measuredangle PQA = \measuredangle PRA = \measuredangle CRA.\] Notice $CQRK$ is cyclic, as \[\measuredangle KRC = \measuredangle BRP = \measuredangle BAC = \measuredangle ADC = \measuredangle KQC.\] Hence we get the desired from \[\measuredangle RCK = \measuredangle RQK = \measuredangle RPA. \quad \blacksquare\]

I am not going to read $81$ posts, so this is probably nothing new, but here it goes. Let $E=BR\cap AD$. Observe that$$\measuredangle QAC=\measuredangle QDC=\measuredangle QPA,$$so $CA$ is tangent to $(AQRP)$, which gives$$CB^2=CA^2=CR\cdot CP.$$Hence $CB$ is tangent to $(BPR)$. Observe now that$$\measuredangle EAP=\measuredangle EAB=\measuredangle CBA\underset{(1)}{=}\measuredangle BRP=\measuredangle ERP,$$where $(1)$ holds because $CB$ is tangent to $(BRP)$. Hence $AQRPE$ is cyclic. Finally, we have$$\measuredangle EDC=\measuredangle EAP=\measuredangle ERP=\measuredangle ERC,$$so $CDER$ is cyclic. Hence the three lines in question are the radical axes of $(AQCD)$, $(AQRPE)$ and $(CDER)$, so they are concurrent (*). (*) If $AQ\parallel CD$ then $Q=P$, so $PA$ is the exterior angle bisector of $\angle CPD$, so $PC=PD$, which gives $P=A$; similarly, if $BR\parallel AD$ then $P=B$; both of these contradict the fact that $P$ lies on the extension of ray $AB$ past $B$.

Heres the angle-chase: Let $X= AQ \cap BR$. Claim: $ABCR$ is cyclic. Proof: Notice that: $$\angle ARC = 180^\circ - \angle ARP = 180^\circ - \angle AQP = \angle AQD = \angle ACD = \angle ABC.$$ Thus, we have: $$\angle CRX = \angle CAB =\angle ADC = \angle CQX$$which implies $CQRX$ to be cyclic. Therefore, by Reims, $AP \parallel CX$ and we conclude.

There´s nothing new, just to record my solution somewhere. Let $K=AQ\cap CD, \angle QDC=\alpha, \angle QDA=\beta$ and $\angle CPA=\theta$ . Since $\angle QPA=\alpha=\angle QAC$, $CA$ is tangent to $(APQ)$. Therefore $CA^2=CB^2=CR\cdot CP$, so $CB$ is tangent $(BPR)$ and $\angle RBC=\theta=\angle RCK$. Now, note that $\angle CAB=\angle ACD$, so $BA$ is tangent to $(ACD)$. Hence $$\beta=\angle PAQ=\angle QKC=\angle QRC \Rightarrow \text{$QCKR$ is cyclic} \Rightarrow \angle RKQ=\angle RCQ=180-(\alpha+2\beta+\theta)$$But $$\angle RBC+\angle BCK+\angle BKR=(\theta +(\alpha +\beta)+(180-(\alpha+\beta))=180 \Rightarrow \text{B,R,K are collinear} \blacksquare$$

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