I recall failing to solve this problem in DC. The key lies in finding

$ \angle CXD = \angle CDB - \angle XCD$ and $ \angle AQB = \angle BQP + \angle AQP = (180 - \angle CDB)+ \angle XCD$ so $ \angle CXD + \angle AQB = 180$ so $ AQBX$ is cyclic quadrilateral. $ \angle QYP = \angle QAP = \angle QXB$ and $ PY\parallel BX$ so $ Q,Y,X$ are collinear. Analogously $ Q,Z,X$ are collinear

K81o7 wrote: the cyclic quadrilateral, $ ABQX$ This follows directly from the fact that $ Q$ is the Miquel point

I think I've got a slightly different solution. $ \angle XCQ = \angle ACQ = \angle BPQ = \angle BDQ$, so $ DXCQ$ is a cyclic quad. Then $ \angle QXD = \angle QCD = \angle QCP = \angle QYP$. Thus $ \angle QXD = \angle QYP$, and since $ XB$ and $ YP$ are parallel, it follows that $ Q,X,Y$ are collinear. Similarly, $ Q,X,Z$ are collinear.

$Q$ is the unique center of the spiral similarity sending $CD$ to $AB$, so it is also the unique center of the spiral similarity sending $CA$ to $DB$; hence $DXQC$ and $AXQB$ are concyclic. Let $Z' = XQ \cap \omega_2$; then \[ \measuredangle CAB = \measuredangle XAB = \measuredangle XQB = \measuredangle Z'QB = \measuredangle Z'PB \implies PZ' \parallel AC \implies Z=Z' \] and $Y=Y'$ is similar. [asy][asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-2.0, 2.0); pair B = (4.031165686870916, 3.133944005249443); pair C = (1.0, -2.0); pair D = (1.0, 0.0); pair P = IntersectionPoint(A--B,Line(C,D,0,lisf)); pair X = IntersectionPoint(A--C,Line(B,D,0,lisf)); pair O_1 = circumcenter(A,P,C); pair O_2 = circumcenter(P,D,B); pair Q = 2*foot(P,relpoint(O_1--O_2,0.5-10/lisf),relpoint(O_1--O_2,0.5+10/lisf))-P; pair Y_prime = (2)*(foot(O_1,X,Q))-Q; pair Z_prime = (2)*(foot(O_2,Q,X))-Q; /* Draw objects */ draw(circumcircle(A,P,C), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(circumcircle(B,P,D), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(A--B, rgb(0.4,0.4,0.4)); draw(C--P, rgb(0.4,0.4,0.4)); draw(A--C, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(B--X, rgb(0.5,0.5,0.5) + linetype("4 4")); draw(P--Y_prime, rgb(0.5,0.5,0.5) + linetype("4 4")); draw(P--Z_prime, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(Y_prime--Z_prime, rgb(0.4,0.8,1.0)); draw(circumcircle(C,Q,D), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(circumcircle(Q,A,B), rgb(0.7,0.8,0.9) + linewidth(1.0)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(D); dot(P); dot(X); dot(Q); dot(Y_prime); dot(Z_prime); /* Label points */ label("$A$", A, lsf * dir(150)); label("$B$", B, lsf * dir(45)); label("$C$", C, lsf * dir(-90)); label("$D$", D, lsf * dir(60) * 1.414); label("$P$", P, lsf * dir(100)); label("$X$", X, lsf * dir(225)); label("$Q$", Q, lsf * dir(-45)); label("$Y'$", Y_prime, lsf * dir(225)); label("$Z'$", Z_prime, lsf * dir(-45)); [/asy][/asy]

v_Enhance wrote: ... Let $Z' = XQ \cap \omega_2$; then \[ \langle CAB = \langle XAB = \langle XQB = \langle Z'QB = \langle Z'PB \implies PZ' \parallel AC \implies Z=Z' \] ... I've noticed that some people on the forum use Reim's theorem in particular if we look at circles $AXBQ$ and $PBQZ'$ we have $PZ'||AX$.

$Q$ is the Miquel point of $CXA, CDP, XDB, APB$, so $CXDQ$ is cyclic. By Reim's theorem on $\omega_1$ and $(CXDQ)$, $Q,X,Y$ are collinear; by Reim's theorem on $\omega_2$ and $(CXDQ)$, $Z,Q,X$ are collinear, and the result follows.

Let $ J= YP \cap \omega_2$. Let $ L= YP \cap \omega_1$. Now, \[ \measuredangle ZQB = \measuredangle ZPB = - \measuredangle BPZ = \measuredangle ZPA = \measuredangle LPA =\measuredangle PAC =\measuredangle XAP \]Now, \[ \measuredangle YQB = \measuredangle YQP + \measuredangle PQB = \measuredangle YCP + \measuredangle PJB = \measuredangle YCP + \measuredangle DPJ = \measuredangle YCP + \measuredangle CPY = - \measuredangle PYC = - \measuredangle PAC = \measuredangle XAP \]Finally observe that $Q$ is the $Miquel$ $Point$ of the complete quadrilateral formed by the vertices $X,D,P,A$. \[ \implies XAQB concyclic. \]. So, \[ \measuredangle XQB = \measuredangle XAB = \measuredangle XAP \]. So , \[ \measuredangle ZQB = \measuredangle YQB = \measuredangle XQB \]This leads us to the desired conclusion.

By spiral similarity $Q$ maps $DB$ to $CA$ so $QCXD$ is cyclic. If $QX$ intersects $\omega_1$ at $Y'$, $PY'$ and $DX$ are both antiparallel to $QC$ so $PY'\parallel DX$, $Y'=Y$, and $Y,Z$ lie on $QX$.

$Q$ is the Miquel point of complete quadrilateral $PBXC$, so $QDXC$ and $ADBX$ are cyclic. Let $QX \cap w_1 \equiv Y'$. Angle chasing, we have $\angle Y'QC = \angle Y'AC = \angle Y'PD$. By the above and $QDXC$ cyclic, we have that $\angle Y'QC = \angle CDX = \angle Y'PD$ so $Y'P \parallel DX$, and similarly for $Z'$. The conclusion follows.

Maybe this solution is already given above, but still I'm writing it down. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.716666666666658, xmax = 17.6, ymin = -1.7266666666666675, ymax = 8.073333333333334; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen ffqqff = rgb(1.,0.,1.); /* draw figures */ draw(shift((3.185458267893191,3.1111558596963573)) * scale(3.1559227963577987, 3.1559227963577987)*unitcircle, linewidth(1.6) + red); draw(shift((5.7380030468626,4.431102969733766)) * scale(2.628399180397597, 2.628399180397597)*unitcircle, linewidth(1.6) + blue); draw((3.8,6.206666666666669)--(5.453214154262895,0.9163593004666906), linewidth(1.6) + red); draw((3.8,6.206666666666669)--(2.4181126438879694,0.04994214780186379), linewidth(1.6) + red); draw((5.155831433496555,1.8679879773650003)--(3.2270742084802806,3.6541096492381664), linewidth(1.6) + blue); draw((2.4181126438879694,0.04994214780186379)--(5.453214154262895,0.9163593004666906), linewidth(1.6) + red); draw((3.8,6.206666666666669)--(6.446607586458077,6.962182158139266), linewidth(1.6) + xdxdff); draw((6.224733389568109,3.9612472849787848)--(3.8,6.206666666666669), linewidth(1.6) + ffxfqq); draw((5.453214154262895,0.9163593004666906)--(6.011390333566628,1.075699412915823), linewidth(1.6) + linetype("4 4") + red); draw((6.011390333566628,1.075699412915823)--(5.155831433496555,1.8679879773650003), linewidth(1.6) + linetype("4 4") + blue); draw((6.011390333566628,1.075699412915823)--(6.446607586458077,6.962182158139266), linewidth(1.6) + dotted + green); draw((5.155831433496555,1.8679879773650003)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + ffqqff); draw((3.2270742084802806,3.6541096492381664)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + ffqqff); draw(shift((3.916839652865845,1.6064204600289003)) * scale(2.1607423224144324, 2.1607423224144324)*unitcircle, linewidth(1.6) + linetype("2 2") + ffqqff); draw(shift((5.593442797870414,1.4824607634596496)) * scale(0.5832108871504142, 0.5832108871504142)*unitcircle, linewidth(1.6) + linetype("2 2") + ffqqff); draw((5.453214154262895,0.9163593004666906)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + red); label("X = X'",(6.133333333333323,1.0566666666666664),SE*labelscalefactor,green); draw((2.4181126438879694,0.04994214780186379)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + ffqqff); /* dots and labels */ dot((3.8,6.206666666666669),red); label("$P$", (3.783333333333327,6.506666666666668), NE * labelscalefactor,red); dot((6.06666666666666,1.823333333333333),red); label("$Q$", (6.2,1.6233333333333333), NE * labelscalefactor,red); label("$\omega_1$", (1.1333333333333315,5.906666666666667), NE * labelscalefactor,red); label("$\omega_2$", (5.4,7.24), NE * labelscalefactor,blue); dot((5.453214154262895,0.9163593004666906),red); label("$A$", (5.4,0.573333333333333), NE * labelscalefactor,red); dot((5.155831433496555,1.8679879773650003),blue); label("$B$", (4.833333333333325,1.69), NE * labelscalefactor,blue); dot((2.4181126438879694,0.04994214780186379),red); label("$C$", (2.2833333333333297,-0.2933333333333338), NE * labelscalefactor,red); dot((3.2270742084802806,3.6541096492381664),blue); label("$D$", (2.966666666666662,3.6233333333333335), NE * labelscalefactor,blue); dot((6.446607586458077,6.962182158139266),xdxdff); label("$Z$", (6.516666666666656,7.123333333333334), NE * labelscalefactor,xdxdff); dot((6.224733389568109,3.9612472849787848),ffxfqq); label("$Y$", (6.4,3.923333333333334), NE * labelscalefactor,ffxfqq); dot((6.011390333566628,1.075699412915823),green); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.1386054421768685) * currentpicture; /* end of picture */[/asy][/asy] First, we prove that $X,Q,Z$ are collinear $\iff XAQB$ is cyclic. Let $ZQ\cap AC=X'.$ Now, $\measuredangle X'AB=\measuredangle BAC=\measuredangle APZ=\measuredangle BQX'.$ So $X'ABQ$ is cyclic. Thus we need to prove $X=X'$ Note that $\measuredangle ABX=\measuredangle PBD=\measuredangle BPY=\measuredangle APY=\measuredangle AQX'=\measuredangle ABX'$ This means $X=X'\implies X,Q,Z$ are collinear. So it remains to prove $X,Q,Y$ are collinear. Note that $\measuredangle AQX=\measuredangle APY=\measuredangle AQY\implies X,Q,Y$ are collinear. Thus, $Q,X,Y,Z$ are collinear. $\blacksquare$

My proof is literally the exact same as everyone else's, but here goes. Let $QX\cap w_1=Y'$. It suffices to show that $PY' \parallel BD$. We have $$\measuredangle{APY'}=\measuredangle{APC}-\measuredangle{Y'PC}=\measuredangle{BPC}-\measuredangle{Y'QC}=\measuredangle{BPD}-\measuredangle{XQC}$$Now observe that $Q$ is the Miquel Point of complete quadrilateral $AXDP$, since $Q=w_1 \cap w_2$, so $DXCQ$ is cyclic. We now have $$\measuredangle{BPD}-\measuredangle{XQC}=\measuredangle{BPD}-\measuredangle{XDC}=\measuredangle{BPD}-\measuredangle{BDP}=-\measuredangle{DPB}-\measuredangle{BDP}=\measuredangle{PBD}$$Hence $PY' \parallel BD$. An identical argument shows that $PZ' \parallel AC$, where $Z'$ is defined analogously to $Y'$. It follows that $Q,X,Y$ and $Z$ are collinear, as desired. $\square$

Note that $\ Q$ is the Miquel Point of the quadrilateral $\ PDXA$, therefore $\ Q$ is the center of the unique spiral similarity sending $\ PD$ to$\ AX$, hence $\angle{AQX}=\angle{PQD}=\angle{PBD}=\angle{ABX}=\angle{APY}=\angle{AQY}$, implying that $\ Q,X$ and$\ Y$ are collinear. In addition, $\angle{BQZ}=\angle{BPZ}=\angle{PAZ}=\angle{PAQ}+\angle{QAC}=\angle{QCP}+\angle{QPC}=\angle{QXD}+\angle{QBD}=180\ -\angle{BQX} $, thus $\ X,Z,Q$ are collinear. Finally, the points $\ X,Y,Z$ and$\ Q$ are collinear.

Since $Q$ is the miquel point of complete quadrilateral $PDXA,$ quadrilateral $ABQX$ is cyclic. But since $\measuredangle AQY=\measuredangle APY=\measuredangle ABX=\measuredangle AQX;$ we get that $Q,X$ and $Y$ are collinear. Similarly $Q,X$ and $Z$ are collinear, and the result follows. $\square$

Note that $Q$ is the miquel point of $APDX$, so $AXQB$ and $CXDQ$ are cyclic. $$\angle DQZ = \angle DPZ = \angle CPZ = \angle ACP = \angle XCD = \angle DQX,$$so $X,Z,Q$ are collinear. $$\angle AQX = \angle ABX = \angle BPX = \angle APY = \angle AQY,$$so $Y,X,Q$ are collinear.

Identify $Q$ as the Miquel point of $ACDB$. Let $Y'$ denote the second intersection of $\omega_1$ and $XQ$, note that by the spiral similarity circles, \[\measuredangle Y'PC=\measuredangle Y'QC=\measuredangle XQC=\measuredangle XDC,\]hence $Y'=Y$. Then, $Q,X,Y$ are collinear, so we are done by symmetry.

N.T.TUAN wrote: Circles $ \omega_1$ and $ \omega_2$ meet at $ P$ and $ Q$. Segments $ AC$ and $ BD$ are chords of $ \omega_1$ and $ \omega_2$ respectively, such that segment $ AB$ and ray $ CD$ meet at $ P$. Ray $ BD$ and segment $ AC$ meet at $ X$. Point $ Y$ lies on $ \omega_1$ such that $ PY \parallel BD$. Point $ Z$ lies on $ \omega_2$ such that $ PZ \parallel AC$. Prove that points $ Q,X,Y,Z$ are collinear. Invert the problem about $P$. Let $T'$ denote the image of point $T$ under the given inversion. Then our problem is Inverted form of given problem wrote: $A',B',C',D',P$ are points in a plane such that $P=C'D' \cap A'B'$. Let the lines $A'C'$ and $B'D'$ intersect at $Q'$. Let $X'=(A'C'P) \cap (B'D'P)$. Let the tangents from $P$ to $(A'C'P), (B'D'P)$ intersect $B'D',A'C'$ at $Z',Y'$ respectively. Prove that $Y'X'PZ'Q'$ is cyclic Observe that $$\angle X'C'Y' =\angle X'P'B' =\angle X'D'B'$$Hence $X'C'Q'D'$ is cyclic. Note that \begin{align*} \angle PZ'D' =\angle Z'PC' -\angle Z'D'P &=\angle PX'C' -\angle Z'D'P \\ &= \angle PX'C' -\angle Q'X'C' \\&= \angle PX'Q' \end{align*}Hence $PX'Q'Z'$ is cyclic. Similarly we can conclude that $PX'Y'Q'$ is cyclic. Hence we are done.

Obviously $Q$ is the Miquel Point, which means that $BAXQ$ and $CXDQ$ are cyclic. I will first show that $Q, Z, X$ are collinear. Note that \[\measuredangle QZP=\measuredangle QDP=\measuredangle QDC=\measuredangle QXC=\measuredangle QXA.\]Since $PZ$ and $AC$ are parallel, $Q, Z, X$ must be collinear. Now I will show that $Q, X, Y$ are collinear. We have \[\measuredangle QXB=\measuredangle QAB=\measuredangle QAP=\measuredangle QYP.\]Since $PY$ and $BD$ are parallel, $Q, X, Y$ must be collinear. Combining the facts that $Q, Z, X$ are collinear and $Q, X, Y$ are collinear, we have that $Q,X,Y,Z$ are collinear.

Solution. Claim 1. $XCQD$ is cyclic. Proof. We show that $\angle XCQ + \angle XDQ = \pi \iff \angle XCQ = \pi - \angle XDQ$. $$\angle XCQ = \angle ACQ = \pi-\angle APQ = \angle BPQ = \angle BDQ = \pi -\angle XDQ$$ Claim 2. $Y,X,Q$ collinear. Proof. We show that $\angle XQP = \angle YQP$. $$\angle XQP = \angle XQD + \angle DQP = \angle XCD + \angle DBP$$$$=\angle ACP + \angle YPA = \angle YQP$$ Claim 3. $X,Q,Z$ collinear. Proof. We show that $\angle XQP+\angle PQZ = \pi\iff \angle PQZ = \pi - \angle XQP$. $$\angle PQZ = \pi - (\angle PBD + \angle DBZ)=\pi -( \angle APY + \angle DPZ)$$$$=\pi -( \angle APY + \angle DCA)=\pi -( \angle APY + \angle PCA)=\pi -\angle XQP$$

Consider complete quadrilateral $APDX$ with $AP \cap DX = B$ and $AX \cap DP = C$. Since $Q = (CAP) \cap (BDP)$, we know it's the Miquel Point of $APDX$. Thus, $CXDQ$ and $BAXQ$ are cyclic. Claim: $X, Q, Z$ are collinear. Proof. Notice $$\measuredangle XQD = \measuredangle XCD = \measuredangle ACP = \measuredangle ZPC = \measuredangle ZPD = \measuredangle ZQD$$which obviously suffices. $\square$ Claim: $AY \parallel BZ$. Proof. Let $BX \cap AY = E$. Then, $$\measuredangle AEB = \measuredangle AYP = \measuredangle ACP = \measuredangle ZPC = \measuredangle ZPD = \measuredangle ZBD = \measuredangle ZBE$$as desired. $\square$ Thus, $$\measuredangle PQY = \measuredangle PAY = \measuredangle BAY = \measuredangle ABZ = \measuredangle PBZ = \measuredangle PQZ$$implying $Q, Y, Z$ are collinear. Combining the two collinearities yields the desired result. $\blacksquare$

Notice that $Q$ is the Miquel point of complete quadrilateral $BPCAXD.$ Let $Y'=\overline{PQ}\cap\overline{ACP}.$ Since $CAPQ$ and $BXAQ$ are cyclic, $$\measuredangle BPY'=\measuredangle APY'=\measuredangle AQY'=\measuredangle ABX,$$so $\overline{PY'}\parallel\overline{BD}.$ Thus, $Y'=Y$ and $Y$ is on $\overline{PQ}.$ Similarly, $X$ is also on $\overline{PQ}.$ $\square$

We have $XCQD$ cyclic because $\angle XCQ=\angle ACQ=\angle QPB=\angle QDB.$ We have $\angle DXQ=\angle DCQ=\angle PCQ=\angle PYQ$ and $PY \parallel BD$ implies $Y,X,Q$ collinear. Similarly, $\angle QXC=\angle QDC=\angle PZQ$ and $XC\parallel PZ$ implies $X,Q,Z.$ The result follows.

[asy][asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-2.0, 2.0); pair B = (4.031165686870916, 3.133944005249443); pair C = (1.0, -2.0); pair D = (1.0, 0.0); pair P = IntersectionPoint(A--B,Line(C,D,0,lisf)); pair X = IntersectionPoint(A--C,Line(B,D,0,lisf)); pair O_1 = circumcenter(A,P,C); pair O_2 = circumcenter(P,D,B); pair Q = 2*foot(P,relpoint(O_1--O_2,0.5-10/lisf),relpoint(O_1--O_2,0.5+10/lisf))-P; pair Y_prime = (2)*(foot(O_1,X,Q))-Q; pair Z_prime = (2)*(foot(O_2,Q,X))-Q; /* Draw objects */ draw(circumcircle(A,P,C), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(circumcircle(B,P,D), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(A--B, rgb(0.4,0.4,0.4)); draw(C--P, rgb(0.4,0.4,0.4)); draw(A--C, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(B--X, rgb(0.5,0.5,0.5) + linetype("4 4")); draw(P--Y_prime, rgb(0.5,0.5,0.5) + linetype("4 4")); draw(P--Z_prime, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(Y_prime--Z_prime, rgb(0.4,0.8,1.0)); draw(circumcircle(C,Q,D), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(circumcircle(Q,A,B), rgb(0.7,0.8,0.9) + linewidth(1.0)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(D); dot(P); dot(X); dot(Q); dot(Y_prime); dot(Z_prime); /* Label points */ label("$A$", A, lsf * dir(150)); label("$B$", B, lsf * dir(45)); label("$C$", C, lsf * dir(-90)); label("$D$", D, lsf * dir(60) * 1.414); label("$P$", P, lsf * dir(100)); label("$X$", X, lsf * dir(225)); label("$Q$", Q, lsf * dir(-45)); label("$Y'$", Y_prime, lsf * dir(225)); label("$Z'$", Z_prime, lsf * dir(-45)); [/asy][/asy] (diagram stolen from @vEnhance) I tried removing the point $X$ first to show that $\overline{Y-Q-Z}$ were collinear and my goodness, how badly stuck was I back then... But fortunately, spiral sim gotchme! Firstly note that $\measuredangle QCA=\measuredangle QPA=\measuredangle QPB=\measuredangle QDB$ and $\measuredangle CAQ=\measuredangle CPQ=\measuredangle DPQ=\measuredangle DBQ$, and so, $Q$ is the center of the spiral similarity taking $CA\mapsto DB$. So we thus get that $QCXD$ and $QXAB$ are cyclic. Now we have $\measuredangle AQX=\measuredangle ABX=\measuredangle PBD=\measuredangle ABD=\measuredangle APY=\measuredangle AQY$ which gives that $\overline{Q-X-Y}$ are collinear and similarly $\overline{Q-X-Z}$ are also collinear and we are done.

first, by spiral similarity $CQD$ and $AQB$ are similar from that, we also get that $AQC$ and $BQD$ are similar, and angles $QAX$ and $QBX$, so $ABQX$ is a cyclic quadrilateral. then, if $BQZ=BAX$ then $XQZ$ are collinear, but $BQZ=BPZ=BAX$ because $PZ||AC$, so $XQZ$ are collinear lastly, we need $PYQ=BXQ$ to prove $Y$ is collinear with $XQZ$, but because we have $PYQ=PAQ=BXQ$, $Y$ is collinear with $XQZ$ therefore, $Q,X,Y$, and $Z$ are collinear

Take the quadrilateral $APDX$. We know that $Q$ is the Miquel point of this quadrilateral, since $Q$ is the second intersection of $(BPD)$ and $(CAP)$. Thus, $AXQB$ is cyclic. Then, we have (using directed angles) \[ \angle QYP = \angle QAP = \angle QAB = \angle QXB. \]Since $\overline{XB} || \overline{YP}$, we must have that $Q$, $X$, and $Y$ are collinear. Similarly, \[ \angle QZP = \angle QBP = \angle QBA = \angle QXA, \]so points $Q$, $X$, and $Z$ are collinear.