Extend $BQ$ and $DP$ to each hit $(ABCD)$ again at $B'$ and $D'$. We will prove that\[BD' \parallel XY \parallel B'D\]which will imply $BD'DB'$ is an isosceles trapezoid, telling us that those three segments share a common perpendicular bisector (passing through circumcenter $O$), finishing the problem. Indeed, note\[\angle DPQ = \angle DPX = 180^{\circ} - \angle DAB = \angle DD'B\]\[\angle BQP = \angle BQY = 180^{\circ} - \angle BCD = \angle BB'D\]proving $BD' \parallel PQ$ and $B'D \parallel PQ$, as desired.

Define $P_1=\overline{DP}\cap (ABCD)\not = D$, and $Q_1=\overline{BQ}\cap (ABCD)\not=B$. By Reim's Theorem, $\overline{BP_1}$ is antiparallel to $\overline{AD}$, which is antiparallel to $\overline{XP}$, so $\overline{BP_1}\parallel \overline{XP}=\overline{PQ}$. Similarly, $\overline{DQ_1}\parallel \overline{PQ}$. Let $G=\overline{BQ_1}\cap \overline{DP_1}$. This is the intersection of two parallel chords in a circle, so $\triangle GDQ_1$ and $\triangle GBP_1$ are isosceles. Also, since $\overline{PQ}$ is parallel to both chords, $\triangle GPQ$ is isosceles. Furthermore, any pair of these three triangles are homothetic at $G$, so the perpendicular bisectors of $\overline{DQ_1}$, $\overline{BP_1}$, and $\overline{PQ}$ coincide. Since $O$ lies on the former two, it also lies on the third.

oops Let \(T=\overline{BQ}\cap\overline{DP}\). From \(\measuredangle TPQ=\measuredangle DAX=\measuredangle YCB=\measuredangle YQT\), we know \(TP=TQ\), so it suffices to show \(\overline{OT}\perp\overline{PQ}\). To this end we use complex numbers with \((ABCD)\) the unit circle and \(\overline{XY}\) aligned with the real axis. Since \(\measuredangle TDA=\measuredangle YXA\) and \(\measuredangle TBC=\measuredangle XYC\), we know \[\frac{t-d}{a-d}\cdot\frac{a-b}1\in\mathbb R\quad\text{and}\quad\frac{t-b}{c-b}\cdot\frac{c-d}1\in\mathbb R.\]That is, \begin{align*} (t-d)\cdot\frac{a-b}{a-d} &=\left(\overline t-\frac1d\right)\cdot\frac{\frac1a-\frac1b}{\frac1a-\frac1d} =\left(\overline t-\frac1d\right)\frac{a-b}{a-d}\cdot\frac db\\ \implies b(t-d)&=d\left(\overline t-\frac1d\right)\\ \implies bt-d\overline t&=bd-1. \end{align*}Analogously, \begin{align*} (t-b)\cdot\frac{c-d}{c-b}&=\left(\overline t-\frac1b\right)\frac{c-b}{c-d}\cdot\frac bd\\ \implies dt-b\overline t&=bd-1. \end{align*} Combining these, we get \(bt-d\overline t=dt-b\overline t\), implying \(t+\overline t=0\). Therefore \(t\) is imaginary, as desired.

Let $DP$ and $BQ$ meet $(ABCD)$ at $P',Q'$ respectively. By Reim's $BP' \parallel PQ \parallel DQ'$ so $BQ'DP'$ is an isosceles trapezium (because it's cyclic) $\implies O$ lies on the perpendicular bisector of $BP',PQ,DQ'$ done. [asy][asy] size(230); pair A=(-3.5247,1.8911), B=(-3.6018, -1.7399), C=(3.6018,-1.7399), D=(1.2105,3.8125); pair X=(-3.5484,0.7764), Y=(2.9994,-0.3414), P=(2.1826,-0.2019), Q=(-2.126,0.5336); pair P1=(2.8206,-2.8362), Q1=(0.1228,3.9981), O=origin; draw(circumcircle(A,B,C), black); draw(A--B--C--D--cycle, black); draw(D--P1, red); draw(D--Q1, dashed+red); draw(B--Q1, red); draw(B--P1, dashed+red); draw(circumcircle(A,X,D), dotted+black); draw(circumcircle(B,Y,C), dotted+black); draw(X--Y, red); dot("$A$",A,NW); dot("$B$",B,W); dot("$C$",C ,E); dot("$D$",D ,NE); dot("$Q'$",Q1 ,N); dot("$P'$",P1,SE); dot("$X$", X,W); dot("$Y$",Y ,E); dot("$P$",P ,SW); dot("$Q$",Q ,NW); dot("$O$", O, S); [/asy][/asy]

Extend $BQ, DP$ to meet the circle at $Q', P'$. Since $\angle BQ'D = \angle DAB = 180 - \angle BCD = \angle BQP$, we have $DQ' || PQ$. Analogously, we have $BP' || PQ$, so $BP'DQ'$ is an isosceles trapezoid and $PQ$ is parallel to the parallel sides. So by symmetry, we have $OP = OQ$, done. $\blacksquare$

[asy][asy] size(7cm); defaultpen(fontsize(9pt)); pair A,B,C,D,X,Y,P,Q,M,N; A=dir(120); B=dir(80); D=dir(210); C=dir(-30); X=0.6A+0.4B; Y=0.55C+0.45D; path w1,w2; w1=circumcircle(A,D,X); w2=circumcircle(B,C,Y); P=intersectionpoint(Y--(2Y-X),w1); Q=intersectionpoint(X--((X+Y)/2),w2); M=intersectionpoint((10Q-9B)--Q, unitcircle); N=intersectionpoint((2P-D)--P,unitcircle); draw(A--B--C--D--cycle); draw(unitcircle); draw(w1); draw(w2); draw(B--M,gray); draw(D--N,gray); draw(D--M,orange+linewidth(0.9)); draw(P--Q,orange+linewidth(0.9)); draw(B--N,orange+linewidth(0.9)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$X$",X,dir(90)*2); dot("$Y$",Y,dir(45)); dot("$P$",P,dir(P)*1.5); dot("$Q$",Q,dir(0)); dot("$M$",M,dir(M)*2); dot("$N$",N,dir(N)); [/asy][/asy] Let $M=\overline{BQ} \cap (ABCD), N=\overline{DP} \cap (ABCD)$ so that by Reim,$$\overline{DM} \parallel \overline{XYPQ} \parallel \overline{BN}.$$So $BNDM$ is an isosceles trapezoid with sides parallel to $\overline{PQ}$, implying the desired.

Reim's Theorem gives $DF\parallel PQ\parallel EB$. Since: $$\angle DPQ=\angle DPX=180^\circ-\angle DAX=180^\circ-\angle DAB=\angle DCB=\angle YCB=180^\circ-\angle YQB=\angle YQF=\angle PQF$$$BEDF$ is an isosceles trapezoid, so the perpendicular bisectors of $BF$ and $ED$ are $OQ$ and $OP$ respectively. Thus $OP=OQ$.

let $(DCY)\cap AD=E$ and $(ABX)\cap BC=F$ also let $R,S= EF \cap (ABX) , (DCY)$ we have $XEYF$ is cyclic and so $QRPS$ is now notice that $EY \parallel AB \parallel RP$ If $O_1$ is the center of $AXB$ we have that $O_1A=O_1B$ nad $OA=OB$ so $OO_1 \perp AB$ hence $OO_1 \perp RP$ so $O$ is on the perpendicular bisector of $PR$ is the same way it is on the perpendicular bisector of $QS$ so $OR=OP=OS=OQ$

ahhhh can't believe wasted so much time just because of overthinking Anyways, Let $DP \cap (ABCD)=B'$ and $BQ \cap (ABCD)=D'$ Claim 1:-$BB'||PQ||DD'$ Simple angle chasing, $$\angle BD'D=\angle ABD=-\angle DCB=-\angle D'QP \implies D'D||PQ $$Similarly $PQ||BB'$ and hence $DD'||PQ||BB'$ and it also implies $BB'DD'$ is an isosceles trapezium and thus $O$ lies in the perpendicular bisector of $PQ$ which implies $OP=OQ$ Edit:-Apparently the solution i was working on works but i left it feeling that it's lengthy and might not work but anyways

I hate myself Let $DP \cap (ABCD)=T$ and $BQ \cap (ABCD)=S$. Claim: $PD=QS$ Proof: $\measuredangle SQP = \measuredangle BQY = \measuredangle BCY = \measuredangle BCD = \measuredangle BAD = \measuredangle XAD = \measuredangle XPD = \measuredangle QPD$. $\blacksquare$ Claim: $BT || PQ$. Proof: $\measuredangle BTD = \measuredangle BAD = \measuredangle XAD = \measuredangle XPD$. $\blacksquare$ Now its easy to see that $DPQS$ is an isosceles trapezium, so $O$ lies on the perpendicular bisector of $PQ$, therefor $OP=OQ$. $\blacksquare$

Storage from the storage (I'm so sorry, i just saw this and wanted to put this sol (thanks to @k12byda5h for reminding me about Reim's xD)) Let $BP \cap (O)=E$ and $CQ \cap (O)=F$ by Reim's we have $DE \parallel PQ\parallel BF$ hence $DEBF$ is an OP isosceles trapezoid which gives that their perpendicular bisector is the same and it passes through $O$, hence $OP=OQ$ as desired.

My problem Posts #2, #3, #5, #7, #8, #9, #10, #11, #13, #14, and #15 are all the same as Solution A. Post #4 is analogous to solution B, but with a different finish. Post #12 is analogous to solution F, but with a different finish.

I am too lazy to find better solution. Let $R$ be midpoint of $\overline{PQ}$. Let $E=\overline{AB}\cap \overline{CD}$. Let $T=\overline{AB}\cap (BCY)$ and $U=\overline{CD}\cap (ADX)$. Claim. $YD\cdot UC=XB\cdot AT$. Proof. Indeed, by PoP, \begin{align*} YD\cdot UC=(ED-EY)(EU-EC)=(EX-EB)(EA-ET)=XB\cdot AT, \end{align*}where we used the fact that $XUYT$ is cyclic as $\measuredangle YTX=\measuredangle DCB=\measuredangle DAX=\measuredangle DUX=\measuredangle YUX$. $\square$ Then, \begin{align*} XR^2-YR^2&=XY\cdot (XP-YQ)=XY\cdot XP-XY\cdot YQ\\&=XB\cdot XT-YU\cdot YD=XA\cdot XB+YC\cdot YD\\&=XO^2-YO^2, \end{align*}which yields that $\overline{OR}\perp\overline{XY}$, we conclude that $OP=OQ$. $\blacksquare$

sad my solution was Holden's solution D

Not the kind of person to post solutions on AoPS usually, but the solution I executed on contest was so unreasonably ugly that I felt the need to defile the thread with it. Let $X' = (ADX)\cap CD, Y' = (BCY) \cap AB$. Let $(ABCD) = \Gamma$, $(ADX) = \omega_1, (BCY) = \omega_2$. Let $f(Z)$ denote $Pow_{\Gamma}(Z) - Pow_{\omega_1}(Z)$. It's well known that $f(Z)$ is linear. Thus, $f(Y) = \frac{YPf(X) + XYf(P)}{XP}$, or $f(P) = \frac{XPf(Y) - YPf(X)}{XY} = \frac{XYf(Y) + YP(f(Y) - f(X))}{XY} = f(Y) + \frac{YP(f(Y) - f(X)}{XY}$. $f(Y) = YC\cdot YD - YX'\cdot YD = X'C\cdot YD$. Note that $X'C\cdot YD = XB \cdot Y'A$ by Reim. Thus, $f(Y) - f(X) = XB \cdot Y'A - XB \cdot XA = XB \cdot Y'X = - XY \cdot XQ$, where the last step is by PoP on $\omega_2$. Thus, $\frac{YP(f(Y) - f(X)}{XY} = -YP\cdot XQ$, and indeed $f(P) = Pow_{\Gamma}(P) - 0 = X'C \cdot YD - YP \cdot XQ$, which is clearly symmetric(again, recall $X'C\cdot YD = XB \cdot Y'A$ by Reim), so indeed $Pow_{\Gamma}(P) = Pow_{\Gamma}(Q)$ or $OP = OQ$.

Let $(ADX)$ intersect $CD$ at $E$ and $(BCY)$ intersect $AB$ at $F$. Let $EF$ intersect $(ADX)$ again at $P'$ and $(BCY)$ again at $Q'$. Note $\measuredangle EXF = \measuredangle EXA = \measuredangle EDA = \measuredangle CDA = \measuredangle CBA = \measuredangle CBF = \measuredangle CYF = \measuredangle EYF$, so $EXFY$ is a cyclic quadrilateral. Note $\measuredangle PP'E = \measuredangle PXE = \measuredangle YXE = \measuredangle YFE$ so $FY\parallel PP'$. Similarly, $EX\parallel QQ'$, so $\measuredangle Q'P'P = \measuredangle EFY = \measuredangle EXY = \measuredangle Q'QP$, meaning $P'PQ'Q$ is a cyclic quadrilateral. Note $\measuredangle FYE = \measuredangle FXE = \measuredangle AXE = \measuredangle ADE$, so $PP'\parallel FY\parallel AD$. Thus the perpendicular bisector of $PP'$ is the same as the perpendicular bisector of $AD$, so the center of the circumcircle of $P'PQ'Q$ lies on the perpendicular bisector of $AD$. It also lies on the perpendicular bisector of $BC$, so either $O$ is the center of $P'PQ'Q$ or $AD\parallel BC$. But the condition is an identity in $X$ by the theory of moving points, so this case does not matter.

Terribly inefficient, but I think it's cool. [asy][asy] size(13cm); defaultpen(fontsize(10)); pair A,B,C,D,X,Y,P,Q,U,V,UU,VV,O,O1,O2,O3; A = dir(100); B = dir(210); C = dir(330); D = dir(40); X = 0.35*A+0.65*B; Y = 0.8*D+0.2*C; P = 2*foot(circumcenter(A,D,X),X,Y)-X; Q = 2*foot(circumcenter(B,C,Y),X,Y)-Y; U = extension(A,D,X,Y); V = extension(B,C,X,Y); UU = 2*foot(circumcenter(A,Q,B),B,C)-B; VV = 2*foot(circumcenter(C,P,D),A,D)-D; O = circumcenter(A,B,C); O1 = circumcenter(C,D,P); O2 = circumcenter(A,B,Q); O3 = circumcenter(U,V,UU); draw(A--B--C--D--cycle); draw(D--U--V--B); draw(C--UU--U); draw(A--VV--V); draw(circumcircle(A,B,C),red); draw(circumcircle(A,D,X),green); draw(circumcircle(B,C,Y),green); draw(circumcircle(U,V,UU),blue+dashed); draw(circumcircle(A,B,Q),magenta+dashed); draw(circumcircle(C,D,P),magenta+dashed); draw(O--O1--O3--O2--cycle,cyan); dot("$A$",A,dir(A)); dot("$B$",B,dir(225)); dot("$C$",C,dir(315)); dot("$D$",D,dir(D)); dot("$X$",X,dir(45)); dot("$Y$",Y,dir(230)); dot("$P$",P,dir(45)); dot("$Q$",Q,dir(45)); dot("$U$",U,dir(U)); dot("$V$",V,dir(V)); dot("$U'$",UU,dir(UU)); dot("$V'$",VV,dir(VV)); dot("$O$",O,dir(90)); dot("$O_1$",O1,dir(270)); dot("$O_2$",O2,dir(0)); dot("$O_3$",O3,dir(270)); [/asy][/asy] Let $XY$ meet $AD$ again at $U$ and $BC$ again at $V$. Note that $$\measuredangle BQU = \measuredangle BQY = \measuredangle BCY = \measuredangle BCD = \measuredangle BAD = \measuredangle BAU,$$so $AQBU$ is cyclic. Similarly, $CPDV$ is cyclic. Let $AD$ intersect $(CPDV)$ at $V' \neq D$ and let $BC$ intersect $(AQBU)$ at $U' \neq B$. By Reim's $UU' \parallel CD$ and $VV' \parallel AB$. Thus, $UU'VV'$ is also cyclic. Let the centers of $(CPDV)$, $(AQBU)$, and $(UU'VV')$ be $O_1$, $O_2$, and $O_3$. We know $OO_1 \perp CD$ and $O_2O_3 \perp UU'$. Since $CD \parallel UU'$, we get $OO_1 \parallel O_2O_3$. Similarly, $OO_2 \parallel O_1O_3$. This means $OO_1O_3O_2$ is a parallelogram, so $$\overrightarrow{O}+\overrightarrow{O_3}=\overrightarrow{O_1}+\overrightarrow{O_2}.$$ Let $\mathcal{P}(\bullet)$ be the projection of $\bullet$ onto $XY$. Then $$\overrightarrow{\mathcal{P}(O)}+\overrightarrow{\mathcal{P}(O_3)}=\overrightarrow{\mathcal{P}(O_1)}+\overrightarrow{\mathcal{P}(O_2)}.$$This means $$\overrightarrow{\mathcal{P}(O)} + \frac{\overrightarrow{U}+\overrightarrow{V}}{2}=\frac{\overrightarrow{V}+\overrightarrow{P}}{2}+\frac{\overrightarrow{Q}+\overrightarrow{U}}{2},$$so $\overrightarrow{\mathcal{P}(O)} = \frac{\overrightarrow{P}+\overrightarrow{Q}}{2}$, which means $OP=OQ$. (The solution may have left out cases when lines are parallel, but these work by continuity.)

Let $S =\overline{AB} \cup \odot{(BCY)}$, $T = \overline{CD} \cup \odot{(XAD)}$. Let $\overline{ST}$ meet $\odot{(XAD)}$ and $\odot{(BCY)}$ at $R$ and $U$, respectively. Claim. $\overline{BC} \parallel \overline{QU} \parallel \overline{XT}$ and $\overline{AD} \parallel \overline{RP} \parallel \overline{SY}$.

From claim we conclude, that pairs $\{\overline{AD}, \overline{RP}\}$ and $\{\overline{BC}, \overline{QU}\}$ share perpendicular bisectors, which means $O$ is the circumcenter of $RQUP$. Hence $$\overline{OP}=\overline{OQ}$$as desired. $\blacksquare$

Very nice problem with a construction that pretty much kills it! Let $DQ \cap \odot(ABCD)= E$ and $BP \cap \odot(ABCD) =F$ Trivial to see that $PQFD$ and $PQEB$ are both cyclic so by Reim's we get that $BE \parallel DF$ Now combining the cyclic condition and the parallel condition $$\measuredangle FDQ = \measuredangle BED = \measuredangle BCD = \measuredangle BAD = \measuredangle XAD = \measuredangle XQD$$which implies $BE \parallel PQ \parallel DF$ Now it is obvious that $O$ lies on the perpendicular bisector of $DF$ and $BE$, and because $PQ \parallel BE$ we get that $PQEB$ is an isosceles trapezium which implies $O$ lies on the perpendicular bisector of $PQ$ which implies $OP=OQ \ \blacksquare$

Let $BQ$ hit the circle at $E$. Now,$\angle DEQ=\angle DAB=180-\angle DPX$,so $DEPQ$ is cyclic.Now, $\angle BQP=180-\angle DCB=\angle DAB=\angle DEQ$,so $DE \parallel PQ$. Thus,so $O$ lies on the perpendicular bisector of $PQ$.

Proof. Let $Z$ be the second intersection of $(AXD)$ and $CD$, $T$ be the second intersection of $(BCY)$ and $AB$, $W$ be the intersection of $XY$ and $ZT$, $R$ be the second intersection of $(AXD)$ and $ZT$, $S$ be the second intersection of $(BCY)$ and $XY$, $V$ be the intersection of $AB$ and $CD$. By PoP $$VT \cdot VB=VC \cdot VY,$$$$VA \cdot VX=VD \cdot VZ,$$$$VC\cdot VD=VA \cdot VB.$$Multiplying all three, we get $$VT \cdot VX=VY \cdot VZ \implies XYZT \; \text{is cyclic.}$$Again by PoP $$WS\cdot WT=WP \cdot WY,$$$$WR \cdot WZ=WX \cdot WQ,$$$$WX \cdot WY=WT \cdot WZ.$$Multiplying all three, we get $$WS \cdot WR=WP\cdot WQ \implies PQRS \; \text{is cyclic}.$$Now, by sine rule for $\triangle BXP$: $BP=\frac{BX}{\sin \angle BPX}\cdot \sin \angle PXB=\frac{BX}{\sin \angle BCD}\cdot \sin \angle YXT$. Similarly, $CS=\frac{CZ}{\sin \angle ABC}\cdot \sin \angle YZT$. But notice that $\angle YXT=\angle YZT$, and $\frac{BX}{\sin \angle BCD}=\frac{CZ}{\sin \angle ABC}$ from the difference of sine rules in $\triangle VBC$ and $\triangle VXZ$ (since by simple angle chasing $BC \parallel XZ$). Thus $BP=CS \implies BPSC$ is an isosceles trapezoid $\implies O$ lies on the perpendicular bisector of $PS$. Similarly, $O$ lies on the perpendicular bisector of $QR$, and hence is the circumcenter of $(PQRS)$, so $OP=OQ$. $\blacksquare$

Let $E= DP \cap (ABCD), F= BQ \cap (ABCD)$, with $E \neq D, F \neq B$. By Reim's Theorem on $(ADXP),(ABCD)$ with the lines $AX,DP$, we have that $BE \parallel XY$. Similarly, we have that $DF \parallel XY$. Therefore, $BE \parallel DF$, so $BEFD$ is an isosceles trapezoid, as well as $BEQP$ and $PQFD$. Hence $PE=BQ,PD=QF \implies QB.QF=PD.PE \implies |Pow_{(ABCD)}P|=|Pow_{(ABCD)}Q|$, but since both $P,Q$ are inside $(ABCD)$, if $R$ is the radius of $(ABCD)$, we have that $OP^2-R^2=OQ^2-R^2 \implies OP=OQ$, as desired. $\blacksquare$

Am I missing something or do we not need both $E$ and $F$? I originally had both because I was trying to prove that the powers of $P$ and $Q$ wrt $(ABCD)$ are equal (same approach as above), which led to a solution that used both, but I realized that it was overcomplicated and could be simplified to only use one point. Let $\overline{DP}$ intersect $(ABCD)$ at $E$. Then we have $$\angle BFP=\angle BCD=180^\circ-\angle XAD=\angle XPD=180^\circ-QPF,$$so $\overline{BF} \parallel \overline{PQ}$. Further, $$\angle BQP=180^\circ-\angle BCY=\angle XAD=\angle QPF,$$so in fact $BFPQ$ is an isosceles trapezoid. Then the perpendicular bisector of $\overline{PQ}$ is also the perpendicular bisector of $\overline{BF}$, which thus passes through $O$. This implies that $OP=OQ$ as desired. $\blacksquare$

Idk why, but I don't like doing latex writeups of contests. Jk I do know why. Let $R=BQ\cap (ABCD)$. Claim: $RDPQ$ is an isosceles trapezoid. Proof: Firstly, by Reim's on $(QBCY),(ABCD)$, we get that $RD \parallel QY\parallel PQ$. Next, using directed angles we have that \begin{align} \measuredangle DPQ = \measuredangle DPX = \measuredangle DAX = \measuredangle DAB = \measuredangle DCB \\ = \measuredangle YCB =\measuredangle YQB = \measuredangle YQR = \measuredangle PQR \end{align}which finishes the proof of the claim. $\square$. Now, note that the perpendicular bisector of $RD$ passes through $O$ since $OR=OD$. But, since $RDPQ$ is an isosceles trapezoid, the perpendicular bisector of $RD$ is the same as the perpendicular bisector of $PQ$, so we also have that $OP=OQ$ and we're done. $\blacksquare$.

Let BQ and DP Meet circumcircle of ABCD at E and F. ∠BQY = 180 - ∠C = ∠BED ---> QP || ED. ∠DPX = 180 - ∠A = ∠DFB ---> QP || BF so EDFB is isosceles trapezoid and O it's center and EQ/QB = DP/PF so OP = OQ

Let $R=BQ\cap (ABCD)$ $\bullet$ $\angle{BRD}=\pi-\angle{DCB}=\angle{BQY}$ $\bullet$$\angle{DPQ}=\pi-\angle{DAX}=\angle{DCB}=\angle{RQP}$ So $PQRD$ is isosceles trapezoid and $O$ lies on the perpendicular bisector of $RD$ which is the same as the perpendicular bisector of $PQ$ $\implies OP=OQ$

Solved with The_Turtle [well, he was playing Ondine in the background, so it would be unfair not to credit him]. Solution writeup also with The_Turtle [this time, he was playing the Ravel Toccata.] Let line $DP$ intersect the circumcircle of quadrilateral $ABCD$ again at $R$. Note that $\angle DRB = 180^{\circ} - \angle BAD$. Moreover, $\angle BQP = 180^{\circ} - \angle BCY = \angle BAD$ and $\angle RPQ = 180^{\circ} - \angle QPD = \angle BAD$. Thus quadrilateral $BRPQ$ is an isosceles trapezoid, from which $OP = OQ$ follows.

Same. Let $U=DP\cap (ABC), V=BQ\cap (ABC)$, $W=BQ\cap DP$. $$ \measuredangle DVW=\measuredangle DCB=\measuredangle YCB=\measuredangle YQW \Longrightarrow DV\parallel XY $$Analogously, $BU,\parallel XY$. Also, $$ \measuredangle WPX=180^o-\measuredangle XAD=\measuredangle DCB=\measuredangle YQW $$Therefore $\triangle WPQ$ is isosceles, and so the perpendicular bisector lines of segments $\overline{DV}, \overline{PQ}, \overline{BU}$ all coincide, and so $OP=OQ$.

Rather long and convoluted, but hopefully more intuitive than the above solutions. The main idea is to introduce symmetry in the diagram. Let the circumcircles of $\triangle AXD$ and $\triangle BYC$ intersect lines $CD$ and $AB$ at points $E$ and $F$, respectively. Claim: $YF \parallel AD$ and $EX \parallel BC$. Proof: $\angle YFB = \angle YCB = \angle DCB = 180^{\circ} - \angle BAD = 180^{\circ} - \angle YAD$ so $YF \parallel AD$ and the other result follows similarly. Claim: $EXFY$ is cyclic. Proof: $\angle XFY = \angle AFY = 180^{\circ} - \angle FAD = 180^{\circ} - \angle XAD = \angle XED = 180^{\circ} - \angle YEX$. Now the pith of the problem is the following: Let line $EF$ meet the circumcircles of $\triangle AXD$ and $\triangle BYC$ at points $M$ and $N$ respectively. We will prove that $O$ is the circumcenter of quadrilateral $PQMN$. This takes two steps: by angle chasing involving the fact that $EXFY$ is cyclic, we can prove $QNCB$ and $PMDA$ are cyclic isosceles trapezoids which implies $ON = OQ$ and $OM = OP$. Furthermore, we can prove that $QPNM$ is cyclic as well, which finishes the problem.

Let $R$ and $S$ be the intersections of $\overline{YX}$ with $(ABCD)$, such that $R$ is closer to $Y$ and $S$ is closer to $X$. It is equivalent to show that $RP = QS$. We have $\triangle APS \sim \triangle ARC$, since $\angle APS = 180^{\circ} - \angle ADS = \angle ABC$ and $\angle ASP = \angle ASR = \angle ACR$. Therefore, it follows that $\triangle ACS \sim \triangle ARP$, so \[ RP = \frac{CS \cdot AR}{AC}. \]Applying symmetric reasoning on $\triangle CQR \sim \triangle CSA$ gives us this exact same expression for $QS$, so the two lengths are equal.

W question Extend $DP$ and $BQ$ and realize that we have three pairwise parallel lines (one of which is $XY$) and we're done. (The motivation is to try and make powers equal so we should obviously extend some lines)

Let $BQ$ intersect $(ABCD)$ again at $R$ and $DP$ intersect $(ABCD)$ again at $S$. We claim that $RDSB$ is a cyclic isosceles trapezoid with $PQ$ is another line parallel to the bases. Let $\angle BAD=\alpha$ and $\angle BCD=\gamma. $ We have from $ADPX$ cyclic that $$\angle SPQ=\alpha.$$However, since $$\angle BRD=\alpha$$due to $ARBD$ cyclic, we have that $RDPQ$ is cyclic. Similarly, $BSPQ$ is cyclic. Hence, by Reim, $RD$ is parallel to $BS$, so $RDSB$ is an isosceles trapezoid. Thus, $$\angle PDR=\angle QRD=\alpha$$as well, and since $\angle SPQ=\alpha$, $PQ$ is parallel to the bases as well. The claim easily solves the problem (e.g. $PS\cdot PD=QB\cdot QR$).

Let $E = \overline{BP} \cap (ABCD)$ and $F = \overline{DQ} \cap (ABCD)$. Note that $$\angle PQF = \angle BAD = 180^\circ - \angle BFQ = \angle BPQ$$hence $BFQP$ is an isosceles trapezoid. Similarly, $DEPQ$ is an isosceles trapezoid, so $OP=OQ$ follows by symmetry.

Let $B'=BQ\cap \omega$,$D'=DP\cap \omega$. Claim: $DB'\parallel PQ\parallel BD'$ Proof: We can say: \[\angle APX=180-\angle DAX=\angle DD'B,\]\[\angle CBQ=180-\angle YCB=\angle DB'B,\]as desired $\square$ Thus, $DD'BB'$ is an isosceles trapezoid, with $PQ$ parallel to the bases. As $O$ passes through the perpendicular bisector of the bases, it must also pass through the perpendicular bisector of $PQ$, as desired $\blacksquare$

Extend $BQ$ past $Q$ to intersect $(ABCD)$ at point $B'$ and $DP$ past $P$ to intersect $(ABCD)$ at $D'$. It is clear that $BDB'D'$ is cyclic so Reim's Theorem gives $\overline{BD'} \parallel \overline{B'D}$. We also have \[\angle QPD' = 180^\circ-QPD = \angle DAB = \angle PD'B\]\[\implies \overline{PQ} \parallel \overline{BD'}.\] Hence, $BD'$, $B'D$, and $PQ$ share a common perpendicular bisector, which passes through $O$. $\square$

Let $K = BQ \cap (ABCD)$ and $L = DP \cap (ABCD)$. Reim's gives us $BL \parallel PQ \parallel DK$, from which it follows that $BLDK$, $BLPQ$, and $KDPQ$ are each isosceles trapezoids. Thus \[PD \cdot PL = QB \cdot QL \implies \operatorname{Pow}_{(ABCD)} P = \operatorname{Pow}_{(ABCD)} Q \implies OP = OQ. \quad \blacksquare\]

Very easy if you have seen IMO 2009 P2. POP is key to show $OP=OQ$.

Spent 2 hours on this one line solution Let $BQ$ hit $(ABCD)$ at $E$, let $DP$ hit it at $F$, then by reims, $BF \parallel QP \parallel DE$ then since $HDBC$ is cyclic $O$ lies on the perpendicular bisector of $BF$, $QP$, and $DE$ implying the result.

Let $DP \cap (ABCD) = T$. We have $\angle DPQ = \angle DPX = 180^\circ - \angle A = \angle DTB$ so $PQ \parallel BT$ and similarly $PQ \parallel SD$ so $BTDS$ is an isosceles trapezoid $\implies$ the perpendicular bisector of $SD$ bisects $PQ$ and passes through $O$, as desired.

Let $R$ be the midpoint of $PQ$ and $(ADX)\cap CD=Y'$, $(BCY)\cap AB=X'$ then $(X'XYY')$ by Reim's Theorem. To Prove: $OR \perp PQ$ Now let the centers $(BCY), (AXD), (X'XYY')$ be $O_1, O_2, O'$. Then we have the following claim: Claim I: $O_1OO'O_2$ is a parallelogram because $BC \parallel XY'$ and $AD \parallel X'Y$. Note that the projection of $O'$ is the midpoint of $XY$, similarly the projections of $O_1$ and $O_2$ are midpoints of $QY$ and $PX$. Because $O_1OO'O_2$ is a parallelogram, by projecting $O$ we get the midpoint of $PQ$