There were just lines, decided to coordinate bash Resulting with "...and substitute the letter and simplifiy." 32+ terms, 8+ variable, doomed.

Let $ML$ and $AB$ intersect at $J$ and let $MN$ intersect $AD$ at $I$. Because the isogonal of $G$ wrt $ABD$ lies on the perpendicular bisector of $BD$ and because $DML \sim BXY$ and $BMN \sim DZY$ we have that $YZIXJM$ is cyclic, call $Q$ the center of this circle. Let $U$ and $V$ be the Miquel points of $MYZIDQ$ and $XJMYBP$ respectively, then $UG \perp DQ$, hence $UG$ passes through $Q$, analogously $VG$ passes through $Q$, thus $\overline {UGQV}$ is perpendicular to both $DQ$ and $BP.$

<Solution 1-Using isogonal conjugates> Let the point $O'$ be $OM=O'M$ and $O,O',M$ is collinear. By angle chasing, we can get $O'$ and $G$ are isogonal conjugate in triangle $ABD$. So if $ML\cap AB=S, MN\cap AD=T$, $O'S\perp AB$ and $O'T\perp AD$. So $(X,S,M,Y,Z,T)$ are concyclic, with center $O_1$, which is midpoint of segment $O'G$. Let $MO_1$ and $SO_2$ meets $(O_1)$ in $U,V$. Then $UG\cap XV=M$, So by the pascal's theorem, we get $G,O_1,XM\cap SY$ is collinear. And the line is perpendicular with $BP$ ans $DQ$ too. So we can get $BP//DQ$. <Solution 2>

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Tbh I don't see any beauty in the configuration with 16 points of this problem...

My solution similar to isogonal conjugates sol above but with different ending. Let $O',N',L'$ be the reflection of $O,N,L$ over point $M$. Note that $N'\in AD,L'\in AB$. By angle chasing, $\angle{GBD}=\angle{FOD}/2=90^{\circ}-\angle{ODF}=\angle{LDO}=\angle{O'BA}$. Similarly, $\angle{GDB}=\angle{O'DA}$. So $O'$ and $G$ are isogonal conjugates w.r.t. triangle $ABD$. Since $\triangle{BO'L}\sim \triangle{BGY}$ and $\triangle{BO'M}\sim \triangle{BGX}$, we get $BL'/BY=BO/BG=BM/BX$. So, $BX\times BL'=BM\times BY\implies XMYL'$ cyclic with center at $K$, the midpoint of $O'G$. Similarly, we get $MYZN'$ is also cyclic with center $K$. Hence $XMYZN'L'$ is cyclic. Let $R$ be the second intersection of $(XL'P),(MYP)$. Consider the radical axis of those two circles, we get $B,P,R$ collinear. But by the property of Miquel point, $R$ is also the intersection of $(BMO'L'),(BXGY)$. Hence, their radical axis, $BR$, is perpendicular to the line connecting their centers, which is parallel to $O'G$. Now, consider the radical axis of $(O'MDN'),(GYDZ)$. Since $MYZN'$ is cyclic, we get that both $D$ and $Q$ lies on the radical axis. And so $DQ$ is perpendicular to the line connecting their centers, which is parallel to $O'G$. Hence, $BP\perp O'G$ and $DQ\perp O'G\implies BP\parallel DQ. \quad \blacksquare$

Collaborative solution with many many others (Gopal Goel, Robin Son, William Wang, Tim Qian, and like 14 other people?): We denote by $V$ and $U$ the reflections of $L$ and $N$ over $M$. We also let $S$ be the reflection of $O$ across $M$. Consequently, $\triangle MVU$ is the pedal triangle of $S$ with respect to $\triangle ABD$. Claim: The points $G$ and $S$ are isogonal conjugates with respect to $\triangle BAD$. Proof. We have $ \angle DBS = \angle OBD = 90^{\circ} - \frac{1}{2} \angle BOD = 90^{\circ} - \angle BFA = \angle FBA$. Similarly $\angle SDB = \angle ADE$. $\blacksquare$ This implies that the points $X$, $Y$, $Z$, $U$, $V$, $M$ are concyclic on the so-called six-point circle of $\{G,S\}$ with respect to $\triangle BAD$. From $PX \cdot PY = PU \cdot PM$ we conclude $\overline{BP}$ is on the radical axis of $(BXGY)$ and $(BUSM)$. These circles have diameter $\overline{BG}$ and $\overline{BS}$ so we conclude $\overline{BP} \perp \overline{GS}$. Analogously $\overline{DQ} \perp \overline{GS}$ solving the problem. [asy][asy] size(12cm); pair E = dir(140); pair F = dir(100); pair B = dir(160); pair D = dir(20); pair A = extension(B, E, F, D); pair G = extension(B, F, D, E); pair Y = foot(G, B, D); pair X = foot(G, B, A); pair Z = foot(G, A, D); pair M = midpoint(B--D); pair O = origin; pair S = 2*M; pair C = B+D-A; filldraw(A--B--C--D--cycle, invisible, red); draw(B--D, red); pair L = foot(O, C, D); pair N = foot(O, B, C); pair V = foot(S, A, D); pair U = foot(S, A, B); draw(V--N, lightblue); draw(L--U, lightblue); draw(CP(O, B), grey+dashed); filldraw(circumcircle(X, Y, Z), invisible, deepgreen); draw(X--G--Z, heavycyan); pair P = extension(X, Y, U, M); draw(U--S--V, lightblue); draw(S--M, lightblue); draw(P--U, lightblue); draw(X--Y--Z, heavycyan); draw(G--Y, heavycyan); draw(X--P, heavycyan); draw(B--P, blue); draw(B--F, orange); draw(D--E, orange); draw(B--S--D--O--cycle, lightcyan); dot("$E$", E, dir(160)); dot("$F$", F, dir(270)); dot("$B$", B, dir(B)); dot("$D$", D, dir(D)); dot("$A$", A, dir(A)); dot("$G$", G, dir(G)); dot("$Y$", Y, dir(Y)); dot("$X$", X, dir(120)); dot("$Z$", Z, dir(Z)); dot("$M$", M, dir(315)); dot("$O$", O, dir(270)); dot("$S$", S, dir(S)); dot("$C$", C, dir(C)); dot("$L$", L, dir(L)); dot("$N$", N, dir(N)); dot("$V$", V, dir(V)); dot("$U$", U, dir(U)); dot("$P$", P, dir(P)); /* TSQ Source: !size(12cm); E = dir 140 R160 F = dir 100 R270 B = dir 160 D = dir 20 A = extension B E F D G = extension B F D E Y = foot G B D X = foot G B A R120 Z = foot G A D M = midpoint B--D R315 O = origin R270 S = 2*M C = B+D-A A--B--C--D--cycle 0.1 lightred / red B--D red L = foot O C D N = foot O B C V = foot S A D U = foot S A B V--N lightblue L--U lightblue CP O B grey dashed circumcircle X Y Z 0.1 lightcyan / deepgreen X--G--Z heavycyan P = extension X Y U M U--S--V lightblue S--M lightblue P--U lightblue X--Y--Z heavycyan G--Y heavycyan X--P heavycyan B--P blue B--F orange D--E orange B--S--D--O--cycle lightcyan */ [/asy][/asy]

Here's a solution that's basically the same as above, but we replace big brain angle chase with a nice application of moving points. [asy][asy] unitsize(1.5inches); pair D=0*dir(0); pair C=2.1*dir(0); pair A=dir(90); pair B=A+C; pair M=(B+D)/2; pair O=0.86*M+0.14*(M-(B-D)*dir(90)); pair OO=2*M-O; pair L=foot(O,D,C); pair N=foot(O,B,C); pair U=foot(O,A,B); pair V=foot(O,A,D); pair E=2*U-B; pair F=2*V-D; pair G=extension(B,F,D,E); pair LL=2*M-L; pair NN=2*M-N; pair X=foot(G,A,B); pair Y=foot(G,B,D); pair Z=foot(G,A,D); draw(D--A--B--C--cycle); draw(D--B); draw(O--OO,dotted); draw(U--L,dotted); draw(V--N,dotted); draw(OO--LL,dotted); draw(OO--NN,dotted); draw(B--F); draw(D--E); draw(G--X,dotted); draw(G--Y,dotted); draw(G--Z,dotted); dot("$A$",A,dir(135)); dot("$B$",B,dir(45)); dot("$C$",C,dir(-45)); dot("$D$",D,dir(-135)); dot("$O$",O,dir(45)); dot("$O'$",OO,dir(-135)); dot("$M$",M,dir(160)); dot("$U$",U,dir(90)); dot("$N$",N,dir(180)); dot("$L$",L,dir(-90)); dot("$V$",V,dir(180)); dot("$E$",E,dir(90)); dot("$F$",F,dir(180)); dot("$G$",G,dir(-45)); dot("$L'$",LL,dir(90)); dot("$N'$",NN,dir(180)); dot("$X$",X,dir(90)); dot("$Y$",Y,dir(-45)); dot("$Z$",Z,dir(135)); [/asy][/asy] Let $O'$ be the reflection of $O$ over $M$. The key claim of the problem is that $O'$ and $G$ are isogonal conjugates in $\triangle ABD$. This can be shown by some big brain angle chasing, but here is a small brain approach. Our goal is to show that $\angle O'DB=\angle ADG$ (directed angles). We show this by moving points. Animate $O$ on the perpendicular bisector of $BD$. The pencil $DO'$ is clearly projective. The map $O\mapsto U$ is projective where $U$ is the foot from $O$ to $AB$, and the map $U\mapsto E$ is projective, since $E$ is the reflection of $B$ over $U$. Thus, the pencil $DE$ or $DG$ is projective, so verifying that $DG$ and $DO'$ are isogonal requires checking it for three values of $O$. First, we take $O=M$. In this case, $\angle O'DB=0$, and $E=A$, so $\angle ADG=0$, verifying the claim in this case. Second, we take $O$ to be on $BC$. In this case, $O'$ is on $AD$, so $\angle O'DB=\angle ADB$. Furthermore, we see that $U=B$, so $E=B$, so $\angle ADG=\angle ADB$, verifying the claim in this case. Finally, we take $O$ to be on $AB$. In this case, $O'$ is on $CD$, so $\angle O'DB=\angle CDB$. Furthermore, $O=E$, so $E$ is the reflection of $B$ in $O$, and since $D$ is the reflection of $B$ in $M$, we have $OM\parallel ED$, so $ED\perp BD$. Thus, $\angle \angle ADG=\angle ADE=\angle ABD$, verifying the claim in this case. Thus, we see that $O'$ and $G$ are isogonal conjugates. Now, we see that the foot from $O'$ to $AB$ is $L'$, the reflection of $L$ in $M$, so $P=XY\cap ML'$. Note that $ML'XY$ is cyclic by six-point circle, so by radical axis on it and $(BO')$ and $(BG)$, we see that $P$ is on the radical axis of $(BO')$ and $(BG)$. Thus, $BP$ is the radical axis of the two, so $BP\perp O'G$. Similarly we may derive $DQ\perp O'G$, finishing the proof.

I’m confused. What allows you to just intuitively tell from looking at a diagram that the isogonal conjugate of $G$ lies on the perpendicular bisector of $BD$.

A fantastic problem. $\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}$ [asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.0928491069498, xmax = 13.03423453766258, ymin = -12.363610466554292, ymax = 5.583051984245472; /* image dimensions */pen zzttff = rgb(0.6,0.2,1); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen ffvvqq = rgb(1,0.3333333333333333,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); /* draw figures */draw((-4.345361448763373,3.2634423422149696)--(7.771394705044084,3.765668444923977), linewidth(1) + blue); draw((-4.345361448763373,3.2634423422149696)--(-3.9550295483692564,-6.153743324990597), linewidth(1) + blue); draw((-3.9550295483692564,-6.153743324990597)--(8.1617266054382,-5.651517222281589), linewidth(1) + blue); draw((8.1617266054382,-5.651517222281589)--(7.771394705044084,3.765668444923977), linewidth(1) + blue); 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[/asy][/asy] Let $O',J,K$ be the reflection of $O,L,N$ in $M$ respectively. CLAIM 1. $G$ and $O'$ are isogonal conjugates w.r.t. $\triangle ABD$ Proof. we have $\angle MOD=\angle AED$, hence $$\angle ADE=90^{\circ}-\angle AED=90^{\circ}-\angle MOD=90^{\circ}-\angle MO'D=\angle O'MD$$similarly $GBD=\angle ABO'$ $\blacksquare$ Corollary. $K,F,Y,M,X,J$ are conyclic. Proof. This follows from the fact that the pedal triangles of isogonal conjugates $G,O'$ share the same circumcircle. $\blacksquare$ $\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 0.9309021989595917, xmax = 7.787932705836236, ymin = -3.716627463876436, ymax = 2.066704315944791; /* image dimensions */pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen qqwuqq = rgb(0,0.39215686274509803,0); pen zzttff = rgb(0.6,0.2,1); pen ffxfqq = rgb(1,0.4980392156862745,0); pen ccqqqq = rgb(0.8,0,0); /* draw figures */draw((2.1677954588028068,0.6231094081805589)--(2.044742617060606,-3.063440403047891), linewidth(1) + blue); draw((2.1677954588028068,0.6231094081805589)--(7.0348029397585545,0.4606542313949273), linewidth(1) + blue); draw((7.0348029397585545,0.4606542313949273)--(2.044742617060606,-3.063440403047891), linewidth(1) + blue); draw((2.1219426348150066,-0.7505989696451015)--(4.1997698325401265,-0.8199544793822966), linewidth(1)+ fuqqzz); draw((4.1997698325401265,-0.8199544793822966)--(4.5397727784095805,-1.301393085826482), linewidth(1) + fuqqzz); draw((4.1997698325401265,-0.8199544793822966)--(4.245622656527927,0.553753898443364), linewidth(1) + fuqqzz); draw(circle((3.4452028199450795,-0.5240533713663383), 1.3425127283009117), linewidth(1)+ qqwuqq); draw((2.6906358073500343,-0.22815226335037878)--(2.1399948077889386,-0.20977249276667953), linewidth(1) + zzttff); draw((2.6906358073500343,-0.22815226335037878)--(2.7184364583639007,0.6047296375968598), linewidth(1) + zzttff); draw((2.6906358073500343,-0.22815226335037878)--(3.8117065538770274,-1.8155700916080453), linewidth(1) + zzttff); draw((3.8117065538770274,-1.8155700916080453)--(4.245622656527927,0.553753898443364), linewidth(1) + green); draw((4.5397727784095805,-1.301393085826482)--(2.7184364583639007,0.6047296375968598), linewidth(1) + green); draw((4.1997698325401265,-0.8199544793822966)--(2.6906358073500343,-0.22815226335037878), linewidth(1) + linetype("4 4") + ffxfqq); draw((3.0786990860131334,0.7674633488753699)--(4.1997698325401265,-0.8199544793822966), linewidth(1) + ccqqqq); draw((4.1997698325401265,-0.8199544793822966)--(4.17196918152626,-1.6528363803295352), linewidth(1) + ccqqqq); /* dots and labels */dot((2.1677954588028068,0.6231094081805589),dotstyle); label("$A$", (2.193718315439543,0.6818769804812693), NE * labelscalefactor); dot((2.044742617060606,-3.063440403047891),dotstyle); label("$B$", (2.071707096455973,-3.00286183282255), NE * labelscalefactor); dot((7.0348029397585545,0.4606542313949273),dotstyle); label("$D$", (7.061965952883993,0.5232623958026281), NE * labelscalefactor); dot((4.5397727784095805,-1.301393085826482),linewidth(4pt) + dotstyle); label("$M$", (4.475328110432306,-1.5021238393246368), NE * labelscalefactor); dot((4.1997698325401265,-0.8199544793822966),dotstyle); label("$O'$", (4.2252051115159865,-0.7578554035248587), NE * labelscalefactor); dot((4.245622656527927,0.553753898443364),linewidth(4pt) + dotstyle); label("$K$", (4.267909038160236,0.6025696881419488), NE * labelscalefactor); dot((2.1219426348150066,-0.7505989696451015),linewidth(4pt) + dotstyle); label("$J$", (2.144913827846115,-0.7029503549822521), NE * labelscalefactor); dot((3.8117065538770274,-1.8155700916080453),linewidth(4pt) + dotstyle); label("$Y$", (3.8347692107685623,-1.7644479601393126), NE * labelscalefactor); dot((2.7184364583639007,0.6047296375968598),linewidth(4pt) + dotstyle); label("$Z$", (2.742768800865609,0.6513741757353768), NE * labelscalefactor); dot((2.1399948077889386,-0.20977249276667953),linewidth(4pt) + dotstyle); label("$X$", (2.1632155106936506,-0.16000043050536486), NE * labelscalefactor); dot((2.6906358073500343,-0.22815226335037878),linewidth(4pt) + dotstyle); label("$G$", (2.7122659961197164,-0.1783021133529004), NE * labelscalefactor); dot((3.4452028199450804,-0.5240533713663377),linewidth(4pt) + dotstyle); label("$H$", (3.4687355538178517,-0.47722959986264735), NE * labelscalefactor); dot((4.007827606431226,-0.7446849321682666),linewidth(4pt) + dotstyle); dot((4.17196918152626,-1.6528363803295352),linewidth(4pt) + dotstyle); label("$Z'$", (4.194702306770094,-1.6058333754606715), NE * labelscalefactor); dot((3.0786990860131334,0.7674633488753699),linewidth(4pt) + dotstyle); label("$Y'$", (3.102701896867141,0.8160893213631966), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] CLAIM 2. The intersection of $ZM$ and $YK$, $I_1$(It is not drawn in the diagram), lies on $GO'$ Proof. It is well-known that the center of the circle is the midpoint $H$ of $GO'$. Let $Z'$ and $Y'$ be the antipode of $Z$ and $Y$ respectively, we justify our claim by applying Pascal's Theorem to $MXJ'JYM'$. $\blacksquare$ Now by symmetry the intersection of $XM$ and $YJ$, $I_2$ lies on $GO'$. Hence $I_1,I_2,H$ are collinear. By Brodcard's theorem, $BP$ and $DQ$ are the polar of $I_1$ and $I_2$ respectively, hence they are parallel as they are both perpendicular to $I_1I_2$, as desired. $\blacksquare$ $\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}$ Plops wrote: I’m confused. What allows you to just intuitively tell from looking at a diagram that the isogonal conjugate of $G$ lies on the perpendicular bisector of $BD$. By angle chasing you can see that $\angle ADE=\angle OMD$, hence their reflection are isogonal conjugates. I actually find out the points $J$ and $K$ first, this is what motivates me to finish the problem.

Pretty tricky problem. Let $O'$ be the reflection of $O$ across $BD$. We claim that $O', G$ are isogonal conjugates wrt $\triangle ABD$. Indeed, \[\angle O'BD = \angle O'DB = 90^{\circ} - \tfrac12 \angle BOD = 90^{\circ} - (180^{\circ} - \angle BED) = \angle ADG = \angle ABG\]proving the isogonality. Then, if we let $L', N'$ be the reflections of $L, N$ over $M$, then we see that $XYZL'MN'$ is cyclic, more specifically the six-point circle $\omega$ of the isogonal conjugate pair $(G, O')$. By PoP wrt $\omega$, note that\[PX \cdot PY = PM \cdot PL' \text{ and } QY \cdot QZ = QM \cdot QN'\]where it follows that $BP$ and $DQ$ are the radical axes of the pairs $(\circ(BG), \circ(BS))$ and $(\circ(DG), \circ(DS))$ respectively. Both are perpendicular to $GS$, and are thus parallel, as desired.

Define $R = LM\cap AB, S = NM\cap AD$. Observe that \[\frac{AZ}{AX} = \frac{GX}{GZ} = \frac{XB}{ZD}\]This means $AZ\cdot ZX = AX\cdot XB$, so the power of $X$ is the same as the power of $Z$ w.r.t $(ABD)$. Since $M$ is the circumcenter of $(ABD)$ (as $M$ is the midpoint of $BD$), we have $MX = MZ$ Now, we have \[\angle ZYD = \angle ZGD = 90 - \angle ZDG = 90 - \angle BGX = \angle XGB = \angle XYB\]Therefore, $BD$ is the external angle bisector of $\angle XYZ$. Since $M$ lies on the angle bisector of $\angle XYZ$ and on the perpendicular bisector of $XZ$, we have $M\in (XYZ)$. Furthermore, \[\angle GDZ = \angle ABG = \angle ZGF\]\[\Rightarrow \angle ZYD = \angle ZGD = \angle ZFG = 180 - \angle BFD = \angle BOM = \angle BNM = \angle MSD\]Therefore, $(YMZS)$ is cyclic. Similarly, $(YMXR)$ is cyclic, so $(YMSZRX)$ is cyclic. Let $O'$ be the reflection of $O$ over $M$, and via a homothety at $M$ with ratio $-1$< we have $M,S,R$ are the feet from $O'$. This means that $G, O'$ are isogonal conjugates (wrt $\triangle ABD$). If $U$ is the center of $(YMSZRX)$, then by pedal circles, $G,P, O'$ are collinear. Now, if we let $T_1 = XM\cap RY, T_2 = ZM\cap YS$, $H_1$ as the miquel point of $RXYM$, and $H_2$ as the miquel point of $ZYMS$, then we have \[\angle GH_1B = \angle O'H_1B = \angle T_1H_1B = \angle UH_1B = 90\]\[\angle GH_2D = \angle O'H_2D = \angle T_2H_2D = \angle UH_2D = 90\]where all of this follows by brocard. Therefore, this means $G,T_1,T_2,O',U$ are collinear, and since $\angle PH_2Q = 90 = \angle PH_1$, we can conclude that $BP || DQ$.