yay this problem made me happy

Who attached a diagram with lots of angle chasing like I did?

huh this problem made me sad. Another way for if part. Let BQ,DC intersect at T. Obviously DTB and QCB similar. Then this gives DGB and QSB similar since BG/BT=BS/BC. So <BDG=<BQS. Since <GPD=<BDC=<BRS so DRGP concyclic. So <BDG=<RPG. So<BQS=<RPG. similar <BQR=<SPG, adding gives <RQS=<RPS. So PQRS cyclic. How to reverse it...

This solution uses pure angle chasing.

greentreeroad wrote: How to reverse it... I did it the same way. To reverse it I just showed $ G$ was unique on $ RS$, (consider the necessary equality of $ \angle GDB$ and $ \angle SQG$).

How many points do you think will be awarded for proving the "Only if" but not the "If"?

greentreeroad wrote: huh this problem made me sad. Another way for if part. Let BQ,DC intersect at T. Obviously DTB and QCB similar. Then this gives DGB and QSB similar since BG/BT=BS/BC. So <BDG=<BQS. Since <GPD=<BDC=<BRS so DRGP concyclic. So <BDG=<RPG. So<BQS=<RPG. similar <BQR=<SPG, adding gives <RQS=<RPS. So PQRS cyclic. How to reverse it... easy to know <AQR = <BPS since both ABPQ and RSPQ are cocyclic. you can get <BQR = <SPG since <AQB = <APB. then <SPQ = <SCG since GSPC are cocyclic(because <GSC = <GPC).then <BQR = <BCG. Now suppose <DBQ bigger than <CBQ. then <BRQ is smaller than <BGC. since they are both bigger than 90 degree, sinBRQ > sinBGC.using sine law in triangle BRQ and BGC and simplify you can get BQ/BC > BD/BM. Now by proving (<BCQ) > (<BMD) > (180 - <BCQ), you know that sinBMD > sinBCQ. now using sine law in triangle BDM and BCQ you can get BQ/BC < BD/BM,which is a contradiction. The other way is similar(but not completely same though, the thinking is the same, just using sine law). Hence <DBQ cannot be bigger than or smaller than <CBQ.

Warning: slightly offtopic post as it contains no relevance to the problem itself On the 'solution' sheet for this problem, all I did was draw a diagram and point out the grammar error: "Let the line through G parallel to AB intersects"

Since $ RS$ and $ DC$ are parallel, the circumcircle of $ BSR$, call it $ \omega'$, is tangent to $ \omega$ at $ B$. i) If $ PQRS$ is cyclic: $ X\in RSG\cap BB\cap QP$ exists since it is the radical center of $ \omega$, $ \omega'$ and $ PQRS$. By pascal on $ BQQPAB$, $ G$, $ X$, and $ QQ\cap AB$ are collinear so $ QQ\cap AB$ lies on $ GX$. But $ GX=RS\parallel AB$ so those lines intersect at infinity so $ QQ\parallel AB\parallel CD$. It follows directly that $ BQ$ bisects $ \angle DBC$. ii) If $ BG$ bisects $ \angle DBC$: Let $ X=BB\cap RS$, $ \{P', Q\}=\omega \cap XQ$. Then $ BX^2=XS\cdot XR=XP'\cdot XQ$ so $ P'QRS$ is cyclic. Let $ \{A', P'\}=P'G\cap \omega$. By pascal on $ BQQP'A'B$, $ G$, $ X$, and $ QQ\cap A'B$ are collinear. But the angle bisector implies that $ QQ\parallel DC\parallel GX$ so $ A'B\parallel DC$ so $ A=A'$, $ P=P'$ so $ PQRS$ is cyclic, as desired.

did anyone else try inverting around B? i think im a little crazy...

did you find a solution with it?

I got inversion to work. Like many other solutions the key was finding the right magic point to mark, after which stuff just works.

So I saw the first line of my proof after inverting. I was like "oh my, $ R'S'$ looks parallel to $ P'Q'$." and then I was like wait that may be trivial. and then I was happy.

This is quite a nice problem. I am a bit disappointed that I didn't see that pair of similar triangles (which essentially crushes the problem). Also to note, the past three USAMOs have had problems which use spiral similarity in a strong way. My approach to geometry pretty much makes this problem routine. I don't have a way to formalize this, but I feel like the iff fact is just a consequence of the fact that you can reverse your proofs by using the fact that a line and circle intersect (at most) two times.

jmerry wrote: If they are, $ P'S'H'$ is a reflection of $ Q'R'B$ across that line, and $ G$ is also on that line by symmetry. I fail to see the symmetry here. How is G proved to be on that line? edit: thread necro was necessary because the solution did not make sense

Let $RS$ intersect $\omega$ at $U$ and $V$, so that $U,R,S,V$ are collinear in that order. We observe that $\angle GSC = \frac{\stackrel{\frown}{VB} + \stackrel{\frown}{UC}}{2} = \frac{\stackrel{\frown}{AU} + \stackrel{\frown}{UC}}{2} = \frac{\stackrel{\frown}{AC}}{2} = \angle GBQ$, so $GSPC$ is cyclic. If: Let $BQ$ intersect $DC$ at $H$. $\angle BDH = \angle BDC = \angle BQC$ and $\angle CBQ = \angle QBD$, so $\triangle BDH \sim \triangle BQC$. Since $RG || DC$, $\triangle BRG \sim \triangle BDH \sim \triangle BGC$. Hence, $\frac{BR}{BG} = \frac{BQ}{BC}$, so $\frac{BR}{BQ} = \frac{BG}{BC}$. Since $\angle CBG \cong r \angle QBR$, $\triangle RQB \sim \triangle GCB$. Hence, $\angle RQP = \angle RQG + \angle GQP = \angle GCB + \angle BQP = \angle GCB + \angle BCP = \angle GCP = 180^{\circ} - \angle PSG = 180^{\circ} - \angle PSR$ since $GSPC$ is cyclic, so $PSRQ$ is cyclic as well. Only if: Since $GSPC$ and $RSPQ$ are cyclic, $\angle RQG = \angle RQP - \angle GQP = (180^{\circ} - \angle PSG) - \angle BQP = \angle GCP - \angle BCP = \angle GCB$. We now claim that if we fix $A,B,C,D,R$, and $S$, there will be at most one $G'$ on segment $SR$ such that the point $Q' = BG' \cap \omega$ satisfies $\angle RQG' = \angle G'CB$. For any such $G'$, let $T' = Q'R \cap \omega$. As $G'$ varies along segment $SR$ from $S$ to $R$, $\angle GCS$ increases strictly. Furthermore, as $G'$ varies along segment $SR$, $Q'$ varies along minor arc $\stackrel{\frown}{CD}$ from $C$ to $D$, whence $T'$ varies along minor arc $\stackrel{\frown}{AB}$ from $A$ to $B$, so $\angle GQ'R = \angle BQ'T'$ decreases strictly as $G'$ varies from $S$ to $R$, whence our claim is proven. From the if part of the problem, we find that $\angle RQ'G' = \angle G'CB$ holds when $G'$ is the intersection of the bisector of $\angle RBS$ with $RS$. By our claim, this is the only possible $G'$ for which the equality can hold, so $G$ must lie on the bisector of $\angle DBC$.

Let $RS\cap PQ=X$, and $RS$ intersect the circumcircle at $M,N$ with $M$ closer to $A$. If: We have that $\widehat{QD}=\widehat{QC}$. Then \[\angle MXQ-\angle CBP=\widehat{MQ}-\widehat{PN}-\widehat{PC}=\widehat{DQ}=\angle DBQ=\angle QBC\rightarrow \angle MXQ=\angle CBP+\angle QBC=\angle GBP\] Thus $BGPX$ is cyclic angle $\angle XBP=\angle PGX=\angle PAB$, and $BX$ is a tangent. But the circle of $BRS$ is tangent to the circle $BDC$, thus $XB^2=XR\cdot XS=XP\cdot XQ$. Only If: Since $BX$ is tangent by PoP and the above tangency of circles, $\angle XBP=\angle BAP=\angle PGX$, thus $BGPX$ is cyclic. Then \[\angle QBC=\angle MXP-\angle CBP=\widehat{MQ}-\widehat{NP}-\widehat{PC}=\widehat{DQ}=\angle DBQ\] Thus $G$ lies on the angle bisector of $\angle DBC$.

If part: $\angle RBG=\angle QBC$ and $\angle BRS=\angle BDC=\angle QBC$ so $\triangle BRQ~\triangle BQC$ and thus $\triangle BRQ~\triangle BGC$ therefore $\angle BQR=\angle GQR=\angle BCG$. Now we have \[\angle SGP=\angle RGA=\angle GAB=\angle PAB=\angle PQB=180^\circ-\angle PCB \\ =180^\circ-\angle PCS\] so quadrilateral $GSCP$ is cyclic and thus $\angle BCG=\angle SCG=\angle SPG$ so $\angle SPG=\angle GQR$. Now $\angle BQP=\angle BAP$ as they intercept the same arc and $\angle BAP=\angle SGP$ since $AB||RS$, we have $\angle BQP=\angle SGP$ and $\angle RQG=\angle SPG$ and so $\angle GSP=180^\circ-\angle SGP-\angle SPG=180^\circ-\angle BQP-\angle GQR=180^\circ-\angle PQR$ and thus quadrilateral $PQRS$ is cyclic. Only if part: Let the radical center of $\odot PQRS$, $\odot BRS$, and $\odot BCD$ be point $F$. Then $F$ is the point of concurrency of line $RS$, $PQ$, and the tangent to $\odot ABCD$ at $B$. Additionally, $\angle FBP=\angle BQP=\angle BAP=\angle FGP$ and so $GBFP$ is cyclic and thus $\angle FBG=\angle APQ=\angle ABG$. Additionally, $\angle FBC=\angle BDC=\angle ABD$ so combining $\angle FBG=\angle ABG$ and $\angle FBC=\angle ABD$, we get $\angle DBQ=\angle dbp=\angle CBP$ so $BQ$ bisects $\angle DBC$ as desired. $\blacksquare$

Let $X$ denote the intersection of the tangent to $\omega$ at $B$ and line $RS$. Remark that the tangent to $\omega$ at $B$ must be the radical axis of $(BRS)$ and $\omega$ because $(BRS)$ is homothetic to $B$ by a homothety about $B$ taking $R$ to $D$, $S$ to $C$. Hence, $X$ lies on line $PQ$ iff $PQRS$ is cyclic by the radical center theorem. Now, we claim that $XBGP$ is concyclic. Remark that \[\measuredangle BPG = \measuredangle BPA = \measuredangle BCA = \measuredangle BDA = \measuredangle CDA - \measuredangle CDB.\]However, it is clear that because $ABCD$ is a trapezoid inscribed in a circle (and hence an isosceles trapezoid), we have \[\measuredangle BPG = \measuredangle CDA - \measuredangle CDB = \measuredangle BCD - \measuredangle CDB = -\measuredangle CDB - \measuredangle DCB =\]\[-\measuredangle CBX - \measuredangle RSB = -\measuredangle SBX - \measuredangle XSB = \measuredangle BXS = \measuredangle BXG.\]Hence, $XBGP$ is concyclic as claimed. Now, remark that $X$ lies on $PQ$ iff \[\measuredangle ABQ = \measuredangle APQ = \measuredangle GPQ = \measuredangle GPX = \measuredangle GBX = \measuredangle QBX,\]or equivalently, line $QB$ bisects $\angle ABX$. It isn't hard to finish from here: note that \[\measuredangle XBC = \measuredangle BDC = \measuredangle BAC = \measuredangle ABD,\]so $QB$ bisects $\angle ABX$ iff it bisects $\angle DBC$. To summarize, $PQRS$ is cyclic iff $X$ lies on line $PQ$ iff $QB$ bisects $\angle ABX$ iff it bisects $\angle DBC$, so we are done.

First, we look at the if case, where $BG$ bisects $\angle{DBC}.$ Also, let the circle with circumcenter $O$ be called circle $O.$ First, we will prove points $G, S, P, C$ are concyclic. We will call this circle $O_1$ for short once we proved it. This is true because $\angle{SGP} = \angle{AGR} = \angle{BAG} = \angle{BAP} = \angle{BCP}$. Next, let $X$ be the second intersection of $QP$ and circle $O_1,$ and we'll prove that $SX$ and $BQ$ are parallel. This is true because $\angle{SXP} = \angle{SCP} = \angle{BCP} = \angle{BQP}.$ After that, let $E$ be the second intersection of $GC$ and circle $O$. We want to prove points $E, R, G, B$ are concyclic, which is true because $\angle{BEG} = \angle{BEC} = \angle{BDC} = \angle{BRG}$. Finally, we will prove that $E, R,$ and $Q$ are collinear. $\angle{QEC} = \angle{QBC}.$ $\angle{REC} = \angle{REG} = \angle{RBG} = \angle{QBC}$. With that, we can prove that $GX$ and $RQ$ are parallel because $\angle{QEC} = \angle{QBC} = \angle{XSC} = \angle{XGC}.$ Because of this, we have $180 = \angle{SPX}+\angle{SGX} = \angle{SRQ} + \angle{SPQ},$ which means points $Q, R, S, P$ are concyclic. Next, we will prove for the only if case. (could write this better but it's basically the same thing) Everything we have proven is true except that we need to prove $E, R, G$ are collinear, as all the other steps did not involve $\angle{DBQ} = \angle{CBQ}.$ We also can use the fact $QR$ is parallel to $GX$ (easily proven). Let $E'$ be the intersection of $QR$ and $CG$. We want to prove $E' = E$, which is true because $\angle{QE'C} = \angle{XGC} = \angle{XSC} = \angle{QBC} = \angle{QEC}$, and both $E$ and $E'$ lie on $CG.$ Finally, $\angle{RBG} = \angle{REG} = \angle{QEC} = \angle{QBC}$ which completes both parts of the proof.

Let $X=BB\cap PQ$. Then by Pascal on $BBQQPA$, we know that $$BB\cap PQ=X, BQ\cap PA=G, AB\cap QQ$$ are colinear. Now by radical axis theorem on $(BSR)$, $(PQRS)$, $\omega$, we have that $PQRS$ is cyclic iff $BB$, $SR$ and $PQ$ concur. So this is iff $XG$ parallel to $AB$, which by the above result is iff $QQ\cap AB$ is the point on infinity on $AB$, i.e $\infty_{AB}$. This is turn is equivalent to $Q$ being the midpoint of arc $CD$, so this is iff $BG$ bisects $\angle CBD$. $\square$

Invert about $B$ with arbitrary radius, denoting $X'$ is the inverse of $X.$ We see $C',D'$ are points on $\overline{S'B},\overline{R'B},$ such that $\overline{C'D'}\parallel\overline{S'R'}.$ Also, $A'=\overline{BB}\cap\overline{C'D'},$ $G'$ is on $(BR'S'),$ and $P'=(A'BG')\cap\overline{C'D'}.$ Since $P'Q'R'S'$ is a trapezoid, it suffices that $P'S'=Q'R'.$ Notice $$\measuredangle A'P'G'=\measuredangle A'BG'=\measuredangle BS'G',$$so we can let $P'=(C'G'S')\cap\overline{C'D'}.$ Notice \begin{align*}\measuredangle G'Q'P'=\measuredangle Q'P'G'&\iff \measuredangle Q'BR'+\measuredangle BD'Q'=\measuredangle C'S'G'\\&\iff \measuredangle G'BR'+\measuredangle BR'S'=\measuredangle S'BG'+\measuredangle BG'S'\\&\iff \measuredangle R'BG'=\measuredangle G'BS'\end{align*}so $\measuredangle R'BG'=\measuredangle G'BS'$ is equivalent to $G'R'=G'S'$ and $G'Q'=G'P'.$ Since $$\measuredangle BGR=\measuredangle BSR=\measuredangle BCD=\measuredangle SCP=\measuredangle SGP,$$we have $$G'R'=G'S'\land G'Q'=G'P'\iff\triangle Q'G'R'\cong\triangle P'G'S'\iff Q'R'=P'S'.$$$\square$

Optx wrote: Anyone tried to invert from point Q? I got some interesting things but cant solve it fully. I gotchu homie . NOOO, why is no one Inverting at $Q$ in any of the earlier posts and Inverting with $\sqrt{bc}$ at $B$ T__T . Am I too dumb for not having thought about $\sqrt{bc}$ T__T . Beautiful problem . Before starting, we prove that for any point $G$, if we let the line parallel to $CD$ through $G$ meet $BD$ and $BC$ at $R$ and $S$ repectively and also let $P=\odot(QRS)\cap\odot(BCD)$, then if $T=PQ\cap RS$, then we will have that $T$ lies on the tangent through $B$ to $\odot(BCD)$. This simply follows by applying Radax to $\left\{\odot(BRS),\odot(BCD),\odot(PQRS)\right\}$. We first adress the if direction that is when $BG$ is the angle bisector. Redefine $P=\odot(QRS)\cap\odot(BCD)$. Let $X=CG\cap QR$ and $Y=DG\cap QS$. Now we have, \begin{align*} \measuredangle GXQ &=\measuredangle CXQ=\measuredangle XQC+\measuredangle QCX\\ &=\measuredangle RQC+\measuredangle QCG\\ &=\measuredangle QRS+\measuredangle RSC+\measuredangle SCQ+\measuredangle QCG\\ &=\measuredangle QRS+\measuredangle DCG\\ &=\measuredangle QRG+\measuredangle RGX\\ &=\measuredangle RXG .\end{align*} So we get that $X\in\odot(BCD)$ and similarly we also get $Y\in\odot(BCD)$. Now $\measuredangle GXR=\measuredangle CXQ=\measuredangle CBQ=\measuredangle QBD=\measuredangle GBR$ which gives that $XRGB$ is cyclic and similarly $BGSY$ is also cyclic. Thus we have the following final POP relations $QR\cdot QX=QG\cdot QB=QS\cdot QY$. Now we perform an Inversion with center $=Q$ and radius $=\sqrt{QG\cdot QB}$ which swaps $\left\{R,X\right\}$, $\left\{G,B\right\}$ and $\left\{S,Y\right\}$. Also notice that by Radax on $\left\{\odot(BCD),\odot(PQRS),\odot(XRSY)\right\}$, we get that $T=PQ\cap RS\cap XY$. Thus we finally derive that $\left\{RS,\odot(BCD)\right\}$ also get swapped. We also deduce that $\left\{P,T\right\}$ also get swapped under this Inversion. Thus now it just suffices to prove that $K=QA\cap RS$, then $TBKQ$ is cyclic which simply follows from $\measuredangle TBQ=\measuredangle BDQ=\measuredangle BAQ=\measuredangle GKQ=\measuredangle TKQ$ and we are done by Inverting back . Now for the other direction, we have that $\overline{A-G-P}$ are collinear. So now Reim's on the the lines $QG$ and $PG$ with the circles $\odot(PQR)$ and $\odot(BCD)$, we get that the tangent at $G$ to $\odot(PGQ)$ is parallel to $AB\parallel CD$. Now letting $T=RS\cap PQ$, we get that $TG^2=TP\cdot TQ=TB^2$ from the initial relation we deduced at the beginning of the solution. This clearly gives $TG=TB$ which immediately finishes as then $\measuredangle DBQ=\measuredangle BDQ+\measuredangle DQB=\measuredangle TBG+\measuredangle DCB=\measuredangle BGS+\measuredangle GSB=\measuredangle GBS=\measuredangle QBC$ and we are finally done .

Taking inversion at $B$ gives the following problem, Inverted Problem wrote: Let $\omega$ be the circumcircle of $\triangle ABC$ and $G$ be a point on arc $BC$ of $\omega$ not containing $A$. Let $P\in AB$ and $Q\in AC$ such that $PQ\parallel BC$. Let $T$ be the intersection $PQ$ and tangent at $A$ of $\omega$. Let $R=PQ\cap (ATG)$ other than $T$ and $S=AG\cap PQ$. Prove that $AG$ is the bisector of $\angle BAC$ if and only if $BCRS$ is cyclic. Let $(ATG)=\gamma$. Define the function \[ f(X)=P(X, \gamma)-P(X, \omega) \]It is known that $f$ is linear. So by linearity we have \[ \frac{ST}{SR}=\frac{f(S)-f(T)}{f(S)-f(R)}=\frac{TA^2}{P(R, \omega)} \implies P(R, \omega)=\frac{TA^2\cdot SR}{ST} \]Let $O$ be the center of $\omega$. Since $PQ\parallel BC$ and $OB=OC$ therefore is sufficient to show $AG$ is the angle bisector of $\angle BAC$ iff $OR=OS$ or equivalently $P(R, \omega)=P(S, \omega)$. Therefore we can say $AG$ is the angle bisector iff \begin{align*} ST\cdot SR=SA\cdot SG=P(S, \omega)=P(R, \omega)=\frac{TA^2\cdot SR}{ST}\iff TA=ST \end{align*}which is easy angle chase.

Kind of an arbitrary geometry problem, but enjoyable nonetheless. Invert about $B$. In the resulting picture, $\overline{AB}$ is tangent to $(BCD)$ and $(BRS)$; $P = \overline{AD} \cap (ABG)$, and $Q = \overline{AD} \cap \overline{BG}$. Observe that $\angle APG = \angle ABG = \angle BRG$. For one direction, assume that $\overline{BG}$ is an angle bisector. Then $$\angle GQP = \frac{\widehat{BS}+\widehat{RG}}2 = \angle BRG = \angle GPQ$$hence $GP = GQ$, and the result follows by symmetry. On the other direction, setting $G'$ to be the arc midpoint, notice that $\angle G'QP = \angle BQC$ by same computation, hence $G'$ lies on $\overline{BQ}$.

Invert about $B,$ and let $X'$ be the image of a point $X$ under this inversion. First, there is a homothety at $B$ taking $DC$ to $RS,$ so $(BCD)$ and $(BSR)$ are tangent so $C'D' \parallel R'S'.$ Next, let $R'S'$ intersect $G'P'$ at $Z.$ We get $\angle R'ZG'=\angle A'P'G'=\angle A'BG'=\angle BS'G'$ and $\angle G'R'Z=\angle G'BS',$ so $\angle R'G'B=\angle ZG'S',$ and letting $G'P'$ intersect $(BR'S'G)$ again at $K$ we get that $BK \parallel R'S'.$ We find that $P'Q'R'S'$ is cyclic iff it is an isosceles trapezoid. However, it is easy to see now that this is true iff $G$ is the midpoint of arc $R'S'$ by considering the perpendicular bisector of $R'S',$ so this is true iff $BG$ bisects $\angle CBD.$

Invert at $B$ with arbitrary radius. We preserve $CE \parallel RS$, but now we know $AB$ is tangent to $(BCD)$ and $(BRS)$. Points $P$, $Q$, $R$, and $S$ are mapped to the points of a trapezoid. Then \begin{align*} BG \text{ bisector} &\iff \measuredangle GSR = \measuredangle SRG = \measuredangle SBG = \measuredangle GBR \\ &\iff \measuredangle GSB = \measuredangle GBA = \measuredangle AQB = \measuredangle PQR = \measuredangle GPR \\ &\iff PQRS \text{ isosceles trapezoid.} \quad \blacksquare \end{align*}

nice and cute. it doesn't fall immediately Let $\ell$ be the line through $G$ parallel to $AB$. Also let $U$ and $V$ be the intersections of $QR$ and $PS$ with $\omega$. If $PQRS$ is cyclic, then by Reim's $UV\parallel AB\parallel CD$. By Pascal's on $BQVPAC$ we then find $QV$ passes through $AC\cap \ell=T$. Here's the thing, though: $R$ and $T$ should be symmetric about the perpendicular bisector of $AB$. Since $U$ and $V$ are also symmetric, it follows that $UR\cap VT=Q$ is on the perpendicular bisector, i.e., $BG$ bisects $\angle CBD$. If $BG$ bisects $\angle CBD$, then $Q$ is on the perpendicular bisector of $AB$. It suffices to have $UV\parallel AB$ (points are defined the same way). For the most part, everything here is identical to the previous. By Pascal's on $BQVPAC$ we again find $QV$ passes through $AC\cap \ell=T$. Now, instead of $UR$ and $VT$ being given as symmetric, we instead have $QR$ and $QT$ symmetric. Thus $U=QR\cap \omega$ and $V=QT\cap \omega$ are clearly symmetric, so $UV\parallel AB$.

Also we can draw $PQ\cap RS=M$ and reflect $G$ across $M$ to $N$. Then we need to show something like $(R,S;G,N)=-1$ which isn't hard if you notice the radical center of $(BRS),(BDC),(PQRS)$ (I didn't oops)

Nice problem! I think I came up with a wonderful solution(not really humble there ). Here is a sketch of it: Let $RS$ intersect $ \omega$ and $PQ$ at $X, Y, Z$. $CG$ intersect $\omega$ at $F$. Then our idea is to use Desargues' Involution theorem on $FPCD$ and line $RS$. It is not hard to verify that then $PQRS$ is concyclic is equivalent to $P, R, F$ lie on the same line, which is equivalent to tangent at $P$ being parallel to $CD$ by Pascal's theorem, which is exactly what we wanted.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -25.251594460355093, xmax = 24.927192375407017, ymin = -12.702874719263233, ymax = 7.819465053019219; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((-1.3452036517273975,0.28788835564809817),0.5880326582315872,-39.87646176125591,-0.33902432445145547)--(-1.3452036517273975,0.28788835564809817)--cycle, linewidth(1) + qqwuqq); draw(arc((-4.069482163222376,2.56383778197193),0.5880326582315872,-39.87646176125591,-0.33902432445144914)--(-4.069482163222376,2.56383778197193)--cycle, linewidth(1) + qqwuqq); draw(arc((-0.7080729112973836,3.584814203364297),0.5880326582315872,-57.110838967275384,-17.57340153047094)--(-0.7080729112973836,3.584814203364297)--cycle, linewidth(1) + qqwuqq); draw(arc((3.2753562200269513,0.26054776469097185),0.5880326582315872,179.66097567554854,285.6547838267051)--(3.2753562200269513,0.26054776469097185)--cycle, linewidth(1) + blue); draw(arc((2.6410188261351224,2.5241306755260275),0.5880326582315872,179.66097567554854,285.6547838267051)--(2.6410188261351224,2.5241306755260275)--cycle, linewidth(1) + blue); draw(arc((-0.7080729112973836,3.584814203364297),0.5880326582315872,-163.10464711843196,-57.110838967275384)--(-0.7080729112973836,3.584814203364297)--cycle, linewidth(1) + blue); /* draw figures */ draw((-6.192115171650039,-4.679590254706536)--(4.677785160575875,-4.7439091915836125), linewidth(1)); draw((4.677785160575875,-4.7439091915836125)--(2.6410188261351224,2.5241306755260275), linewidth(1)); draw(circle((-0.7431525425540128,-2.343643479006383), 5.928561467353529), linewidth(1)); draw((-4.069482163222376,2.56383778197193)--(2.6410188261351224,2.5241306755260275), linewidth(1)); draw((-4.069482163222376,2.56383778197193)--(-6.192115171650039,-4.679590254706536), linewidth(1)); draw((-4.069482163222376,2.56383778197193)--(4.677785160575875,-4.7439091915836125), linewidth(1)); draw((-0.7080729112973836,3.584814203364297)--(4.677785160575875,-4.7439091915836125), linewidth(1)); draw((-6.192115171650039,-4.679590254706536)--(3.6084618748954225,1.6826815245216786), linewidth(1)); draw((-0.7080729112973836,3.584814203364297)--(9.912245917184492,0.2212762280214005), linewidth(1)); draw((-11.367451861549748,0.3471915994932009)--(-0.7080729112973836,3.584814203364297), linewidth(1)); draw((-11.367451861549748,0.3471915994932009)--(9.912245917184492,0.2212762280214005), linewidth(1)); draw((-11.367451861549748,0.3471915994932009)--(-6.192115171650039,-4.679590254706536), linewidth(1)); draw((4.677785160575875,-4.7439091915836125)--(9.912245917184492,0.2212762280214005), linewidth(1)); /* dots and labels */ dot((-6.192115171650039,-4.679590254706536),dotstyle); label("$A$", (-6.552155928590619,-5.215258871114335), NE * labelscalefactor); dot((4.677785160575875,-4.7439091915836125),dotstyle); label("$B$", (4.5812624005941,-5.293663225545214), NE * labelscalefactor); dot((2.6410188261351224,2.5241306755260275),dotstyle); label("$C$", (2.71915898286074,2.7231820150121155), NE * labelscalefactor); dot((-4.069482163222376,2.56383778197193),linewidth(4pt) + dotstyle); label("$D$", (-4.592047067818661,2.6447776605812368), NE * labelscalefactor); dot((-0.7080729112973836,3.584814203364297),linewidth(4pt) + dotstyle); label("$Q$", (-1.0442500298214181,3.7816407998289754), NE * labelscalefactor); dot((1.4345559109606933,0.27144007421207395),dotstyle); label("$G$", (1.072667539812296,-0.39339107361530595), NE * labelscalefactor); dot((3.6084618748954225,1.6826815245216786),linewidth(4pt) + dotstyle); label("$P$", (3.836421033500756,1.4295101669026198), NE * labelscalefactor); dot((-1.3452036517273975,0.28788835564809817),linewidth(4pt) + dotstyle); label("$R$", (-1.2598620045063336,0.44945573651663817), NE * labelscalefactor); dot((3.2753562200269513,0.26054776469097185),linewidth(4pt) + dotstyle); label("$S$", (3.346393818307767,0.4102535593011989), NE * labelscalefactor); dot((9.912245917184492,0.2212762280214005),linewidth(4pt) + dotstyle); label("$R'$", (9.991162856324703,0.37105138208575966), NE * labelscalefactor); dot((-11.367451861549748,0.3471915994932009),linewidth(4pt) + dotstyle); label("$S'$", (-11.922854207105782,0.48865791373207745), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We consider inversion $\Psi$ centered at $G$ with radius $r=-\sqrt{GB\cdot GQ}.$ We know that $\Psi(Q)=B$ and $\Psi(P)=A.$ Let $QC\cap RS = R'$ and let $QD\cap RS = S'$. Claim: $R'\equiv \Psi(R).$ Proof: Angle chasing gives $$\angle BQR' = \angle BQC = \angle BDC = \angle BRS = \angle BRR'.$$Hence, $BRQR'$ is cyclic, so $GR\cdot GR' = r^2$ and $G$ is on segment $RR'$, so we are done. In a similar fashion we can prove that $S'\equiv \Psi(S).$ Now $PQRS$ is cyclic iff $ABR'S'$ is cyclic. However, since $AB\parallel R'S'$, we get that $ABR'S'$ is cyclic iff $AS'=BR'.$ Denote with $S_{XY}$ the perpendicular bisector of segment $XY$. Clearly, we have that $S_{AB}\equiv S_{DC}.$ Direction 1: If $G$ lies on the angle bisector, then $Q\in S_{AB}$ and since $\hspace{2mm}$ $RS\perp S_{AB}$, we get that $S'$ and $R'$ are symmetric with respect to $S_{AB}$. However, $A$ and $B$ are also symmetric wrt to $S_{AB}$, so $AS'=BR'$ and we are done. Direction 2: If $PQRS$ is cyclic, then $ABR'S'$ is cyclic. Thus, $S'$ and $R'$ are symmetric wrt to $S_{AB}.$ Since $D$ and $C$ are also symmetric wrt to $S_{AB}$, we get that $Q=DS'\cap CR'$ lies on $S_{AB}$. Hence, $Q$ is the midpoint of the arc $CD$ and hence $G$ lies on the angle bisector of $\angle CBD$.

after inversion, the key insight really was trying to remove points from the diagram lol

Great problem. I don't think a solution similar to mine has been posted yet so here it is, First the if condition. Observe that as $RS \parallel CD$ $(BCD),(BSR)$ are tangent. Claim: $PQ \cap RS \in BB$ Proof: Let $X = BB \cap RS$ and let $XQ \cap \omega = P' \neq Q$ Observe $$\angle QBX = \angle QBC + \angle CBX = \angle DBQ + \angle ABD = \angle ABG \dots (1)$$Additionally by sine law in $ \triangle BGX$ $$ \frac{BG}{BX} = \cos(\angle GBX) \dots (2)$$Similarly by applying the extended sine law on $\omega$ we get $$ \frac{AB}{BQ} = \cos(\angle DBQ + \angle ABD) = \cos(\angle ABG) = \cos(\angle GBX) \dots (3)$$With $1,2,3$ we conclude $$ \triangle ABG \sim \triangle BQX $$So $$ \angle BAG = \angle BQX = \angle BQP' $$This implies $A-G-P'$ and hence $P' \equiv P$. Finally by pop we get that $$ XP\cdot XQ = XB^2 = XR \cdot XS $$Hence $PQRS$ cyclic. Now the only if part: By radical axis we get $PQ \cap RS \in BB$. Next \begin{align*} \angle XGP &= \angle RGA \\ &= \angle GAB \\ &= \angle PQG \end{align*}Hence $XG$ is tangent to $(PQG)$. So by many pops $$XG^2 = XP \cdot XQ = XB^2$$Hence $XB = XG$. Now as its well known that the tangent from a vertex, the opposing side and the perpendicular bisector of the angle bisector of the angle at vertex are concurrent, we are done.