For this one I have found the following solution. In fact,the bisectors of $\angle{AOB},\angle{BOC}$ are the perpendicular bisectors of $AB,BC$ respectively. Lemma:If $R,S$ are the second intersection points of the perpendicular bisectors of $AB,BC$ with $\Gamma$,then $PQRS$ is an isosceles trapezium. Proof:Suppose that $X,Y$ are the intersections of the perpendicular bisectors of $AB,BC$ respectively,with $BM$.Also,let $K$ be the center of $\Gamma$. Obviously $OK\perp BM$ because $K$ is the midpoint of $BM$.Also $\angle{OXK}=\angle{BXP}=90^{\circ}-\frac{\angle{B}}{2}$. Similarly $\angle{OYB}=90^{\circ}-\frac{\angle{B}}{2}$,which means that $\triangle{OXY}$ is isosceles. Hence,$OK$ is also median of this triangle which means that $KX=KY\Leftrightarrow BX=MY$. Power of point thorem gives $PX\cdot XR=BX\cdot XM$ and $SY\cdot YQ=MY\cdot BY$,thus $PX\cdot XR=SY\cdot YQ$. Also $PR,YQ$ intersect each other at $O$ thus $PO\cdot OR=SO\cdot OQ$. This is equivalent to $(PX+OX)(RX-OX)=(SY+YO)(QY-YO)\Leftrightarrow $ $PX\cdot RX+OX(RX-PX)-OX^2=SY\cdot YQ+OY(QY-SY)-OY^2$ Since $PX\cdot XR=SY\cdot YQ$ and $OX=OY$,the latter becomes $PX-RX=SY-YQ$. This gives $(PX-RX)^2=(SY-YQ)^2\overset{PX\cdot RX=SY\cdot YQ}{\Leftarrow \!=\!=\! \Rightarrow} (PX-RX)^2+4PX\cdot RX=(SY-YQ)^2+4SY\cdot YQ$. This is equivalent to $(PX+XR)^2=(SY+YQ)^2\Leftrightarrow PR=SQ$. Every pair of chords that don't share starting and ending points form an isosceles trapezium,thus the lemma is proved. $\rule{420pt}{1pt}$ Back to our problem,we know that in an isosceles trapezium,the segment connecting its circumcentre with the intersection of the diagonals is the perpendicular bisector of its parallel sides.Thus $OK\perp PS$.However $OK\perp BM$ thus $BM\parallel PS$ which gives that $PSMB$ is isosceles trapezium.Thus $\angle{PSQ}=\angle{BYQ}=90^{\circ}-\frac{\angle{B}}{2}$. Since $\angle{BKQ}=2\angle{BYQ}$,we get $\angle{BKQ}=180^{\circ}-\angle{B}=\angle{POQ}$ thus $OKQP$ is cyclic. $\rule{430pt}{1pt}$. Because I have already defined a point $R$,let $T$ be the point corresponding to $R$. Thus,we have to prove that $BT\parallel AC$.By definition,$T\equiv KO\cap PQ$. From the cyclic $OKQP$ we get $\angle{OPK}=180^{\circ}-\angle{KPQ}-\angle{TPO}=180^{\circ}-\angle{KQP}-\angle{OKQ}=\angle{PTO}$. Thus $\triangle{OPK}\simeq \triangle{KPT}\Rightarrow KP^2=KO\cdot KT\overset{KP=KB}{\Leftarrow \!=\!=\!=\! \Rightarrow} KB^2=KO\cdot KT$ $\triangle{KOB}\simeq \triangle{KBT}\Rightarrow \angle{KTB}=\angle{OBK}$. Hence,$\angle{KBT}=90^{\circ}-\angle{OBK}=90^{\circ}-\left(\angle{C}+\frac{\angle{B}}{2}-90^{\circ}\right)=\angle{A}+\frac{\angle{B}}{2}$ Thus $\angle{ABT}+\angle{ABK}=\angle{A}+\frac{\angle{B}}{2}\Leftrightarrow \angle{ABT}=\angle{A}\Leftrightarrow BT\parallel AC$ which is the desired result. (There are three attachments,corresponding to the three parts of the proof,defined by the black lines)

My solution : Let $ X \equiv MO \cap \Gamma $ and $ Y \in OR $ be the midpoint of $ BM $ . From $ \angle BXO=\angle BYO=90^{\circ} \Longrightarrow B, O, X, Y $ lie on a circle with diameter $ BO $ . From $ OY \perp BM, OQ \perp BC, OP \perp AB \Longrightarrow \angle QOY=\angle CBM=\tfrac{1}{2} \angle CBA=90^{\circ}-\tfrac{1}{2} \angle POQ $ , so $ Y $ lie on the external bisector of $ \angle POQ \Longrightarrow Y $ is the midpoint of arc $ POQ $ in $ \odot (POQ) $ ( $\because YP=YQ $ ) , hence $ R $ is the radical center of $ \Gamma, \odot (BO) ,\odot (POQ) \Longrightarrow R \in BX \Longrightarrow BR \parallel AC $ ( $ \because BR \perp OM, OM \perp AC $ ) . Q.E.D

Let $M_a, M_c$ be the midpoints of $\overline{BC}, \overline{BA}$, respectively, and let $O'$ be the midpoint of $\overline{BM}$ (the center of $\Gamma$). Let $X$ be the second intersection of $OM$ with $\Gamma.$ Because $\overline{BM}$ is a diameter of $\Gamma$, we have $BX \perp OX.$ Meanwhile, $OO'$ is the perpendicular bisector of $\overline{BM} \implies BO' \perp OO'.$ Hence, $B, X, O, O'$ are inscribed in the circle $\omega$ of diameter $\overline{BO}.$ Furthermore, From $BX \perp OX$, it follows that $BX \parallel AC.$ Observe that $M_a$ and $M_c$ also lie on $\omega.$ Therefore, \[\measuredangle O'M_aQ = \measuredangle O'M_aO = \measuredangle O'M_cO = \measuredangle O'M_cP.\] Meanwhile, from $\angle M_aBO' = \angle M_cBO'$, it follows that $O'M_a = O'M_c$ (equal arcs subtend equal angles). Combining these relations with $O'Q = O'P$, we obtain $\triangle O'M_aQ \cong \triangle O'M_cP.$ Hence, \[\measuredangle O'QO = \measuredangle O'QM_a = \measuredangle O'PM_c = \measuredangle O'PO.\] Therefore $O, O', P, Q$ are concyclic. It follows that $BX, OO', PQ$ concur at the radical center $R$ of $\Gamma, \omega,$ and $\odot (OO'PQ).$ Thus, $BR \parallel AC$, as desired. $\square$

I think that this solution is very similar to the TelvCohl's solution. Let $X$ $=$ $\Gamma$ $\cup$ $MO$, and let $D,E$ be the midpoints of $BC$ and $AB$ respectively. Let $T$ be the midpoint of $BM$. Since $BM$ is a diameter of $\Gamma$ $\implies$ $MOX$ $\perp$ $BX$ $\implies$ $BX$ $\parallel$ $AC$ . Observs that $E,T,D$ are the midpoints of chords of $\Omega$ with center $O$ .$\implies$ $OE$ $\perp$ $BE$ , $OT$ $\perp$ $BT$ and $BD$ $\perp$ $OD$ Hence $E,O,T,D$ and $B$ are cyclic From the above result $\angle$$EOR$$=$$\angle$$EBT$$=$$\angle$$TBC$$=$$\angle$$TOD$ $\implies$ $T$ lies on the external angle bisector of $POQ$. On the other hand, $T$ $\epsilon$ perpendicular bisector of $PQ$. Hence $P,O,T,Q$ are cyclic Hence $R$ is the radical center of $\odot$($POTQ$), $\odot$($BXEOTD$) and $\Gamma$. $\implies$ $B,X,R$ are collinear Hence $BR$ $\parallel$ $AC$ and we are done.

[asy][asy] size(8cm); pair B = dir(120); pair A = dir(230); pair C = dir(310); pair M = dir(270); pair M_B = midpoint(A--C); pair M_C = midpoint(A--B); pair M_A = midpoint(B--C); pair O = circumcenter(A, B, C); pair S = extension(B, B+C-A, O, M_B); draw(circumcircle(B, M_A, M_C), heavygreen); pair T = midpoint(B--M); path w = CP(T, B); draw(w, blue); pair P = IP(M_C--(100*M_C-99*O), w); pair Q = IP(M_A--(100*M_A-99*O), w); draw(B--A--C--cycle, heavyred); draw(unitcircle, red); draw(B--O--M, blue); draw(P--O--Q, blue); draw(B--M, heavyred+dashed); draw(T--O, blue+dashed); dot("$B$", B, dir(B)); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$M_C$", M_C, dir(210)); dot("$M_A$", M_A, dir(335)); dot("$O$", O, dir(310)); dot("$S$", S, dir(45)); dot("$T$", T, dir(240)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); /* Source generated by TSQ */ [/asy][/asy] Let $T$ be the midpoint of $BM$ and $S$ the point such that $SA = SC$, $BS \parallel AC$. Let $\omega$ be the circumcircle of $BSOT$, which passes through the midpoints of $AB$ and $AC$. Since $TP = TQ$ and $OT$ is the external angle bisector of $\angle POQ$, the quadrilateral $TOQP$ is cyclic as well, and we're done by radical axis.

v_Enhance, I was wondering how you noticed that $OT$ is the external angle bisector of $\angle POQ$ and how you chose to go this path? e.g. what was the thought process that you went through to get the answer. I spent hours trying to prove that $T, O, Q, P$ are cyclic, but I just couldn't find anything useful.

@MathPanda1: Because I already knew that if you have a cyclic quad with two equal lengths, that's equivalent to having an angle bisector (internal or external). The "internal" configuration is more common, viz http://www.mit.edu/~evanchen/handouts/Fact5/Fact5.pdf.

wait why is OT the external angle bisector of $\angle POQ$? (sorry if this is obvious)

AkshajK wrote: wait why is OT the external angle bisector of $\angle POQ$? (sorry if this is obvious) Don't apologize. We have $AM = MC$, so by the homothety with ratio $\tfrac12$ at $B$ we obtain $M_CT = M_AT$. On the other hand $T$, $M_C$, $M_A$, $O$ are concyclic on the circle with diameter $BO$.

Let $S$ be the center of $\Gamma$ and let $l$ be the line through $B$ parallel to $AC$. Let $T=l\cap \Gamma$, and note that $BS\perp SO$. We have $l\perp TM$ because $BM$ is a diameter, and $OM\perp AC$ is clear, so because $l\parallel AC$ we have $BT\perp TO$ and thus $BTOS$ is cyclic. As $R=OS\cap PQ$ and $BTPQ$ is cyclic, by radical axis it suffices to prove $QPOS$ cyclic. Let $E=OQ\cap BC, F=OP\cap AB$. As $E,F,S$ are the midpoints of $BC,AB, BM$, respectively, taking the homothety from $B$ shows that $SE=SF$ (because $MC=MA$). Also, $S$ is the center of $\Gamma$ so $SQ=SP$. We have $BESOF$ cyclic with diameter $OB$, so $\angle SFO=\angle SEO\implies \angle SFP=\angle SEQ$, so we have $\triangle SFP\cong \triangle SEQ\implies \angle SPF=\angle SQE$ so $QPOS$ is cyclic and we are done.

We will show that $R$ is in fact the orthocenter of the triangle $OBM$. Let's begin with an auxiliary result: Lemma: In a triangle $XYZ$, if $W$ is the foot of perpendicular from $X$ to $YZ$ and $ST$ is a chord in the circle with diameter $YZ$ such that $X,W,S,T$ are concyclic, then $ST$ passes through the orthocenter of $\triangle XYZ$. Proof : By radical axes theorem on $(YZ),(XSWT),(XWZ)$, we are done. $\square$ Now coming back to the problem at hand, let $BA$ and $BC$ intersect $\Gamma$ again at $A',C'$ respectively. Also, suppose $D,F,N$ are the midpoints of $BC,AB,BM$ respectively. Since $OP,OQ$ are really the perpendicular bisectors of the respective sides, they pass through $F,D$ respectively. From the cyclic quadrilateral $ODBF$, it's easy to get $\angle QOP=\angle DOF=180^{\circ}-\angle B.$ Next, note that the triangles $BC'M$ and $BA'M$ are congruent, so $BC'=BA'.$ On the other hand, the congruency of the triangles $MCC'$ and $MAA'$ gives $CC'=AA'$. Now $$BD=\frac{BC}{2}=\frac{BC'-CC'}{2}=\frac{BA'-AA'}{2}=BA'-\frac{BA'+AA'}{2}=BA'-\frac{BA}{2}=BA'-BF=A'F.$$Similarly $C'D=BF$. Considering the isometry that maps $\overarc{BC'}\to\overarc{BA'}$ and $D\to F$, we see that $Q$ must map to $P$, so the triangles $QBC'$ and $PA'B$ are congruent, with vertices in this order. We can now infer \begin{align*} \angle QNP& =2\angle QMP=2(\angle QMB+\angle BMP)=2(\angle QC'B+\angle BA'P)\\ &=2(\angle QC'B+\angle QBC')=2(180^{\circ} - \angle BQC')=2 \angle BMC'=2(90^{\circ}-\angle MBC')\\ &=2\left(90^{\circ}-\frac{\angle B}{2}\right)=180^{\circ}-\angle B.\end{align*} This gives $\angle QNP=\angle QOP\implies QNOP$ is cyclic. Now applying the Lemma with the triangle $OBM$ and chord $PQ$ gives $PQ$ passes through the orthocenter of $\triangle OMB$. The fact that $RN$ is an altitude confirms that $R$ is indeed the orthocenter. This means $BR\perp MO$; but since $MO\perp AC,$ we get $BR||AC$, as desired. $\blacksquare$

It is very similar to the G2 2013, the same idea and just a bit harder proof.

Here's another solution: Suppose $P $ lies between $R $ and $Q $. Let $OP\cap AB = D$ and $OQ\cap BC = E$. Also, let $O'$ be the center of $\Gamma $. It is simple enough to note that $O'$, $O $, and $R $ are collinear. Claim: $OO'QP $ is cyclic. Proof of the claim: It is easy to see that $D $ and $E $ are the midpoints of $AB $ and $BC $ respectively and $BDOO'E $ is cyclic. As, $O'P = O'Q $, $O'D = O'E$, and $\measuredangle O'DO = \measuredangle O'EO $, so, $O'$ is the center of the spiral similarity carrying $DP $ to $EQ $. Hence, $OO'QP $ is cyclic. Main problem: From the claim, we have, $Pow_R (\odot (POO'Q)) = Pow_R (\Gamma) $. As $BM $ is a diameter of $\Gamma$, so, $O $ is the orthocenter of $\Delta RBM $. But, $RB = RM $. Thus, $R $ and $O $ are inverses of each other under that inversion centered at $O'$ with radius $O'B $ or $O'M $. Hence, $\measuredangle MBR = \measuredangle O'OM = \measuredangle BAM\implies BR || AC$.

After two years, I suddenly feel like writing this up hajimbrak wrote: Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$. (Here we always assume that an angle bisector is a ray.) Proposed by Sergey Berlov, Russia Let $D$ be the mid-point of $\overline{BM}$. Then Claim. $D,P,Q,O$ are concyclic. (Proof) Note that $DP=DQ=DB$ and $\overline{DO} \perp \overline{BM}$, $\overline{OP} \perp \overline{AB}$, $\overline{OQ} \perp \overline{CB}$, so $\overline{DO}$ bisects $\angle POQ$ since $\overline{BM}$ bisects $\angle ABC$. $\blacksquare$ Redefine $R$ as the radical center of $\odot(OB), \odot(MB), \odot(POQ)$; then $R$ lies on $\overline{PQ}, \overline{DO}$ and the parallel from $B$ to $\overline{AC}$. So $BR=MR$ and $R$ lies on $\overline{PQ}$, as desired.

My proof is different from the ones above, so am including it here hajimbrak wrote: Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$. (Here we always assume that an angle bisector is a ray.) Let $K$ be the midpoint of $BM$ and $S=\Gamma \cap BR$. Since $BM$ is a diameter of $\Gamma$, hence we have that $MS \perp BR$.

to show that $OS \perp BR \Leftrightarrow S$ lies on $\odot(BOK)$ (since $BO$ is a diameter of this circle). The key observation is that $PKOQ$ is cyclic. (this is proved by everyone above)

Thus, by Power of a Point, $RO \cdot RK \overset{\odot(PKOQ)}{=} RQ \cdot RP \overset{\odot(PBSQ)}{=} RS \cdot RB$ and thus $S$ lies on $\odot(BOK)$, as desired. $\blacksquare$

Here is my solution. I had found the steps in this order only. Let $OR \cap BM = N, BM \cap AC = T$. Then $RP \cdot RQ = NR^2 - NB^2$. Now, We have to show that \[\angle RBM = \angle BTC = 180^{\circ} - \angle ACB - \frac{1}{2}\angle ABC = 180^{\circ} - \angle BCM = \angle BAM = \frac{1}{2} \angle BOM = \angle BON \Leftrightarrow \angle BRO = \angle NBO\]\[\Leftrightarrow \text{NB is tangent to} \text{ } \odot (BRO) \Leftrightarrow NB^2 = NO \cdot NR \Leftrightarrow NR^2 - RP \cdot RQ = NO \cdot NR \Leftrightarrow RP \cdot RQ = RN \cdot RO \Leftrightarrow \text{PONQ is cyclic}\] And, As $NP =NQ$, it suffices to show that $NO$ is the external angle bisector of $\angle POQ$, i.e. of $\angle M_AOM_C$, where $M_A$ is the midpoint of $BC$ and $M_C$ is the midpoint of $AB$. But as $\angle OM_AB = \angle OM_CB = \angle ONB = 90^{\circ}$ so $B,M_A,M_C,O,N$ are concyclic. Also, $BN$ bisects $\angle M_ABM_C$, and so $NM_A = NM_C$, giving that $NO$ is the external angle bisector of $\angle M_AOM_C$. Hence, done.

ISL 2014 G3 wrote: Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$. (Here we always assume that an angle bisector is a ray.) Solution: Let $N$ be midpoint of $BM$ and $M_A,M_C$ be midpoints of $BC,AB$ $\implies$ $BM_CONM_A$ is cyclic. Now, $\tfrac{NP}{NM_C}=\tfrac{NQ}{NM_A}$ $\implies$ $\Delta NQM_A$ $\sim$ $\Delta NPM_C$ $\implies$ $QPON$ is cyclic. Now, Finally Apply Radical Axes Theorem $\qquad \blacksquare$

ISL 2014 G3 wrote: Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$. (Here we always assume that an angle bisector is a ray.) Proposed by Sergey Berlov, Russia Lemma:- Let $\ell$ be the A- external bisector of $\angle BAC$ and $X\in \ell$ such that $BX=CX$. Then $X\in\odot(ABC)$. (Well known) Let $OR\cap BM=K$, so, $OK\perp BM$. Let $OQ\cap BC=X$ and $OP\cap AB=Y$. So, $\angle QOK=B/2\implies OK$ is the O-external bisector of $\angle POQ$ and as $KQ=KP$. So, by our Lemma we get that $K,O,P,Q$ are concyclic. Let $RB\cap\odot(BM)=T$. So, $RO\cdot RK=RP\cdot RQ=RT\cdot RB\implies T\in\odot(BOK)\implies\angle OTB=90^\circ$ also $\angle MTB=90^\circ$. So, $\overline{M-O-T}$. So, $TM\perp AC$ also $TM\perp BR\implies BR\| AC$.

Nice problem! $D=$center of $\Gamma$ and $A_0,C_0$ are midpoints of $BC$ and $BA$,respectively. First of all note that $RB=RM$ means that $R\in OD$. Let $R'\in OD$ such that $BR'||AC$ and $Q'=R'P\cap \Gamma$. Now, that's enough to prove that $Q'\in A_0O$. $\rule{430pt}{1pt}$ Claim 1 : $\angle DBO=\angle BR'O$. Proof : Let $E=R'O\cap AC$ and $F=OM\cap AC$. 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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $\rule{430pt}{1pt}$ Claim 2 : $Q'POD$ is cyclic. Proof : From Claim 1 we get $DB^2=DR'\cdot DO$ $\implies$ $R'D\cdot R'O=R'D^2-DO\cdot DR'=R'D^2-DB^2=Pow(R',\Gamma)=R'P\cdot R'Q$ $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14.cm); 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 = -2.4758125297833034, xmax = 2.098966346474646, ymin = 0.3008081282625627, ymax = 3.6590903628872407; /* image dimensions */ pen ttttff = rgb(0.2,0.2,1); pen ffwwqq = rgb(1,0.4,0); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((-0.30400072362596164,2.2820813058443474)--(-0.3625554570975552,2.2014666233637774)--(-0.2819407746169855,2.142911889892184)--(-0.223386041145392,2.2235265723727538)--cycle, linewidth(1) + qqwuqq); draw(arc((-1.3253520822907845,3.0239431447455085),0.14090694690732905,-99.41822363775223,-67.70556175061617)--(-1.3253520822907845,3.0239431447455085)--cycle, linewidth(1) + qqwuqq); 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draw(arc((-1.3253520822907845,3.0239431447455085),0.14090694690732905,-67.70556175061616,-35.99289986348005), linewidth(1) + qqwuqq); draw((-1.2491895790459495,2.9269865585347485)--(-1.2274288638331396,2.899284676760246), linewidth(1) + qqwuqq); draw((-1.3253520822907845,3.0239431447455085)--(1.9226657063760997,2.984682738126643), linewidth(1)); draw((-1.4556510411453925,2.2384215723727547)--(-0.3462343719959593,2.054396280120791), linewidth(1)); draw((-2.017738388706867,2.331658217613929)--(-1.4556510411453925,2.2384215723727547), linewidth(1) + linetype("2 2")); draw(arc((-0.3462343719959593,2.054396280120791),0.14090694690732905,170.58177636224778,202.29443824938386), linewidth(1) + qqwuqq); draw((-0.46875040718264416,2.0405713883563306)--(-0.5037549886645539,2.0366214192807703), linewidth(1) + qqwuqq); draw(arc((-0.3462343719959593,2.054396280120791),0.14090694690732905,22.294438249383838,54.00710013651993), linewidth(2) + qqwuqq); draw((-0.2492777857851994,2.1305587833656263)--(-0.22157590401069657,2.152319498578436), linewidth(1) + qqwuqq); draw((-0.8441204848350525,1.8502547611449036)--(-2.017738388706867,2.331658217613929), linewidth(1)); draw((-0.8441204848350525,1.8502547611449036)--(0.11145789966944286,2.6845198408747066), linewidth(1)); draw(arc((0.11145789966944286,2.6845198408747066),0.14090694690732905,-170.5901662312376,-138.87750434410154), linewidth(1) + qqwuqq); draw((-0.00004076942654478582,2.63189519550785)--(-0.03189753202539798,2.6168595825458905), linewidth(1) + qqwuqq); /* dots and labels */ dot((-1.58595,1.4529),dotstyle); label("$C$", (-1.5693111713461532,1.4985171769748604), NE * labelscalefactor); dot((-1.3253520822907845,3.0239431447455085),dotstyle); label("$B$", (-1.3062848704524723,3.0719780841067026), NE * labelscalefactor); dot((0.87858,1.42311),dotstyle); label("$A$", (0.8965603995321052,1.4703357875933947), NE * labelscalefactor); dot((-0.3462343719959593,2.054396280120791),linewidth(4pt) + dotstyle); label("$O$", (-0.33872383502214615,1.9447225088480695), NE * labelscalefactor); dot((-0.36288888737932057,0.6765663775442985),linewidth(4pt) + dotstyle); 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/* end of picture */ [/asy][/asy] Since $BDOC_0$ , $BA_0DO$ are cyclic and $DQ'=DP$ $\implies$ $=\angle A_0OD=\angle A_0BD=\angle DBC_0=\angle C_0OR'=\angle DQ'P=\angle Q'PO=\angle Q'OD$ $\implies$ $\angle A_0OD=\angle Q'OD$ $\implies$ $Q'\in A_0O$. So we are done

Let $N,M_A,$ and $M_C$ be the midpoints of $\overline{BM},\overline{BC},$ and $\overline{AB}.$ Let $D=\Gamma\cap\overline{PO},$ $E=\Gamma\cap\overline{QO},$ and $F=\Gamma\cap\overline{MO}.$ We know $F$ lies on $(ONM_ABM_C)$ as $\angle BFO=90.$ Hence, $\overline{BF}\parallel\overline{AC}$ since $$\measuredangle M_CBF=\measuredangle M_COF=\measuredangle(\overline{PD},\overline{FM})=\measuredangle(\overline{BA},\overline{AC}).$$Also, $\overline{ON}$ is the perpendicular bisector of $\overline{BM}$ so $N,O,$ and $R$ are collinear. Lemma: The perpendicular bisectors of $\overline{BC}$ and $\overline{AB}$ are reflections over $\overline{ON}.$ Proof. Let $\ell_a$ and $\ell_c$ be the perpendicular bisectors of $\overline{BC}$ and $\overline{AB}.$ Notice \begin{align*}\measuredangle(\ell_a,\overline{ON})&=90-\measuredangle(\overline{BM},\ell_a)\\&=\measuredangle CBM\\&=\measuredangle NBA\\&=90-\measuredangle(\overline{AB},\overline{ON})\\&=\measuredangle(\overline{ON},\ell_c).\end{align*}$\blacksquare$ Hence, $DEPQ$ is a cyclic isosceles trapezoid so $R,E,$ and $D$ are collinear. We also see $DEON$ is cyclic for $$180-\angle DNO=\angle(\overline{DN},\overline{ON})=\tfrac{1}{2}\angle QND=\angle QED.$$Radical Axis on $(BNOF),(BFED),$ and $(DONE)$ implies that $R$ lies on $\overline{BF}.$ $\square$

[asy][asy] size(200); pair A = (-1,2); pair B = (-2,-3); pair C = (2,-3); draw(A--B--C--cycle , green); draw(circumcircle(A,B,C) , orange); pair O = circumcenter(A,B,C); pair M = (2.55 , 0.73); draw(B--M , olive); pair D = midpoint(B--M); draw(circumcircle(B,C,M) , fuchsia); pair P = (-2.54 , -0.29); draw(O--P); pair Q = (0 , -3.77); draw(O--Q); pair R = (-4.21 , 1.99); draw(Q--R); draw(circumcircle(B,R,M) , purple); draw(B--R , olive); draw(M--R); draw(circle((0.27 , -1.14),2.94) , orange+dashed+linewidth(0.5)); pair F = midpoint(B--C); pair U = midpoint(A--B); draw(R--M , olive); dot("$F$" , F , SE); dot("$U$" , U , NE); dot("$A$" , A , N); dot("$B$" , B , S); dot("$C$" , C , S); dot("$D$" , D , S); dot("$O$" , O , N); dot("$Q$" , Q , S); dot("$M$" , M , NE); dot("$R$" , R , NW); dot("$P$" , P , NE); [/asy][/asy] Let $D$ be the midpoint of $BM$ and let $F$ and $U$ be the midpoints of the sides $BC$ and $AB$ repectively. Therefore we see that $BUODF$ is cyclic. Now we have that \begin{align*} \frac{DP}{DU} &= \frac{DQ}{DF} \\ \implies \triangle DQF &\sim \triangle DPU \\ \implies QPOD \; &\text{is cyclic} \\ \end{align*}Apply Radical Axis Lemma to finish the problem. $\blacksquare$

Let $X$ be the midpoint of $BM$. Let $N$ be the intersection of $MO$ with $\Gamma$. Since $BM$ is a diameter, $\angle ONB=\angle MNB=90=\angle OXB$. So, $BNOX$ is cyclic. Since $B, N, Q, P$ all lie on $\Gamma$, $BNQP$ is cyclic. Let $O'$ and $Q'$ be the reflections of $O$ and $Q$ around $BM$. Note $O'Q'\parallel OP$. Let $E$ be the midpoint of $Q'P$. Since $X$ is the midpoint of $OO'$ and $O'Q'\parallel OP$, $XE\parallel OP\parallel O'Q'$. Also, since $X$ is the center of $\Gamma$, $\angle XEP=90$. So, $\angle O'\angle OPQ'=90$. Let $U$ be the intersection of $PQ'$ and $BM$. Then, $\angle UQO=\angle UQ'O'=90=\angle UPO=\angle UXO$. So, $UPXOQ$ is cyclic. Since $QPOX$, $BNOX$, and $BNQP$ are all cyclic, by radical axis concurrence, $QP$, $OX$, and $BN$ concur. Assume they concur at a point $V$. Since $V$ is on $OX$, $BV=MV$. Also, $V$ is on $PQ$. So, $V=R$. So, $R$ is on $BN$. Since $OM$ is perpendicular to $AC$ and is also perpendicular to $BN$, $BN\parallel AC$. So, $BR\parallel AC$.

Obviously $K=MO\cap B\infty_{AC}\in$ $\Gamma.$ Denote by $N$ midpoint of $BM$, so we get $N\in \odot (BOK).$ By $$|NP|=|NQ|,\text{ } \measuredangle NOP=\measuredangle MBA=-\measuredangle MBC=-\measuredangle NOQ$$we redefine $N$ as midpoint of arc $POQ.$ Radical center theorem on $\Gamma ,\odot (BOKN),\odot (OPQN)$ gives $R\in BK,$ done.

If we let $L$ be the center of $(BPQ)$ then the main idea is to show that $POLQ$ is cycilc. Denote with $Z$ and $S$ the midpoints of $AB$ and $BC$ respectively. It is not hard to show that $BSLOZ$ is cyclic. Indeed $OZ\perp AB$ and $OS\perp BC$ give that $BSOZ$ is cyclic, and also $LZ || AM$ and $LS||MC$ (since they are midpoints), implies that $\angle SLZ=180^{\circ}-\angle BAC$. SInce $L$ is center we have that $LP=LQ$ and since $BZLS$ is cyclic and $BL$ is the angle bisector of $\angle ZBS$ we have that $LZ=LS$ and $\angle LSQ=180^{\circ}-\angle LSO=180^{\circ}-\angle LZO=\angle LZP$. Now we work with law of sines in triangles $LZP$ and $LSQ$. $\frac{SL}{\sin\angle LQS}=\frac{LQ}{\sin\angle LSQ}=\frac{LP}{\sin\angle LZP}=\frac{ZL}{\sin\angle LPZ}\Rightarrow \sin\angle LQS=\sin\angle LPZ$. One can easily check that points $P$ and $Q$ are always outside triangle $ABC$, forcing $\angle LSQ$ and $\angle LZP$ to be obtuse, hence forcing $\angle LQS$ and $\angle LPZ$ to be acute. Since, $\sin\angle LQS=\sin\angle LPZ$ we have that $\angle LPZ=\angle LQS$ and we have proved that $PQLO$ is cyclic. Now let $BR$ intersect $(BPQ)$ again at $X$ and we have that $RB\cdot RX=RP\cdot RQ=RO\cdot RL$ implying that $BLOX$ is cyclic. $\angle MOL=\angle BOL=\angle BXL=\angle XBL=180^{\circ}-\angle LOX$, implying that $X$, $O$, and $M$ are collinear. Use the fact that $BM$ is diameter and $OM$ is the perpendicular bisector of $AC$ and the conclusion follows. Note: Proving $P$ and $Q$ are outside triangle $ABC$ is simple. Assume wlog that $P$ is inside triangle $ABC$ then by $LP=LB\Rightarrow \angle LBP=\angle LPB=90^{\circ}+\angle ZBP>90^{\circ}$ which leads to contradiction. Also $L$ is always inside triangle $ABC$. It is neccessary to check these cases because there are two possible triangles for an "AngleSideSide" similarity, so we need to make sure the configuration eleminates one of them, and leaves us with the proper triangle.

The perpendicular bisector of $BM$ bisects the perpendicular bisectors of $BA, BC$ at $O$. So extend $OP, OQ$ to intersect $\Gamma$ again at $OP', OQ'$. So by Brokard's theorem on $PQ'QP'$, $R$ is the inverse of $O$ in $\Gamma$ so $BR \perp OM \perp AC$, finishing.

Absolutely amazing problem. Let $O_1$ be the center of $\Gamma$. Then notice that $OO_1$ is the external angle bisector of $\angle POQ$ which actually means that $P,O,Q,O_1$ are concyclic. Let $BR$ intersect $\Gamma$ at $K$. Since $$RK\cdot RB=RP\cdot RQ=RO\cdot RO_1$$it follows that $OKBO_1$ is cyclic. Therefore $\angle MKB=\angle OKB=90^{\circ}$ which obviously means that $M,O,K$ are collinear. At this point both $BR$ and $AC$ are perpendicular to $OM$ and we are done. $\blacksquare$

Let $BO$ and $MO$ intersect $\Gamma$ at $D$ and $E$ respectively. Let $F$ be the midpoint of $BM$. Note that $OB=OM$ so $OF\perp BM$. Further since $BM$ is diameter, $\angle BEM=90^\circ$ so $BEOF$ is cyclic. Since $FP=FQ$ and $\angle FOQ=180^\circ-\angle POF$, $QPOF$ is cyclic. Since $BEPQ$ is also cyclic, $BE,PQ,OF$ concur at $R$. Thus, $BR\parallel BE\perp BM \perp AC$ and we're done.

??????? We first prove the following lemma. Lemma: Let $\ell$ be a diameter of a circle $\omega$ and let $P \in \ell$. Let $A,B$ lie on $\omega$ such that lines $\overline{PA}$ and $\overline{PB}$ are symmetric with respect to $\ell$. Then $\overline{AB}$ intersects $\ell$ at the inverse of $P$ with respect to $\omega$. Proof: Let $A',B'$ be the reflections of $A,B$ over $\ell$, so $A,B',P$ and $A',B,P$ are collinear. Then by Brokard on $AA'BB'$, $\overline{AB} \cap \overline{A'B'}$ is the pole of $P\infty_{\perp \ell}$. This clearly lies on $\ell$ as well so we're done. $\blacksquare$ Let $N$ be the center of $\Gamma$ and the midpoint of $\overline{AM}$. Let $X$ be the point on $\overline{MO}$ such that $\overline{BX} \parallel \overline{AC}$, so $X$ lies on $\Gamma$. Let $D$ and $E$ be the midpoints of $\overline{AB}$ and $\overline{BC}$. It suffices to show that $\overline{BX},\overline{NO},\overline{PQ}$ concur. Since $\angle BDO=\angle BEO=\angle BNO=\angle BXO=90^\circ$, $BXDENO$ is cyclic. Obviously we have $NB=NX$, and by a homothety at $B$ we have $ND=NE$. Thus $\overline{ON}$ is a bisector of both $\angle DOE$ and $\angle BOX$. Thus from our lemma, the desired concurrency point is the inverse of $O$ with respect to $\Gamma$. $\blacksquare$

My solution is same as @above's

Let $K$ be the center of $\Gamma$ (so it is the midpoint of $BM$). The heart of this problem is the following: Claim: $P,O,K,Q$ are concyclic. Proof: Since $OK \perp BM$ ($OK$ is the perpendicular bisector), we see that $\angle (AB, OK) = \angle (BC, OK).$ But $\angle QOK = 90^\circ - \angle(BC, OK)$ and $\angle POK = 90^\circ + \angle(AB, OK).$ Thus $\angle QOK + \angle POK = 180^\circ,$ so $OK$ is the external bisector of $\angle POQ.$ But $PK = KQ,$ so this evidently implies that the four points are concyclic, as claimed. Say these four points are concyclic on circle $\omega.$ Now construct $\gamma,$ the circle with diameter $BO$ (which $K$ lies on). The radical axis of $\Gamma$ and $\gamma$ is perpendicular to the line connecting their centers, which by homothety is parallel to $OM.$ Thus the radical axis of these two circles is parallel to $BC.$ The problem then follows by applying the Radical Center Theorem on $\Gamma, \omega, \gamma.$

With some angle chasing, showing $\overline{BR} \parallel \overline{AC}$ reduces to showing $\angle OBM = \angle ORB$. Let $D$ be the midpoint of $\overline{BM}$, and extend $OP$ and $OQ$ to meet $(BM)$ again at $P'$ and $Q'$, respectively. Since $QP'Q'P$ is an isoscels trapezoid with $\overline{QP'} \parallel \overline{Q'P}$, it follows that $O$ and $R$ are polars WRT $(BM)$. Therefore, $\triangle RBD \sim \triangle BOD$ and $\angle OBM = \angle ORB$ follows.

Pretty standard problem. Felt rather contrived and not natural. We denote by $B'$ the intersection of the line through $B$ parallel to $\overline{AC}$ with $\Gamma$ and by $O_1$ the center of $\Gamma$. The entirety of the problem, is the following two concyclicity claims. Claim : Points $B$ , $B'$ , $O$ and $O_1$ are concyclic. Proof : Note that since $OM \perp AC$ and $BB' \parallel AC$ , $OM \perp BB'$ which implies that points $B'$ , $O$ and $M$ are collinear. Now, it is clear that, \[\measuredangle OB'B = 90^\circ = \measuredangle OO_1B \]from which the claim follows. Claim : Points $P$ , $Q$ , $O$ and $O_1$ are concyclic. Proof : First note that $O_1$ lies on the external $\angle POQ-$bisector since, \[\measuredangle QOP = \measuredangle CBA = 2\measuredangle O_1BA = 2 \measuredangle O_1OP\]Further, since $O_1$ is the center of $\Gamma$, it lies on the perpendicular bisector of segment $PQ$. Thus, by the Incenter-Excenter Lemma, it must be the major $PQ$ arc midpoint in $(POQ)$ and thus must lie on this circle, proving the claim. To finish note that by the Radical Center Theorem on circles $(BB'OO_1)$ , $(PQOO_1)$ and $\Gamma$, it follows that their pairwise Radical axes, $\overline{BB'}$ , $\overline{PQ}$ and $\overline{OO_1}$ concur. Now, the point $R$ is the point on line $\overline{PQ}$ which lies on the perpendicular bisector of segment $BM$ (which is simply line $\overline{OO_1})$. Thus, $R$ is the above concurrence point and it indeed lies on the line through $B$ parallel to $\overline{AC}$, as desired.

ok if we go one step at a time we are cool Let the angle bisectors of $\angle AOB$ and $\angle BOC$ meet $BC$ and $CA$ in $X$ and $Y$. Let $N$ be the midpoint of $BM$ and let $BR$ meet $\Gamma$ again in $Z$. CLAIM 1: $O,X,B,Y,N$ are concyclic with $XN=XY$ PROOF: Indeed, $\angle BXO=\angle BYO=\angle BNO=90$. Also, $\angle XNB=\frac{\angle B}{2}=\angle YBN \implies XN=YN$ CLAIM 2: $O,N,P,Q$ are concyclic PROOF: Note that $\angle NYQ=180-\angle OYN=180-\angle OXN=\angle NPX$, $NY=NX$ and $NP=NQ$. We see that $\triangle NYQ \equiv \triangle NXP$, so $N$ is the Miquel point of $PXYQ$ and $PONQ$ is cyclic. To finish the problem, note that $RO \cdot RN=RP \cdot RQ=RZ \cdot RB$, so $ZONB$ is cyclic, so $\angle OZB=90=\angle MZB \implies Z-O-M$. Finally, $ZM \perp AC $ and $ZM \perp BR$, so $BR \parallel AC$ and we are done. $\blacksquare$