[asy][asy] defaultpen(fontsize(9pt)); size(9cm); pair A,B,C,K,O,X,M,P,V,Q,U,R; A=dir(120); B=dir(210); P=dir(135); C=dir(330); K=-(circumcenter(A,B,C))+2*foot((circumcenter(B,circumcenter(A,B,C),C)),circumcenter(A,B,C),(circumcenter(B,circumcenter(A,B,C),C))); O=circumcenter(A,B,C); X=-A+2*foot(O,A,K); M=(A+X)/2; V=extension(B,P,X,A); Q=-C+2*foot(O,C,V); U=extension(B,Q,C,P); R=-P+2*foot(O,P,M); draw(A--B--C--cycle,blue); draw(circumcircle(A,B,C)); draw(A--X); draw(O--M); draw(B--M--C); draw(M--P); draw(B--V--A); draw(C--V); draw(M--Q); draw(B--Q); draw(C--P); draw(circumcircle(B,M,P),dotted+red); draw(circumcircle(C,M,Q),dotted+red); draw(M--R); draw(Q--R,dotted); draw(B--K--C); draw(K--X); draw(circumcircle(B,O,C),dotted); dot("$A$" , A , dir(65)); dot("$B$" , B , dir(B)); dot("$C$" , C , dir(C)); dot("$X$" , X , dir(240)); dot("$M$" , M , dir(160)); dot("$O$" , O , dir(300)); dot("$P$" , P , dir(P)); dot("$V$" , V , dir(90)); dot("$Q$" , Q , dir(75)); dot("$U$" , U , dir(10)); dot("$R$" , R , dir(270)); dot("$K$" , K , dir(270)); [/asy][/asy] The problem is valid even for any two Isogonal Pairs $\{\overline{MP},\overline{MQ}\}$ WRT $\Delta BMC$. It is well known that $M$ is the midpoint of the $A-$ Symmedian Point or the $A-$ Dumpty Point of $\Delta ABC$. ( See Definition 3 from the Section 2.2 here.) Let $O$ be the circumcenter of $\Delta ABC$. So, $\overline{OM}\perp\overline{AM}$. If $K$ is the point where the tangents at $B,C$ to $\odot(ABC)$ meet then $\angle OMK=\angle OBK=\angle OCK\implies M\in \odot(ABC)$ and $\angle BMK=\angle CMK$ as $KB=KC$. Let $\overline{CP}\cap\overline{AM}=\{U\}$. Then $$\angle BPU=\angle BAC=\frac{\angle BOC}{2}=\frac{\angle BMC}{2}=\pi-\angle UMB\implies P,U,M,B \text{ are concyclic.}$$Let $\overline{BU}\cap\odot(ABC)=\{Q^*\}$ and $\overline{PM}\cap\odot(ABC)=\{R\}$. Then, $\angle UMP=\angle Q^*BP=\angle Q^*RP\implies\overline{RQ^*}\parallel\overline{AM}\perp\overline{OM}$. So, $MQ^*=MR\implies\angle OMR=\angle OMQ^*\implies\angle KMR=\angle Q^*MA=\angle PMA\implies Q^*\equiv Q$. Hence, we get that $\overline{BQ},\overline{CP},\overline{AM}$ are concurrent. Now again, $$\angle UQC=\angle BAC=\frac{\angle BOC}{2}=\frac{\angle BMC}{2}=\angle KMC\implies Q,U,M,C\text{ are concyclic.}$$Now taking pairwise radical axis and applying radical axis theorem on $\odot(ABC),\odot(BMUP)$ and $\odot(CMUQ)$ we get that $\overline{BP},\overline{AM},\overline{CQ}$ are concurrent. $\blacksquare$

[asy][asy] size(10cm); import geometry; pair A = dir(125), B = dir(205), C = dir(-25), O1 = extension(A, rotate(90, A)*C, (A+B)/2, rotate(90, (A+B)/2)*A), O2 = extension(A, rotate(90, A)*B, (A+C)/2, rotate(90, (A+C)/2)*A); pair M = intersectionpoints(circle(O1, abs(O1-A)), circle(O2, abs(O2-A)))[1], P = intersectionpoints(line(M, rotate(90, M)*B), unitcircle)[0], Q = intersectionpoints(line(M, rotate(90, M)*C), unitcircle)[0], N = extension(B, Q, P, C), R = extension(B, P, C, Q), O = (0, 0); //draw(circle(O1, abs(O1-A)), lightblue); draw(circle(O2, abs(O2-A)), lightblue); draw(unitcircle, royalblue); draw(circumcircle(B, M, N), purple); draw(circumcircle(C, M, N), purple); draw(arc(circumcenter(B, O, C), circumradius(B, O, C), 10, 170), fuchsia); draw(A--B--C--cycle, orange); draw(B--Q^^C--P, lightmagenta); draw(P--M--B, deepgreen+linewidth(1.2)); draw(Q--M--C, deepgreen+linewidth(1.2)); draw(B--R--C, blue); draw(M--R, dashed); string[] names = {"$A$", "$B$", "$C$", "$M$", "$N$", "$O$", "$P$", "$Q$", "$R$"}; pair[] points = {A, B, C, M, N, O, P, Q, R}; pair[] ll = {dir(170), dir(260), dir(-10), dir(165), dir(20), dir(-90), dir(135), dir(85), R}; for (int i=0; i<names.length; ++i) {dot(names[i], points[i], dir(ll[i]));} [/asy][/asy] Claim: $BMOC$ is cyclic. Proof. We have \begin{align*} \angle AMP &= \angle AMB - 90^\circ \\ &= (180^\circ - \angle MBA - \angle BAM) - 90^\circ \\ &= 90^\circ- \angle MAC - \angle BAM \\ &= 90^\circ - \angle BAC, \end{align*}and similarly $\angle QMA = 90^\circ - \angle BAC$, so $\angle QMP = 180^\circ- 2\angle BAC$. Therefore, \[\angle BMC = 2\angle BAC=\angle BOC,\]indicating that $BMOC$ is cyclic as claimed. Claim: $M$ is the Miquel point of quadrilateral $BPCQ$. Proof. We proceed indirectly. Let $M'$ be the Miquel point of $BPCQ$. We will show that $M \equiv M'$. Since $BPCQ$ is cyclic, it is well known that $M'$ lies on the circle $(BOC)$. Moreover, since $M'$ lies inside the circumcircle of $BPCQ$, it is located somewhere on the arc $\widehat{BC}$ of $(BOC)$. Furthermore, since it is the center of spiral similarity sending $\overline{BP}$ to $\overline{QC}$, it must satisfy $\angle PM'B = \angle CM'Q$. Suppose that $M'$ lies in the interior of arc $\widehat{BM}$ of $(BOC)$. Since $M'$ is inside $(PMB)$ (which has diameter $\overline{PB}$), $\angle PM'B < 90^\circ$. Since $M'$ is outside $(CMQ)$ (which has diameter $\overline{CQ}$), $\angle CM'Q > 90^\circ$. However, this would cause $\angle PM'B \neq \angle CM'Q$, which is a contradiction. Thus, $M'$ cannot lie in the interior of arc $\widehat{BM}$ of $(BOC)$. Similarly, $M'$ cannot lie in the interior of arc $\widehat{CM}$. It follows that $M'$ must coincide with $M$, as desired. By the above claim it follows, by definition of the Miquel point, that $BPNM$ and $CQNM$ are cyclic. Claim: $M, N, A$ are collinear. Proof. We have seen already that $\angle QMA = \angle AMP = 90^\circ - \angle BAC$. So it suffices to show that $\overline{NM}$ also bisects $\angle QMP$. Indeed, \[ \angle NMP = \angle NBP = \angle QCN = \angle QMN, \]as needed. By radical axis on $(BPNM)$, $(CQNM)$, and $(BPQC)$, we obtain the concurrence of $\overline{BP}$, $\overline{CQ}$, and $\overline{NM}$ (the last line of which contains $A$), as desired.

Cute question. [asy][asy] size(13cm); defaultpen(fontsize(11pt)); import olympiad; pair A = dir(120), B = dir(220), C = dir(-40), O = (0,0); pair K = 2*foot(O,A,2*circumcenter(O,B,C)) - A; pair M = (A+K)/2; pair P = dir(146); pair Z = extension(A,M,C,P); pair Q = 2*foot(O,B,Z) - B; draw(A--B--C--cycle, royalblue); draw(unitcircle, fuchsia); draw(A--K, fuchsia); draw(C--P, orange); draw(B--Q, dashed + orange); draw(Q--M--P, brown); draw(B--M--C, brown); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$M$", M, dir(20)); dot("$K$", K, dir(K)); dot("$T$", Z, dir(20)); [/asy][/asy] Choose $(KA; BC) = -1$ and let $CP$ intersect $AM$ at $T$. Let $QT$ intersect $(ABC)$ again at $B'$. Note that $M$, the dumpty point, is known to be the midpoint of $AK$. Observe that since $\angle PMB = \angle PMC$, $MA$ is the angle bisector of $PMQ$, and we have that $(AK; PQ) = -1$. $$(AK; CB) = -1 = (KA; PQ) \stackrel{T}{=} (AK; CB')$$implies that $QTB$ are collinear, which is what we wanted to show. $\blacksquare$ Remark: We never need $\angle PMB = 90$.

NJOY wrote: Let $APBCQ$ be a cyclic pentagon. A point $M$ inside triangle $ABC$ is such that $\angle MAB = \angle MCA$, $\angle MAC = \angle MBA$ and $\angle PMB = \angle QMC = 90^\circ$. Prove that $AM$, $BP$, and $CQ$ concur. Proposed together with Navilarekallu Tejaswi

let $N=CP \cap BQ$ and $R=BP \cap QC$ note that $M$ is the $A-$Dumpty point so $AM$ is symmedian note that $MO , BC , AA$ are concurrent so $MA$ is angle bisector of $BMC$ claim:$M$ is the miquel point of $BPCQ$ note that $MB.MC=MA^2$ so invert around $M$ with radius $MA$ and reflect across $AM$ note that this fixes $(ABC)$ so $P \longrightarrow Q$ so $MP.MQ=MA^2=MB.MC \implies \triangle MBP \sim \triangle MQC $ $\blacksquare$ now since $\{MB,MC\} \& \{MP,MQ\}$ are pairs of conjugates so $\{MN,MR\}$ are isogonal but $M-N-R$ are collinear since they on the radical axis of $(PMC) , (QMB)$ so $MR$ is the angle bisector of $\angle BMC$ so $M-N-A-R$ are collinear and we win

We note that $M$ is not any random point, but the $A -$Dumpty Point . Therefore, we get that $M\in AX$ (where $X$ is the intersection of tangents to $ABC$ at $B$ and $C)$ and $M\in (OBCX).$ Therefore $\angle XBC = \angle XMC$ and $\angle XCB = \angle XMC.$ Therefore, we can say that $\overline{AM}$ bisects $\angle BMC.$ $\newline$ Lemma: Consider two circles $\omega_1$ and $\omega_2$. Let $\overline{MI}$ be their radical axes. Construct $P$ and $Q$ on $\omega_1$ and $\omega_2$, respectively; such that $\angle PMC =\angle QMB = 90^{\circ}$ and there exists a non-degenerate circle passing through points $B, P, Q, C$.Then, the following are equivalent: $\newline$ $(1)$ ${MI}$ bisects $\angle BMC.$ $\newline$ $(2)$ ${BQ}$ and ${CP}$ pass through $I$ $\newline \underline{\textbf{Proof:}}$We assume that ${MI}$ bisects $\angle BMC$. Hence, $$\angle QIC = \angle QMC = 90^{\circ} -\angle PMQ = \angle BMP =\angle BIP$$. As $\angle QIC$ and $\angle BIP$ are vertically opposite, we conclude that ${BQ}$ and ${CP}$ pass through $I$. $\newline$ Now we flip the condition. Hence, $$\angle PMI =\angle PBI =\angle PCQ =\angle ICQ = \angle IMQ.$$and $\angle BMP =\angle QMC.$ Hence we conclude that ${MI}$ bisects $\angle BMC.$ $\newline \rule{\textwidth}{0.5pt} \newline$ Now, just apply this lemma to the main figure with refrence triangle as $MBC$ and hence $BP', CQ'$ and $M-$angle bisector of $\angle BMC,$ where $(P'= MP \cap (ABC)$ and $Q' = MQ \cap ABC)$ concur at say $I.$ Now apply Pascal's theorem on hexagon $BP'PCQ'Q,$ which means that $BP' \cap CQ'= I , P'P \cap Q'Q = M $ and $CP \cap BQ ($say $I'$) are collinear. Hence $AM, BQ, CP$ concur. By Isogonal lines lemma, on pairs $(MP , MQ), (MB, MC)$ we get $BP \cap CQ$ and $I'$ are isogonal; but $I' \in AM$ which is the angle bisector of $\angle BMC.$ Therefore, $BP, CQ, AM$ concur.

[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 = -21.746537070883942, xmax = 18.69832730969797, ymin = -23.631067886310635, ymax = 10.841831550365722; /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */draw((-7.458892967532918,4.374655659216323)--(-8.77469001886136,-6.191593389330231), linewidth(0.8)); draw((-8.77469001886136,-6.191593389330231)--(7.134492510837027,-6.829555596034926), linewidth(0.8) + blue); draw((7.134492510837027,-6.829555596034926)--(-7.458892967532918,4.374655659216323), linewidth(0.8)); draw(circle((-0.6328249701472005,-1.8404345075499673), 9.231606040444344), linewidth(0.8)); draw(circle((-0.9981201321161531,-10.949982609150762), 9.116869373350514), linewidth(0.8)); draw((-8.77469001886136,-6.191593389330231)--(-5.599362812090376,-3.0794125554176643), linewidth(0.8)); draw((-5.599362812090376,-3.0794125554176643)--(7.134492510837027,-6.829555596034926), linewidth(0.8)); draw((-7.458892967532918,4.374655659216323)--(-1.3634152940851074,-20.059530710751567), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-8.77469001886136,-6.191593389330231)--(-1.3634152940851074,-20.059530710751567), linewidth(0.8)); draw((-1.3634152940851074,-20.059530710751567)--(7.134492510837027,-6.829555596034926), linewidth(0.8)); draw((-9.460586038983477,0.8601551683369683)--(-5.599362812090376,-3.0794125554176643), linewidth(0.8)); draw((-2.576713612699046,7.184189973907777)--(-5.599362812090376,-3.0794125554176643), linewidth(0.8)); draw((-16.786860968084987,-5.870303326203471)--(-2.576713612699046,7.184189973907777), linewidth(0.8) + linetype("4 4") + blue); draw((-16.786860968084987,-5.870303326203471)--(-0.6328249701472002,-1.8404345075499673), linewidth(0.8) + linetype("4 4") + blue); draw((-16.786860968084987,-5.870303326203471)--(-8.77469001886136,-6.191593389330231), linewidth(0.8) + blue); draw((-8.77469001886136,-6.191593389330231)--(-2.576713612699046,7.184189973907777), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-9.460586038983477,0.8601551683369683)--(7.134492510837027,-6.829555596034926), linewidth(0.8) + linetype("4 4") + qqwuqq); /* dots and labels */dot((-7.458892967532918,4.374655659216323),dotstyle); label("$A$", (-8.586758032816618,4.941387144710537), NE * labelscalefactor); dot((-8.77469001886136,-6.191593389330231),dotstyle); label("$B$", (-10.088689336074301,-7.145583819601299), NE * labelscalefactor); dot((7.134492510837027,-6.829555596034926),dotstyle); label("$C$", (7.970246572143085,-6.859501666599835), NE * labelscalefactor); dot((-1.3634152940851074,-20.059530710751567),linewidth(4pt) + dotstyle); label("$T$", (-1.613505553405944,-21.52121200792484), NE * labelscalefactor); dot((-3.7398326566478333,-10.533480770051652),linewidth(4pt) + dotstyle); label("$D$", (-3.1869573949139935,-10.185206695241849), NE * labelscalefactor); dot((-5.599362812090376,-3.0794125554176643),linewidth(4pt) + dotstyle); label("$M$", (-6.083539194053812,-4.284762289586663), NE * labelscalefactor); dot((-0.6328249701472002,-1.8404345075499673),linewidth(4pt) + dotstyle); label("$O$", (-0.8267796326519191,-2.7828309863289795), NE * labelscalefactor); dot((-9.460586038983477,0.8601551683369683),linewidth(4pt) + dotstyle); label("$P$", (-10.374771489075766,0.9004767335648639), NE * labelscalefactor); dot((-2.576713612699046,7.184189973907777),linewidth(4pt) + dotstyle); label("$Q$", (-3.2942382022895424,7.6234073290992574), NE * labelscalefactor); dot((-16.786860968084987,-5.870303326203471),linewidth(4pt) + dotstyle); label("$E$", (-17.49106504498717,-5.607892247218432), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] It is well-known that $M$ is the $A$-dumpty point of $\triangle ABC$. Hence $M$ lies on the $A$-symmedian. Suppose $AM$ intersect $(ABC)$ again at $D$. Suppose the $P$-symmedian of $\triangle APD$ intersect $(ABC)$ again at $Q'$. Then $M$ is the midpoint of the symmedian chord in $\triangle APQ'$, hence it is the $A$-dumpty point of $\triangle APQ'$. Therefore, $\angle AMP=\angle AMQ'$. On the other hand, we have $\angle AMB=\angle AMC$, hence $$\angle AMP=\angle AMB-90^{\circ}=\angle AMC-90^{\circ}=\angle AMQ$$Hence $Q=Q'$ since they are both the unique point on minor arc $AC$ satisfying $\angle AMP=\angle AMQ$ Hence $M$ is the $A$-dumpty point of $\triangle APQ$, so $PQMO$ is cyclic. Applying radical axis theorem to $(PQMO),(BMOC),(ABC)$ we have $PQ,MO,BC$ are concurrent at a point $E$. By a well-known fact, $BC$ is the $C$-symmedian in $\triangle ADC$. Therefore, $E=MO\cap BC$ is the intersection of tangent at $A$ and $E$, hence $AD$ is the polar of $E$ w.r.t. $(ABC)$. \newline \newline By Brokard's theorem $BP\cap CQ$ lies on the polar of $E$, that is $AM$ as desired.

Let $O$ be circumcentre of $ABC$. Invert wrt $M$ with power $-MA^2$ and reflect everything in line $OM$. $(P,Q),(A,A)$ and $(B,C)$ are swapped, hence they are pairs of involution. So tangent from $A$, lines $PQ$ and $BC$ concur, say at $X$.Let $T=PC \cap QB$. So $AT$ is the polar of $X$, and so is $AM$ (bcoz $AO\perp AM$ at $M$). We are done.

We only need that $\angle APM = \angle AMQ$. Let $AD \cap \odot(ABC) = K$. $M$ is just the $A$-dumpy point of $\triangle ABC$, which is also midpoint of segment $AK$. Our problem is clearly equivalent to showing $$ (A,K ; P,Q) = (K,A ; B,C) $$As $(K,A ; B,C)= -1$ (another property of dumpy point), thus our problem is equivalent to $$ (A,K ; P,Q) = -1 $$Recall $\angle APM = \angle AMQ$. But this is just a well known configuration in EGMO (converse of part (g), Lemma 4.26).