This is the inverted(at $A$ by $\sqrt{bc}$ then reflected by bisector) statement. Note that $P$ is the intersection of the circle with diameter $AC$ and the circle which passes through $B,C$ and tangent to $AC$. Issac Newton doesn't wrote: Let $ABC$ be a triangle with circumcircle $\omega$ and $S$ be the antipode of $A$. $BR,CR$ intersect the line perpendicular to $BC$ at $C,B$ at $Q,P$ respectively. Let $R$ be the intersection of $\odot(ACQ)$ and $\odot(ABP)$. Prove that center of $\odot(APQ)$ lies on $AR$ Claim 1: $\angle PAB = \angle CAQ$.

Let the center of $\odot(ACQ)$ and $\odot(ABP)$ be $O_1,O_2$ respectively. Claim 2: $O_1,O_2,S$ are collinear. Proof: Let $M_1,M_2,$ be the midpoint of $CQ,BP$ respectively. Obviously, $M_1,M_2,S$ are collinear. By Claim 1, $\triangle BPO_2 \sim \triangle CQO_1$ are similarly. $\frac{O_1M_1}{O_2M_2} = \frac{CQ}{BP} = \frac{M_1S}{M_2S}$. The claim is proven. Since $\measuredangle PRQ = \measuredangle PRA + \measuredangle ARQ = \measuredangle PBA + \measuredangle ACQ = \measuredangle CAB = \measuredangle PSQ$, $P,Q,R,S$ are concyclic. From Claim2, we get $AS=SR$ implies $\measuredangle SAR = \measuredangle ARS$. The rest part is angle chasing. \begin{align*} \measuredangle PAR &= \measuredangle PAS + \measuredangle SAR\\ &=\measuredangle PAB + \measuredangle BAS + \measuredangle ARS\\ &=\measuredangle CAQ + \measuredangle BAS + \measuredangle ARP + \measuredangle PRS\\ &=\measuredangle CAQ + \measuredangle QCA + \measuredangle ABP + \measuredangle PQS\\ &=\measuredangle CQA + \measuredangle ABQ + \measuredangle BQC + \measuredangle PQS\\ &=\measuredangle BQA + \measuredangle PQS + \measuredangle ABQ\\ &=\measuredangle PQA + 90^\circ

Define $Q'$ as the isogonal conjugate of $P$. Then a bit of angle chasing gives that $Q' = Q$, (since $Q$ is unique) hence $P,Q$ are isogonal conjugates. Let $S = BQ \cap CP$. Note that $PAQS$ is cyclic and $R,S$ are isogonal conjugates. Now angle chasing shows that $AR \perp PQ$: $$\angle APQ + \angle PAR = \angle ASQ + \angle SAQ = 90^\circ$$.

Let $D$ be the foot from $A$ to $BC$, then $\angle{ADC}=\angle{ADB}=90^o$. and $APDC$ and $AQDB$ are cyclic quadrilaterals. Since $\angle{ACP}=\angle{PBC}$ and $\angle{PAC}=\angle{PDB}$ because of the cyclic, $\triangle{APC}$ and $\triangle{DPB}$ are similar, and therefore there is a spiral of similarity with center at $P$ sending $AD$ to $CB$. Similarly, there is a spiral of similarity with center $Q$ sending $AD$ to $BC$. Also note that because of the similarities we know, $\angle{DPC}=\angle{DQC}=90^o$ and so $PQRD$ is cyclic. Claim 1: $\angle{QCB}=\angle{PBC}$ Proof: Let's consider $M$ and $N$ the midpoints of $BC$ and $AD$ respectively. We have that the similarity with center P will send $ND$ to $MB$ so $\angle{PND}=\angle{PMB}$, analogously, $\angle{QND}=\angle{QMC}$ so: $$\angle{PMQ} =180^o-\angle{QMC}-\angle{PMB}\\ =180^o-\angle{QND}-\angle{PND}\\ =180^o-\angle{QNP}$$ Therefore, $QMPN$ is cyclic. But $\angle{NDM}=90^o$ and because of the similarities, $\angle{NQM}=\angle{NPM}=90^o$, so $D$ lies in the circumcircle of $QMNP$ and finally $PQRMDN$ is cyclic. Specially, $\angle{RMD}=90^o$ so $R$ lies in the perpendicular bisector of $BC$ thus, $\angle{QCB}=\angle{PBC}$ which is what we wanted. Claim 2: $NM$ is perpendicular to $PQ$ Proof: $\angle{NMQ}=\angle{ADQ}=\angle{ABQ}=\angle{QCB}=\angle{PBC}=\angle{ACP}=\angle{NDP}=\angle{NMP}$ this implies that $NM$ is angle bisector of $\angle{PMQ}$, but because $NM$ is also a diameter in the circumcircle of $PQRMDN$, $P$ and $Q$ are symmetric respect to the line $NM$, and therefore, $NM$ is perpendicular to $PQ$, what we wanted. Finally, we note $RMDN$ is a rectangle, this is because $RMDN$ is cyclic and $\angle{RMD}=\angle{NDM}=90^o$. So $RM$ is parallel to $AD$ and $RM=ND=NA$ therefore, $ARMN$ is a parallelogram. And because $NM$ is perpendicular to $PQ$, $PQ$ will be perpendicular to $AR$ as well, which is what we wanted

Let $D$ be the foot of the altitude from $A$ to $BC$. Claim: $AD$ is tangent to both $(DPB)$ and $(DQC)$. Proof: Note that $D$ lies on $(ACP),(ABQ)$. Hence \[\measuredangle PDA=\measuredangle PCA=\measuredangle PBC=\measuredangle PBD\]\[\measuredangle ADQ=\measuredangle ABQ=\measuredangle BCQ=\measuredangle DCQ\]These yield $(DPB),(DQC)$ are tangent to $AD$.$\square$ Claim: $D,P,Q,R$ are concyclic. Proof: \[\measuredangle RPD=180-\measuredangle DPB=\measuredangle PBD+\measuredangle BDP=\measuredangle PDA+\measuredangle BDP=90\]\[\measuredangle DQR=180-\measuredangle CQD=\measuredangle QDC+\measuredangle DCQ=\measuredangle QDC+\measuredangle ADQ=90\]Thus, $D,P,Q,R$ lie on the circle with diameter $DR$.$\square$ Now take the inversion centered at $D$ with any radius. New Problem Statement: $PBCQ$ is a trapezoid with $\measuredangle PBC=90=\measuredangle BCQ$. $D,R$ are the feet of the altitudes from $A,D$ to $BC,PQ$ respectively. $(DPQ)$ intersects $BC$ at $X$. Prove that $AX,(DAR),(DPQ)$ are concurrent. Proof:Let $K=(ADR)\cap (DPQ), \ QD\cap BR=S$. Since $(PQ\cap BC,D;B,C)=-1$ and $\measuredangle (BC,DA)=90,$ we have $\measuredangle PDA=\measuredangle ADQ$. Denote $D'$ as the reflection of $D$ with respect to $A$ which lies on $PQ$ because of homothety centered at $PQ\cap BC$. Since $\measuredangle D'RD=90$ and $A$ is the midpoint of $DD',$ we observe that $AD=AD'=AR$. Also $(RBDP)$ and $(PDQX)$ are concyclic which gives that $BR\parallel QX$. \[\measuredangle AKD=\measuredangle ARD=\measuredangle RDA=\measuredangle RDP+\measuredangle PDA=\measuredangle RDP+\measuredangle ADQ=\measuredangle RBP+\measuredangle ADQ=\measuredangle RSQ=\measuredangle XQS\]\[\measuredangle XQS=\measuredangle XQD=\measuredangle XKD\]Subsequently $\measuredangle AKD=\measuredangle XKD$ which implies $A,X,K$ are collinear as desired.$\blacksquare$