Easy man. Drop perpendicular to $AC$ from $R$ to cut it at Y. Produce $YR$ to meet $AC$ at $X$. We have to prove $A,R,X,B$ are concyclic. By angle chase and noting that triangle $XCB$ is isosceles, we get $\angle AXB=2*\angle C=\angle ARB$. So, they are concyclic. We can also do this the straight way by producing $YR$ to $X'$ and show that $X=X'$.

$\angle AXB=\angle ARB=2\angle ACB\implies\angle CBX=\angle BCX$, done. Best regards, sunken rock

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -3.52066848231, xmax = 8.63034242359, ymin = -1.71772531254, ymax = 4.13589211518; /* image dimensions */ pen qqwuqq = rgb(0.,0.392156862745,0.); draw(arc((2.01126809201,2.76332957195),0.316982893197,-147.856657128,-134.307084653)--(2.01126809201,2.76332957195)--cycle, qqwuqq); draw(arc((2.62,0.14),0.316982893197,146.885565468,160.435137943)--(2.62,0.14)--cycle, qqwuqq); draw(arc((-2.1,0.18),0.316982893197,18.5937703973,32.1433428719)--(-2.1,0.18)--cycle, qqwuqq); /* draw figures */ draw((2.01126809201,2.76332957195)--(2.62,0.14)); draw((2.62,0.14)--(-2.1,0.18)); draw((-2.1,0.18)--(2.01126809201,2.76332957195)); draw(circle((0.266917536204,0.976269272091), 2.49726718171)); draw(circle((1.71116417062,1.31140025174), 1.48261967967)); draw((xmin, 118.*xmin-30.52)--(xmax, 118.*xmax-30.52)); /* line */ draw((0.272804746579,1.67096009628)--(2.62,0.14)); draw((2.62,0.14)--(0.266917536204,0.976269272091)); draw((0.266917536204,0.976269272091)--(2.01126809201,2.76332957195)); draw((0.266917536204,0.976269272091)--(-2.1,0.18)); /* dots and labels */ dot((2.01126809201,2.76332957195),dotstyle); label("$A$", (2.05823043796,2.82569615663), NE * labelscalefactor); dot((2.62,0.14),dotstyle); label("$B$", (2.66049793504,0.205304239528), NE * labelscalefactor); dot((-2.1,0.18),dotstyle); label("$C$", (-2.0625471736,0.247568625288), NE * labelscalefactor); dot((0.266917536204,0.976269272091),dotstyle); label("$R$", (0.304258428937,1.04002585828), NE * labelscalefactor); dot((0.272804746579,1.67096009628),dotstyle); label("$X$", (0.314824525377,1.73738822332), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $ABRX$ is cyclic so $\angle XBR=\angle XAR=\angle XCR$ thus $\angle XCB=\angle XBC$ and $RX$ is the perpendicular bisector of $BC$.

Lemma(not really a lemma as it's too simple) $\longrightarrow$ In a $triangle$ $ABC$ the angle made by $BC$ at the circumcentre is $twice$ $angleA$. Proof-- Just chase some angles. Using this the problem is dead. Let $\angle RBA$$=x$. As $R$ is the circumcentre $RA=RB$.$\longrightarrow$$\angle RAB=x$.$\longrightarrow$$\angle ARB=180-2x$.. By the lemma we have $\angle C=90-x$ Also $\angle AXB$$=$$\angle ARB$(angles in the same segment) This implies $\angle XBC=180-2x=$$\angle C$ which further implies that $triangle BCX$ is isosceles from which the result follows.

Here, we have a triangle $ABC$ with circumcenter $O.$ Given that $\odot AOB$ intersect the side $AC$ of $\Delta ABC$ again at $X.$ Let the line $RX$ produced meet the side $BC$ of $\Delta ABC$ at $A'.$ Now, we know that $\angle RBA=90-\angle C.$ And we know that the quadrilateral $ABRX$ is cyclic. So $-$ $$\angle RBA+\angle AXR=180\rightarrow \angle AXR=180-\angle RBA=90+\angle C.$$So $\angle RXC=90-\angle C,$ as $A,X,C$ are collinear. Thus, $\angle XA'C=180-\angle C-(90-\angle C)=90.$ So $XA'\perp BC,$ and thus $RX\perp BC$ as $X,O,A'$ are collinear. $\Square$

In most of the above solutions , people have proved that $\triangle$ BCX is isosceles , and then they say that the result follows ; but I don't see why . Just because it is isosceles , why does RX have to be perpendicular to the base ? Can someone please explain me ?

kk108 wrote: In most of the above solutions , people have proved that $\triangle$ BCX is isosceles , and then they say that the result follows ; but I don't see why . Just because it is isosceles , why does RX have to be perpendicular to the base ? Can someone please explain me ? \(X\) and \(R\) are both equidistant from \(B\) and \(C\), so the line joining them is the perpendicular bisector of \(BC\).

The easiest solution I found, do check it and point out mistakes if any: Let RX meet BC at O. $\angle BRO = \angle BAX = \angle BAC = A.$ (since $ABRX$ is cyclic.) Now since R is circumcenter of ABC $ \angle BRC = 2A.$ This implies that $\angle CRO = \angle BRO = A.$ and the result follows.