Hello Leart! I hope you can figure out who I am Solution: We begin by noting that if $E$ is the second intersection of the circumcircles of triangle $XAC$ and $XBD$, then the line passing through the circumcenters of those triangles bisects $XE$ (as it is the radical axis), so it suffices to show that $\angle XEO = 90^{\circ}$. Moreover, we claim that $E, X,$ and $M$ are collinear, where $M$ is the midpoint of $AC$; this follows by $MA \cdot MC = MB \cdot MD$, so $M$ lies on the radical axis. Let $H$ be the foot of the perpendicular from $X$ to $OM$, and let $F$ and $G$ be the midpoints of arc $AC$ not containing $D$ and arc $AC$ containing $D$, respectively. It is obvious that $X$ lies on $BG$ and $FD$. Let $Y$ be the intersection of $BF, XH,$ and $GD$, which we know exists by the Radical Center theorem; then $F$ is the orthocenter of triangle $GXY$, and as $M$ is the intersection of $BD$ and $GH$ it follows by a well-known lemma that $(H,M; F,G)$ is harmonic. Thus $OAHC$ is cyclic, so $ME \cdot MX = MA \cdot MC = MO \cdot MH$ so $OEHX$ is cyclic and $\angle OEX = 90^{\circ}$, as desired.

wu2481632 wrote: Hello Leart! I hope you can figure out who I am Good job, Andrew!

wu2481632 wrote: Hello Leart! I hope you can figure out who I am Solution: We begin by noting that if $E$ is the second intersection of the circumcircles of triangle $XAC$ and $XBD$, then the line passing through the circumcenters of those triangles bisects $XE$ (as it is the radical axis), so it suffices to show that $\angle XEO = 90^{\circ}$. Moreover, we claim that $E, X,$ and $M$ are collinear, where $M$ is the midpoint of $AC$; this follows by $MA \cdot MC = MB \cdot MD$, so $M$ lies on the radical axis. Let $H$ be the foot of the perpendicular from $X$ to $OM$, and let $F$ and $G$ be the midpoints of arc $AC$ not containing $D$ and arc $AC$ containing $D$, respectively. It is obvious that $X$ lies on $BG$ and $FD$. Let $Y$ be the intersection of $BF, XH,$ and $GD$, which we know exists by the Radical Center theorem; then $F$ is the orthocenter of triangle $GXY$, and as $M$ is the intersection of $BD$ and $GH$ it follows by a well-known lemma that $(H,M; F,G)$ is harmonic. Thus $OAHC$ is cyclic, so $ME \cdot MX = MA \cdot MC = MO \cdot MH$ so $OEHX$ is cyclic and $\angle OEX = 90^{\circ}$, as desired. Leartia wrote: wu2481632 wrote: Hello Leart! I hope you can figure out who I am Good job, Andrew! Elegant proof!