Let the circumcircle of triangle $ABC$ be $\omega$. Let the intersection point of the tangent line of $A$ at $\omega$ and the line $BC$ be $T$. Let the intersection point of $AY$ and $\omega$ be $M$. Since $OM \perp BC$ and $XD \perp BC$, $OM \parallel XD$. Therefore, since $AO=OM$, $AX=XD.$ So, since $Y$ is the midpoint of segment $AD$, $XY \perp AD$. By some angle-chasing, $TA=TD$ can be easily proved. Therefore, $TY \perp AD$, so $T,Y,X$ are collinear. So $TY \times TX = TA^2 = TB \times TC$. $\therefore X,Y,B,C$ are cyclic.

Let $M$ be midpoint of arc $BC$ not containing $A$, $D' = 2M - D$. Infamous pop trick at $D$ gives $YD \cdot DD' = 2YD \cdot DM = AD \cdot DM = BD \cdot DC$ implying $BYCD'$ cyclic. Since $XD \parallel OM$, $AX:XD = AO:OM = 1$ hence $XY$ is the perpendicular bisector of $\overline{AD}$. Let parallel line to $BC$ through $D'$ cut $XD$ at $E$ so $\measuredangle D'ED = 90^\circ$ and since $M$ is the center of $(D'ED)$ we have $E$ is the reflection of $D'$ over $OM$ so $BCD'E$ is an isosceles trapezium, as a result $BYCD'E$ is cyclic. To finish just note that $\measuredangle D'YX = 90^\circ = \measuredangle D'EX$ so infact $B,C,Y,E$ lie on the circle with diameter $D'X$.

used as claim to 2012 G4 ISL

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