Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $CI$ intersect the segment $BC$ and the arc $BC$ (not containing $A$) of $\Omega$ at points $U$ and $V$ , respectively. Let the line passing through $U$ and parallel to $AI$ intersect $AV$ at $X$, and let the line passing through $V$ and parallel to $AI$ intersect $AB$ at $Y$ . Let $W$ and $Z$ be the midpoints of $AX$ and $BC$, respectively. Prove that if the points $I, X,$ and $Y$ are collinear, then the points $I, W ,$ and $Z$ are also collinear.

First, let $U'=AB\cap UX.$ Observe that $\measuredangle BIV = \measuredangle BIC - 90^\circ = \measuredangle BAI = \measuredangle BYV\implies B\in (YIV),$ so we also have $\measuredangle BU'U=\measuredangle BYV = \measuredangle BIU,$ giving $BU'IU$ cyclic as well; since $\tfrac{U'X}{AI}=\tfrac{YX}{YI}=\tfrac{VU}{UI}=\tfrac{VX}{AI},$ $X$ is the midpoint of $U'U,$ so this implies that $XI$ is the perpendicular bisector of $U'U$ as $I$ is the midpoint of arc $\widehat{U'IU}$ on $(BU'IU).$ This means that $AI\parallel U'X \perp IX,$ so $V,Y$ are the $A$-mixtilinear touch points with $(ABC)$ and $AB.$ Next, let $S$ be the $C$-mixtilinear touchpoint with $AC.$ It's well-known that $VY$ passes through the midpoint $M_C$ of arc $\widehat{BC}$ on $(ABC)$ not containing $C,$ so we must then have $\measuredangle AVM_C=\measuredangle ACI=\measuredangle ICU$ and we already have $\measuredangle AIX=90^\circ=\measuredangle UIC,$ so \[\triangle ASI\cong \triangle AXI\sim\triangle CUI\implies \measuredangle ASI=\measuredangle IUB=\measuredangle VXI=\measuredangle AXI\implies S\in (AXI).\]This means that if $T$ is the $B$-mixtilinear incircle touch point with $(ABC),$ which is well-known to lie on $(BU'IU),$ then we must also have $ATXIS$ cyclic. Finally, let $L$ and $M_A$ be the respective midpoints of arcs $\widehat{BAC}$ and $\widehat{BC}$ on $(ABC)$ not containing $A;$ obviously $AI,LZ$ pass through $M_B,$ and it's well-known that $TU$ does too. On the other hand, if $AI\cap BC=D,$ then $\measuredangle DAL=90^\circ=\measuredangle DZL\implies D\in (AZL)$ and $\measuredangle UTA=\measuredangle ALM_A=\measuredangle UDA\implies D\in (AUT),$ so by radical center $TUZL$ is cyclic. We also have $\measuredangle TPU=\measuredangle TAV=\measuredangle TLU\implies P\in (TUZL),$ but combined with the well-known fact that $L,I,V$ are collinear, this means that $\measuredangle TPI=\measuredangle TAI=\measuredangle TLM_A=\measuredangle TPZ,$ so $P,I,Z$ collinear, but clearly $P,W,I$ must be collinear as $PI$ and $AX$ are diameters of $(ATXIS),$ so we're done. $\square$.

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