My solution : Let $ T \equiv XU \cap AB $ . Since $ \angle BIV=\angle BIC-90^{\circ}=\tfrac{1}{2} \angle BAC=\angle BAI=\angle BYV=\angle BTU $ , so we get $ B, I, V, Y $ are concyclic ... (1) and $ B, I, T, U $ are concyclic ... (2) From $XT:IA=YT:YA=VU:VI=XU:IA \Longrightarrow XT=XU $ , so from (2) and notice $\angle UBI=\angle TBI \Longrightarrow XI $ is the perpendicular bisector of $ TU $ , hence $ \angle AIY=90^{\circ} \Longrightarrow $ combine (1) we get $ V $ is the tangency point of A-mixtilinear circle with $ \Omega $ , so $ AV $ is the isogonal conjugate of the A-Nagel line of $ \triangle ABC $ WRT $ \angle BAC $ (well-known) . ... (3) Since $ IZ $ is parallel to the A-Nagel line of $ \triangle ABC $ (well-known) , so combine (3) $\Longrightarrow \angle ZIA=180^{\circ}-\angle VAI=180^{\circ}-\angle AIW \Longrightarrow I, W, Z $ are collinear . Q.E.D

Since $X,Y,I$ are collinear and $UX \parallel YV \parallel AI$ so $\tfrac{UI}{UV}= \tfrac{AI}{VI} \left( = \tfrac{XI}{XY} \right)$. If $IU \cap AC \equiv K$ then $IK=IU$. We then have $\tfrac{IK}{UV}= \tfrac{AI}{VI}$. Note that $\angle AIK= \angle YVU$ so $\triangle AIK \sim \triangle YVU$. This follows $YU \parallel AC$. $YI \cap AC=L$ then since $YU \parallel AC$ and $IU=IK$ so $IY=IL$. Hence, $AI \perp YL$. Since $VY \parallel AI$ so $\angle YVI= \angle AIK= \tfrac 12 \angle B= \angle IBY$. This follows $Y,I,V,B$ are concyclic. Therefore $\angle YIV= \angle LCI (=180^{\circ}- \angle ABV)$. This means $I,L,C,V$ are concyclic. Therefore $\angle ICL= \angle IVL$. Note that $\angle IVL= \angle YVA= \angle VAI$ so $\angle VAI= \tfrac 12 \angle C$. Since $XI \perp AI$ and $W$ is the midpoint of $AX$ so $\angle VAI= \angle WIA= \tfrac 12 \angle C$. since $\angle YIB= \tfrac 12 \angle C$ so $\angle BIV= \angle ICV (= \angle ACV- \tfrac 12 \angle C)$. Therefore $\angle BVI= \angle IVC (=90^{\circ}- \angle BIV)$ or $VI$ passes through the midpoint $R$ of arc $BAC$. We have $MB^2=MI^2=MZ \cdot MR$ so $\angle MIZ= \angle MRI= \angle MAV= \angle WIA$ or $\angle WIA= \angle MIZ$. This follows $W,I$ and $Z$ are collinear.

A slightly different finish: you can get $BIVY$ cyclic as others have shown, and it is clear that $B(I,V;U,IV\cap AB)$ is harmonic, so $(BA,BC;BI,BV)$ is harmonic and $V$ is the midpoint of arc $ABC$. So $\angle AIX=\angle VYI=\angle VBI=90$. Then angle chasing yields it suffices to show that $\angle BIZ=90$, or that $Z$ is the $B$-mixtilinear touchpoint on $BC$. This is equivalent to the midpoint $T$ of arc $BAC$ being the $B$-mixtilinear touchpoint, but $VT\perp CI$ so $V,I,T$ are collinear. But the fact that $V$ is the midpoint of arc $ABC$ implies that the intersection of $VI$ and the circumcircle is indeed the $B$-mixtilinear touchpoint, so we are done.

Let $R$ be the midpoint of $AC$. After proving the concyclic of 4 points $B,Y,I,V$ we get $\angle IBV=\angle IYV=90^\circ$, then $V$ is the midpoint of arc $ABC$. Then applying problem from here, we get $AC+BC=3AB$. This follows that the $C$-midline $ZR$ of triangle $ABC$ is tangent to $(I)$. From trapezoid $ABZR$ we get angle bisectors of $\angle ABZ$ and $\angle BZR$ are perpendicular. Therefore $\angle BIZ=90^\circ$. On the other side, since $VA=VC$, $\angle VAC=90^\circ-\frac{1}{2}\angle ABC$. Then $\angle WIA=\angle WAI=\angle VAC-\angle IAC=\frac{1}{2}\angle ACB=\angle XIB$. Thus $\angle BIW=\angle XIA=90^\circ$. This means $WI$ and $ZI$ are perpendicular to $BI$ or $Z,I,W$ are collinear.

a different approach: -Let $T$ the midpoint of arc $BAC$, $D$ the midpoint of minor arc $BC$, $N=XI\cap AC$ Lemma 1: $V,I,T$ are collinear proof: using the fact that $DI=DC$ and $AIB=90+B/2$ one gets by straightforward angle chase: $AIV=180-VID=90+DIC=90+DCI=90+C/2+A/2=180-B/2=180-(AIC-90)=180-AIT$ as desired. Next suppose $I,X,Y$ are collinear by lemma 1, $V$ is nothing but the intersection of $(ABC)$ and the $A$_mixtilinear incircle. also, it is clear that $IN=IY$ since $AI\perp NY$, therefore $(VNY)$ is the $A$_mixtilinear incircle. lemma 2: $CIZ\sim CVD$ proof: $ITZ=ITD=ATD-ATI=C+A/2-(90-B/2)=C+A/2+B/2-90=C/2=ICZ$ and therefore $T,I,Z,C$ are concyclic. Henceforth and since $T,Z,D$ are collinear, the center of spiral similarity sending $D\rightarrow Z$ and $V\rightarrow I$ is $(TIZ)\cap (TVD)=C$,i.e $CIZ\sim CVD$. Now we angle chase the final result: $DIZ=180-IZD-IDZ=IZT-ADT=ICT-ACT=ICA=VCD=VAD$ QED.

houssam9990 wrote: a different approach: -Let $T$ the midpoint of arc $BAC$, $D$ the midpoint of minor arc $BC$, $N=XI\cap AC$ Lemma 1: $V,I,T$ are collinear proof: using the fact that $DI=DC$ and $AIB=90+B/2$ one gets by straightforward angle chase: $AIV=180-VID=90+DIC=90+DCI=90+C/2+A/2=180-B/2=180-(AIC-90)=180-AIT$ as desired. Hey! It's wrong , since $\angle CIT$ isn't invariant($\equiv 90^{\circ}$). Your proof just regard $T$ as $VI\cap \Omega$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11.86294706749860cm); 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 = -8.690708564946124, xmax = 20.96665910380037, ymin = -11.46353533749292, ymax = 4.454711287577997; /* image dimensions */ pen ttzzqq = rgb(0.2000000000000002,0.6000000000000006,0.000000000000000); pen ccqqqq = rgb(0.8000000000000009,0.000000000000000,0.000000000000000); pen qqwuqq = rgb(0.000000000000000,0.3921568627450985,0.000000000000000); draw((7.175322333511941,-1.764700526050644)--(7.574874362682265,-2.162387381421302)--(7.972561218052923,-1.762835352250978)--(7.573009188882599,-1.365148496880320)--cycle, qqwuqq); /* draw figures */ draw((0.9702148095989103,-7.998910354506586)--(6.430296993760646,3.418297901304598)); draw(circle((6.559280219098369,-3.657585108605876), 7.077058502115000), linewidth(1.200000000000002) + ttzzqq); draw((6.430296993760646,3.418297901304598)--(13.43160946335172,-1.967659725360100)); draw((13.43160946335172,-1.967659725360100)--(0.9702148095989103,-7.998910354506586)); draw((0.9702148095989103,-7.998910354506586)--(7.573009188882599,-1.365148496880320)); draw((5.248949493451049,0.9480621102315469)--(12.94380691547388,-6.710874466925294)); draw((6.430296993760646,3.418297901304598)--(12.94380691547388,-6.710874466925294)); draw((12.94380691547388,-6.710874466925294)--(11.44576247608541,-0.4399907671596416)); draw((6.430296993760646,3.418297901304598)--(7.573009188882599,-1.365148496880320)); draw((8.600596784346335,1.748733407500978)--(9.897068884314148,-3.678359103992187)); draw((7.573009188882599,-1.365148496880320)--(13.43160946335172,-1.967659725360100)); draw((7.573009188882599,-1.365148496880320)--(11.44576247608541,-0.4399907671596416), linetype("4 4")); draw(circle((10.25840805217824,-4.038011481902800), 3.788873657393818), linewidth(1.600000000000002) + dotted + blue); draw(circle((10.60785065336160,-0.6401585914770015), 3.120236061175142), linewidth(1.600000000000002) + dotted + ccqqqq); draw((5.248949493451049,0.9480621102315469)--(8.600596784346335,1.748733407500978)); draw((7.200912136475315,-4.983285039933342)--(7.839564914045445,1.226742526529496), linetype("4 4")); /* dots and labels */ dot((6.430296993760646,3.418297901304598),dotstyle); label("$A$", (6.536595802609200,3.577746114577428), NE * labelscalefactor); dot((7.573009188882599,-1.365148496880320),dotstyle); label("$I$", (7.679307997731153,-1.205700283607490), NE * labelscalefactor); dot((3.700255901679778,-2.290306226600995),dotstyle); label("$S$", (3.799401474758941,-2.135814861032334), NE * labelscalefactor); dot((0.9702148095989103,-7.998910354506586),dotstyle); label("$C$", (1.088781849120820,-7.849375836642096), NE * labelscalefactor); dot((13.43160946335172,-1.967659725360100),dotstyle); label("$B$", (13.52574248440161,-1.816918434486673), NE * labelscalefactor); dot((9.897068884314148,-3.678359103992187),dotstyle); label("$U$", (9.991307090187197,-3.517699376063533), NE * labelscalefactor); dot((12.94380691547388,-6.710874466925294),dotstyle); label("$V$", (13.04739784458312,-6.547215428247314), NE * labelscalefactor); dot((9.248832834330242,-0.9648128482456046),dotstyle); label("$X$", (9.353514237095874,-0.8070797504254130), NE * labelscalefactor); dot((11.44576247608541,-0.4399907671596416),dotstyle); label("$Y$", (11.55921452070337,-0.2755857061826445), NE * labelscalefactor); dot((7.839564914045445,1.226742526529496),dotstyle); label("$W$", (7.945055019852537,1.398620533182076), NE * labelscalefactor); dot((7.200912136475315,-4.983285039933342),dotstyle); label("$Z$", (7.307262166761215,-4.819859784458316), NE * labelscalefactor); dot((5.248949493451049,0.9480621102315469),dotstyle); label("$T$", (5.367308905275110,1.106298808848554), NE * labelscalefactor); dot((8.600596784346335,1.748733407500978),dotstyle); label("$K$", (8.715721384004551,1.903539875212707), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We denote $\alpha=\angle{CAB}$, $\beta=\angle{ABC}$, $\gamma=\angle{BCA}$. Let $T=IU\cap AC$, $K=UX\cap AB$ and let $S$ be the midpoint of side $AC$. First observe that $$\angle{AIT}=\angle{AIB}-\frac{\pi}{2}=(\pi-\frac{\alpha}{2}-\frac{\gamma}{2})-\frac{\pi}{2}=\frac{\beta}{2}$$hence $\angle{KUI}=\frac{\beta}{2}$ (because $KU$ is parallel to $AI$); thus $\angle{KUI}=\angle{KBI}$ and quadrilateral $BUIK$ is cyclic. Since $I$ lies on the angle bissector of $\angle{KBU}$, $I$ is the midpoint of arc $UK$, hence $IK=IU$. On the other hand, triangles $CIU$ and $CIT$ are congruent, yielding $IT=IU$. Thus $IK=IU=IT$, yielding that $K$ lies on the circle with diameter $TU$; it follows that $TK\perp KU$. Since $KU$ is parallel to $AI$, $TK\perp AI$, hence due to symmetry $AT=AK$. Since $I, X, Y$ are collinear, triangles $AXI$ and $BXY$ are similar, as well as triangles $UXV$ and $IAV$. It follows that $$\frac{IA}{YV}=\frac{AX}{XV}=\frac{IU}{UV}=\frac{IT}{UV}$$or, equivalently, $\frac{IA}{IT}=\frac{VY}{VU}$. Since $AI$ and $YV$ are parallel, triangles $AIT$ and $YVU$ are similar; hence $UY$ is parallel to $AC$. Thus $\angle{UYB}=\alpha$, and $\angle{KYU}=\pi-\alpha$; moreover, since $AK=AT$, $\angle{AKT}=\frac{\pi}{2}-\frac{\alpha}{2}$ and $\angle{UKY}=\pi-\frac{\pi}{2}-(\frac{\pi}{2}-\frac{\alpha}{2})=\frac{\alpha}{2}$. This implies $\angle{YUK}=\pi-\frac{\alpha}{2}-(\pi-\alpha)=\frac{\alpha}{2}$, and $YK=YU$. Since $IK=IU$, the line containing $I$, $X$ and $Y$ is actually the perpendicular bissector of $KU$, and hence is perpendicular to $KU$; it follows that it is also perpendicular to $AI$ and $YV$. Now note that $$\angle{YVI}=\angle{KUI}=\frac{\beta}{2}=\angle{YBI}$$implying that quadrilateral $BVIY$ is cyclic. Since $\angle{IYV}=\frac{\pi}{2}$, we also have $\angle{IBV}=\frac{\pi}{2}$; since $BI$ is the internal angle bissector of $\angle{ABC}$, it follows that $BV$ is the external angle bissector of the same angle, and thus $V$ is the midpoint of the arc $AC$ (containing $B$) of $\Omega$. Since $VA=VC$ and $\angle{AVC}=\beta$, $\angle{CAV}=\frac{\pi}{2}-\frac{\beta}{2}$. Since $\angle{CAI}=\frac{\alpha}{2}$, $\angle{IAV}=\frac{\pi}{2}-\frac{\beta}{2}-\frac{\alpha}{2}=\frac{\gamma}{2}=\angle{ICA}$, hence $VA$ is tangent to the circumcircle of $AIC$. Similarly, $VC$ is tangent to such circle as well. It follows from a well-known lemma that $VI$ is the $I$-symmedian of $AIC$. Since $VI\perp CI$, the corresponding median $IS$ is perpendicular to $AI$; that is, $\angle{AIS}=\frac{\pi}{2}$. Consider the homothety centered at $C$ mapping the excenter $I_C$ to $I$. Since $AI_C\perp AI$, this homothety maps $A$ to $S$. Thus it maps $B$ to the midpoint of $BC$, that is, to $Z$. Hence $IZ$ is parallel to $BI_C$; since $BI_C$ is perpendicular to $BI$, it follows that angle $\angle{ZIB}$ is right. Since $\angle{XIA}$ is right, point $W$ is actually the circumcenter of $XIA$. We have previously noticed that $\angle{IAW}=\frac{\gamma}{2}$, hence $\angle{WIA}=\frac{\gamma}{2}$. On the other hand, $\angle{BIY}=\angle{BIA}-\frac{\pi}{2}=(\pi-\frac{\alpha}{2}-\frac{\beta}{2})-\frac{\pi}{2}=\frac{\gamma}{2}=\angle{WIA}$, hence $\angle{BIW}=\angle{XIA}=\frac{\pi}{2}$. It follows that $IW\perp BI$. Since $IZ\perp BI$, it follows that $I$, $W$ and $Z$ are collinear, as desired.

Let $N,N_1$ be midpoint of arc $BC$ not containing $A$ and midpoint of arc $BAC$. After proving that $BYIV$ is cyclic with $VI$ as its diameter (angle and length chase), we get $\angle NVI=\angle NVB-\angle BVI=180-\angle A/2-(90-\angle A/2)=90$ so $V,I,N_1$ are collinear. Since $N_1C$ and $N_1B$ are tangents to circumcircle of $BIC$, we get that $IN_1$ is symmedian to triangle $BIC$ and thus $\angle BIZ=\angle CIU=90$. On the other hand, easy angle chase yealds $\angle WIB=90$, so $W,I,Z$ are collinear.

Astounding.

official proposal

My solution : First we prove that quadrilateral $IYBV$ is cyclic $IV$ cuts $(O)$ again at $S$ then : $\angle YVI = \angle AIS = \angle YBI \Rightarrow IYBV$ is cyclic $XU$ cuts $AB$ at $T$ then : $\angle TUI = \angle YVI = \angle YBI \Rightarrow $ quadrilateral $TIUB$ is cyclic Three points $I,X,Y$ are collinear , by Thales : $\frac{XT}{IA} = \frac{YX}{YI} = \frac{VX}{VA} = \frac{XU}{IA}$ $\Rightarrow$ $X$ is the midpoint of $UT$ $(1)$ Quadrilateral $TIUB$ is cyclic and $BI$ is the bisector of $\angle TBU$ then $IT = IU$ $(2)$ Combine $(1)$ with $(2)$ $\Rightarrow$ $YI$ is the perpendicular bisector of $UT $ $\Rightarrow$ $UT \perp YI$ Since $UT $ is parallel with $AI$ then we have $YI$ is perpendicular with $IA$ $\Rightarrow$ $\angle YIB = \angle ICB $ But we have $Y,I,V,B$ are concyclic , therefore $\angle YIB = \angle YVB$ then $\angle YVB = \angle ICB$ $\Rightarrow$ $YV$ passes through $R$ which is the midpoint of arc$AB$ not containing $C$ Let $S'$ be the midpoint of arc$BC$ containing $A$ so $\angle RVS' = \frac{1}{2} \angle B = \angle ABI = \angle RVS$ then we have $S \equiv S'$ $IU$ passes through $S$. Therefore, $IU$ is the symmedian line of $\triangle IBC$ ; it follows that $\angle BIZ = \angle CIV = 90^{\circ}$ $\Rightarrow$ $IZ$ is perpendicular to $IB$ $(3)$ Note that $VR$ is the bisector of $\angle BVA$ then $\angle RVB = \angle RVA = \angle VAI = \angle WIA$ Hence , $BYIV$ is cyclic so $\angle BVY = \angle BIY$ and it follows that : $\angle BIY = \angle AIW$ $\Rightarrow$ $\angle WIB = \angle AIY = 90^{\circ}$ then $WI$ is also perpendicular with $BI$ Combine $(3)$ with $(4)$ , we come to a conclusion that $W,I,Z$ are collinear

An interesting (probably not, yeah..) property about $IX$: Define $M_c,M_a$ by midpoint of $\overarc{AB}, \overarc{BC}$. \Claim: $IX,AB,M_cV$ are always concurrent. Define $E=IX\cap M_aV$. Applying Desargues' involution theorem on quadrilateral $EXUV$ and line $\ell = AI$, we get an involution $\sigma:\ell\rightarrow \ell$ with fixed point $I$ and $\sigma(M_a)=\infty_\ell$, which is harmonic conjugation wrt $(I,I_A)$ where $I_A$ is $A$-excenter. Therefore $EU\cap\ell=\sigma(A)=BC\cap AI$ which tells $E\in BC$. Now applying Pascal's theorem on $ABCM_cVM_a$ we get $IE,AB,M_cV$ are concurrent, hence \End Claim. As a direct consequence, we get $Y\in VM_c$ in the problem. \return;

Saki wrote: Applying Desargues' involution theorem Hi, What is this! Thanks

\Above: Desargues' involution theorem states: "The eight points of intersection of a line $\ell$ with the three pairs of opposite sides of a complete quadrangle and a conic section circumscribed about the quadrangle form four pairs of an involution on $\ell$". where involution means projectivity of period 2. For some proof, try this paper. \return;

My solution: Notations Let $J$ be the intersection of $XU, AB$. $L$ is the second intersection of $IU, \Omega$. $M_a$ is the midpoint of the minor arc $\overarc{\textit{BC}}$ not containing $A$. Solution Notice that $\angle YVU = \angle AIL = \angle AIC - 90^\circ = \frac{B}{2} = \angle IBA$, so $IYBV$ is cyclic. Also note that $\frac {XJ}{AI} = \frac{YJ}{YA} = \frac{VU}{VI} = \frac {XU}{AI}$, so $XJ=XU$. Also, $\angle BJU = \angle BYV = \angle BIV = \angle BIU$ so $BIUJ$ is cyclic. Now that $BI$ bisects $\angle B$, so $JI = UI$, so $\angle AIY = 90^\circ$. So $Y$ is actually the $A-$mixtilinear incircle touch point with $AB$, and as $BVIY$ is cyclic, we come to know that $V$ is the $A-$mixtilinear incircle touchpoint with $\Omega$. So we have $\angle WIU = \angle WIY + \angle YIU = 90^\circ - \angle M_aAV + 90^\circ - \frac{B}{2}$. By the diameter of the incircle lemma and homothety at the incircle touchpoint with $BC$ and ratio $\frac12$, we know that $IZ \parallel A-\text{Nagel cevian}$, so $\angle ZIM_a = \angle VAM_a$ as the $A-$Nagel cevian is isogonal to $AV$ with respect to $\angle A$ and $\angle UIM_a = 90^\circ - \angle M_aIC = \frac{B}{2}$ $\Longrightarrow$ $\angle ZIU = \angle M_aAV + \frac{B}{2} = \angle WIU$, and thus we are done. $\blacksquare$.

I will only post a brief sketch; the details can be verified easily. Let $V = (x : y : z)$. We get $U = (0:bx-ay:cx-az)$, $Y = (cx+(b+c)z : cy - bz : 0)$. From this we deduce that $X = AV \cap IY = (cx + (a+b+c)z : cy : cz)$. We must use the condition that $UX$ is parallel to $AI$: we get \[ (cx+az)(cy-bz) = 0. \]We must have $cx + az = 0$, hence $V$ is midpoint of $I_aI_c$ and $II_c = IC$, yielding $3c = a+b$. We get $V = (-a : \frac{b^2}{a-c} : c)$ and $X = (2c^2 + bc - b^2 : b^2 : 2c^2 - bc)$. Thus $W = (6c^2 + bc - b^2 : b^2 : 2c^2 - bc)$. Combined with $I = (3c-b : b : c)$ and $Z = (0 : 1 : 1)$, we get that $I,W,Z$ are collinear, as desired. $\blacksquare$

IMO ShortList 2014 G7 wrote: 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. Proposed by David B. Rush, USA Nice problem! Quite unexpected to see how neatly everything fits in!

Solution: Let $M_1$ and $M_2$ be the midpoints of smaller and larger arcs $BC $ of $\odot (ABC) $ respectively. Let $UX\cap AB = N $. Claim 1: $BVIY$ and $INBU $ are cyclic. Proof of claim 1: $\angle BIV = \angle BAI = \angle BYV\implies BVIY $ is cyclic. Again, $\angle IUN = \angle IVY = \angle IBN\implies IUBN $ is cyclic. Claim 2: $UX=XN $. Proof of claim 2: Let $AB\cap IV = P $ and $PX\cap AI = Q $. As, $VY||AI $, so, $AQ = IQ$. This gives $XN = XU $, as $UN||AI $. Claim 3: $AI,UN,VY\perp IY $. Proof of claim 3: Follows directly from claim 1 (as $IN=IU$) and claim 2 (as $XN=XU $). Claim 4: $VI\cap \odot (ABC) = M_2$. Proof of claim 4: Let $IV\cap \odot (ABC) = M_2'$. As $BVIY $ is cyclic, application of Reim's Theorem gives $AM_2'||IY $. Let $AM_2'\cap UN = R $. So, $AIXR $ is a rectangle. This gives $\angle IAM_2' = 90$°. But, $\angle M_1AM_2 = 90$° (antipodal points). So, $M_2' = M_2$. Main problem: Observe that $M_1$, $Z$, and $M_2$ are collinear, as $M_1M_2$ is the $\perp$ bisector of $BC$. Also, $I $, $W $, and $R $ are collinear. As $M_2RUZ $ is cyclic, it suffices to show that $\angle URI = \angle UM_2Z $. This follows directly as $ARXI $ and $AVM_1M_2$ are cyclic.

Let $AU\cap XI=\{L\}$ , $VI\cap AB=\{K\}$ and $BI\cap \odot{(ABC)}=\{B,W\}$. Clearly $AL$ bisects the segment $AI$. I claim that $V$ is the midpoint of arc $ABC$ and that $VI$ intersects $\odot{(ABC)}$ at $L$, which happens to be the midpoint of arc $BAC$. To prove this it is enough to show that $(AC,VW)=-1$, the latter claim following from a simple angle-chase. $$(AC,VW)=B(AC,VW)=(KU,VI)=A(KU,VI)=(YL,XI)=V(YL,XI)=V(YL,AI)=-1$$Note that $LIZC$ is cyclic, thus $\angle{ZIC}=0.5\angle{BAC}$, so it is enough to show that $\angle{AIW}=\angle{ACB}$, which is equivalent to showing that $\angle{AIX}=90^{\circ}$. $$\frac{AX}{AI}=\frac{\frac{UI}{VI}\cdot AV}{AI}=\frac{CU}{CI}\cdot\frac{UI}{CU}\cdot\frac{CV}{VI}\cdot\frac{CI}{AI}=\frac{CU}{CI}\sin{0.5\angle{BCA}}\cdot\frac{1}{\sin{0.5\angle{BAC}}}\cdot\frac{CI}{AI}=\frac{CU}{CI} \implies \triangle{AXI} \sim \triangle{AUI} \ \ \ \blacksquare$$

Wow, really nice problem! Extend $UX$ to meet $AB$ at $P$ and let $AI$ meet $BC$ at $D$. Since $\angle PUI = \angle UID = 90 - \angle DIC = 90 - (\frac{A}{2} + \frac{C}{2}) = \frac{B}{2} = \angle PBI$ and so $PIUB$ is cyclic. Also since $\angle YVI = \angle XUI = \angle PUI = \angle PBI = \angle YBI$, $BVIY$ is also cyclic. Observe that since $BUIP$ is cyclic, and $BI$ is angle bisector, $IP = IU$ by Fact 5. Due to the parallel lines, $\frac{AI}{PX} = \frac{YI}{YX} = \frac{VI}{VU}$ and since $UX || AI$ in $\triangle AVI$, we have $\frac{VI}{VU} = \frac{AI}{XU}$ and so this means $\frac{AI}{PX} = \frac{AI}{XU} \implies PX = XU$. Now, since $P,X,U$ are collinear and $PX = XU$ in isosceles $\triangle IPU$, we have that $\angle IXP = 90^\circ$, which means $\angle AIY = \angle IYV = \angle IXU$ are all $90^\circ$. Obviously, since $\angle YIA = 90^\circ$, $Y$ is the point where the mixtilinear incircle touches $AB$ and so since $BYIV$ is cyclic, $V$ is the mixtilinear touchpoint. Let $Q$ be the ex-touch point. Its well known that $IZ || AQ$ and $AQ, AV$ are isogonal. So, $\angle DIZ = \angle DAQ = \angle DAV = \angle IAW = \angle WIA$ and so $I,W,Z$ are collinear, as required

Nice problem. First of all note that $\angle BAI=\angle BIV$ Let $U'=UX \cap AB$, Then because of the given conditions $BUIU'$ and $BYIV$ are cyclic. Notice that due to similarity we have $UX=\frac{XI}{YI} \times VY$ and $U'X=\frac{XY}{YI} \times AI$. Also $\frac{AI}{VY}=\frac{XI}{XY} \implies UX=U'X$ Clearly in $\triangle BUU'$, $BI$ is the angle bisector of $\angle UBU' \implies XI$ is the perpendicular bisector of $UU'$ $$\implies YX \perp VY \implies XY \perp AI \implies \angle BIY=\angle BVY=\frac{C}{2} \implies VY \cap CI \in (ABC)$$But it is well known (trivial angle chase) that $(BYI) \cap (ABC)$ is the mixtilinear intouch point. Also note that because of concyclicity of $B,Y,I,V$, we get $\angle YVI =\frac{B}{2} \implies \angle VBI=90$ By tangent secant theorem, since $VI \perp CI$, $CI$ is tangent to $(BYVI)$ Let $D= VY \cap CI$. Then we have $$\angle DVA=\angle VAI=\angle WAI=\angle AIW \implies BI \perp WI$$ Note that it suffice to show that $BI \perp IZ \iff \angle ZIC=\frac{A}{2} \iff IU$ is symmedian of $\triangle BIC$. Clearly $V$ is the centre of spiral similarity sending $BI$ to $CI$, hence we can say that $IV$ is the symmedian.

Proposed by USA=-10 mohs (jk dont kill me lmao) The proposer is already a hint as u can think on spamming usa config, and more especificaly mixtilinear incircle config. These days i dont count with so much time so yeah, me typin fast. Let $XU \cap AB=D$, $M,N$ the midpoint of the arcs $BC, ABC$ on $(ABC)$, respectivily also let $IC \cap (ABC)=M_C$ also let $VM \cap BC=G$, so now we being with the most important claim Claim 1: $V$ is the touch point of the $A$-mixtilinear incircle with $(ABC)$ Proof: Its enough to show that $\overline{XYI} \perp AI$, clearly $\angle BYV=\angle BAI=\angle BIV$ so $BYIV$ is cyclic and now $\angle DBI=\angle YVI=\angle DUI$ so $BDIU$ is cyclic, now since $BI$ bisects $\angle DBU$ we have $\triangle DIU$ isosceles and now by Thales spam on those parallels we can get $DX=XU$ so indeed $\overline{XYI} \perp AI$ which means that $Y$ is the touch point of the $A$-mixtilinear incircle with $AB$ and by well known mixtilinear config that means $V$ is the touch point of the $A$-mixtilinear incircle with $(ABC)$ as desired. Finishing: Since $\angle AIX=90$ we have that $W$ is center of $(AIX)$ so $\triangle AWI$ is isosceles, now by well known usa mixtilinear config we have $V,U,I,N$ colinear, $G,Y,X,I$ colinear and $M_C,Y,V$ colinear. So using these colinearities by angle chase $\angle GVN=90=\angle GZN$ so $GVZN$ is cyclic and now since $\angle GIM=90=\angle GZM$ we have $GIZM$ cyclic so now by the final angle chasing: $$\angle WIA=\angle IAW=\angle VNM=\angle VGZ=\angle MIZ \implies W,I,Z \; \text{colinear}$$Thus we are done

Finally did something worthy in this holiday,so yayyy!(Even tho I might have glanced on Telv Cohls diagram,yes taking a small hint is ok right?).Also it seems like I have lost every skill except angle chasing ISL 2014 G7 wrote: 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. Proposed by David B. Rush, USA Say $UX \cap AB=D$.Notice that by angle chasing we have $\{B,Y,I,V \},\{B,D,I,X\}$ as cyclic quadraples.Also by Fact 5 $IU=ID$ and as we have $VY || UX || AI$ by similarity ratios,$\frac{XU}{YV}=\frac{IU}{IV}=\frac{AX}{AV}=\frac{XD}{YV} \implies XU=XD$(i.e. $X$ is the midpoint of $UD$).So we get $IX \perp UD \implies YI \perp AI$.Now we present a claim which is the crux of the problem. Claim :$V$ is the $\text{A-Mixtillinear touch point}$ Proof : By angle chasing $\angle IBV=\angle IYV=\angle YIA=90$.So, \begin{align*} \angle BVI &=90-\angle BIV \\ &=90-(\angle BIC-\angle CIV) \\ &=90-(\frac{1}{2}\angle A+90-90)\\ &=90-\frac{1}{2}\angle A \end{align*}So $VI$ bisects $\angle BVC$ so if $M$ is the midpoint of $\widehat{BAC}$ we have $V-I-M$ which implies our claim $\blacksquare$ Now say $G$ is the midpoint of $\widehat{BC}$,note that we have $\angle MIC=90=\angle MZC \implies \text{ M,I,Z,C are concyclic}$ So now as $\triangle AIX$ is right angled, \begin{align*}\angle WIX &=\angle WXI \\ &=90-\angle XAI \\ &=90-\angle VAI \\ &=90-(\angle VAC-\angle CAI)\\ &=90+\frac{1}{2}\angle A-\angle VBC \\ &=\angle IBC+\frac{1}{2}\angle A \\ &=\frac{1}{2}\angle B+\frac{1}{2}\angle A \end{align*}Again by angle chasing, $$\angle XIU=180-(\angle ABV)=180-(\frac{1}{2}\angle B+90)=90-\frac{1}{2}\angle B \implies \angle WIU=90+\frac{1}{2}\angle A$$Again as $MIZC$ is cyclic we get, $$\angle CIZ=\angle CFZ=\angle CFG=\angle CAG=\frac{1}{2}\angle A \implies \angle ZIU=90-\frac{1}{2}\angle A$$Finally $$\angle ZIU+\angle WIU=90+\frac{1}{2}\angle A+90-\frac{1}{2}\angle A =180$$Which means $I,W,Z$ are collinear.The End $\blacksquare$

Let $T=UX\cap AB.$ Step 1. By $\angle BTU=\angle BYV=\frac{1}{2}\angle BAC=\angle BIV=\angle BIU$ sets $(BIYV),(BITU)$ are concyclic. Step 2. Since $BI$ bisects angle $TBU$ we conclude $|IT|=|IU|.$ Now observe that $$|TX|:|AI|=|YT|:|YA|=|VU|:|VI|=|UX|:|AI|\implies TU\perp IY\implies AI\perp IY.$$ Step 3. Since $\angle BVI=\angle IYA=90^\circ -\frac{1}{2}\angle BAC=\frac{1}{2}\angle BVC$ point $V$ is the $A-\text{Mixtilinear touch point}.$ Step 4. From two previous steps $IW$ is parallel to isogonal of $AV$ in $BAC,$ which is $A-\text{Nagelian} \text{ } \ell.$ But well-known that $IZ\parallel \ell,$ thus we are done.

Missing configs... Did I mention I HATE BARY? Let $V = I + t(U-I)$ Then $(ABCV)$ gives $(a+b+c)c + (a+b+c)(a-b)t + (a^2-2ab+b^2-c^2)t^2 = 0$ And $\overline{IXY}$ gives $2c + (a-b-c)t = 0$ So taking a polynomial gcd gives $a + b = 3c$ Notice $\overline{IWZ}$ is equivalent to $(a+b+c)(c-b) + (a^2-ab-ac+2b^2-2c^2)t = 0$ which follows from substitution, so we are done. Thank goodness.

As I have seen so far, nobody had my solution, so here we go Let $I_a$ be the $A-$excentre of $ABC$. Claim 1: $V\in BI_a$ Proof: Consider the point $\{N\}=XU\cap AB$ \[\angle{BYV}=\angle{BAI}=\angle{BIV}\]resulting in $BYIV$ being cyclic. On the other hand, \[\angle{BNU}=\angle{BYV}=\angle{BIU}\]so $BUIN$ is cyclic as well. Now a simple angle chasing leads to $XI\perp AI$ and $\angle{ABV}=\angle{ABI_a}\Rightarrow V\in BI_a$. Now, consider $YV\cap CI=\{T\}$ Claim 2: $T\in (ABC)$ Proof: \[\angle{BVY}=\angle{BI_aI}=\angle{BCI}=\angle{BCT}\] Claim 3: $WI\parallel VI_a$ Proof: Angle chase again: \[\angle{TVC}=\angle{TBC}=\angle{ABC}+\frac{\angle{ACB}}{2}\Rightarrow \angle{YVA}=\frac{\angle{ACB}}{2}=\angle{VI_aA}\]Since $WI=WA=WX$, we obtain $\angle{WIA}=\angle{VI_aA}\Rightarrow WI\parallel VI_a$. Finally, construct the following points: $S$ the midpoint of $\overline{CI_a}$, $M$ the midpoint of $\overline{II_a}$ and $\{K\}=IV\cap (BIC)$ Claim 4: $V$ is the midpoint of $KI$ Proof: $KI\perp IC$ and $I_aC\perp IC$, thus $ICI_aK$ is a rectangle centred at $M$. So it is enough to prove that $MV\perp IK$ and the claim would be proved. But just use another angle chase, THIRD TIME: \[\angle{MVI}=180-\angle{MIV}-\angle{VMI}=180-\angle{VCA}-\frac{\angle{ABC}}{2}=\angle{ABV}-\frac{\angle{ABC}}{2}=90\] So I have that $VI=SI_a=\frac{KI}{2}$ and $VI\parallel I_aS\Rightarrow VISI_a$ is a parallelogram. This means nothing else than $SI\parallel VI_a$. In conclusion, $ZS\parallel BI_a$, $IS\parallel BI_a$, and $WI\parallel BI_a\Rightarrow W, I$ and $Z$ are collinear. 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Absolute beauty.

oops I forgot to post this one. Suppose that $I$, $X$, and $Y$ are collinear. We are required to show that $I$, $W$, and $Z$ are collinear. Let $P=\overline{UX} \cap \overline{AB}$. Claim: $BUIP$ is cyclic. Proof. Angle chase: \[ \angle BPU = \angle BAI = \angle A/2 = \angle BIC - 90^{\circ} = \angle BIU, \]as desired. Claim: $X$ is the midpoint of $UP$. Proof. Since $\overline{AI} \parallel \overline{PU} \parallel \overline{YV}$, and noting that $X = \overline{AV} \cap \overline{YI}$, we readily have $XU=XP$, as desired. By the Incenter-Excenter lemma on $(BUIP)$, we have $\overline{IX} \perp \overline{XU} \parallel \overline{AI}$, so $\angle AIY = 90^{\circ}$. Let $M$ denote the midpoint of arc $BAC$. Claim: $M$ lies on $\overline{IUV}$. In particular, $V$ is the $A$-mixtillinear touchpoint of $\triangle ABC$. Proof. It suffices to show that $\overline{VI}$ is an angle bisector in $\triangle BVC$. First, I contend that $BYIV$ is cyclic. This is easy to show by angle chase: \[ \angle YBI = \angle PBI = \angle PUI = \angle XUI = \angle YVI, \]which implies the claimed concyclicity. The proof of the claim is easy from here: write \[ \angle BVI = \angle AYI = 90^{\circ} - \angle A/2 = \frac{1}{2}\cdot (180^{\circ}-\angle A) = \frac{1}{2} \angle BVC, \]as desired. Note that $V$ is the center of the spiral similarity that maps $\overline{BI} \mapsto \overline{IC}$. Thus $\overline{IV}$ is the symmedian of $\triangle BIC$. Angle chasing shows that $\angle WIB = 90^{\circ}$ so $WI$ and $IU$ are isogonal conjugates in $\triangle BIC$. Thus $W$, $I$, and $Z$ are collinear, and we are done.

Pretty sure I, like many others, just used angle-chasing. 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Define $T = \overline{UX} \cap \overline{AB}$ and $E = \overline{UV} \cap \overline{AC}$. Note that we easily find $(BVYI)$ and $(BUIT)$ as, \begin{align*} (\angle YVI = \angle XUI) &=\angle EIA\\ &= \angle AIC - 90\\ &= (180 - \angle A/2 - \angle C/2)\\ &= \angle B/2 \end{align*}from the parallel condition. Claim: $(UTE)$ has center $I$. Proof. Indeed note that to show $IU = UT$ we have, \begin{align*} \angle IUT &= \angle B/2\\ &= \angle IBU\\ &= \angle ITU \end{align*}Then to show that $IU = IE$ note that $CU = CE$ as $\overline{EU} \perp \overline{IC}$ and $\overline{IC}$ bisects $\angle ACB$. Thus our claim follows. $\square$ Claim: $X$ is the midpoint of $\overline{TU}$ and equivalently $\overline{IXY} \perp \overline{UXT}$ Proof. To show this let $D \neq B$ be the intersection of $(BVIY)$ with $\overline{BC}$. Then note that $\triangle TIY \overset{+}{\sim} \triangle IUD$ and $\triangle TIU \overset{+}{\sim} \triangle YID$. Now some standard angle chasing gives, \begin{align*} \angle IXT &= 180 - \angle XTI - \angle XIT\\ &= 180 - \angle DYI - \frac{1}{2}(\angle YIZ)\\ &= 180 - \angle DBI - \frac{1}{2}(180 - \angle YBD)\\ &= 180 - \angle B/2 - 90 + \angle B/2\\ &= 90 \end{align*}Note then that $(XIA)$ has center $W$. Moving forward, denote by $F$ the second intersection of $(TDU)$ with $\overline{AB}$. Now we aim to show that $\overline{EF} \perp \overline{IW}$ so we can rewrite $\overline{WI}$ as the perpendicular bisector of $\overline{EF}$ and delete $W$ from the picture. To do this we will first show, Claim: $E, F \in (XIA)$ Proof. Note that, \begin{align*} \angle EAV &= \angle CBV\\ &= \angle VID\\ &= \angle VIY\\ &= \angle VIX \end{align*}where the last step follows from noting that under the aforementioned spiral similarities we have $\overline{IV} \perp \overline{YD}$ in isosceles $ \triangle YID$. Thus $E \in (XIA)$. Now we show the claim for $F$: \begin{align*} \angle FDI &= \angle FDU\\ &= \angle FTU\\ &= \angle BIU\\ &= \angle BIC - 90\\ &= \angle A/2 \end{align*}as desired. Then we find that $\overline{WZ}$ is the perpendicular bisector of $\overline{EF}$ as claimed. $\square$ Now to show the collinearity we may focus on just $Z$. We will show it lies on the perpendicular bisector of $\overline{EF}$. To do this first note, Claim: $\overline{EF} \perp \overline{BI}$. Proof. Simply note that, \begin{align*} \angle AFE &= \angle AIE\\ &= \angle AIC - 90\\ &= \angle B/2 \end{align*}as desired. $\square$ Then all that remains is to show that $\overline{IZ} \perp \overline{IB}$. This is then equivalent to showing that $\overline{IZ}$ and $\overline{IU}$ are isogonal with respect to $\angle BIC$. Then as $\overline{IZ}$ is a median, it suffices to show that $\overline{IU}$ is the symmedian of $\triangle IBC$. Then noting that the tangents to $(IBC)$ at $B$ and $C$ meet at the midpoint of arc $\widehat{BAC}$, say $K$, it suffices to show that $\overline{IVK}$ collinear. This is easy as, Claim: $\overline{IVK}$ collinear. Proof. Simply note that, \begin{align*} \angle AIV &= 90 + \angle C/2 + \angle BIV\\ &= 90 + \angle C/2 + \angle BYV\\ &= 90 + \angle C/2 + BAI\\ &= 90 + \angle C/2 + \angle A/2\\ &= 180 - \angle B/2 \end{align*}but we also have, \begin{align*} \angle AIK &= \angle AIC - \angle CIK\\ &= 90 + \angle B/2 - 90\\ &= \angle B/2 \end{align*}so we are done. $\square$ This completes the proof.

We begin with the following claim: Claim: The points $B, Y, I, V$ are concyclic. Proof. Note that $\angle BYV = \angle BAI = \frac{\angle BAC}{2} = \angle BIV$, so $BYIV$ is cyclic. $\blacksquare$ Now let $N = XU \cap AB$. Then $XU \parallel AI \implies \angle BNU = \angle BAI = \angle BIU$, hence $BNIU$ is cyclic. Since $\angle IUN = \angle INU = \frac{\angle ABC}{2}$, so $IN = IU$. Now since the lines $AI, NU, YV$ are parallel, thus $\frac{XU}{AI} = \frac{XV}{VA} = \frac{XY}{YI} = \frac{XN}{AI}$, so we get $XU = XN$. Combined with the fact that $IN = IU$, we see that $XI \perp NU$, i.e $\angle AIX = \angle AIY = 90^{\circ}$. Therefore $A-$mixtilinear circle touches $AB$ at $Y$ and since $BYIV$ are cyclic, so $A$-mixtilinear circle touches $\Omega$ at $V$. Let $M = AI \cap \Omega$ and let $I_A$ be the $A$-excenter and assume $A$-excircle touches $BC$ at $D$. Claim: $IZ \parallel AD$. Proof. Let $L = AI \cap BC$. Then $\frac{LI}{LA} = \frac{\frac{BL \cdot LC}{LI_A}}{\frac{BL \cdot LC}{LM_A}} = \frac{LM_A}{LI_A}$, so $\frac{IM_A}{AI_A} = \frac{ZM_A}{DI_A}$ and $\angle IM_AZ = \angle AI_AD \implies \triangle IM_A$ is similar to $\triangle AI_AD$. Therefore $\angle LIZ = \angle LAD$, so $IZ \parallel AD$. $\blacksquare$ Now it's well-known that $AV$ and $AD$ are isogonal. Combining with the fact that $\angle AIX = 90^{\circ}$ and $\angle WAI = \angle WIA$, we see that $I, Z, W$ are collinear, as desired. $\blacksquare$

Let $IV$ intersect $AB$ at $T$ and $M_B$ be the midpoint of arc $AC$ not containing $B$. Clearly, we have \[\frac{TI}{TV}=\frac{AI}{YV}=\frac{AX}{XV}=\frac{UI}{UV}\]so $(T,U;I,V)=-1$. Taking perspectivity through $B$, $(A,C;M_B,V)=-1$. This implies that $V$ is the midpoint of arc $ABC$. Now, observe that \[\angle BIV=\angle BIC-90^\circ=\frac12 \angle BAC=\angle BAI=\angle BYV\]so $BYIV$ is cyclic. Thus, $\angle VYI=\angle VBI=90^\circ$. Since $AI\parallel YV$, $\angle AIY=90^\circ$. Since $VA=VC$ and $\angle AIC=90^\circ+\frac12 \angle AVC$, $VA$ and $VC$ are tangent to $(AIC)$. Thus, $VI$ is symmedian of $\triangle AIC$. Let $Y$ be the midpoint of $AC$ then $\angle AIY'=\angle CIT=90^\circ$. Therefore, $Y,I,Y'$ are collinear. Furthermore, $Y'$ is the reflection of $Y$ over $I$, and $Y'$ is on $VM_B$. We have \[\angle YVI=\angle YBI=\angle ABM_B=\angle AVM_B=\angle XVY'\]so $VX$ is symmedian in $\triangle YVY'$. This implies $AY$, $AY'$ tangent to $(YVY')$. Thus, $(YVY')$ is the mixtillinear incircle of $\triangle ABC$. Let $E$ be the point where the $A$-excircle is tangent to $BC$. Using a $\sqrt{bc}$ inversion, we can see that $AV$ and $AE$ are isogonal. Let $X'$ be the reflection of $X$ over $I$. We know that $X'$ is on $AE$, so $WI\parallel AE$. We know that $AE\parallel IZ$ so we are done.

Define $R = UI \cap AC$, $T = UX \cap AB$ and suppose $I-X-Y$. $\frac{VY}{IA} = \frac{VX}{XA} = \frac{VU}{UI} = \frac {VU}{IR}$ $\implies \frac{VY}{VU} = \frac {IA}{IR}$ Since $VY \parallel IA$, $\angle YVI = \angle AIR \implies\triangle YVU \sim \triangle AIR$ (because the angle + above equality). From this we get $\angle VYU = \angle IAC \implies AC \parallel YU$! $\implies \angle AIY = 90$ (Ez to prove with trig). Now, notice that $\angle BIV = \angle VYB \implies BYIV$ cyclic. $\implies 90 = \angle AIY = \angle IYV = \angle IBV = \frac{\angle B}{2} + \angle VAC = \frac{\angle B}{2} + \frac{\angle A}{2} + \angle VAI$ $\implies 90 = \frac{\angle B}{2} + \frac{\angle A}{2} + \angle VAI \implies \angle VAI = \frac{\angle C}{2}$ With simple angle chasing we can get from this: $\cdot$ Line $WI$ bissects the segment $VC$; $\cdot$ And $WI \parallel BV$; So $WI$ bissects $BC$ too, as desired.

Let $M,N$ be the midpoint of the arcs $BC,BAC$. $VM \cap BC=T$. Claim: $V$ is the midpoint of the arc $ABC$. Proof: Let $AB\cap VI=K$. \[\frac{KV}{KI}=\frac{VY}{IA}=\frac{XV}{XA}=\frac{UV}{UI}\implies -1=(K,U;V,I)\overset{B}{=}(A,C;V,BI\cap (ABC))\]Since $BI\cap (ABC)$ is the midpoint of the arc $AC$, we get that $VA=VC$.$\square$ Claim: $V,I,N$ are collinear. Proof: \[\measuredangle MIV=\measuredangle ADC-\measuredangle IUC=\frac{\measuredangle A}{2}+\measuredangle B-\frac{\measuredangle A+\measuredangle B}{2}=\frac{\measuredangle B}{2}\]\[\measuredangle VMA=\measuredangle VCA=\frac{\measuredangle A+\measuredangle C}{2}\]Hence $\measuredangle IVM=180-\measuredangle MIV-\measuredangle VMA=90=\measuredangle NVM$ yields $V,I,N$ are collinear.$\square$ Claim: $\measuredangle XIA=90$. Proof: Since $\measuredangle UXV=\measuredangle IAV=\frac{\measuredangle C}{2}=\measuredangle CBM-\measuredangle VMB=\measuredangle UTV,$ we conclude that $T,X,U,V$ are concyclic. $\measuredangle TVU=90$ gives $\measuredangle UXT=90$. \[\frac{TU}{TD}=\frac{MI}{MD}.\frac{VU}{VI}=\frac{AI}{ID}.\frac{XU}{AI}=\frac{XU}{ID}\]Thus, $T,X,I$ are collinear. $180-\measuredangle XIA=\measuredangle DIT=\measuredangle UXT=90$, this gives us $\measuredangle XIA=90$ which is the desired angle condition.$\square$ $MI^2=MZ.MN$ yields $\measuredangle ZIM=\measuredangle MNI=\measuredangle MNV=\frac{\measuredangle C}{2}$. Also since $\measuredangle XIA=90$ and $W$ is the midpoint of $AX,$ we have $WA=WI=WX$. \[\measuredangle WAI=\measuredangle IAW=\frac{\measuredangle C}{2}=\measuredangle ZIM\]Thus, $I,Z,W$ are collinear as desired.$\blacksquare$