Firstly, after inversion wrt $N$ and radius $NA,$ note that $\odot(ANB)$ goes to $\overleftrightarrow{AB}$, and vice-versa. Therefore, the circle $(I)$ remains constant. So, $NA=NB=NC=ND,$ leading to the fact that $ABCD$ is a cyclic quadrilateral with centre at $N.$ Now, let $\omega$ be the circle $(N, NA);$ then note that $\angle ICD=\angle IND,$ therefore $IC$ and $ID$ are tangents to $\omega,$ and so $I$ is the pole of $CD$ wrt $\omega.$ Similarly, note that $\angle MAC=\angle ANM,$ and therefore $M$ is the pole of $AB$ wrt $\omega.$ So, $IM$ is the polar of $AB\cap CD$ wrt $\omega,$ which is the same as $XY.$ Therefore, $X,Y,I,M$ are collinear.$\Box$

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http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=474644

Let $ \omega$ touch $AB$ and big circle at $Z$ and $T$ respectively. Because <ATZ=<ATN=<NAB we have that $NA$ is tangent to circle $AZT$ so $NA^2=NZ*NT=ND^2$ we have $NA=ND=NC$ and in similar way $NC=NB$ that means $N$ is circumcenter for $ABDC$. Circles $(ABDC)$ and $ \omega$ are orthogonal so $I$ is intersection of tangents at $C$ and $D$ on $(ACDB)$ , also circles $(ABDC)$ and $(M,MA)$ are also orthogonal so $M$ is point of intersections of tangents at $A$ and $B$ on $(ABDC)$. Now what is left to prove is well-known lemma (follows from Pascal's theorem)

$M,N$ are the midpoints of the called and complementary arc. $G=AX\cap BY, H=AY\cap BX$.Let the smaller circle which is inscribed in the circular arc and the chord as $\omega$. Let the circle containing the arc and the chord be $\Gamma$. Let $\omega$ is tangent to $\Gamma$ at $P$ and tangent to the chord $AB$ at $Q$.$PQ$ is the internal angle bisector of $\angle APB$. And $PM$ is the external bisector of this angle. Let $AB\cap XY=Z$. Then the $Z$ is the pole of $PQ$ wrt $\omega$. This gives that $ZP$ is tangent to $\Gamma$ and $\omega$ at $P$ so we get $ABXY$ is cyclic. Also $N$ is the centre of the circle $ABXY$. $Z$ is also the pole of $GH$ wrt $\odot ABXY$.$NZ$ is the radical axis of the circles $(M,MA)$ and $\omega$. Now so $I,M,G,H$ are collinear.

By Power of Point $N$ WRT $\omega,$ we have $NC^2=NQ\cdot NP.$ But this quantity is also $NA^2$ by the Shooting Lemma, so $NA=NC$ and analogously we obtain that $N$ is the circumcenter of $ACDB.$ Now from cyclic quadrilateral $ICND,$ we have $\angle ICD=\angle IND$ so $IC,ID$ are tangents to $(ACDB)\implies CD$ is the polar of $I$ WRT $(ACDB).$ Similarly $AB$ is the polar of $M$. Now by La Hire we have that $IM$ is the polar of $CD\cap AB,$ but the polar of $CD\cap AB$ is also $XY$ by Brokard, implying the desired collinearity.

Truly Projective! Sharygin 2012/22 wrote: A circle $\omega$ with center $I$ is inscribed into a segment of the disk, formed by an arc and a chord $AB$. Point $M$ is the midpoint of this arc $AB$, and point $N$ is the midpoint of the complementary arc. The tangents from $N$ touch $\omega$ in points $C$ and $D$. The opposite sidelines $AC$ and $BD$ of quadrilateral $ABCD$ meet in point $X$, and the diagonals of $ABCD$ meet in point $Y$. Prove that points $X, Y, I$ and $M$ are collinear. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.182cm); 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 = -4.26, xmax = 16., ymin = -5.2, ymax = 6.34; /* image dimensions */ /* draw figures */ draw(circle((4.732328757277367,1.562979228060088), 2.8976577985276055)); draw((7.42,0.48)--(2.82,3.74)); draw((2.82,3.74)--(1.96,0.72)); draw(circle((3.9910461468119207,2.4068653196747483), 1.774428683787815)); draw((2.82,3.74)--(4.605082053223209,-1.3318832891719834)); draw((1.96,0.72)--(5.672801008275183,1.8409167886282969)); draw((2.5786488371805976,1.3327466141327784)--(7.42,0.48)); draw((4.2279602060716375,2.9663227177336378)--(1.96,0.72)); draw((4.2279602060716375,2.9663227177336378)--(7.42,0.48)); draw((3.9910461468119207,2.4068653196747483)--(2.5786488371805976,1.3327466141327784)); draw((2.5786488371805976,1.3327466141327784)--(4.605082053223209,-1.3318832891719834)); draw((4.605082053223209,-1.3318832891719834)--(5.672801008275183,1.8409167886282969)); draw((5.672801008275183,1.8409167886282969)--(3.9910461468119207,2.4068653196747483)); draw((2.82,3.74)--(-0.49515241066053656,0.8279187872817818)); draw((-0.49515241066053656,0.8279187872817818)--(4.605082053223209,-1.3318832891719834)); draw((-0.49515241066053656,0.8279187872817818)--(7.42,0.48)); draw(shift((4.605082053223209,-1.3318832891719834))*xscale(3.3476385857297597)*yscale(3.3476385857297597)*arc((0,0),1,15.637236155935852,170.289472327311), linetype("2 2")); draw((4.859575461331523,4.457841745292159)--(2.7419121873807155,-0.5428846316537113), linetype("2 2")); draw((-0.49515241066053656,0.8279187872817818)--(5.672801008275183,1.8409167886282969)); /* dots and labels */ dot((2.82,3.74),dotstyle); label("$E$", (2.9,3.94), NE * labelscalefactor); dot((1.96,0.72),dotstyle); label("$A$", (1.6,0.46), NE * labelscalefactor); dot((7.42,0.48),dotstyle); label("$B$", (7.6,0.46), NE * labelscalefactor); dot((4.859575461331523,4.457841745292159),linewidth(3.pt) + dotstyle); label("$M$", (5.02,4.52), NE * labelscalefactor); dot((4.605082053223209,-1.3318832891719834),linewidth(3.pt) + dotstyle); label("$N$", (4.52,-1.68), NE * labelscalefactor); dot((3.913124522765808,0.6341483726256787),linewidth(3.pt) + dotstyle); label("$G$", (4.,0.76), NE * labelscalefactor); dot((3.9910461468119207,2.4068653196747483),linewidth(3.pt) + dotstyle); label("$I$", (4.08,2.16), NE * labelscalefactor); dot((2.5786488371805976,1.3327466141327784),linewidth(3.pt) + dotstyle); label("$C$", (2.58,1.52), NE * labelscalefactor); dot((5.672801008275183,1.8409167886282969),linewidth(3.pt) + dotstyle); label("$D$", (5.76,1.96), NE * labelscalefactor); dot((-0.49515241066053656,0.8279187872817818),linewidth(3.pt) + dotstyle); label("$T$", (-0.48,1.04), NE * labelscalefactor); dot((4.2279602060716375,2.9663227177336378),linewidth(3.pt) + dotstyle); label("$X$", (4.38,2.98), NE * labelscalefactor); dot((3.4697231623008404,1.1757944352612038),linewidth(3.pt) + dotstyle); label("$Y$", (3.22,1.3), NE * labelscalefactor); dot((3.2526459147705995,0.6631803993507428),linewidth(3.pt) + dotstyle); label("$F$", (3.36,0.3), NE * labelscalefactor); dot((2.7419121873807155,-0.5428846316537113),linewidth(3.pt) + dotstyle); label("$Z$", (2.66,-0.92), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $E$ be the tangency point of $\omega$ with the arc and $G$ be the tangency point of $\omega$ with $AB$. Let $EF$ ($F$ lies on $AB$) be a symmedian in triangle $AEB$. Let $T$ be a point on line $AB$ with $TE$ tangent to the circumcircle of $AEB$. Dilation at $E$ sending $\omega$ to the circumcircle of $ABE$ maps $G$ to $N$. Note that $$NC^2=ND^2=NG\cdot NE=NA^2,$$so $A, B, C, D$ lie on a circle (call it $\Gamma$) with center $N$. Since $\angle ICN=\angle IDN=90^{\circ}$, lines $IC$ and $ID$ are tangent to $\Gamma$. From the radical axis theorem for the circles $\odot (AEB), \omega,$ and $\Gamma,$ it follows that $T$ lies on line $CD$. Evidently, line $XY$ is the polar of $T$ w.r.t. $\Gamma$, so $XY \perp TN$. By Pascal's Theorem on the cyclic hexagon $(ACCBDD)$ we conclude $X, I, Y$ are collinear. From $(A, B; F, T)=-1$, we get $F$ lies on $XY$. It suffices to show that $MF \perp TN$ in order to prove that $M$ lies on $XY$. Let $MF$ meet the circumcircle of $ABE$ at $Z$. Projecting from $Z,$ $$-1=(ABMN)=(ZA, ZB; ZM, ZN)=(A, B; F, ZN \cap AB),$$so $Z, N, T$ are indeed collinear. The result is established. $\, \square$

anantmudgal09 wrote: Truly Projective! Sharygin 2012/22 wrote: A circle $\omega$ with center $I$ is inscribed into a segment of the disk, formed by an arc and a chord $AB$. Point $M$ is the midpoint of this arc $AB$, and point $N$ is the midpoint of the complementary arc. The tangents from $N$ touch $\omega$ in points $C$ and $D$. The opposite sidelines $AC$ and $BD$ of quadrilateral $ABCD$ meet in point $X$, and the diagonals of $ABCD$ meet in point $Y$. Prove that points $X, Y, I$ and $M$ are collinear. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.182cm); 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 = -4.26, xmax = 16., ymin = -5.2, ymax = 6.34; /* image dimensions */ /* draw figures */ draw(circle((4.732328757277367,1.562979228060088), 2.8976577985276055)); draw((7.42,0.48)--(2.82,3.74)); draw((2.82,3.74)--(1.96,0.72)); draw(circle((3.9910461468119207,2.4068653196747483), 1.774428683787815)); draw((2.82,3.74)--(4.605082053223209,-1.3318832891719834)); draw((1.96,0.72)--(5.672801008275183,1.8409167886282969)); draw((2.5786488371805976,1.3327466141327784)--(7.42,0.48)); draw((4.2279602060716375,2.9663227177336378)--(1.96,0.72)); draw((4.2279602060716375,2.9663227177336378)--(7.42,0.48)); draw((3.9910461468119207,2.4068653196747483)--(2.5786488371805976,1.3327466141327784)); draw((2.5786488371805976,1.3327466141327784)--(4.605082053223209,-1.3318832891719834)); draw((4.605082053223209,-1.3318832891719834)--(5.672801008275183,1.8409167886282969)); draw((5.672801008275183,1.8409167886282969)--(3.9910461468119207,2.4068653196747483)); draw((2.82,3.74)--(-0.49515241066053656,0.8279187872817818)); draw((-0.49515241066053656,0.8279187872817818)--(4.605082053223209,-1.3318832891719834)); draw((-0.49515241066053656,0.8279187872817818)--(7.42,0.48)); draw(shift((4.605082053223209,-1.3318832891719834))*xscale(3.3476385857297597)*yscale(3.3476385857297597)*arc((0,0),1,15.637236155935852,170.289472327311), linetype("2 2")); draw((4.859575461331523,4.457841745292159)--(2.7419121873807155,-0.5428846316537113), linetype("2 2")); draw((-0.49515241066053656,0.8279187872817818)--(5.672801008275183,1.8409167886282969)); /* dots and labels */ dot((2.82,3.74),dotstyle); label("$E$", (2.9,3.94), NE * labelscalefactor); dot((1.96,0.72),dotstyle); label("$A$", (1.6,0.46), NE * labelscalefactor); dot((7.42,0.48),dotstyle); label("$B$", (7.6,0.46), NE * labelscalefactor); dot((4.859575461331523,4.457841745292159),linewidth(3.pt) + dotstyle); label("$M$", (5.02,4.52), NE * labelscalefactor); dot((4.605082053223209,-1.3318832891719834),linewidth(3.pt) + dotstyle); label("$N$", (4.52,-1.68), NE * labelscalefactor); dot((3.913124522765808,0.6341483726256787),linewidth(3.pt) + dotstyle); label("$G$", (4.,0.76), NE * labelscalefactor); dot((3.9910461468119207,2.4068653196747483),linewidth(3.pt) + dotstyle); label("$I$", (4.08,2.16), NE * labelscalefactor); dot((2.5786488371805976,1.3327466141327784),linewidth(3.pt) + dotstyle); label("$C$", (2.58,1.52), NE * labelscalefactor); dot((5.672801008275183,1.8409167886282969),linewidth(3.pt) + dotstyle); label("$D$", (5.76,1.96), NE * labelscalefactor); dot((-0.49515241066053656,0.8279187872817818),linewidth(3.pt) + dotstyle); label("$T$", (-0.48,1.04), NE * labelscalefactor); dot((4.2279602060716375,2.9663227177336378),linewidth(3.pt) + dotstyle); label("$X$", (4.38,2.98), NE * labelscalefactor); dot((3.4697231623008404,1.1757944352612038),linewidth(3.pt) + dotstyle); label("$Y$", (3.22,1.3), NE * labelscalefactor); dot((3.2526459147705995,0.6631803993507428),linewidth(3.pt) + dotstyle); label("$F$", (3.36,0.3), NE * labelscalefactor); dot((2.7419121873807155,-0.5428846316537113),linewidth(3.pt) + dotstyle); label("$Z$", (2.66,-0.92), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $E$ be the tangency point of $\omega$ with the arc and $G$ be the tangency point of $\omega$ with $AB$. Let $EF$ ($F$ lies on $AB$) be a symmedian in triangle $AEB$. Let $T$ be a point on line $AB$ with $TE$ tangent to the circumcircle of $AEB$. Dilation at $E$ sending $\omega$ to the circumcircle of $ABE$ maps $G$ to $N$. Note that $$NC^2=ND^2=NG\cdot NE=NA^2,$$so $A, B, C, D$ lie on a circle (call it $\Gamma$) with center $N$. Since $\angle ICN=\angle IDN=90^{\circ}$, lines $IC$ and $ID$ are tangent to $\Gamma$. From the radical axis theorem for the circles $\odot (AEB), \omega,$ and $\Gamma,$ it follows that $T$ lies on line $CD$. Evidently, line $XY$ is the polar of $T$ w.r.t. $\Gamma$, so $XY \perp TN$. By Pascal's Theorem on the cyclic hexagon $(ACCBDD)$ we conclude $X, I, Y$ are collinear. From $(A, B; F, T)=-1$, we get $F$ lies on $XY$. It suffices to show that $MF \perp TN$ in order to prove that $M$ lies on $XY$. Let $MF$ meet the circumcircle of $ABE$ at $Z$. Projecting from $Z,$ $$-1=(ABMN)=(ZA, ZB; ZM, ZN)=(A, B; F, ZN \cap AB),$$so $Z, N, T$ are indeed collinear. The result is established. $\, \square$ Perfect approach by Anant.I liked this proof.

it easy to prove that $NB=ND=NC=NA$ so the quadrilateral $ABCD$ is cyclic with circumcenter $N$ . let $Z=the intersection of CD and AB (may be the point in the infiny)$ the polar of $Z$ wrt (ACDB) is $XY$ and since $NB=NA$ and $MA=MB$ ; $MA\perp NA$ and $MB\perp NB$ $\implies$ so $Z\in the polar of M wrt (ACBD)$ so $N\in XY$ The tangents from $N$ touch $\omega$ in points $C$ and $D$ means that The tangents from $I$ touch $(ACBD)$ in points $C$ and $D$ ; $Z\in the polar of I wrt (ACBD)$ so $I\in XY$ sincerely

Observe by Fact 5 that $A,B, C,D$ are concyclic with center $N$. So $I$ is the intersection of the tangents to its circumcircle at $C, D$. So $X , Y, I$ are collinear on the polar of $T'=AB\cap CD$ wrt $(N)$. If $T = AB \cap \text{common tangent of (I), original circle}$, we can easily see that $DCT$ is collinear by radical axes, so $T\equiv T'$. Also, $TN \perp XY$ due to polarity. As the circle centered at $M$ and passing through $A$ is orthogonal to $(N)$, $M$ is the pole of $BC$ wrt $(N)$. So it suffices to show that $M \in XY$ which is equivalent to $\text{pole of XY} \in BC$, which is obvious as $T$ is the pole of $XY$ by Brokard's theorem.