Suppose that $ABCDGE$ is our complete quadrilateral with Miquel point $\mathcal{M}$ and $F$ the intersection of diagonals. Drop the incircles and only use that $EP=EQ$. Animate $P$ on $\overline{AD}$ and let $P_{\infty}$ be the point at infinity on the external bisector of $\angle{AEB}$. Clearly $P\mapsto Q$ is projective because $Q=PP_{\infty} \cap BC$. Analogously $P\mapsto M$ and $P\mapsto N$ are projective. Note that $P\mapsto PY$ is projective, because it is a rotation by $90^{\circ}$ about $P$ of line $AD$. Because $Y$ is the intersection of line $PY$ and the bisector of $\angle{AEB}$ we have that $P\mapsto Y$ is projective, analogously $M\mapsto X$ is projective, thus $P\mapsto M \mapsto X$ is projective. It is enough to check the assertion of the problem for three positions of $P$. Let $P_1$ be a point on $AD$ such that $FP_1 \mid \mid l$, then if $P\equiv P_1$, we have that $EPXQ\mathcal{M}$ is cyclic by Miquel, following that $XF\perp EG$, thus $X,F$ and $O$ are colinear, but $F\equiv Y$ in this case. Analogously we do the case $P\equiv E$. Now if $P\equiv R_{\infty}$, where $R_{\infty}$ is the point at infinity on $AD$, because the bisector of $\angle{AEB}$ and $\angle{AFB}$ are parallel, we have that $X\equiv Y$.

Denote by $\omega$ the circle that touches $\Gamma$ internally, $AE, BE$ where $E$ is the intersection of the diagonals. By one consequence Sawayama-Thebault for $\Delta ABD$ and $E$, the chord that $\omega$ makes with $AE,BE$ passes through $J$, analogiously through $I$ when we look from $\Delta ABC$ and $E$, so $M,N$ are those points and $X$ is center of $\omega$. Analogiously, let $k$ touches $\Gamma$ internally at a point on the arc $AB$ without $C,D$, and touches $AD,BC$. We similarly prove $Y$ is the center of $k$ (this needs a bit more explanation because of the positions of the points since the intersection of $AD, BC$ lies "outside"). So we only need to prove that $k$ and $\omega$ touch $\Gamma$ at the same point. Let $\omega$ touches $\Gamma$ at $K$. Another result from Sawayama-Theabault gives us that $J,M,A,K$ are concyclic, so $\angle KBQ=\pi-\angle KAM=\angle KJM=\pi-\angle KJQ$, so $J,K,B,Q$ are also concyclic. But the same Sawayama for $k$ also says $J,L,B,Q$ are concyclic where $L$ is the touchpoint of $k$ and $\Gamma$, so $K=L$ since both of them lie on the same arc $AB$, that we wanted to prove. To conclude, those lemmas are used from "Sawayama configuration": Let $ABC$ be a triangle inscribed in $k$, $I$ is the incenter and $D$ lies on the line $BC$. Circle $\omega$ touches $k$ at $T$ internally, $AD$ at $Y$ and $BC$ at $X$. $1) A,Y,I,T$ are concyclic; $2) I$ lies on $XY.$

Lemma (w.l.o.g suppose the situation is in the figure below) Let $ABCD$ be a quadrilateral inscribes $(O),$ a line $\ell$ cuts $CD, AD, BD, AC, BC, AB$ at $E, F, G, H, I, J,$ resp A point $S \in (O),$ $SE, SF, SG, SH, SI, SJ$ cuts $(O)$ again at $R, M, P, Q, N, T$ Then $\ell, TR, MN, PQ$ are concurrent Solution: Let $\ell \cap (O) = K, L$ By Desargues Involution theorem, $(E, H);(K, L);(F,I);(G,H)$ are reciprocal pair, so $(R,K;D,P)=S(E,K;F,G)=S(J,L;I,H)= T, L, N, Q \Rightarrow \ell, TR, MN, PQ$ are concurrent Remark: Let $X = AP \cap DT, X'=TD \cap BM$ Apply Pascal theorem for $\binom{A,D,S}{T,P,B} \Rightarrow X \in GJ = \ell$ for $\binom{T,M,A}{B,D,S} \Rightarrow X' \in FJ = \ell$ $\Rightarrow X \equiv X',$ so $AP, DT, MS$ are concurrent at $X \in \ell$ Anagously we can prove $AN, TC, BQ$ are concurrent at $Y \in \ell$ The locus of $T$ when $S$ moves is $(O),$ so we get this problem Let $ABCD$ be a quadrilateral inscribes $(O),$ $T$ is a point on $(O),$ let $X, Y \in DT, CT$ $XY$ cuts $AD, BD, AC, BC$ at $, F, G, H, I$ resp $BX, AY, AX, BY$ cuts $(O)$ again at $M, N, P, Q,$ then $MF, PG, QH, NI$ are concurrent on $(O)$ Back to the main problem Let $E, F, G, H, K$ be the midpoint of arc $\overarc{AB}, \overarc{BC}, \overarc{AD}, \overarc{AC}, \overarc{BD}$ of $(O)$ By Lemma+Remark, $MH, NK, PG, QF$ are concurrent at $T \in (O)$ Note that $(YP \parallel OG) \perp AD; (YQ \parallel OF) \perp BC \Rightarrow Y \in OT$ Similarly, $X \in OT$ So $X, O, Y$ are collinear $\blacksquare$ P/s: I forgot two more green line in the first figure

Firstly, use Sawayama-Thebault in $\triangle ABD$ to obtain that the circle centered at $X$ and going through $M, N$ is simply the circle tangent to $AC, BC$ and $\Gamma$ which is on the opposite side of $AC$ as $D$ and the opposite side of $BD$ as $C$. Analogously, $Y$ is the center of the circle which is tangent to $AD, BC$ and arc $AB$ (not containing $C, D$). Now, let $Z$ be the point where the circle centered at $X$ is tangent to $\Gamma.$ Let $M_{ac}, M_{ad}$ be the midpoints of arcs $AC, AD$ not containing $B$. Then, by Pascal's Theorem on $M_{ac}XM_{ad}CAD,$ we know that $M, XM_{ad} \cap AD, E$ are collinear, where $E$ is the $C-$excenter of $\triangle ACD.$ It would suffice to show that $E \in IJ,$ which would imply that $XM_{ad} \cap AD$ and hence the result. However, this is easy when applying the Japanese Rectangle Theorem and also applying Fact 5 in $\triangle ABD, \triangle ACD.$ $\square$

Let $\cal C$$_1$: the mixtlinear incircle that is tangent to $ AD,BC,\Gamma$ at $P',Q',T_1$ ; $\cal C$$_2$: the mixtlinear incircle that is tangent to $ AC,BD,\Gamma$ at $M,N,T_2$ By Sawayama we get $P=P',Q'=Q,M'=M,N'=N$ whence $X,Y$ are the centers of $\mathcal{C}_1,\mathcal{C}_2$ so by the lemma from here we deduce that $T_1=T_2$ hence $T_1,X,Y,O$ are collinear.

This is problem 27198 for 9th graders in Gazeta Matematica nr. 3/2016.

This is actually the famous Poncelet Coaxial Theorem. It claims that the following three circles: 1. The circle $\Gamma$ 2. The circle centered at $X$ and tangent to $AC,BD$ at $M,N$ 3. The circle centered at $Y$ and tangent to $AD,BC$ at $P,Q$ are coaxial. Hence their centers $O,X,Y$ are colinear.