Any solution

I asumed that $\omega$ is the incircle of $\triangle ADC$ Let $I$ the incenter of $\triangle ADC$, $N=\omega\cap CD$, $Z=\omega\cap MX$, $W=\omega\cap MY$ and $E=MD\cap XY$, by angle-chasing we get: $\measuredangle BIC$ $=$ $\measuredangle BDC$ $=$ $2(\measuredangle IAC+\measuredangle ICA)$ $\Longrightarrow$ $BDCI$ is cyclic and $\overline{KNM}$ is the Simson's line of $I$ WRT $\odot (BDIC)$, hence $K$, $N$ and $M$ are collinear. So from $D$ belongs in the polar of $M$ WRT $\omega$, by la Hire's theorem we get: the polar of $M$ through $D$ $\Longrightarrow$ $D$, $Z$ and $W$ are collinear, hence $M(D,N,Z,W)=-1$ $\Longrightarrow$ $M(E,K,X,Y)=-1$ $\Longrightarrow$ $D(E,K,X,Y)=-1$, so from $\measuredangle EDK=90^{\circ}$ we get $DE$ is a external bisector of $\measuredangle XDY$ and $MX=MY$ combining both results we obtain that: $X$, $D$, $M$ and $Y$ are cyclic.

Nice solution! Mine is the same except from a different proof for the collinearity of $K,N,M$. Note that since $ABC$ is isosceles,$KL \parallel BC$ and $KB=LC=NC$.So $\frac{CM}{BM}\frac{DN}{NC}\frac{KB}{KD}=1$ and by the reverse of Menelaus theorem for the triangle $DBC$ we obtain that $K,N,M$ are collinear.

Really great solution FabrizioFelen and john111111

I have a new solution for this question which is a simple angle-chasing and using symmetry in the question ( I draw the other line just like CD)

Note that $KL$ is parallel to $BC$ and $MA$ bisects $\angle{XMY}$. Let $U$ be the contact point of $\omega$ and $DC$. If $KU \cap AM = M'$, then it is well-known (see page 8) that $\angle{CM'A}=90^{\circ}$, thus $M' \equiv M$, which means that $M , U , K$ are colinear. Applying dDIT

on $ADUC \cap \omega$ with perspectivity at $M$, we have that $\{MX, MY; MD, MC; MU, MA\}$ are reciprocal pairs of an involution, $f$ of the plane. Taking perspectivity at $M$, we have: $$(MX, MY; MD, MK)=(MX, MY; MD, MU)=(Mf(X), Mf(Y); Mf(D), Mf(U))=(MY, MX; MC, MA)=-1$$Now projecting this onto line $\overline{XY}$ we see that $DM$ bisect $\angle{XDY}$ and because $MX \equiv MY$, we have that $M$ is the midpoint of the arc $XDY$ in $\odot(XDY)$, thus $M, D, X, Y$ are concyclic.

FabrizioFelen wrote: I asumed that $\omega$ is the incircle of $\triangle ADC$ Let $I$ the incenter of $\triangle ADC$, $N=\omega\cap CD$, $Z=\omega\cap MX$, $W=\omega\cap MY$ and $E=MD\cap XY$, by angle-chasing we get: $\measuredangle BIC$ $=$ $\measuredangle BDC$ $=$ $2(\measuredangle IAC+\measuredangle ICA)$ $\Longrightarrow$ $BDCI$ is cyclic and $\overline{KNM}$ is the Simson's line of $I$ WRT $\odot (BDIC)$, hence $K$, $N$ and $M$ are collinear. So from $D$ belongs in the polar of $M$ WRT $\omega$, by la Hire's theorem we get: the polar of $M$ through $D$ $\Longrightarrow$ $D$, $Z$ and $W$ are collinear, hence $M(D,N,Z,W)=-1$ $\Longrightarrow$ $M(E,K,X,Y)=-1$ $\Longrightarrow$ $D(E,K,X,Y)=-1$, so from $\measuredangle EDK=90^{\circ}$ we get $DE$ is a external bisector of $\measuredangle XDY$ and $MX=MY$ combining both results we obtain that: $X$, $D$, $M$ and $Y$ are cyclic. I didn't solve this, but a slight correction: $M(D, N, Z, W) = -1$ should be $M(D, Q, Z, W) = -1$, where $Q = KN \cap ZW$. In addition, I found it easier to show that $BDIC$ was cyclic by showing that angles $DBC$ and $DIC$ were supplementary.

Another way to prove the collinearity of $K, N, M$ is the following pretty well-known lemma. $\textbf{Lemma}$ Let the incircle of $\triangle ABC$ centered at $I$ touches $BC, AC, AB$ in $D, E, F$. Denote $P = EF \cap BI$. Then $\angle BPC = 90 ^ \circ $. $\textit{Proof:}$ Let $S = EF \cap BC$. Assume $S, B, D, C$ are in this order. Then $(S, D; B, C) = -1$. Now notice $\triangle FBP \equiv \triangle DPB$ so $BP$ is the internal bisector $\angle SPD$, thus $CP$ must be its external bisector. Now apply the lemma for the $\triangle ADC$, line $KN$ and bisector $AM$ to get the desired property.