let $M'$ be the midpoint of $CD$ $\omega $ touches $CD ,AB,AC,BC$ at $K',N,Q,P$ and $O$ is the center of $\omega$ note that $O-P-Q$and $M-,M'-O$ are collinear and $PQ \perp MM'$ $MK=MN=M'K'$ so $K'$ is the reflection of $K$ across $PQ$ so $MK \cap PQ \cap DC=R$ since $(R,K';C,D)$ that is from the triangle $CD\infty AD$ since $MM'K'K$ is an isosceles trapezoid $RM'.RK'=RC.RD=RK.RM$ and we win

Let $O$ be centre of $\omega$. Let $\omega$ touch $AB$ at $D$. Let $R$ and $S$ be feet of $\perp$ from $C$ and $D$ onto $AD$ and $BC$ respectively. Then reflecting in $OM$ we get that problem is equivalent to proving that $MDRS$ is cyclic. Note that $\triangle OAM\sim\triangle CAD$. And $OD$ and $CR$ are altitudes of $\triangle OMA$ and $\triangle CDA$. So $\frac{AR}{RD}=\frac{AD}{DM}$ which means $MD\parallel DR$. Now $\angle RSD=\angle CDS$ as $CRDS$ is rectangle. Also $MD=MS\implies\angle MDS=\angle MSD\implies\angle MSR=\angle MDC=\angle AMD=\angle ADR\implies MDRS$ is cyclic as desired. Q.E.D.

Any solution using complex numbers?

Ali3085 wrote: let $M'$ be the midpoint of $CD$ $\omega $ touches $CD ,AB,AC,BC$ at $K',N,Q,P$ and $O$ is the center of $\omega$ note that $O-P-Q$and $M-,M'-O$ are collinear and $PQ \perp MM'$ $MK=MN=M'K'$ so $K'$ is the reflection of $K$ across $PQ$ so $MK \cap PQ \cap DC=R$ since $(R,K';C,D)$ that is from the triangle $CD\infty AD$ since $MM'K'K$ is an isosceles trapezoid $RM'.RK'=RC.RD=RK.RM$ and we win May I ask what does $(R,K’;C,D)$ mean?

It is a cyclic quadrilateral, and the notation denotes the quadrilateral's circumcircle.

$MK\cap AD=E$. Let the midpoint of $CD$ be $M^{\prime}$ and $\omega$ be tangent to $CD$ at $K^{\prime}$. Let the other tangent from $M^{\prime}$ to $\omega$ cut $AB$ at $Y$ and $MK\cap CD = X$. We see that $\omega$ is also incircle of $MXM^{\prime}Y$ and it's a parallelogram, so $MXM^{\prime}Y$ is a rhombus. We conclude that $KK^{\prime} \parallel MM^{\prime} \parallel AD$ and $MK=M^{\prime}K^{\prime}$, so by Thales's, we see that $ME=M^{\prime}D$. Writing $MK=b, KE=a, MB=a+b$, etc. We can easily verify that $XD\cdot XC = XK^{\prime}\cdot XM^{\prime}$, hence these points are cyclic

Here is my bashy solution: Let $L=AD\cap KM$, $G=CD\cap KM$. Let $F=AB\cap w$, $H=CD\cap w$, $J=AD\cap w$. Firstly note that just $BMLDC$ is a tangential pentagon with incircle $w$, thus we have $FB=HD=DJ=JL=LK$ and $KM=MF$ and $GK=GH$. Claim. $AL=2MF$. $$AL=AD-LD=AD-2DH=2MB-2BF=2MF.$$We need to show that $GK\cdot GM=GD\cdot GC$, i.e. $$GH^2+ GH\cdot KM=GD^2+ GD\cdot CD\Longleftrightarrow$$$$GH(GD+DH)+ GH\cdot MB-GH\cdot FB=GD^2+ GD\cdot CD\Longleftrightarrow$$$$GH\cdot GD+ GH\cdot MB=GD^2+ GD\cdot CD\Longleftrightarrow$$$$\frac{GH\cdot MB}{GD}=GD+CD-GH=AF\Longleftrightarrow$$$$1+\frac{HD}{GD}=1+\frac{KM}{AM}\Longleftrightarrow$$$$\frac{AM}{GD}=\frac{KM}{HD}$$ Now we use the fact that $\triangle ALM\sim\triangle DLG$, thus we have $$\frac{AL}{DL}=\frac{AM}{GD}$$Hence, we need $\frac{AL}{DL}=\frac{KM}{HD}$, but that is true since $DL=2HD$ and $AL=2KM$. We are done.

Here is a finish with complex numbers. Obviously the incircle will be the unit circle and since $ABCD$ is a rhombus we can set the coordinates in the following way: $$d=-b \in \mathbb{R}$$$$c=-a \in i\mathbb{R}$$Let $X,Y,Z,W$ be the touch points of the incircle with $AB,BC,CD,DA$, respectively. In a similar manner we set $x,y,z,w$ in the following way: $$z=-x$$$$w=-y$$But we can take wlog that $y=1$, since we can get any rhombus we want by just moving $X$ around. Thus we have that $a=-\frac{2x}{x-1}$, $b=\frac{2x}{x+1}$, $c=\frac{2x}{x-1}$, $d= -\frac{2x}{x+1}$. This implies that $m=\frac{1}{2}\left( a+b \right)=-\frac{2x}{x^2-1}$. Since $K$ is inside $ABCD$, then we know that the coordinates of $k$ is equal to $\overline{m}=\frac{1}{m}=k$. Thus we just need to check if: $$\frac{c-d}{k-d} : \frac{c-m}{k-m} \in \mathbb{R}$$but we see that this is true when we plug in what we got. Thus we have that $CDKM$ is a cyclic quadrilateral.

Let $KM$ meet $AD$ at $T$ and $\omega$ meet $AB,CD,DA$ at $G,F,E$ and $IE$ meet $MK$ at $S$. Claim $: S,D,C$ are collinear. Proof $:$ Note that we only need to prove $TE = ED$. Note that $TE = TK , ED = DF$ so we need to prove $KF || TD$. Note that $\angle TKF = \angle KGF = \angle KGI = \angle KMI \implies MI || KF$ and Note that $MI || AD$ cause $BI = ID , BM = MA$. Note that $\angle MIS = \angle MIA + \angle AIS = \angle IAE + AIE = \angle 90 = \angle MKI \implies SK.SM = SI^2$ and $\angle ICD = \angle 90 - \angle IDC = \angle 90 - \angle IDE = \angle SID \implies SD.SC = SI^2$ so $SK.SM = SD.SC$ so $CDKM$ is cyclic.

EulersTurban wrote: But we can take wlog that $y=1$, since we can get any rhombus we want by just moving $X$ around . How would you take $y=1$ while $b,d \in \mathbb{R}$ and clearly $Y$, $D$ and $B$ are not collinear?