Let $N$ be the projection of $I$ onto $XY$. $\rightarrow ND=NX+XC=XC+\frac{MX+XY-MY} {2}=XC+\frac{XD+XY-YC}{2}=\frac{CD}{2}$....Done.

$\textrm{Easy for Iran 3rd round}$ Let tangents of $M$ to $\omega_1$ and $\omega_2$ be $S,R$ respectively. $Q$ be exsimilicenter of $\omega_1 , \omega_2$ Notice $M$ is on radical axis of $\omega_1 , \omega_2$ so $R,S,Q$ are collinear and in fact the inversion centered at $Q$ sending $\omega_1 \rightarrow \omega_2$ sends $R \rightarrow S$. By PoP $QR \cdot QS = QA \cdot QB = QC \cdot QD \implies (RSCD)$ is cyclic (or easily by the same inversion) now notice $I$ is on $\overline{YO_1} , \overline{XO_2} \implies I$ is on perpendicular bisector of $CS , RD$, so obviously $I$ is the center of $\odot(RSCD) \implies IC=ID$ $\boxed{\mathcal{Q.E.D}}$

If we get $E , F$ be the intersection of $MX , MY$ with $\omega_1 , \omega_2$ . $ XE = XC \implies XI \perp EC \implies IE = IC $ $ YF = YD \implies YI \perp DF \implies ID = IF $ $ \angle IME = \angle IMF , ME=MB=MA=MF \implies IE=IF \implies IC=ID \implies \blacksquare $

$\textrm{Beautiful Problem}$ We need to prove that : $IC=ID$ $\implies$ $I$ is perpendicular on the midpoint of $CD$. Let the midpoint of $CD$ be $N$ and let the tangent from $M$ to $\omega_1 , \omega_2$ touch the two circles at $Q$ and $P$ respectively. Now since $M$ is the midpoint of $AB$ $\implies$ $MA=MB$ $\implies $ $MP=MQ$. also $YP=YD$ and $XQ=XC$. Now we mark points $S$ and $T$ on segments $MY$, $MX$ such that $PS$ = $QT$= $\frac{CD}{2}$. Now we note that, $$MS=MT$$$$YS=YN=(YD-ND)$$$$XT=XN=(XC-CN)$$ Now its easy to see that $S$ , $N$ and $T$ are the tangency points of the incircle of triangle $MXY$ implying that $I\perp CD$ on $N$.$\blacksquare$