Let $\overline{IO}$ intersect $O_{1}$ at another point $E$. Note that the tangent to $O_{1}$ at $I$ is parallel to $\overline{CD}$ by Reim's theorem. Let $P_{\infty}$ be the point at infinity along line $CD$. Let $CD$ interest $IO$ at $M'$. Note that $AEBI$ is harmonic any EGMO lemma 9.9, thus taking perspectivity at $I$ yields $-1=(A,B;E,I)\stackrel{I}{=}(C,D;M',P_{\infty})$. Thus $M'$ is the midpoint of $\overline{CD}$ and $M=M'$, as desired. $\blacksquare$

$IO$ is the $I$-symmedian of $IAB$. $CD$ is antiparallel to $AB$, therefore $IO$ is the median of $ICD$. Done

Let $J=IO \cap (O_1)$ and $M'=IO\cap DC$, we will sjow that $M'=M$. Observe that $IO$ is the $I$-symmedian of triangle $\triangle IAB$ and $ID\cdot IB= IC\cdot IA$, so \[\frac{DM'}{M'C}=\frac{ID\sin \angle DIM}{IC \sin \angle AIM}=\frac{IA}{IB}\cdot \frac{\sin \angle BIJ}{\sin \angle AIJ}= \frac{AJ}{BJ}\cdot \frac{\sin \angle BAJ}{\sin \angle ABJ}=1.\]Hence $M'=M. \blacksquare$