parmenides51 wrote: A quadrilateral $ABCD$ is inscribed into a circle $\omega$ with center $O$. Let $M_1$ and $M_2$ be the midpoints of segments $AB$ and $CD$ respectively. Let $\Omega$ be the circumcircle of triangle $OM_1M_2$. Let $X_1$ and $X_2$ be the common points of $\omega$ and $\Omega$ and $Y_1$ and $Y_2$ the second common points of $\Omega$ with the circumcircles of triangles $CDM_1$ and $ABM_2$. Prove that $X_1X_2 // Y_1Y_2$. Let $AB \cap CD = T$. Note that $T \in (OM_1M_2)$. Claim: $M_2Y_2, AB, (CDM_1)$ are concurrent. Proof : Let $(CDM_1) \cap AB = Z_1$. Now we will prove that $Z_1$ lies on $M_2Y_2$. By easy length chasing, we get $\text{Pow}(Z_1, (ABM_2)) = \text{Pow}(Z_1,(ABM_2))$. So, $Z_1 \in (M_2Y_2)$. $\square$. Now, by applying converse of radical axis theorem (or using PoP), we get $M_1,M_2,Z_1,Z_2$ to be concyclic. So, $\measuredangle{Z_1M_1Z_2} = \measuredangle{TZ_2Y_2} = \measuredangle{Z_2M_2Z_1}$. This gives $TY_1 = TY_2$ which combined with $OY_1 = OY_2$ gives the desired result. $\blacksquare$.

We present two completely different solutions, one by inversion, and one by radical axis. Solution 1: Let $P=AB\cap CD$, and let $\ell$ be its polar with respect to $\omega$. Let $Z_1=\ell\cap AB$, $Z_2=\ell\cap CD$, and let $T_1$ be the intersection of the tangents at $A$ and $B$, $T_2$ the intersection of the tangents at $C$ and $D$. Clearly we have $T_1,T_2\in\ell$. We invert about $\omega$. Note that $M_i$ inverts to $T_i$, so $\Omega$ inverts to $T_1T_2=\ell$. Let $Y_i'$ be the inverse of $Y_i$, and since $OP\perp\ell$, it suffices to show that $X_1Y_1'=X_2Y_2'$. We see that $Y_1'=(CDT_1)\cap\ell$, so \[Z_2Y_1'\cdot Z_2T_1 = Z_2C\cdot Z_2D = Z_2X_1\cdot Z_2X_2,\]and similarly \[Z_1Y_2'\cdot Z_1T_2 = Z_1X_1\cdot Z_1X_2.\]The problem is now reduced to a 1D coordinate bash since $Z_i$ and $T_i$ are harmonic conjugates in $X_1X_2$. Let $X_1=1$, $X_2=-1$. Note that $t_i=1/z_i$, so \[(1-z_2)(z_2+1)=(1/z_1-z_2)(z_2-y_1'),\]so \[y_1'=\frac{z_2-z_1}{1-z_1z_2}.\]A similar calculation gives \[y_2'=\frac{z_1-z_2}{1-z_1z_2},\]so $y_1'=-y_2'$, as desired. Solution 2: Adopt notations from the previous solution. Note that by radical axis on $\omega$, $\Omega$, and $(CDM_1)$, we have that \[Y_1=M_1Z_2\cap\Omega,\]and similarly \[Y_2=M_2Z_1\cap\Omega.\]By Pascal on $M_1Y_1Y_2M_2PP$, we have that $Y_1Y_2\cap\ell_P\in\ell$, where $\ell_P$ is the tangent to $\Omega$ at $P$. However, $OP$ is a diameter of $\Omega$, so this says that $Y_1Y_2\parallel \ell$, as desired.

Let $X = \overline{AB}\cap \overline{CD}$; then $\Omega$ has diameter $\overline{OX}$, and $\overline{X_1X_2}$ is the polar of $X$ with respect to $\omega$, which we denote by $\ell$. Furthermore, let $(CDM_1)$ and $(ABM_2)$ intersect $\overline{AB}$ and $\overline{CD}$ again at $R$ and $S$. [asy][asy] size(300); defaultpen(fontsize(10pt)); pair A, B, C, D, O, M1, M2, R, S, Z1, Z2, M, X, X1, X2; A = dir(150); B = dir(70); C = dir(340); D = dir(200); X = extension(A, B, C, D); M1 = midpoint(A--B); M2 = midpoint(C--D); O = (0,0); X1 = IP(CP(midpoint(X--O), O), circumcircle(A, B, C), 0); X2 = IP(CP(midpoint(X--O), O), circumcircle(A, B, C), 1); R = extension(X1, X2, A, B); S = extension(X1, X2, C, D); Z1 = IP(circumcircle(C, D, M1), Line(X1, X2, 20), 1); Z2 = IP(circumcircle(A, B, M2), Line(X2, X1, 20), 0); M = foot(O, R, S); draw(A--B--C--D--cycle^^A--X--D, orange); draw(circumcircle(A, B, C), red); draw(CP(midpoint(X--O), O), lightblue); draw(circumcircle(A, B, M2), heavycyan+dashed); draw(circumcircle(C, D, M1), heavycyan+dashed); draw(Z1--Z2, heavygreen); dot("$A$", A, dir(120)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(240)); dot("$X$", X, dir(200)); dot("$O$", O, dir(0)); dot("$R$", R, dir(130)); dot("$S$", S, dir(225)); dot("$M_1$", M1, dir(80)); dot("$M_2$", M2, dir(300)); dot("$Z_1$", Z1, dir(290)); dot("$Z_2$", Z2, dir(90)); dot("$M$", M, dir(190)); [/asy][/asy] First, by Power of a Point at $X$, we have $(\overline{XR}; \overline{AB}) = (\overline{XS}; \overline{CD}) = -1$, so $R$ and $S$ lie on $\ell$. Now consider an inversion $\Gamma$ at $X$ with power $XA\cdot XB$; let $Z_1$ and $Z_2$ be the images of $Y_1$ and $Y_2$. Note that: $\Gamma$ swaps $\Omega$ and $\ell$, and so $O$ is sent to $M$, the midpoint of $\overline{X_1X_2}$. $\Gamma$ fixes $(ABM_2)$ and $(CDM_1)$, and so $Z_1 = (CDM_1)\cap \ell$ and $Z_2 = (ABM_2)\cap \ell$. It suffices to show that: Claim: $M$ is the midpoint of $\overline{Z_1Z_2}$. Proof. Since $\overline{OM}\perp \overline{RS}$, we have \begin{align*} RS\cdot (MR-MS) &= MR^2-MS^2 = OR^2 - OS^2 \\ &= (R^2- RA\cdot RB) - (R^2 - SC\cdot SD) \\ &= SC\cdot SD - RA\cdot RB \\ \implies MR - MS &= \frac{1}{RS}(SC\cdot SD - RA\cdot RB) = SZ_1 - RZ_2, \end{align*}which forces $M$ to be the midpoint, as desired. $\blacksquare$