Lemma: $ I$ is incenter of a $ \triangle ABC,$ $ M$ is midpoint of $ BC,$ and $ E \in AB, F \in AC$ are points on its sidelines, such that $ \overline{EB} = \overline {FC}.$ Parallel $ m \parallel AI$ through $ M$ cuts $ EF$ at its midpoint $ D.$ (Posted before.) Parallel $ m \parallel AI$ through $ M,$ cutting the lines $ AB, AC$ at $ K, L,$ is perimeter cleaver of the $ \triangle ABC$ $ \Longrightarrow$ $ \overline{KB} = \overline {LC} = \frac {_1}{^2} ({\overline{AB} + \overline{AC}),}$ $ \overline{AK} = \overline{LA} = \frac {_1}{^2} (\overline{AB} - \overline{AC})$ $ \Longrightarrow$ $ \overline{KE} = \overline{KB} - \overline{EB} = \overline{LC} - \overline{FC} = \overline{LF}.$ By Menelaus for the $ \triangle AEF$ cut by transversal $ DKL,$ we have $ \frac {\overline{DF}}{\overline{DE}} = \frac {\overline{LF}}{\overline{LA}} \cdot \frac {\overline{KA}}{\overline{KE}} = - 1$ $ \Longrightarrow$ $ D$ is midpoint of $ EF.$ Problem: $ (O_1) \cong (O_2)$ and $ \angle APC = \angle BPD$ $ \Longrightarrow$ $ AC = BD.$ Let $ AB, CD$ meet at $ X.$ From cyclic $ PQAC, PQDB,$ $ \angle CAQ = \angle DPQ = \angle DBQ$ and $ \angle BDQ = \angle APQ = \angle ACQ$ $ \Longrightarrow$ $ \triangle QAC \cong \triangle QBD$ by ASA, their Q-altitudes are equal $ \Longrightarrow$ $ XQ$ bisects the angle $ \angle AXD = \angle CXB.$ (Also posted before.) By the above lemma, $ MN \parallel XQ.$ Midpoint $ K$ of $ XP$ is on the Newton line $ MN$ of complete quadrilateral $ (AB, CD, AC, BD)$ with diagonals $ AD, BC, XP.$ But $ KO \parallel XQ$ is midline of the $ \triangle XPQ$ $ \Longrightarrow$ $ O \in MN.$

My solution: W.L.O.G CD is between P,Q. Let $ AP \cap CQ = \{ I \},BP \cap DQ = \{ J \}$.$ L$ is midpoint of $ IJ$.Applying Gauss's theorem to the complete quadrilateral $ PCQDIJ$ we conclude that $ O,N,L$ are collinear.Then it's suffice to prove that $ N,M,L$ are collinear. The spiral similarity that took $ AC$ to $ DB$ has center $ Q$ and $ R_(O_1) = R_(O_2)$ then $ \triangle ACQ \cong \triangle DBQ$,then $ QA = QD$ and $ QB = QC$.It follows that $ \triangle QAI \cong \triangle QBJ$ then $ IA = JB$,similar $ ID = JC$.So $ \frac {\overline{IA}}{\overline{ID}} = \frac {\overline{JB}}{\overline{JC}}$.Use the E.R.I.Q theorem (posted here:http://www.mathlinks.ro/viewtopic.php?p=1534951#1534951) to the tranversals $ \overline{IAD},\overline{JBC}$with$ \frac {\overline{LI}}{\overline{LJ}} = \frac {\overline{MA}}{\overline{MB}} = \frac {\overline{ND}}{\overline{NC}} = - 1$ we have $ N,M,L$ collinear,which ends the proof. Image not found

$ \vec{OM} = \frac {1}{2}(\vec{O_1A} + \vec{O_2D}), \vec{ON} = \frac {1}{2}(\vec{O_1C} + \vec{O_2B})$ So we need to prove that $ \vec{OM}//\vec{ON}$ Construct 2 lines which pass through $ O_2$ and parallel to $ O_1A$, pass through $ O_2$ and parallel to $ O_1C$, they cut $ (O_2)$ at $ L,K$. We only need to show $ \angle DO_2K = \angle BO_2L$ $ \Leftrightarrow \angle DO_2O_1 + 180^o - \angle CO_1O_2 = \angle AO_1O_2 + 180^o - \angle BO_2O_1$ $ \Leftrightarrow \angle AO_1C + \angle PO_1Q = \angle BO_2D + \angle PO_2Q$ $ \Leftrightarrow \angle AO_1C = 2\angle APC = 2\angle DPB = \angle DO_2B$ (right!) $ \rightarrow$ QED

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Let $ OB$ and $ OD$ intersect circle $ (O_1)$ at $ B'$ and $ D'$, respectively (notice this is a reflection over $ O$). As $ \angle APC = \angle BPD$, arcs $ AC$ and $ BD$ are equal. This implies arcs $ AC$ and $ B'D'$ are equal, so $ AD' || B'C$. Now, as $ M$ and $ O$ are the midpoints of $ AD$ and $ D'D$ respectively, $ MO || AD'$ Similarly, as $ N$ and $ O$ are the midpoints of $ CB$ and $ B'B$, respectively, $ NO || B'C$ Then we get $ MO || NO$, and the result follows.