Solution posted by Zarif_Ibragimov: We have to prove that: Intersection point of circumcircles of the triangles $MO_{2}C$ and $MO_{1}D$ lies on the line $O_{1}O_{2}$. Let $N_{1}$ - intersection point of the line $O_{1}O_{2}$ with circumcircle of triangle $MO_{2}C$. and let $\angle{BAD}=\alpha$. => $\angle{ABC}=\pi-\alpha, \angle{MO_{1}D}=2\alpha$. => $\angle{O_{1}MD}=\angle{O_{1}DM}=90-\alpha$. $\angle{ABC}=\pi-\alpha$ => $\angle{MO_{2}C=2\alpha}$ => $\angle{O_{2}MC=\angle{O_{2}CM}=90-\alpha}$. The quadrilateral $MO_{2}CN_{1}$ is cyclic. => $\angle{O_{2}CM}=\angle{O_{2}N_{1}M}=90-\alpha$. => $\angle{O_{1}DM}=\angle{O_{1}N_{1}M}=90-\alpha$. => the quadrilateral $MO_{1}DN_{1}$ is cyclic. It follows circumcircle of triangles $MO_{2}C$ and $MO_{1}D$ meets at point $N_{1}$ which lies on the line $O_{1}O_{2}$. => $N=N_{1}$. Easy one!

Denote ${O_1}{O_2} \cap CD = Y$, $\left( {{O_1}} \right) \cap CD = \left\{ {X,D} \right\}$, then $\angle X = \angle A = 180^\circ - \angle B$, so $X,M,B,C$ are concyclic. Because ${O_1}{O_2} \bot MX$, $\angle Y = 90^\circ - \angle X$. $\angle {O_2}MC = 90^\circ - \frac{1}{2}\angle M{O_2}C = 90^\circ - \angle X = \angle {O_2}YC$, so $Y,M,{O_2},C$ are concyclic. On the other hand, $\angle DYM = \angle XYM = 180^\circ - 2\angle X = 180^\circ - \angle M{O_1}D$, so $Y,M,{O_1},D$ are concyclic. Hence $Y=N$, we done.

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From $\hat A+\hat B=180^\circ\implies \angle MO_1D=\angle MO_2C $ and there is a spiral similarity mapping the triangles $MO_1O_2$ and $MDC$, and if $N'=CD\cap O_1O_2$, then $\angle DN'O_1=\angle CMO_2=\angle DMO_1$ and, clearly $N'$ is the second intersection point of the 2 circles, i.e. $N'\equiv N$. Best regards, sunken rock

The points D,C and N are collinear, so from the O1M=O1D, O2M=O2C to be we have NO1 and NO2 are bisectors of angle MND. Therefore, O1, O2 and N are collinear. Quote:

The points D,C and N are collinear, so from the O1M=O1D, O2M=O2C to be we have NO1 and NO2 are bisectors of angle MND. Therefore, O1, O2 and N are collinear. Quote:

Thanks sina!

sina wrote: Solution posted by Zarif_Ibragimov: We have to prove that: Intersection point of circumcircles of the triangles $MO_{2}C$ and $MO_{1}D$ lies on the line $O_{1}O_{2}$. Let $N_{1}$ - intersection point of the line $O_{1}O_{2}$ with circumcircle of triangle $MO_{2}C$. and let $\angle{BAD}=\alpha$. => $\angle{ABC}=\pi-\alpha, \angle{MO_{1}D}=2\alpha$. => $\angle{O_{1}MD}=\angle{O_{1}DM}=90-\alpha$. $\angle{ABC}=\pi-\alpha$ => $\angle{MO_{2}C=2\alpha}$ => $\angle{O_{2}MC=\angle{O_{2}CM}=90-\alpha}$. The quadrilateral $MO_{2}CN_{1}$ is cyclic. => $\angle{O_{2}CM}=\angle{O_{2}N_{1}M}=90-\alpha$. => $\angle{O_{1}DM}=\angle{O_{1}N_{1}M}=90-\alpha$. => the quadrilateral $MO_{1}DN_{1}$ is cyclic. It follows circumcircle of triangles $MO_{2}C$ and $MO_{1}D$ meets at point $N_{1}$ which lies on the line $O_{1}O_{2}$. => $N=N_{1}$. Easy one! Can you say me form where did you get that <MO1D=2\alpha?

Let $N' \equiv CD \cap O_1O_2.$ Since $\triangle MO_1D$ and $\triangle MO_2D$ are both isoceles and $\measuredangle MO_1D = 2\measuredangle MAD = 2\measuredangle MBC = \measuredangle MO_2C$, it follows that $\triangle MO_1D \sim \triangle MO_2C.$ Therefore, $M$ is the center of spiral similarity that sends $\overline{O_1D} \mapsto \overline{O_2C}.$ Hence, it is well-known that $M$ is the second intersection of $\odot(O_1DN'), \odot(O_2CN') \implies N' \equiv N.$ The desired result follows. $\square$

By angle chasing we get triangles $MO_2C$ and $MO_1D$ are similar, thus M is a Miquel point of a quadrilateral $DO_1O_2C$. And the result follows from general properties of Miquel point.

Actually my solution is similar but I did without define any Phantom point.. We need to show that $O_1 , O_2 , N$ are collinear Let $\angle O_1DM = a \implies \angle O_1MD = a\implies \angle MO_1D = 180-2a\implies \angle MAD = 90-a\implies \angle ABC = 90+a \implies \angle MO_2C = 180-2a \implies \angle O_2MC = a\implies \angle O_2CM = a$ Thus $\triangle MO_1D \sim \triangle MO_2C\implies $ $M$ is a center of Spiral Similarity which sends $O_1D$ to $O_2C(*)$ Also since $MO_2CN$ is cyclic $\implies \angle MNC = 2a(1)$ and $MO_1DN$ is cyclic $\implies \angle MND = 2a(2)$ From $1$ and $2$ we get $D,C,N$ are collinear $(**)$ Hence from $(*)$ and $ (**) \implies O_1,O_2,N$ are collinear $\blacksquare$(it is well-known property) so we are done

No need of phantom point. Because $AD \parallel BC, \angle{A}+\angle{B}=180^{\circ}$ and note that $\angle{MO_1D}=\angle{MO_2C} \implies \triangle{MO_1D} \sim \triangle{MO_2D}$. Now notice that $M,O_1,D,N$ and $M,O_2,C,N$ are cylic which implies that $\angle{MND}=\angle{MNC} \implies D-C-N$. Just note that $\angle{O_2NC}=\angle{O_2MC}=\angle{O_1MD}=\angle{O_1ND}$ and we're done. $\square$