Let O be the midpoint of AB, the center of the semicircle (O) with the points C, D forming a convex quadrilateral ABDC. $OM = OA\ \cos \widehat{AOM}$ and $ON = OB\ \cos \widehat{BON}$. $ON \parallel AD$ is a midline of the triangle $\triangle ABD$ and $OM \parallel BC$ is a midline of the triangle $\triangle ABC$. Let $E, F$ be the midpoints of PC, PD. Denote $\alpha = \angle (OO_A, PE) = \angle (OM, PC)$ and $\beta = \angle (OO_B, PF) = \angle (ON, PD)$. From the trapezoid $OO_AEP$, $OO_A = \frac{PE}{\cos \alpha}$ and from the trapezoid $OO_BFP$, $OO_B = \frac{PF}{\cos \beta}$. Since $OA = OB = \frac{AB}{2}$ and $PE = \frac{PC}{2} = \frac{CD}{4} = \frac{PD}{2} = PF$, we get $\frac{OM}{ON} = \frac{\cos \widehat{AOM}}{\cos \widehat{BON}},\ \ \ \frac{OO_A}{OO_B} = \frac{\cos \beta}{\cos \alpha}$ But from the circle (O) and form $AD \parallel ON, BC \parallel OM$, $\alpha = \angle (OM, PC) = \angle (BC, DC) = \angle BCD = \angle BAD = \angle BON$ $\beta = \angle (ON, PD) = \angle (AD, CD) = \angle ADC = \angle ABC = \angle AOM$ Consequently, $\frac{OM}{ON} = \frac{OO_A}{OO_B}$, which means $O_AO_B \parallel MN$.

Attachments:

Let $ L$ be the midpoint of $ AB$ and $ Q \equiv \odot(ACP) \cap \odot(BDP),$ different from $ P.$ Clearly $ O_AO_B \perp PQ$ and lines $ CA,DB$ and $ PQ$ concur at the radical center $ E$ of $ \odot(ACP),$ $ \odot(BDP)$ and the circumference with diameter $ AB.$ Since $ CD$ is antiparalell to $ AB$ with respect to $ EA,$ $EB,$ then $EP$ and $ EL$ are isogonals with respect to $ \angle AEB,$ i.e. $ EP$ is the E-symmedian of $ \triangle EAB.$ Since $ EMLN$ is a cyclic quadrilateral on account of $ \angle EML = \angle ENL = 90^{\circ},$ we have $ \angle MEL = \angle MNL = \angle PEN$ $ \Longrightarrow$ $ MN \perp PE.$ Hence $ O_AO_B \parallel MN.$

Let be the point $ Q\equiv (O_{a})\cap (O_{b}),$ the other than $ P,$ where $ (O_{a}),\ (O_{b})$ are the circumcircles of the triangles $ \bigtriangleup ACP,\ \bigtriangleup BDP,$ respectively. The line segments $ AC,\ PQ,\ BD,$ as the radical axis of the circles $ (O),\ (O_{a}),\ (O_{b})$ taken per two of them, are concurrent at one point so be it $ R,$ as their radical center, as well. $ \bullet$ Because of now, $ PM\parallel AD$ and $ PN\parallel BC$ and $ AD\perp BD$ and $ BC\perp AC,$ we conclude that $ P$ is the orthocenter of the triangle $ \bigtriangleup RMN$ and so, we have that $ RP\perp MN$ $ \Longrightarrow$ $ PQ\perp MN$ $ ,(1)$ From $ (1)$ and beause of $ O_{a}O_{b}\perp PQ$ we conclude that $ O_{a}O_{b}\parallel MN$ and the proof of the proposed problem is completed. Kostas Vittas.

Attachments:
t=82947.pdf (9kb)

Take $\{T\}= BD \cap AC$, then easily $P$ is the orthocenter of $\triangle TNM$, $PT \perp MN$, but $P$ and $T$ belong to the radical axis of the 2 circles, hence $TP \perp O_AO_B$. Best regards, sunken rock

Project all 7 points $A$,$M$, $O_A$, $O$, $O_B$, $N$, and $B$ onto $CD$, then it's trivial! Let $h_{X}$ denote the projection of $X$ onto $CD$. Let $O$ be the midpoint of $AB$. Then $O$,$O_A$, and $M$ are collinear, similarly $O$,$O_B$, and $N$ are collinear. Then since $AO=OB$, $h_{A}P=h_{B}P$. Also, $CP=PD$, so $h_{A}C=h_{B}D$. Since $M$ and $N$ are the midpoint of $AC$ and $BD$, $h_{M}P=h_{N}P$. Also, since $O_A$ and $O_B$ are on the perp bisectors of $CP$ and $PD$, $h_{O_A}P=h_{O_B}P$. Thus, $\frac{MO_A}{O_AO}=\frac{h_{M}h_{O_A}}{h_{O_A}P}= \frac{h_{N}h_{O_B}}{h_{O_B}P}= \frac{NO_B}{O_BO}$ So $MN \parallel O_AO_B$.

let AC,BD ntersect at E.circle $O_A,O_B$ intersect at P and T it suffices to prove that PT is perpendicular to MN by equal power theorem,E,P,T collinear then $MP\bot EN,NP\bot EM$ so P is the orthocemter of triangle EMN hence $EP\bot MN$