I will also sketch a simple solution: $ O \in MN \Longleftrightarrow OP^{2} - \frac {AD^{2}}{4} = OQ^{2} - \frac {BC^{2}}{4}$, because $ MN$ is the radical axe of the circles, and where $ P,Q$ are respetively the midpoints of $ AD$, $ BC$. But $ OP^{2} = \frac {2(OA^{2} + OD^{2}) - AD^{2}}{4}$, and the same for $ OQ$, and then we optain : $ O \in MN \Longleftrightarrow OA^{2} + OD^{2} - AD^{2} = OB^{2} + OC^{2} - BC^{2}$, but $ AD^{2} = OA^{2} + OB^{2} - 2 OA \cdot OB \cos{\angle{AOB}}$, and then it is obvious.

my solution is the same of yours, pohoatza! i was posting when i see that you posted first!

This problem is more accurately so : Quote: Let $ABCD$ be a convex quadrilateral. Denote $O\in AC\cap BD$. Prove that the point $O$ belongs to the radical axis of the circles of diameters $AD$ and $BC$ $\Longleftrightarrow$ $AB\parallel CD\ \ \vee\ \ AC\perp BD$.

Let the cirles with diameter $AD,BC$ be $(O1)$ and $(O2)$ $AC,BD$ meets $(O1), (O2)$ at $F,E$ $DF,CE$ meet at $M$ We see $CF\bot MD, BD\bot MC$ so $O$ is the orthecenter of triangle $MCD$.This means $MO\bot CD$ Because of trapezium $ABCD$ and $O1,O2$ are the midpoint of $AD,BC$ we have $O1O2//CD$ so $MO\bot O1O2$ $(1)$ Otherwise $\angle CED=\angle CFD=90$ so $C,E,F,D$ are cyclic So $MF*MD=ME*MC$.Thus $M$ is on the radical axis of $(O1)$ and $(O2)$ $(2)$ from $(1)$ and $(2)$ we have $O$ is on the radical axis of $(O1)$ and $(O2)$ $q.e.d$

Attachments:

It seems my solution does not belong among those found yet around here: Let $\{O\}\in AC\cap BD$, $P$ the projection of $A$ onto $BD$, $Q, R$ the projections of $B, D$ respectively onto $AC$. Obviously $BQPA$ is cyclic, so $\angle ABP=\angle AQP$, but $\angle ABO=\angle ODC$, so $CDPQ$ is cyclic, with $OC\cdot OQ=OD\cdot OP$. From $APDR$ cyclic, $OA\cdot OR=OP\cdot OD$. From the last 2 equalities, $O$ has equal power w.r.t. the two circles. Best regards, sunken rock