Let $M$ be the midpoint of $CD$. Since, $AB$ is the radical axis of the two circles , $M$ must lie on $AB$. $WLOG$, let $A$ be closer to $CD$.. From alternate segment theorem , $\angle DCA=\angle CBA$ and $\angle CDA=\angle DBA$ Adding both these, $\angle CBD=\angle DCA +\angle CDA =180^{\circ}-\angle CAD$ Hence $\triangle CAD$ and $\triangle CBD$ have common base $CD$ , but the respective angles opposite to the common base are supplementary. Hence, midpoint of $O_{a}O_{b}$ is the midpoint of $CD$ , that is point $M$ which lies on $AB$.

It is clear that $AB$ bisects $CD$ and $O_a, O_b$ are symmetric about midpoint of $CD$

There were 2 problems on the same configuration (i.e., the vertex of the D-triangle of ABC corresponding to $A$) i.e. P1 and P5! (Maybe that's why mine was the first solution on both days in 2 minutes ).

Was certainly easy: (Considering $CD $ to be nearer to $A$). It's obvious that $AB $ bisects $CD $. So, it's suffices to show that $\angle CO_aD = \angle CO_bD $. This follows directly from the fact that $\angle CAD + \angle CBD = 180$°, which implies that $ACA'D $ and $BCB'D $ are parallelograms, where $A' = AB\cap \odot(BCD)$ and $B' = AB\cap \odot (ACD) $.

Tumon2001 wrote: Was certainly easy: (Considering $CD $ to be nearer to $A$). It's obvious that $AB $ bisects $CD $. So, it's suffices to show that $\angle CO_aD = \angle CO_bD $. This follows directly from the fact that $\angle CAD + \angle CBD = 180$°, which implies that $ACA'D $ and $BCB'D $ are parallelograms, where $A' = AB\cap \odot(BCD)$ and $B' = AB\cap \odot (ACD) $. i didnt get what actually are A' and B', could someone help?