Solution:

Answer: $82.5$

Lemma: Given a triangle $\triangle ABC$, let $I$ be the incenter, $I_{A}$ be the excenter opposite $A$, and $\check{S}$ be the second intersection of $AI$ with the circumcircle. Then $\check{S}$ is the center of the circle through $B, I, C$, and $I_{A}$.

Proof. First, note
$$\angle IBI_{A} = \angle IBC + \angle CBI_{A} = \frac{\angle ABC}{2} + \frac{180^{\circ} - \angle ABC}{2} = 90^{\circ}.$$
Similarly $\angle ICI_{A} = 90^{\circ}$. Therefore $BICI_{A}$ is cyclic. Now note that $A, I, \check{S}$, and $I_{A}$ are collinear because they are all on the angle bisector of $\angle BAC$. Hence
$$\angle CI\check{S} = 180^{\circ} - \angle CIA = \angle CAI + \angle ACI = \angle BC\check{S} + \angle ICB = \angle IC\check{S}$$
(Note $\angle CAI = \angle BA\check{S} = \angle BC\check{S}$ since $A, B, \check{S}$, and $C$ are concyclic.) Hence $\check{S}C = \check{S}I$. Similarly $\check{S}B = \check{S}I$. Thus $\check{S}$ is the center of the circle passing through $B, I$, and $C$, and therefore $I_{A}$ as well.

Let $BA$ and $CD$ intersect at $E$ and $DA$ and $CB$ intersect at $F$. We first show that $F, O_{1}, O$, and $O_{3}$ are collinear.

Let $O_{1}'$ and $O_{3}'$ denote the intersections of $FO$ with the circumcircles of triangles $FAB$ and $FDC$. Since $O$ is the excenter of triangle $FAB$, by the lemma $O_{1}'$ is the circumcenter of $\triangle ABO$; since $O$ is incenter of triangle $FDC$, by the lemma $O_{3}'$ is the circumcenter of $\triangle DOC$. Hence $O_{1}' = O_{1}$ and $O_{3}' = O_{3}$. Thus, points $F, O_{1}, O$, and $O_{3}$ are collinear, and similarly, we have $E, O_{2}, O$, and $O_{4}$ are collinear.

Now $\angle BEC = 55^{\circ}$ and $\angle DFC = 20^{\circ}$ so considering quadrilateral $EOFC$, the angle between $O_{1}O_{3}$ and $O_{2}O_{4}$ is
$$\begin{aligned} \angle EOF & = \angle OEC + \angle OFC + \angle FCE \\ & = \frac{\angle BEC}{2} + \frac{\angle DFC}{2} + \angle FCE \\ & = 27.5^{\circ} + 10^{\circ} + 45^{\circ} = 82.5^{\circ}. \end{aligned}$$