20. [12] Let $A B C D$ be a quadrilateral circumscribed about a circle with center $O$. Let $O_{1}, O_{2}, O_{3}$, and $O_{4}$ denote the circumcenters of $\triangle A O B, \triangle B O C, \triangle C O D$, and $\triangle D O A$. If $\angle A=120^{\circ}, \angle B=80^{\circ}$, and $\angle C=45^{\circ}$, what is the acute angle formed by the two lines passing through $O_{1} O_{3}$ and $O_{2} O_{4}$ ?
Answer: $\square$
Lemma: Given a triangle $\triangle A B C$, let $I$ be the incenter, $I_{A}$ be the excenter opposite $A$, and $\breve{S}$ be the second intersection of $A I$ with the circumcircle. Then $\breve{S}$ is the center of the circle through $B, I, C$, and $I_{A}$.
Proof. First, note
$$
\angle I B I_{A}=\angle I B C+\angle C B I_{A}=\frac{\angle A B C}{2}+\frac{180^{\circ}-\angle A B C}{2}=90^{\circ} .
$$

Similarly $\angle I C I_{A}=90^{\circ}$. Therefore $B I C I_{A}$ is cyclic. Now note that $A, I, \breve{S}$, and $I_{A}$ are collinear because they are all on the angle bisector of $\angle B A C$. Hence
$$
\angle C I \check{S}=180^{\circ}-\angle C I A=\angle C A I+\angle A C I=\angle B C \check{S}+\angle I C B=\angle I C \breve{S} .
$$
(Note $\angle C A I=\angle B A \breve{S}=\angle B C \breve{S}$ since $A, B, \breve{S}$, and $C$ are concyclic.) Hence $\breve{S} C=\breve{S} I$. Similarly $\breve{S} B=\breve{S} I$. Thus $\breve{S}$ is the center of the circle passing through $B, I$, and $C$, and therefore $I_{A}$ as well.
Let $B A$ and $C D$ intersect at $E$ and $D A$ and $C B$ intersect at $F$. We first show that $F, O_{1}, O$, and $O_{3}$ are collinear.
Let $O_{1}^{\prime}$ and $O_{3}^{\prime}$ denote the intersections of $F O$ with the circumcircles of triangles $F A B$ and $F D C$. Since $O$ is the excenter of triangle $F A B$, by the lemma $O_{1}^{\prime}$ is the circumcenter of $\triangle A B O$; since $O$ is incenter of triangle $F D C$, by the lemma $O_{3}^{\prime}$ is the circumcenter of $\triangle D O C$. Hence $O_{1}^{\prime}=O_{1}$ and $O_{3}^{\prime}=O_{3}$. Thus, points $F, O_{1}, O$, and $O_{3}$ are collinear, and similarly, we have $E, O_{2}, O$, and $O_{3}$ are collinear. Now $\angle B E C=55^{\circ}$ and $\angle D F C=20^{\circ}$ so considering quadrilateral $E O F C$, the angle between $\mathrm{O}_{1} \mathrm{O}_{3}$ and $\mathrm{O}_{2} \mathrm{O}_{4}$ is
$$
\begin{aligned}
\angle E O F & =\angle O E C+\angle O F C+\angle F C E \\
& =\frac{\angle B E C}{2}+\frac{\angle D F C}{2}+\angle F C E \\
& =27.5^{\circ}+10^{\circ}+45^{\circ}=82.5^{\circ} .
\end{aligned}
$$