I am looking for a solution of the problem as follows:
Let $A_1A_2A_3A_4A_5$ be a cyclic pentagon, let $B_i=A_{i-1}A_{i} \cap A_{i+1}A_{i+2}$ for $i=1, 2, 3, 4, 5$, here we take modulo $5$. Let $(O_i)$ is circle through $B_i, A_{i+2}, A_{i+4}$. Let $C_i=O_{i+1} \cap O_{i+4}$ then show that $C_1, C_2, C_3, C_4, C_5$ lie on a circle.
As far as I know, this theorem is called Takada's theorem (高田の定理), and was first discovered in 1989 by Hideyuki Takada, a then high school student, and proved by Tominosuke Otsuki and Kazuo Masuda. You can see their proofs in Tominosuke Otsuki and Kazuo Masuda, "CERTAIN THEOREMS ON PENTAGONS."
Here is another elementary proof.
Let $\mathcal{P_1}=\odot{(A_1A_2C_1)}$ and $\mathcal{P_5}=\odot{(A_5A_1C_5)}$. Let $D_1$ be the second intersection of $\mathcal{P_1}$ and $\mathcal{P_5}$, different from $A_1$.
Observe that
$$\begin{align}\measuredangle{A_1D_1A_2} & =\measuredangle{A_1C_1A_2}\\ & =\measuredangle{A_1C_1A_4}+\measuredangle{A_4C_1A_2}\\ & =\measuredangle{A_1B_2A_4}+\measuredangle{A_4B_5A_2}\\ & =\measuredangle{B_3A_4A_3}\end{align}$$ Likewise, $\measuredangle{A_5D_1A_1}=\measuredangle{A_4A_3B_3}$. Therefore,
$$\begin{align}\measuredangle{A_5D_1A_2} & =\measuredangle{A_5D_1A_1}+\measuredangle{A_1D_1A_2}\\ & =\measuredangle{A_4A_3B_3}+\measuredangle{B_3A_4A_3}\\ & =\measuredangle{A_5B_3A_2}\end{align}$$ This implies that $A_2$, $B_3$, $A_5$, $D_1$ are concyclic, i.e. $D_1\in (O_3)$. Let $E_2$ be the second intersection of $\mathcal{P_1}$ and $\overleftrightarrow{A_2A_3}$, different from $A_2$. Let $E_9$ be the second intersection of $\mathcal{P_5}$ and $\overleftrightarrow{A_4A_5}$, different from $A_5$. By applying a converse of Miquel's theorem to $(O_3)$, $\mathcal{P_1}$, and $\mathcal{P_5}$, we get that $E_2$, $A_1$, $E_9$ are collinear. Hence,
$$\begin{align}\measuredangle{E_9E_2A_3} & = \measuredangle{A_1C_1A_2}\\ & =\measuredangle{B_3A_4A_3}\\ & =\measuredangle{E_9A_4A_3}\end{align}$$ So we find that $E_2$, $A_3$, $A_4$, $E_9$ are concyclic. Furthermore, we have
$$\begin{align}\measuredangle{A_4C_1E_2} & =\measuredangle{A_4C_1A_2}+\measuredangle{A_2C_1E_2}\\ & =\measuredangle{A_4B_5A_2}+\measuredangle{A_2A_1E_2}\\ & =\measuredangle{E_9B_5A_1}+\measuredangle{B_5A_1E_9}\\ & =\measuredangle{A_5E_9A_1}\\ & =\measuredangle{A_4A_3E_2}\end{align}$$ Thus, $C_1$, $E_2$, $A_3$, $A_4$ are concyclic. Similarly, $E_9$, $C_5$, $A_3$, $A_4$ are concyclic. Therefore, $C_1$, $E_2$, $A_3$, $A_4$, $E_9$, $C_5$ lie on a common circle. Denote this circle by $\mathcal{Q_1}$. Let $\mathcal{R}$ be the circle $\odot{(A_1A_2A_3A_4A_5)}$. By applying Miquel's six circle theorem to $(O_1)$, $(O_3)$, $(O_5)$, $\mathcal{Q_1}$ and $\mathcal{R}$, we can see that $C_1$, $C_2$, $C_4$, $C_5$ are concyclic. With similar arguments, $C_2$, $C_3$, $C_5$, $C_1$ are concyclic, and we are done. $\square$
There is another remarkable theorem in this configuration: construct the radical line of $(O_{i})$ and $(O_{i+1})$ (indices are taken $\bmod 5$). Then these five radical lines meet in a point. This holds true for any pentagon $A_1A_2A_3A_4A_5$. It is often called the five-circle incidence theorem and attributed to Larry Hoehn. Seemingly this theorem was re-discovered by Tran Quang Hung; the point of concurrency is listed as 5G-s-P5 1st 5G-Hung’s Point in the Encyclopedia of Polygon Geometry.
I found an interesting property about this configuration: the two circles $\odot{(A_1B_2B_4)}$ and $\odot{(A_1C_1C_5)}$ are tangent to each other at $A_1$.