Taiwan IMO Selection Camp

Perform an inversion with center $A$ and denote inverse point with prime.

Let $P \equiv AC'_1 \cap \odot(AB'B'_2)$, $Q \equiv AD'_1 \cap \odot(AE'E'_2)$, $R \equiv B'D'_1 \cap \odot(AB'D')$, $S \equiv E'C'_1 \cap \odot(AC'E')$. Clearly $E'_1 \in \odot(AB'D')$, $B'_2 \in \odot(AB'C')$.

$B'D'_1$ and $B', C', E'_1, B'_2$ are concyclic, so from Reim's theorem we get $AR \parallel C'B'_2$. Similarly, we can prove that $AS \parallel D'E'_2$.

Note that $A, B', E', C'_1, D'_1$ are concyclic, so from Reim's theorem we get $C'_1D'_1 \parallel B'_2P \parallel E'_2Q \parallel D'R \parallel C'S$, hence
$$\left\{ \begin{array}{l} \dfrac{AD'_1}{C'D'_1} = \dfrac{D'_1R}{D'_1B'_2} = \dfrac{D'C'_1}{C'_1P} \\ \\ \dfrac{AC'_1}{D'C'_1} = \dfrac{C'_1S}{C'_1E'_2} = \dfrac{C'D'_1}{D'_1Q} \end{array} \right\} \Longrightarrow \dfrac{AC'_1}{C'_1P} = \dfrac{AD'_1}{D'_1Q} \Longrightarrow C'_1D'_1 \parallel PQ,$$
which implies that $P, Q, B'_2, E'_2$ are collinear and $C'_1D'_1 \parallel B'_2E'_2$.
From Reim's theorem we get $B', E', B'_2, E'_2$ are concyclic, so $B, E, B_2, E_2$ lie on a circle $\Gamma$, which implies that $AA_2, BB_2, EE_2$ are concurrent at the radical center $T$ of $\Gamma$, $\odot(ABD_1)$, and $\odot(AEC_1)$. Analogously, we can prove $T$ lies on $CC_2$ and $DD_2$, so $AA_2, BB_2, CC_2, DD_2, EE_2$ are concurrent.