Taiwan IMO Selection Camp, 2023

Let $R$ be the intersection of $AE$ and $BC$. To prove the original statement, it suffices to prove that $\odot(EBC)$ and $\odot(ERP)$ are tangent at $E$ (because this tells us that $EP$, $EA$ are antiparallel with respect to $EB$, $EC$, and the latter is preserved with respect to $EQ$, $EQ$ antiparallel).

Let $AE$ intersect $\Omega$, $\odot(EBC)$ at $S$, $T$ respectively, then by Reim's theorem, $P$, $Q$, $R$, $S$ are concyclic. Since $EQ$ bisects $\angle BEC$, the intersection point $M$ of $EQ$ and $\odot(EBC)$ is the midpoint of arc $BC$ on $\odot(EBC)$. From
$$\angle ETM = \angle (EM, BC) = \angle (PX, BC) = \angle (CB, DP) = \angle ADP = \angle ASQ,$$
we obtain $TM \parallel SQ$, hence $\odot(ETM)$ and $\odot(ESQ)$ are tangent at $E$.

Note that the radical center of the three circles $\Omega$, $\odot(EBC)$, $\odot(ESQ)$ is the intersection point $U$ of $BC$ and $SQ$, hence from the tangency of $UE$ to $\odot(ESQ)$ we know $\overline{UE}^2 = US \cdot UQ = UR \cdot UP$, that is, $UE$ is tangent to $\odot(ERP)$. Combining with $UE$ tangent to $\odot(EBC)$, we obtain that $\odot(EBC)$ and $\odot(ERP)$ are tangent. $\square$