Taiwan IMO Selection Camp

Let $T$ be the intersection point of $\Omega_E, \Omega_F$ other than $A$, and let $AT$ meet the circumcircle of $\triangle BTC$ again at point $H_D$. From $\angle^*BCH_D = \angle^*BTA = \frac{1}{2}\angle^*BFA = \frac{1}{2}\angle^*CDB$, we know that $BH_D \perp CD$. Similarly we obtain $CH_D \perp BD$, so $H_D$ is the orthocenter of $\triangle BDC$.

Similarly, we can set $H_E, H_F$ to be the orthocenters of $\triangle CEA, \triangle AFB$ respectively, then $BH_E, CH_F$ are the radical axes of $(\Omega_F, \Omega_D)$ and $(\Omega_D, \Omega_E)$ respectively, that is, $AH_D, BH_E, CH_F$ are concurrent at the radical center $P$ of $\Omega_D, \Omega_E, \Omega_F$.

Let $X, Y, Z$ be the reflections of $D, E, F$ with respect to $BC, CA, AB$ respectively.

From $\triangle BXC \sim \triangle BFA$, we get $\triangle BCA \sim \triangle BXF$, so $\frac{AE}{BF} = \frac{CA}{AB} = \frac{FX}{BF}$, from which we know $AE = FX$. Similarly we obtain $AF = EX$, so $AEXF$ is a parallelogram.

Let $O$ be the circumcenter of $\triangle DEF$, and let $Q$ be the reflection of $P$ with respect to $O$. Then from the above discussion we know that $XQ \perp EF$. Similarly, $YQ, ZQ$ are perpendicular to $FD, DE$ respectively. Let $H$ be the orthocenter of $\triangle DEF$. Note that
$$\frac{DX}{DH_D} = \frac{EY}{EH_E} = \frac{FZ}{FH_F},$$
so $H, P, Q$ are collinear, that is, $P$ lies on the Euler line $OH$ of $\triangle DEF$. This completes the proof.