BMO Shortlist, 2019

We will prove that the desired point of concurrency is the midpoint of $BC$. Assume that $\triangle ABC$ is acute. Let $(ABC)^5$ intersect $(AEF)$ at the point $Y'$; we will prove that $Y = Y'$.

Figure 7: G7
Using the fact that $H$ is the incenter of $\triangle DEF$ we get that $D$, $E'$, $F$ and $D$, $F'$, $E$ are triples of collinear points. Furthermore,
$$90^\circ = \angle^6 AEH = \angle AF'H = \angle AE'H = \angle AFH \Rightarrow F', E', H \in (AEFY').$$
We will now prove that the points $Y'$, $B$, $D$, $F'$ are concyclic. Indeed,
$$\angle Y'BD = \angle Y'BC = \angle Y'AC = \angle Y'AE = \angle Y'F'E \Rightarrow (Y', B, D, F').$$
Now, as
$$\angle F'Y'B = \angle F'DC = \angle EDC = \angle CAB = \angle CY'B,$$
the points $C$, $F'$, $Y'$ are collinear. Similarly we get that $B$, $E'$, $Y'$ are collinear, which implies
$$Y' = Y = (ABC) \cap (AEF).$$
⁵$(XYZ)$ denotes the circumcircle of $\triangle XYZ$
⁶$\angle$ denotes a directed angle modulo $\pi$
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Since we proved this property using directed angles, we know that it is also true for obtuse triangles.
Notice that the points $A$, $B$, $C$, $H$ form an orthocentric system; in other words $H$ is the orthocenter of $\triangle ABC$ and $A$ is the orthocenter $\triangle HBC$. Furthermore, notice that $F'$ is to $\triangle ABC$ as $E'$ is to $\triangle HBC$ and that $E'$ is to $\triangle ABC$ as $F'$ is to $\triangle HBC$. This means that $X$ is to $\triangle HBC$ as $Y$ is to $\triangle ABC$ and, as we know the proven property is also true for obtuse triangles, we get
$$X = (HBC) \cap (AEF).$$
By Reflecting the Orthocenter Lemma we know that in a triangle $ABC$, the reflection of its orthocenter over the midpoint of $BC$ is the antipode of $A$ w.r.t. $(ABC)$. Applying this Lemma on the triangles $ABC$ and $HBC$ we get that $YH$ and $AX$ both go through the midpoint of $BC$, thus finishing the solution. □