, 2023

The two parts may be completed independently, and in the three solutions below we demonstrate different approaches to both parts, though one can create valid solutions combining either first part with either second part.
Let $\omega_B$, $\omega_C$ denote the circumcircles of triangles $ABG$ and $ACG$ respectively, and the points $Y$ and $Z$ the second intersection of the line through $B$ parallel to $AC$ and $\omega_B$ and the second intersection of the line through $C$ parallel to $AB$ and $\omega_C$.
The lines $BY$ and $CZ$ thus intersect at $A'$, the reflection of $A$ across the midpoint of $BC$, and in particular on the $A$-median. Using Power of a Point from $A'$ with respect to the circles $\omega_B$ and $\omega_C$ we obtain

$$|A'B| \cdot |A'Y| = |A'A| \cdot |A'G| = |A'C| \cdot |A'E|$$
implying from the converse of Power of a Point that the quadrilateral $YBCZ$ is cyclic. The perpendicular bisector of $BY$ is orthogonal to $BY \parallel AC$ and passes through $F$ and thus $X$ as well. Similarly, the perpendicular bisector of $CZ$ passes through $Z$. Hence $X$ is the center of circle ($YBCZ$) and thus on the perpendicular bisector of the line $BC$.
Let $M$ and $N$ denote the midpoints of $BC$ and $EF$, respectively. To prove that $N$ lies on the perpendicular bisector of $BC$, let $V$ and $W$ denote the second intersections of $\omega_B$ and $\omega_C$ with the line $BC$, respectively.
From Power of a Point from $M$ with respect to $\omega_B$ and $\omega_C$ we obtain
$$|MV| \cdot |MB| = |MG| \cdot |MA| = |WM| \cdot |CM| \implies |MV| = |WM|$$
so $M$ is the midpoint of the segment $VW$. Let $E'$, $N'$, $F'$ denote the projections of $E$, $N$ and $F$ onto $BC$ respectively. Since $N$ is the midpoint of $EF$, $N'$ will be the midpoint of $E'F'$.
Moreover, from the fact that $E$ and $F$ are the centers of $\omega_B$ and $\omega_C$ we get that $E'$ and $F'$ are the midpoints of $BV$ and $WC$, and hence $M$ is the midpoint $E'F'$ as well, implying $N' = M$ and that $N$ is on the perpendicular bisector of $BC$.

Let $G'$ denote the reflection of $G$ across the midpoint of $BC$. We begin by proving that triangles $ABC$ and $DFE$ are orthological, with orthology centers $G'$ and $X$.
Observe that $G'$ is on the $A$-median and thus $AG' \perp EF$. Furthermore, quadrilateral $BGCG'$ is a parallelogram and hence $BG' \parallel CG \perp DE$ and $CG' \parallel BG \perp DF$. Hence, $G'$ is the first orthology center of $\triangle ABC$ and $\triangle DFE$.

Thus, by the property of orthological triangle, the second orthology center must exist, which is defined as the common intersection of the normal from $D$ to $BC$, $E$ to $AB$ and $F$ to $AC$, i.e. the point $X$. Since $D$ is on the perpendicular bisector of $BC$, by virtue of being the circumcenter of triangle $BGC$, and $XD \perp BC$ so must point $X$.
Moreover, let $O$ denote the circumcenter of triangle $ABC$. Then $EO \perp AC \perp FX$ implies $EO \parallel FX$ and $FO \perp AB \perp EX$ implies $FO \parallel EX$, meaning that quadrilateral $FOEX$ is a parallelogram. Hence, the midpoint of $EF$ lies on the line $XOD$ i.e. the perpendicular bisector of segment $BC$.

Let $M$ be the midpoint of $BC$. Let $N$ be the intersection of $EF$ and $DM$. We claim that $N$ is the midpoint of $EF$.
Namely, we have $\triangle DEN \sim \triangle CGM$ because corresponding pairs of sides are orthogonal. Similarly, $\triangle DFN \sim \triangle BGM$. Hence proving that $|EN| = |FN|$, as desired.

Next, let $X'$ resp. $X''$ denote the intersection of $DN$ with the perpendicular from $E$ to $AB$ resp. the perpendicular from $F$ to $AC$. Just as above we have $\triangle ENX' \sim \triangle AMB$ and $\triangle FNX'' \sim AMC$, thus
This shows that $X = X' = X''$ lies on $DN$.
Remark: That the medians of triangle $DEF$ coincide with the perpendicular bisectors of triangle $ABC$ implies that the centroid