Let $O$ be the circumcenter of $\Delta ABC$. $OX,OY$ are the perp. bisectors of $AB,AC$ and hence external angle bisectors in $\Delta ZXY$ , thus $O$ is the $Z$ excenter in $\Delta ZXY$. Let $A-$ symmedian, $BZ,CZ$ meet $(ABC)$ again at $A',B',C'$. Note that $AY=YC \implies A'C'AC$ is isosceles trapezoid ,thus $A'C' \parallel AC \implies BC',BA'$ are isogonal lines. Let $M$ be the midpoint of $BC$ ,we know isogonal conjugate of $A'$ is $\infty_{AM}$ (since isogonal conjugate of $(ABC)$ is line at infinity) , thus $BC' \parallel AM$ . Similarly $CB' \parallel AM$. In particular $AM$ is the midline of trapezoid $BCB'C'$. Due to tangency points of excircle we know $OE \perp BB' ;OF \perp CC' \implies E,F$ are midpoints of $BB',CC' \implies E,F \in AM$.

GeoKing wrote: Let $M$ be the midpoint of $BC$ ,we know isogonal conjugate of $A'$ is $\infty_{AM}$ (since isogonal conjugate of $(ABC)$ is line at infinity) , thus $BC' \parallel AM$ . Very nice! My solution is almost the same as yours; I just used angle chasing to show that $BC' \parallel AM$ since we have $\measuredangle C'BA = \measuredangle C'CA = \measuredangle CAY = \measuredangle MAB.$