We shall construct $X,Y$ Consider the tangent through $A$ to a circle with diameter $BC$ and call the intersections of that lines with that circle $X',Y'$ Now $BA$ is the $B-symmedian$ in $\triangle BX'Y'$ and $M$ is the center of $(BX'Y')$, but it's well-known that in any triangle , the foot of perpendicular through the circumcenter of $ABC$ to $A-symmedian$ is in fact $A-Dumpty$ point. So $F$ is $B-Dumpty$ point in $\triangle BX'Y'$ and it satisfies that similarity condition. similar for $C,E$. But $X,Y$ are unique(it's just using the similarity formula in complex numbers and by writing 2 equations drived from the similarity condition., we'd have the values of both $x*y$ and $x+y$ which implies the uniqueness) and ${X',Y'}={X,Y}$ . But clearly $AM$ is the perpendicular bisector of $XY$ and we're done.

Let circle with diameter $BC$ intersect $AB$ and $AC$ at $P$ and $Q$respectively. Notice that the angle conditions mean $E$ is $C$-Dumpty point of $CXY$ and $F$ is $B$-Dumpty point of $BXY$. Since dumpty point is midpoint of symedian chord, $BPXY$ and $CXYQ$ are cyclic. However, if we take circles $(BPXY),(CQXY),(BCPQ)$ and look at the radical centeri it implies $A,X,Y$ are collinear but this is not possible since $CQXY$ are cyclic. So those $3$ circles must coincide, hence $(XYPQBC)$ are cyclic. Also since $Q$ is dumpty point of $CQXY$ tangents from $X$ and $Y$ meet at $A$. So $X$ and $Y$ must be the tangents from $A$ to $(BPQC)$ and $XY\perp AM$

Is there a fully computational solution? i.e using complex numbers or barycentric coordinates