Can the $9$ point circle be generalized to $n$-gons of $n\gt3$? All triangles have concyclic vertices and have a $9$ point circle which intersects the triangle's feet and the midpoints of its sides (as well as $3$ other significant points).
Is this special for triangles?  Or can this be generalized in some sense to all irregular $n$-gons (of $n\gt3$) with concyclic vertices?
Can a circle be drawn which intersects the feet and midpoints of the sides of an irregular pentagon with concyclic vertices?
I guess a prerequisite is that the irregular $n$-gon have feet and altitudes which all intersect at a single orthocenter.  So I guess all shapes with even $n$ are eliminated.
 A: Update. This may-or-may-not count as an answer, depending on whether OP admits polygons with crossed edges and/or coincident vertices. Nevertheless, since equations have no bias, it's worth noting that this is the kind of thing that arises from symbol-pushing alone:

Above is cyclic pentagon $ABCDE$ with coincident vertices $B=E$. The midpoints of edges $AB$, $CD$, $EA$ coincide at the circumcenter; these lie on common circle with the midpoints of edges $BC$ and $DE$. Moreover, that circumcenter is the foot of the perpendicular from $A$ to "opposite" edge $CD$, the foot from $C$ to $EA$, and the foot from $D$ to $AB$. Point $B$ is itself the foot of the perpendicular to $DE$ through $B$; and $E$ is the foot of the perpendicular to $BC$ through $E$.
Thus, the smaller circle contains all the midpoints and all the feet of perpendiculars-to-opposite-edges. (Note, however, that the five perpendiculars do not pass through a common point.)
I'm still investigating the general case.
Previous (Non-)Answer. It's easy enough to construct a pentagon in which the perpendiculars-to-opposite-edges meet at a common point, but such that the feet of those perpendiculars do not lie on a common circle:

