I propose an alternative "fourth vertex".

To paraphrase Sherman's result:

The "fourth side" ($w$) of a triangle *is a chord of the circumcircle, is a tangent to the incircle ($\bigcirc I$), and is bisected by the nine-point circle ($\bigcirc N$). (The last aspect is equivalent to $w$ intersecting the nine-point circle at the foot of the perpendicular from circumcenter $K$.)*

Thus, this "fourth side" fits a description that applies to the (standard) three; importantly, there is exactly one such "fourth side" that can do so.

My proposal:

The "fourth vertex" ($W$) of a triangle *determines a line through the orthocenter ($H$) that intersects the nine-point circle at the foot of a perpendicular tangent to the incircle, and is such that the midpoint of $\overline{WH}$ is the "other" intersection with the nine-point circle.* (The last part distinguishes $W$ from the other endpoint of the chord determined by $\overleftrightarrow{WH}$.)

**Note.** As observed in comments, point $W$ is Kimberling's triangle center X(1309).

As before, the ostensible "fourth vertex" fits a description that applies to the (standard) three; also, as it turns out, there is exactly one "fourth vertex" that can do so.

The simplicity of this proposed definition has some appeal, but what makes it particularly relevant is a fact the reader may have suspected from the images: The two "fourth" elements *involve the same construction*. The line of segment $w$ serves as the tangent to $\bigcirc I$ perpendicular to $\overline{WH}$ where they meet on $\bigcirc N$; that is, Sherman could have —and *may* have (I haven't checked)— noted:

$w$ intersects the nine-point circle *again* at the foot of the perpendicular from orthocenter $H$.

Proof isn't complicated (essentially all that is needed is to rotate some elements around the center of the nine-point circle), but getting too caught-up in the *triangle* context is unnecessarily limiting. It's better to see these "three elements and a spare" results as special cases of a broader "four elements" result.

**Lemma.** Suppose tangents at $A$, $B$, $C$ of $\bigcirc P$ meet $\bigcirc Q$ at feet ($A_+$, $B_+$, $C_+$) of perpendiculars from a common point $R_+$. Then there is a unique point $D$ of $\bigcirc P$ such that the tangent at $D$ meets $\bigcirc Q$ at the foot ($D_+$) of the perpendicular from $R_+$.

Moreover, reflecting $R_+$ in $Q$ gives a point $R_-$ such that the feet ($A_-$, $B_-$, $C_-$, $D_-$) of perpendiculars to the tangents lie on $\bigcirc Q$.

(In the triangle context, $\bigcirc P$ and $\bigcirc Q$ are respectively the incircle and nine-point circle; lines $\overleftrightarrow{A_+A_-}$, etc, contain the "four sides"; and points $R_+$ and $R_-$ are the circumcenter and orthocenter.)

The "moreover" follows from recognizing that a $180^\circ$ rotation about $Q$ effectively creates four inscribed rectangles. This rotational symmetry guarantees that sides of these rectangles concur iff the *opposite* sides concur.

The main part of the Lemma can be proven with some light coordinate bashing. For instance, taking $P=(0,0)$ and $R_+=(r,0)$, defining angles $\alpha:=\angle R_+PA$, $\beta:=\angle R_+PB$, $\gamma:=\angle R_+PC$, $\delta:=\angle R_+PD$, and letting the radius of $\bigcirc P$ be $p$, then
$$A_+ = R_+ + (p - r \cos\alpha) (\cos\alpha,\sin\alpha), \qquad B_+=\cdots, \qquad C_+=\cdots, \qquad D_+=\cdots$$
We find that $A_+$, $B_+$, $C_+$, $D_+$ are concyclic iff
$$2 p \sin\sigma = r \left(\;\sin(\sigma-\alpha) +\sin(\sigma-\beta) +
\sin(\sigma-\gamma) + \sin(\sigma-\delta)\;\right) \tag{1}$$ where $\sigma:=\frac12(\alpha+\beta+\gamma+\delta)$. Condition $(1)$ determines any one angle from the other three; eg, we can write
$$\begin{align}
&\;\phantom{-}\cos\tfrac12\delta\left(\;
2 p \sin\tau -
r ( \sin(\tau-\alpha) + \sin(\tau-\beta) + \sin(\tau-\gamma) + \sin\tau)
\;\right) \\[0.5em]
= &-\sin\tfrac12\delta \left(\;
2 p \cos\tau -
r (\cos(\tau-\alpha) + \cos(\tau-\beta) + \cos(\tau-\gamma) - \cos\tau)\;\right)
\end{align} \tag{1'}$$
where $\tau:=\frac12(\alpha+\beta+\gamma)$. This identifies $\delta/2$ up to a half-turn, hence $\delta$ up to a full turn, guaranteeing the unique $D$ claimed in the Lemma. $\square$

It is perhaps worth noting that (barring degeneracies) $(1)$ allows for determining distance $r=|PR_+|$ to make any chosen collection of distinct angles (ie, any chosen collection of points $A$, $B$, $C$, $D$ on $\bigcirc P$) "work" to give the concurrencies shown in the figure.