Here's an extension of Ceva's theorem that incorporates the concurrency of perpendicular bisectors (and, actually, (almost) any trio of lines).
Let distinct lines $p := \overleftrightarrow{P_1P_2}$, $q := \overleftrightarrow{Q_1Q_2}$, $r := \overleftrightarrow{R_1R_2}$ meet the (extended) edges of non-degenerate $\triangle ABC$ such that
$$P_1 = A + p_1 (B-A) \qquad Q_1 = B + q_1 (C-B) \qquad R_1 = C + r_1(A-C)$$
$$P_2 = A + p_2 (C-A) \qquad Q_2 = B + q_2 (A-B) \qquad R_2 = C + r_2(B-C)$$
for some $p_1$, $p_2$, $q_1$, $q_2$, $r_1$, $r_2$. Then $p$, $q$, $r$ concur if and only if
$$\begin{array}{c}
p_1 q_1 r_1 ( 1 - p_2 - q_2 - r_2 ) + p_2 q_2 r_2 ( 1 - p_1 - q_1 - r_1 ) \\[4pt]
+\; p_1 q_1 \; q_2 r_2 \;+\; q_1 r_1 \; r_2 p_2 \;+\; r_1 p_1 \; p_2 q_2 \quad = \quad 0
\end{array} \qquad (\star)$$
(Proof ---for instance, using straightforward vector methods--- is left as an exercise for the reader.)
When $p_1 = q_1 = r_1 = 1$, then $p$, $q$, $r$ are cevians through $B$, $C$, $A$, respectively, and $(\star)$ reduces to
$$( 1 - p_2 )( 1 - q_2 )( 1 - r_2 ) = p_2 q_2 r_2 \qquad (\star\star)$$
which recaptures Ceva's Theorem.
If $p$, $q$, $r$ are perpendicular to $\overline{AB}$ (of length $c$), $\overline{BC}$ (of length $a$), $\overline{CA}$ (of length $b$), respectively, then, via right triangle $\triangle AP_1 P_2$,
$$|AP_1| = |AP_2|\cos A \quad\implies\quad c p_1 = b p_2 \cos A = b p_2 \frac{-a^2+b^2+c^2}{2bc}$$
whence
$$
p_2 = \frac{2 c^2 p_1}{-a^2+b^2+c^2} \qquad
q_2 = \frac{2 a^2 q_1}{a^2-b^2+c^2} \qquad
r_2 = \frac{2 b^2 r_1}{a^2+b^2-c^2}$$
and (for non-degenerate $\triangle ABC$) condition $(\star)$ reduces to
$$a^2 + b^2 + c^2 = 2 \left( a^2 q_1 + b^2 r_1 + c^2 p_1 \right) \qquad (\star\star\star)$$
If $p$, $q$, $r$ are, more-specifically, perpendicular bisectors of the triangle's edges, then $p_1 = q_1 = r_1 = 1/2$. This satisfies $(\star\star\star)$, so the lines are concurrent.
Edit. After bouncing this idea around a bit, I hit upon a better rendition of the Ceva extension, including a more-Ceva-like version of $(\star)$. I'll take this opportunity to rename some points.
Extended Ceva's Theorem. Consider points $D_1$, $D_2$, $E_1$, $E_2$, $F_1$, $F_2$ on the (extended) edges of $\triangle ABC$, with $D_i$, $E_i$, $F_i$ on the (extended) edge opposite vertex $A$, $B$, $C$, respectively.
Lines $\overleftrightarrow{D_1E_2}$, $\overleftrightarrow{E_1F_2}$, $\overleftrightarrow{F_1D_2}$ concur if and only if
$$\begin{align}
1 &= \frac{|BD_1|}{|D_1C|} \frac{|CE_1|}{|E_1A|} \frac{|AF_1|}{|F_1B|}
+ \frac{|D_2C|}{|BD_2|} \frac{|E_2A|}{|CE_2|} \frac{|F_2B|}{|AF_2|} \\[6pt]
&+ \frac{|BD_1|}{|D_1C|} \frac{|D_2C|}{|BD_2|}
+ \frac{|CE_1|}{|E_1A|} \frac{|E_2A|}{|CE_2|}
+ \frac{|AF_1|}{|F_1B|} \frac{|F_2B|}{|AF_2|} \qquad\qquad (\star\star\star\star)
\end{align}$$
Note: The above uses signed lengths, with $\overrightarrow{AB}$, $\overrightarrow{BC}$, $\overrightarrow{CA}$ indicating the direction of a positively-signed segment on each of the triangle's sides.
We get Ceva's Theorem back from the Extended Ceva's Theorem by moving $D_2$, $E_2$, $F_2$ to coincide with $C$, $A$, $B$, respectively, so that $|D_2C| = |E_2A| = |F_2B| = 0$; this eliminates all but the first term of the right-hand side of $(\star\star\star\star)$.
For the perpendicular variant, with the three lines ("orthians"?) perpendicular to the sides of $\triangle ABC$ at $D_1$, $E_1$, $F_1$, we can write
$$|D_2C| = |BC| - |BD_2| \qquad
|E_2A| = |CA| - |CE_2| \qquad
|F_2B| = |AB| - |AF_2|$$
and then also
$$|BD_2| = -\frac{|F_1B|}{\cos B} \qquad
|CE_2| = -\frac{|D_1C|}{\cos C}\qquad
|AF_2| = -\frac{|E_1A|}{\cos A}$$
(the negatives maintain the relationships of the signed lengths). Expressing the cosines in terms of the lengths of the triangles edges, and then expressing those lengths as
$$|BC| = |BD_1|+|D_1C| \qquad |CA| = |CE_1| + |E_1C| \qquad |AB| = |AF_1|+|F_1B|$$
equation $(\star\star\star\star)$ eventually reduces to something much nicer than $(\star\star\star)$; namely,
$$|BD_1|^2 + |CE_1|^2 + |AF_1|^2 = |D_1C|^2 + |E_1A|^2 + |F_1B|^2$$
(In a recent answer, I call the above result "Ortha's Theorem" and provide a stand-alone geometric proof.)
