$A{P^2} + M{Q^2} = P{B^2} + M{Q^2} = (P{M^2} - B{M^2}) + (Q{C^2} + M{C^2}) = P{M^2} + A{Q^2} \Leftrightarrow $ $\boxed{PQ\bot AM}$

Attachments:

I think I solved a different problem...

$\textbf{\sffamily Cartesian Coordinate Solution:}$ Assume that, \begin{align*} B=(0,0)\\ C=(2c,0)\\ A=(2a_1,2a_2) \end{align*}If $M$ is the midpoint of $BC$, then $M=(c,0)$. Let $m_i$ denote the slope of line $i$. We have \begin{align*} m_{AM}=\frac{a_2}{a_1-c} \end{align*}Let $R$ be the midpoint of $AC$, then $R=(a_1+c,a_2)$. We have $m_{AC}=a_2/(a_1-c)$, so the slope of perpendicular bisector of $AC$ will be $(c-a_1)/a_2$ and then the equation of this will be, \begin{align*} y-a_2=\frac{c-a_1}{a_2}(x-a_1-c) \end{align*}And similarly, equation of perpendicular bisector of $AB$ will be, \begin{align*} y-a_2=\frac{-a_1}{a_2}(x-a_1) \end{align*}So the soln of $x=0$ and perp bisec of $AC$ will be the coordinates of $P$, hence \[P=\bigg(0,\frac{a_1^2-c^2+a_2^2}{a_2}\bigg)\]and similarly solution of $x=2c$ and perp bisec of $AB$ is coordinate of $Q$, hence \[Q=\bigg(2c,\frac{a_1^2-2ca_1+a_2^2}{a_2}\bigg)\]Therefore, \begin{align*} m_{PQ}=\frac{c^2-2ca_1}{2ca_2}=\frac{c-2a_1}{2a_2} \end{align*}So, $m_{PQ}\cdot m_{AM}=-1$, which proves that $AM$ is perpendicular ot $PQ$.

Observe $$MP^2 - MQ^2 = (MB^2 + BP^2) - (MC^2 + CQ^2) = BP^2 - CQ^2$$$$= (BP^2 + BC^2) - (CQ^2 + BC^2) = CP^2 - BQ^2 = AP^2 - AQ^2$$which finishes via the Perpendicularity Lemma. $\blacksquare$ Remarks: I spent far too long trying to find the orthocenter of $APQ$ and attempting to use Radical Axes on $(BQ)$ and $(CP)$. Instead, I should've started out by abusing the perpendicular lines which are already in the diagram via Pythag.

Let $\omega_P$ be the circle centered at $P$ with radius $PB$, and define $\omega_Q$ similarly. Then they are both tangent to $\overline{BC}$, so $\overline{AM}$ is their radical axis, and so $\overline{PQ} \perp \overline{AM}$.