Let $ABC$ be a triangle and $M$ be the midpoint of $BC$. Squares $ABQP$ and $ACYX$ are erected. Show that $PX = 2AM$. 
$\textbf{Question:}$ Let $ABC$ be a triangle and $M$ be the midpoint of $BC$. Squares $ABQP$ and $ACYX$ are erected. Show that $PX = 2AM$.

I could solve this problem using computational techniques but I am looking for purely synthetic solution. I tried drawing some extra midpoints, connected them. But still couldn't find the solution.
Any kind of hint or full solution both are appreciated.

Let's first give names to the sides of $\overset{\Delta}{PAX}$ triangle: $|PA| = k$, $|AX| = m$ and $|PX| = b$. Also for $\overset{\Delta}{ABC}$ triangle: $|AB| = k$, $|AC| = m$, $|BC| = 2l$ and the length of the median $|AM| = a$. Writing law of cosines for $\overset{\Delta}{PAX}$ triangle for angle $\widehat{PAX}$, which is the opposite of angle $\widehat{A}$ in $\overset{\Delta}{ABC}$ triangle, along with the fact that $\widehat{PAB} = 90^{\circ}$ and $\widehat{XAC} = 90^{\circ}$ ($\widehat{PAX} + \widehat{A} + \widehat{PAB} + \widehat{XAC} = 360^{\circ}$), we have:
\begin{align*}
b^{2} &= k^{2} + m^{2} - 2\ k\ m\ cos (180^{\circ} - \widehat{A}) \\
b^{2} &= k^{2} + m^{2} + 2\ k\ m\ cos (\widehat{A}) \tag{1}
\end{align*}
Then write the same law for $ABC$ triangle along with Apollonius's theorem and we have:
\begin{align*}
(2l)^{2} &= k^{2} + m^{2} - 2\ k\ m\ cos(\widehat{A}) \\
4 l^{2} &= k^{2} + m^{2} - 2\ k\ m\ cos(\widehat{A}) \tag{2}
\end{align*}
\begin{align*}
k^{2} + m^{2} = 2(l^{2} + a^{2}) \tag{3}
\end{align*}
Summing $(1)$ and $(2)$ and using the right-hand side of $(3)$ in place of $k^{2} + m^{2}$ we get:
\begin{align*}
b^{2} + 4 l^{2} &= 2 (k^{2} + m^{2}) \\
b^{2} + 4 l^{2} &= 4 (l^{2} + a^{2}) \\
b^{2} &= 4 a^{2} \\
b &= 2a \implies |PX| = 2 |AM| 
\end{align*}
