Let $\Delta ABC$ be a right angled isosceles triangle with $\angle A=90^°$. $D$ is the foot of its altitude from $A$, $ M$ the midpoint of $AD$ and $S$ the midpoint of $AM$.
$B'$ is the reflection of $B$ across the line $CM$, $C'$ is the reflection of $C$ across the line $BM$.
Let E be the intersection of $CB'$ and $BC'$.
Construct a parallel to $BC$ through the point $A$, denote by $P$ its intersection with segment $CE$ by. A parallel line to $BC$ through $S$ meets $AC$ at the point $Q$.
Show that the quadrilateral $ASQP$ is a square.
It is easy to see that triangle $\Delta EBC$ is isosceles and that $ED$ coincides with $AD$. From here I managed to prove this by using some trigonometric formulas and I was wondering if there was a synthetic way of showing this fact.
The only problem that I have is showing that $ASQP$ is a parallelogram.
Since there is no circle in your figure, and we are dealing with a right-angled isosceles triangle, placing the points on the coordinate system seems to be an effortless approach.
WLOG, let's assume $A=(0,2), B=(-2,0)$, and $C=(2,0)$. So, $M=(0,1)$ and $S=(0,\frac{3}{2}).$
An easy computation reveals that:
$$B'=(\frac{-2}{5}, \frac{16}{5}), \\ C'=(\frac{2}{5}, \frac{16}{5}).$$
Now, $P$ is the intersection of the line $y=2$ and the line $y=\frac{-4}{3}x+\frac{8}{3}$ (the line passing through $B'$, $E$, and $C$); hence:
$$P=(\frac{1}{2}, 2).$$
On the other hand, $Q$ is the intersection of the line $y=\frac{3}{2}$ and the line $y=2-x$ (the line passing through $A$ and $C$); hence:
$$Q=(\frac{1}{2}, \frac{3}{2}).$$
At this stage, it is easy to note that $|AS|=|AP|=|SQ|=\frac{1}{2}.$
We are done.