Thanks to Andrei of School 205 Bucharest for this well explained solution.

First, I drew a line $HG$ , perpendicular on $P$ , and a line through H, parallel with $PR$ .
On it, I took a segment $HE$ of the same length with $HO$ .
Then I finished drawing the square $HEFO$ , drawing $EF$ perpendicular on $PR$ .
The line PE intersects $QR$ in $A$ .
From $A$ I drew a parallel to $PR$ (let the intersection point with $PQ$ be $D$ ), and a perpendicular to $PR$ (let the intersection point with $PR$ be $B$ ).
This way I found 3 vertices of the rectangle $ABCD$ , and I finished the rectangle finding the vertex $C$ on $PR$ , so that $CD$ is perpendicular to $PR$ .

The rectangle $ABCD$ is a square because:- Triangles $PEF$ and $PAB$ are similar, they are both right-angled triangles, with a common angle. The similarity ratio is:

$\frac{PE}{PA} = \frac{EF}{AB}$ $(1)$
Triangles $PEH$ and $PAD$ are similar, because angles $PEH$ and $PAD$ are equal and $HPE$ is a common angle. The similarity ratio

$\frac{PE}{PA} = \frac{HE}{AD}$ $(2)$
From $(1)$ and $(2)$

$\frac{AB}{EF} = \frac{AD}{HE}$

As $HE = EF$ (sides of a square) Therefore $AB = AD$ $ABCD$ is a rectangle with two adjacent sides equal

Therefore $ABCD$ is a square.

solution

Now, the construction of the inscribed square must be done in the following steps:

Chose a point $H$ on side $PQ$ , near $P$
Draw line $HG$ , perpendicular on $PR$
Take the distance $HG$ as the compass distance, and draw a circle arc, with centre $G$ . $F$ is the point of intersection of this arc with $PR$ .
Construct two circle arcs with centres $F$ and $H$ (with the same radius as before). Their intersection is point $E$ , and $EFGH$ is a square.
Draw line $PE$ . Let the intersection point of this line with side $QR$ be $A$ .
Draw from $A$ parallel to $EF$ and $HE$ . Their intersections with $PR$ and $QP$ are points $B$ and $D$ respectively.
Draw from $D$ a parallel to $AB$ . Its intersection with $PR$ is $C$ .
$ABCD$ is the square to be found.

Note

The choice of point $P$ , so that $E$ is interior to the triangle $PQR$ is not a restrictive condition, the construction is the same if $E$ is exterior to triangle $PQR$ .