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'Squirty' printed from https://nrich.maths.org/
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: $${{PE}\over{PA}} = {{EF}
\over {AB}}$$ Triangles $PEH$ and $PAD$ are similar, because angles
$PEH$ and $PAD$ are equal and $HPE$ is a common angle. The
similarity ratio $$\frac{PE}{PA} = \frac{HE}{AD}$$ 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.
Now, the construction of the inscribed square must be done in the
following steps:
- Choose 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 $PQR$.