Squirty
Problem
Using a ruler and compass only, it is possible to construct a square in any triangle so that one side of the square rests on one side of the triangle, and the other two vertices of the square touch the other two sides of the triangle:
Can you find a way to construct the square, for any triangle?
Can you explain why your method works?
Getting Started
Can you draw a square ABCD with BC on the base PR of the triangle and the vertex D on the side PQ using ruler and compasses?
What happens to the point A as you enlarge the square? The interactivity may help. Move Q to specify your triangle, and then move D to enlarge the square.
How can you use this to help you draw the square you require?
Student Solutions
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 rightangled 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.
Teachers' Resources
Why do this problem
This activity is plain straight forward problem solving. Here with traditional instruments. Working out what the task is asking for, doing something which might help but isn't the whole solution, reviewing the result of that to see what might bethere to be noticed, and then using that observation to close the task. Along the way working into a deeper appreciation of the things we thought we
already knew and could do easily.
Possible approach
This printable worksheet may be useful: Squirty.
Start the group off with the task of drawing a square of any size, using only a ruler and compasses. There's plenty to challenge understanding just in that.
Next have each student draw a triangle, pointing out that because they can each draw the triangle they choose, what follows will be an exploration of any triangle.
Allow some discussion to establish that every student appreciates the task and that many 'first thoughts' about the task can be shared.
Pursue each suggestion, together as a group but with each student actually drawing for themselves. Exploring suggestions to their point of failure deepens understanding, skipping this reduces the value of the activity considerably.
If a prompt towards something fruitful is needed, point out that partially meeting criteria can sometimes be helpful in the problemsolving process. In this case to create a square, based on one side, one of whose 'top' corners touch one of the other two sides of the triangle. Several of these will be needed so that we can think about the special one of this set which touches both sides.
Take time to let students notice how they create their squares. How they choose the base or the 'touch point' and from that go on to create the rest of the square. Once they have three or four squares invite them to stand back, take a look and then share what they notice. Ask how that helps with the actual task.
This is a relatively closed task but follow the extension suggestion to foster a classroom culture that 'plays' with compasses and sparks with geometric reasoning and insight.
Here's a couple of nice problems to start discussion and play.

If I only have the three midpoints from the sides of a triangle can I recreate the triangle ?

If I only have the centres of three circles where each circle touches the other two can I recreate the circles.
As before, these tasks are to use ruler and compasses, but do allow the discussion to wander into methods which depart from that constraint.
Also take time to allow students to appreciate what kind of constraint 'ruler and compasses' imposes.
'Ruler' means tha points can be joined and the line continued indefinitely from both ends, and 'compasses' means that a length can be picked up and transferred somewhere else.
Key questions
 Can you draw a square using only a straight edge and a pair of compasses ?

What, in your own words, is the challenge or task here ?
 How might you start on that ? Can you do it 'straight off'' ?

How can you use that to get the square you want ?
Possible support
Stick with creating squares just using ruler and compasses constructions.
Possible extension
ZigZag explores more of what makes a square.