Draw any two squares which meet at a common vertex $C$ and join the
adjacent vertices to make two triangles $CAB$ and $CDE$.
Construct the perpendicular from $C$ to $AB$, (the altitude of the
triangle). When you extend this line where does it cut $DE$?
Now bisect the line $AB$ to find the midpoint of this line $M$.
Draw the median $MC$ of triangle $ABC$ and extend it to cut $DE$.
What do you notice about the lines $MC$ and $DE$?
Will you get the same results about the two triangles formed if you
draw squares of different sizes or at different angles to each
other? Make a conjecture about the altitude of one of these
triangles and prove your conjecture.
In a right-angled tetrahedron prove that the sum of the squares of the areas of the 3 faces in mutually perpendicular planes equals the square of the area of the sloping face. A generalisation of Pythagoras' Theorem.