Show, without invoking the Pythagorean theorem, that the $3-4-5$ triangle is right The ancient Egyptians knew the $3-4-5$ triangle was a right triangle, but they did not possess the Pythagorean theorem or any equivalent theory.  Can it be shown that the $3-4-5$ triangle is a right triangle without using the Pythagorean theorem or any ideas related to it?  
This problem was shown to me by a fellow peer tutor a while ago.  I gave it some thought initially, but then I gave up trying when I realized it might not be doable.  What do you think?   

This is a proof (without using Pythagoras' theorem) of the following: If a right-angled triangle has a hypotenuse of length 5 and one leg of length 3, the remaining leg is of length 4, i.e. it must be a 3-4-5 triangle. 
Given that the sides of a triangle uniquely determine its angles we should conclude from the above that a 3-4-5 triangle has to be right-angled.
Start by drawing a 5-5-6 rectangle with one of the length-5 sides as its base. Extend the base by a line of length 5:

Draw a circle with the three lines of length 5 as radii:

Draw an altitude of the original triangle to the side of length 6. Also, join the top of the triangle to the end of extended line of length 5:

If we call the equal angles in the isosceles 5-5-6 triangle $\theta$ then we can calculate some of the other angles. In particular the angle between the altitude drawn earlier and the base is $\pi/2 - \theta$ and this equals the angle between the line joining the top of the triangle and the extended line.

So the altitude is parallel to the line joining the top to the right extension. Also, by similar triangles, if the altitude has length $x$ then the second line has length $2x$

If we extend the other radius of length 5 to make it a diameter of the circle, we can complete a cyclic quadrilateral with the edges shown and with its diagonals both being diameters of the circle

Ptolemy's theorem says that, for a cyclic quadrilateral, the 
product of the two diagonal lengths  equals the sum of the products of opposite sides. In our case, this means that
$$
10 \times 10 = 6 \times 6 + 2x \times 2x \Rightarrow
100 = 36 + 4x^2
\Rightarrow
25 = 9 + x^2 
$$
which means that $x=4$ and we have not used Pythagoras to find this value. 
So we have shown that the `half' of the original triangle defined by the base, the altitude and the length of 3 joining the base to the altitude is a 3-4-5 triangle.
There can only be one 3-4-5 triangle, we cannot have a second one with different angles. So we have shown that a 3-4-5 triangle must be right-angled. 

