A Complicated Construction of Something Simple

The 3-4-5 triangle is my desert-island geometric fact: if you have a triangle with a sides of length 3, 4, and 5, it's a right triangle. This is great because it's super easy to mark a rope into 3+4+5=12 equal lengths, and this means it's super easy to make yourself a right triangle. I've used this before to lay out an orchard, making sure all the rows were nice and parallel, and it worked beautifully.

Because it's so easy to make a 3-4-5 triangle directly, it seems pretty silly to go to much greater lengths to make one using a pair of compasses, but let's not let mere uselessness stop us! After all, there's the Taoist saying: "When purpose has been used to achieve purposelessness, the thing has been grasped." ;)

I'm feeling playful, so in honor of the sovereign Sun whose day it is, and his dutiful son Pythagoras, let's pick up our compasses and hop to it:

Given points A and B, draw

1. circle AB,
2. circle BA intersecting circle AB at points C and D,
3. circle CD intersecting circle AB at point E≠D,
4. circle EB intersecting circle BA at points F and G,
5. circle FA,
6. circle GA intersecting circle FA at point H≠A,
7. circle HA intersecting circle EB at point I;

then triangle HIE is a 3-4-5 triangle. I'm not going to write up a full proof right now, but a quick sketch runs like this: let's define AB=2. It can be shown that EAB is collinear, therefore EB=EI=EA+AB=2×AB=4. Suppose line FG intersects line AB at X: it can be shown that AX=3/2. H is the reflection of A about line FG and FG is perpendicular to AB, therefore EABH is collinear and AH=HI=AX+XH=2×AX=3. Finally, EH=EA+AH=AB+AH=5.

I believe this to be the simplest possible construction (that is, using the fewest circles) of such a triangle using only compasses. (There are much easier ways to draw a right angle, though!)