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# Origami proof of the Pythagorean theorem

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You don't need numbers or fancy equations to prove the Pythagorean Theorem, all you need is a piece of paper. There is a ton of ways to prove it, and people are inventing new ones all the time, but I am going to show you my favorite. Only instead of looking at diagrams, we're gonna fold it. First, you need a square, which you can probably obtain from a rectangle if you ask nicely. Step one, fold your square in half one way, then the other way, then across the diagonal. No need to make these creases sharp, we're just taking advantage of the symetries of the square for the next step. But, be precise. Step 2: Make a crease along this triangle, parallel to the side of the triangle that has the edges of the paper. You can make it anywhere you want. This is where you are choosing how long and pointy, or short and fat, your right triangle is going to be, because this is a general proof. Now when you unwrap it, you'll have a square centered in your square. Extend those creases and make them sharp, and now we've got we've got four lines all the same distance from the edges, which will allow us to make a bunch of right triangles that are all exactly the same. Step three: fold from this point to this one. Basically taking a diagonal of this rectangle. Now we've got our first right triangle. Which has the same shape and area as this one. Let's call the sides: "A little leg", "a big leg", and "hypothenus". Rotate ninety degrees, and fold back another triangle, which of course is just like the first. Repeat on the following two sides. The original paper minus those four triangles, gives us a lovely square. How much paper is this ? Well, the length of a side is the hypotenus of one of these triangles. So the area is the hypotenus squared. Step four: unfold, and this time let's choose a different four triangles to fold back. Rip along one little leg, and fold back these two triangles. Then you can fold back another two over here. The area of the unfolded paper, minus four triangles, must be the same, no matter which four triangles you remove. So let's see what we've got. We can divide this into two squares, This one has sides the length of the little leg of the triangle. And this one has sides as long as the big leg. So the area of both together, is little leg squared, plus big leg squared. Which has to be equal to this area, which is hypotenus squared. If you called the sides of your triangle something more abstract, like: a, b, and c, you'd of course have a squared plus b squared equals c squared So quick review: Step 0: Aquire a paper square. OK, Step One: Fold it in half three times. Step 2: Fold parallel to the edges anywhere you choose and extend the crease. Step Three: Fold back four right triangles around the square and admire the area hypothenus squared that is left over. Step Four: unfold and rip along a short side to fold back another four right triangles and admire the area one leg squared plus the other leg squared that is left. And that is all there is to it! Of course, mathematicians are rebels and never believe anything anyone tells them unless they can prove it for themselves. So be sure to not believe me when I tell you things like: This is a square. Think of a few ways you could convince yourself that no matter what the triangles on the outside look like, this will always be a square, and not some kind of a rombus or parallelogram or dolphin or something. Or, you know, maybe it is a dolphin, in which case you should define what a dolphin is and then show that this fits that definition. Also, these edges look like they line up together. Do they always do that? Is it exact?