Geometry – How to Know Common Rearrangement Proofs of the Pythagorean Theorem Work for Any Right Triangle

geometry

I’m a little bit puzzled by geometrical proofs, like the common algebraic proof for the Pythagorean theorem listed Wikipedia's "Pythagorean theorem" entry.

I understand the idea of arranging the right triangles and the area $c^2$ in a neat way to form another square and writing the area of the new square in different terms and going from there.

But I’m a little confused on how you actually know that the right triangle you draw isn’t just the one special right triangle with which you can actually form such a square.

How do we know this way of rearranging the pieces works for any other right triangle?

Best Answer

But I’m a little confused on how you actually know that the right triangle you draw isn’t just the one special right triangle with which you can actually form such a square.

If I am not mistaken, this is the figure you are referring to.

enter image description here

Ask yourself, what is it about this right triangle that is not applicable to every right triangle possible? We are not making any unwarranted assumptions about its side lengths or angles, right? We are not saying something like - one side has a length $4$ units or one acute angle measures $40^{\circ}$. So undoubtedly, this isn't "just one special right triangle".

As for being able to rearrange the triangles into a square, again, imagine any right triangle. Can you make four copies of it? Of course, you can. Can you arrange the four triangles as shown in the figures below? sure, why not?

Perhaps you are not sure if the corners would make right angles. Well, they have to. If you observe any corner made by the triangles, it's either the right angle itself, or the sum of the two acute angles of the right triangle. And remember, the sum of the acute angles in a right triangle is $90^{\circ}$.

Also, the quadrilaterals formed have to be squares. Each angle is $90^{\circ}$ and the sides are all equal in length ($a, b, a+b, \text{ or } c$depending on the square).

Response to OP's comment

How can we know that we can place the $c^2$ square in the middle for any right triangles. It seems a bit hand-wavy due to the square fitting in there while being all slanted like that.

We know the quadrilateral formed in the center (right figure) is a square because the sides are all equal ($c$) and the angles are $90^{\circ}$ (straight angle minus the two acute angles = $180^{\circ} - 90^{\circ}$). Again, this is independent of the shape and size of the right triangle.

Also, as ilkkachu says in the comments the squares are all formed as a result of us arranging the triangles. We are not trying to fit any squares.

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