Proof by Rearrangement of the Pythagorean Theorem

[FlatAssembler]

So, what do you think, is the Proof by Rearrangement of the Pythagorean Theorem valid? I've made a SVG animation presenting it here:
http://flatassembler.000webhostapp.com/p...ences.html
At first, it seems to make perfect sense. However, as I was making that animation, it appeared to me that it's actually a form of circular reasoning. Namely, how is it possible to prove that the angles of the c*c square are actually right angles, without appealing to the Pythagorean Theorem itself and the formula for the area of a parallelogram (P=a*b*sin(alpha))? It appears to me that it isn't.

[Tiberius]

What other angles could they be? The hypotenuses are all the same length, we don’t need to prove that as it’s part of the design of the initial setup. The triangles are all the same, that’s likewise by design. They are all transformed on the plane in the same way.

The only possible shape they can make is a square when all that is taken into account.

[Brian37]

I had a hypotenuse once, but I went to the doctor, got a shot for it, and it cleared up.(Note to readers: Nothing to see here, just being silly) We now return you to your regularly scheduled thread already in progress.

[Angrboda]

"Hypot en use. Hardhat required."

[GrandizerII]

So, what do you think, is the Proof by Rearrangement of the Pythagorean Theorem valid? I've made a SVG animation presenting it here:
http://flatassembler.000webhostapp.com/p...ences.html
At first, it seems to make perfect sense. However, as I was making that animation, it appeared to me that it's actually a form of circular reasoning. Namely, how is it possible to prove that the angles of the c*c square are actually right angles, without appealing to the Pythagorean Theorem itself and the formula for the area of a parallelogram (P=a*b*sin(alpha))? It appears to me that it isn't.

You just need to take a "look" to see it makes sense, right? Not sure where the necessity to appeal to that formula for the area of a parallelogram comes from.

[polymath257]

So, what do you think, is the Proof by Rearrangement of the Pythagorean Theorem valid? I've made a SVG animation presenting it here:
http://flatassembler.000webhostapp.com/p...ences.html
At first, it seems to make perfect sense. However, as I was making that animation, it appeared to me that it's actually a form of circular reasoning. Namely, how is it possible to prove that the angles of the c*c square are actually right angles, without appealing to the Pythagorean Theorem itself and the formula for the area of a parallelogram (P=a*b*sin(alpha))? It appears to me that it isn't.

You need to know the angles of a triangle add up to two right angles. That is enough to show that the central figure is a square.

To show the angles of a triangle add up to two right angles requires results on opposite interior and opposite exterior angles for parallel lines.

Those results require the parallel postulate.

[FlatAssembler]

What other angles could they be? The hypotenuses are all the same length, we don’t need to prove that as it’s part of the design of the initial setup. The triangles are all the same, that’s likewise by design. They are all transformed on the plane in the same way.The only possible shape they can make is a square when all that is taken into account.

But the sides being the same length doesn't mean the angles are the same. That's true only for triangles, not for the polygons with four (or more) angles.

As for the other replies, this is not a trolling question, so please don't answer as if it were.

Edit: Sorry, Polymath, I haven't seen your response, which actually makes some sense.

[GrandizerII]

I was being serious, FTR.

Like Tibs said, the triangles are exactly the same and rearranged in the same way. That could only result in a square, not just any rhombus.

Polymath's answer is more rigorous, fair enough.

[polymath257]

I was being serious, FTR.

Like Tibs said, the triangles are exactly the same and rearranged in the same way. That could only result in a square, not just any rhombus.

Polymath's answer is more rigorous, fair enough.

Something is required since the result fails for non-Euclidean geometry. And, in fact the point that fails is the angles are not right angles. The lines around the (a+b) squares are no longer straight.

[polymath257]

I'd also point out there are other implicit assumptions here. For example, how is the square of side length a+b constructed? Well, we can take a line segment of length a+b, say AB. Do perpendiculars from both ends, giving sides AC and BD, each of length a+b.

iI is an *assumption* that the line CD is perpendicular to both AC and BD and is of length a+b. In non-Euclidean geometry, this line is NOT perpendicular to either AC or BD (the angle is *less* than a right angle) and its length is *more* than a+b. There *are* no squares in Lobachevskian geometry with four equal sides and right angles. They simply don't exist.

So even the original figures have hidden assumptions. To be perfectly rigorous, all these details would need to be addressed.