You might like to visualise this way:

1. Draw your triangle.
2. Draw the triangle you get from rotating the fist triangle 180 degrees about the centre of the circle.
3. You can see, without doing the work of proving it, that what you have constructed can only be a rectangle and all four corners will be of 90 degrees.

This assumes the triangle was already a right triangle. Circluar (pun intended) reasoning. Also it only explains those two specific triangles.

It does not assume that. Maybe it doesn't work for you, but I can see from this thought experiment, without needing the standard proof, that it can only be a rectangle.

If you need some interim steps between looking at the image and concluding that it is a rectangle (getting nearer to the proof that OP specifically said he didn't want) you can imagine trying to deform the quadrilateral to vary the angles while still maintaining the vertices on the circle, you can see that this is impossible. The quadrilateral is fixed. Looking at the symmetry again shows that all the angles are right angles.

Since the starting triangle was a random angle in a semi-circle, the visualisation applies to all such triangles.

Can you not see this without needing an algebraic or verbal proof?

The proof is the "reason and logic" behind it. I understand what you're asking, but that's just not how math works. You want some pretty, intuitive reason for why it is true, but not every proof in math is elegant like that. Plenty of things just work becuase you show they do, and it's not always pretty, clear, or obvious. Sometimes you make a bunch of reductions and inferences, most of them seemingly completely unrelated, until you go from what you know to be true, to what you want to prove to be true. And it's not pretty, and it's hard to follow, and it challenges some people's notion of math being this elygant, aesthetically pleasing set of rules that are somehow hidden in the universe, and we have to discover them.

That's not how math works. It wasn't hidden in the world by God. Humans made it up, and sometimes, even if we start from something simple and pretty, and we end at something simple and pretty, the journey may be complicated and ugly. And we just have to deal with it.

So yeah, you can look up the proofs online, no need for someone to transcribe them here, but there's no guarantee you will like them. They are not meant to be elegant, they are just meant to be correct.

Well, the reason and logic behind it are in the proofs for it, of which there are several. I like this one:

https://upload.wikimedia.org/wikipedia/commons/thumb/7/7c/Thales%27_Theorem.svg/180px-Thales%27_Theorem.svg.png

Since OA = OB = OC, ∆OBA and ∆OBC are isosceles triangles, and by the equality of the base angles of an isosceles triangle, ∠OBC = ∠OCB and ∠OBA = ∠OAB.

Let α = ∠BAO and β = ∠OBC. The three internal angles of the ∆ABC triangle are α, (α + β), and β. Since the sum of the angles of a triangle is equal to 180°, we have:

a + (a + B) + B = 180

2a + 2B = 180

2(a + B) = 180

a + B = 90

Thanks for the answer!

Thank you sm for taking out the time to reply!!

Okay, thank you so much!!

The question is if you make a Thales triangle, is it a right triangle? You start by assuming that if you have a right triangle, then you flip it over the hypothenuse, and you get a rectangle. Well, of course you do, it was right triangle to begin with! If it wasn't a right triangle, and you did your construction, you wouldn't get a rectangle.

Maybe you can't see it, but I can.

I started with a triangle in a semi-circle as requested by OP and did not assume it was a right-angled triangle. However when completing the construction as I described, it is clear (to me at least) that it can only have been a right angle to start with. I filled out a couple of extra steps to help you complete the visualisation, but it seems you need the formal algebra. Maybe OP still needs the proof, but to me it is so obvious as not to need formal proof.

Ok buddy, what happens if it isn't a right triangle? Never heard of a rhombus?

Also, you keep using the words "formal algebra". I don't think they mean what you think they mean.

That's my point; I'm not assuming it is a right triangle, but when the construction is drawn I (even if not you) can see that it must be one. I've even given you a couple of interim steps to seeing why this is so, but you don't seem to grasp it even then.

You ARE assuming the original triangle is a right triangle. If you take a triangle ABC, rotate it 180° and translate it so that AC coincides with the original CA, you're guaranteed to get a PARALLELOGRAM. You're only guaranteed to get a rectangle if the original triangle was a right triangle.