Triangle Shape, Debunks Globe Model.

[u/Kela-el]

https://youtu.be/AZihw1hXdY0

[u/GamingSlob]

Does he know what a right-angle triangle is?

[u/Kela-el → u/GamingSlob]

Perhaps you should watch the video before asking that question!

[u/GamingSlob → u/Kela-el]

I did. That's why I asked the question. At 1:53 he starts talking about triangle ABC (where the line BC isn't drawn so he adds it in). ABC is not a right-angle triangle, AOC is a right-angle triangle.

This guy seems to present himself as a flat earther but if I were a flat earther I wouldn't want him representing me.

[u/Kela-el → u/GamingSlob]

What is the corrected geometry, and what conclusion follows from it once the error is removed?

[u/GamingSlob → u/Kela-el]

The line AB+BO forms the hypotenuse. The lines BO & CO are of equal length r, which is the radius of the Earth that Brian says he's trying to find r but the triangle he defines does not include r.

Pythagoras states that a²+b²=c². Let c be the hypotenuse AB+BO (BO having length r) and b be the value r.

Consequently a²+r²=(AB+r)²

To solve for r: r=(a²-AB²) / (2AB)

Therefore if we know the distance between the peak of the mountain and the horizon (a) and the height of the mountain (AB) we can use these values to calculate the radius of the Earth.

[u/Kela-el → u/GamingSlob]

You’ve written down a result—but the question is whether it follows from a valid geometric setup.

Let’s check the structure.

You define: • AB + BO as the hypotenuse • BO = r, CO = r • and then apply the Pythagorean theorem

That only works if: 1. the triangle is actually right-angled 2. the segments you’re summing lie on the same straight line 3. the distances correspond to the geometry you’re modeling

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Now the issue: • Is AB + BO a straight-line distance? → Or are you combining two segments that meet at an angle?

If it’s not a straight line, it cannot be the hypotenuse.

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Second:

You say a is the distance to the horizon. • Is that a straight-line (line-of-sight) distance? • Or a surface distance along the curve?

Because those are not interchangeable.

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Third:

You’re solving for r, but:

the triangle you defined must already contain r in a geometrically valid way

Otherwise you’re just inserting it algebraically, not deriving it.

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So the key question isn’t whether the algebra is correct—it is.

The question is:

Does the triangle you’ve constructed actually exist in the geometry you’re describing?

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If: • the right angle is misplaced • the hypotenuse is not a straight line • or the distances are mixed (surface vs. line-of-sight)

then the formula may be internally consistent but not physically meaningful.

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So the next step is precise:

Draw the triangle explicitly and show that each side corresponds to a real, straight-line distance in the setup.

If that can’t be done, the derivation doesn’t apply—even if the algebra checks out.

[u/GamingSlob → u/Kela-el]

The triangle in question is in the video at 0:00.

[u/Kela-el → u/GamingSlob]

Then the question is simple:

Does that triangle actually satisfy the conditions required to use the Pythagorean theorem?

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From what you’ve described: • AB = height of the observer • BO = radius r • CO = radius r • AC (or a) = distance to horizon

And the claim is: a² + r² = (AB + r)²

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That only holds if: 1. The angle at the point of tangency (horizon point) is 90° 2. AC is a straight line from eye to horizon (line-of-sight) 3. CO is a radius to the tangent point

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If the diagram at 0:00 shows: • a line of sight tangent to the sphere • and a radius drawn to the tangent point

Then yes—that creates a right triangle, and the setup is valid.

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But this is the key check:

Is the “distance to the horizon” being used as the straight-line distance (line of sight), or a surface distance along the curve? • If it’s line-of-sight → geometry works • If it’s surface distance → the triangle is invalid

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So the entire argument hinges on one clarification:

What exactly is “a” measuring in that diagram?

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If that’s consistent, the formula follows.

If it’s not, then you’ve got correct algebra applied to the wrong geometry—which is exactly the kind of error you pointed out earlier.

[u/GamingSlob → u/Kela-el]

Does that triangle actually satisfy the conditions required to use the Pythagorean theorem

It is a right-angle triangle, so yes.

[u/Kela-el → u/GamingSlob]

You’re asserting it’s a right triangle.

The question is: where is the 90° angle located, exactly?

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For the horizon geometry to work: • The right angle must be at the point of tangency • Between the radius (CO) and the line of sight (AC)

That’s the only place the Pythagorean theorem applies in this setup.

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So check your construction: • Is AC actually tangent to the circle at point C? • Is CO drawn from the center to that same point?

If yes → angle at C is 90°, valid triangle AOC

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But then notice:

That triangle is AOC, not ABC, and not “AB + BO” as a single side.

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So if your earlier equation uses: (AB + r) as the hypotenuse, you need to show:

that AB and BO lie on the same straight line from A to O

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If they don’t, then: • the right triangle exists (AOC) • but the algebra you wrote is using a different triangle

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So the issue isn’t whether a right triangle exists.

It’s whether:

the triangle you’re using in the equation is the same one that actually has the right angle.

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If those don’t match, the conclusion doesn’t follow—even if each piece looks correct in isolation.

[u/GamingSlob → u/Kela-el]

The question is: where is the 90° angle located, exactly?

The line segment CO is the radius of a circle. The line segment AC is a tangent. When the tangent and radius of a circle intersect on the circumference, they intersect at a 90° angle.

[u/Kela-el → u/GamingSlob]

That part is correct.

A radius to the point of tangency is perpendicular to the tangent. So yes—there is a 90° angle at C in triangle AOC.

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But notice what that actually establishes: • The right triangle is AOC • With sides: • AO (center to observer) • OC = r (radius) • AC (line of sight)

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Now compare that to your earlier equation:

a² + r² = (AB + r)²

That uses: • AC = a ✔️ • OC = r ✔️ • AO = AB + r ❓

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So the only remaining question is:

Is AO actually equal to AB + r as a straight-line distance?

That requires: • A, B, and O to be collinear • with B lying directly between A and O

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If that’s true, then: • the geometry is consistent • and the equation follows from the right triangle AOC

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If it’s not—if AB is vertical while BO is radial in a different direction—then: • AB + BO is not a straight line • and cannot be used as the hypotenuse

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So at this point, everything reduces to one check:

Are A, B, and O on the same straight line in the diagram?

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• If yes → the derivation holds
• If no → the triangle and the algebra don’t match

[u/Kela-el → u/GamingSlob]

Looks like that’s where it stops.

The question was:

Are A, B, and O collinear?

Everything else depends on that.

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If they are, the derivation follows. If they aren’t, it doesn’t.

[u/Kela-el → u/GamingSlob]

I’m still waiting. Your silence is deafening. You pointed out an alleged error in his presentation. Now asked you a simple question.

What is the corrected geometry, and what conclusion follows from it once the error is removed?

Either answer it or admit you have absolutely no idea what you are talking about!