To define it, yes he did use theorems. However he first discoved it by playing with pencils. After he realized that the triangels were equal he checked it all with theorems

Are not all theorums just a colaboration of other theorums? There proofs, used to prove another theorum.

 

 

 

 

 

 

 

And that proof just happens to be what tested the theorum. For example, prove Vertical Angles are congruent uses angle addition postulate. Without that postulate, you would not be able to prove that statement is true.

 

 

 

 

 

 

 

In other words; in geometry, you prove theorums by other theorums, not just reword them. Even then, there have been accepted theorums that have been reworded.

 

 

 

 

 

 

 

A theorem is a proposition that has been or is to be proved on the basis of explicit assumptions. Proving theorems is a central activity of mathematicians. Note that "theorem" is distinct from "theory

 

 

 

 

 

 

 

Exactly.

 

 

 

 

 

 

 

A theorem is based on different theorems, and is to be proven true.

 

 

 

 

 

 

 

In this case, he has proven his theorem.

... But to know a traingle is simlar is not realy imporart in geometry.

 

 

 

 

 

 

 

I disagree. The 3rd proof for the pathagrean therom reqires the use of 3 simlar triangles. Friction on a slope in phisics uses a simlar triangle as well. Just 2 examples.

I did check on the internet for this theorum, (I also asked a pre-calculus techer) but I didn't find anything on it. I may do some research later on.

 

 

 

 

 

 

 

Very important to check, that's how some brilliant people waste their time. They spend all of it trying to prove something that's already proven (unless they're checking it), don't waste your time doing that...

 

 

 

Who said I was Brilliant? :mrgreen:

How do you know if you were the first to find this out?

 

 

 

 

 

 

 

Oops. I didn't see the part about you researching. :oops:

Sorry, but I fail to be impressed :P

 

 

 

 

 

 

 

The top drawing that you give is true, simply because there are two 'Z' Angles in it. Observe:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

If angles are in a 'X' shape, the angle on either side of it must be equal. This is true because if you take one side of either of the lines, that side of the line makes up 180 degrees. To bi-sect it at some point leaves two angles, and because both lines are straight, you must get two pairs of identical but opposite angles on each ide of that line.

 

 

 

 

 

 

 

If two parallel lines exist with a line intersecting both, then the opposite angles must also be equal (see drawing). This is true because if you split the diagonal line on the diagram, you get two 'X' shapes. Because the two horizontal lines are parallel, you effectively have two copies of the same thing. This is called the 'Z' rule.

 

 

 

 

 

 

 

Your 'theorem' uses nothing more than this. As someone else has said, it is not really a theorem, it is more of a deviation or derivative of a pre-existing rule. Sorry to disappoint you!

 

 

 

 

 

 

 

Challenge for you: Geometricaly tri-sect an angle. Do that and I'll send you a Mars bar by courier :P

(180 and 0 degrees don't count, presumably).

Sorry, but I fail to be impressed :P

 

 

 

 

 

 

 

The top drawing that you give is true, simply because there are two 'Z' Angles in it. Observe:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

If angles are in a 'X' shape, the angle on either side of it must be equal. This is true because if you take one side of either of the lines, that side of the line makes up 180 degrees. To bi-sect it at some point leaves two angles, and because both lines are straight, you must get two pairs of identical but opposite angles on each ide of that line.

 

 

 

 

 

 

 

If two parallel lines exist with a line intersecting both, then the opposite angles must also be equal (see drawing). This is true because if you split the diagonal line on the diagram, you get two 'X' shapes. Because the two horizontal lines are parallel, you effectively have two copies of the same thing. This is called the 'Z' rule.

 

 

 

 

 

 

 

Your 'theorem' uses nothing more than this. As someone else has said, it is not really a theorem, it is more of a deviation or derivative of a pre-existing rule. Sorry to disappoint you!

 

 

 

 

 

 

 

Challenge for you: Geometricaly tri-sect an angle. Do that and I'll send you a Mars bar by courier :P

 

 

 

 

 

 

 

And I say yet again, most other theorums are like this. For example, the vertical angle theorum is decided by both the angle addition postulate and the reflexive proporty. Therefore, stating that this is not a true therom would not be logical, as you used them in your diagram. Also, you must not know the geometry terms that I used. Not only that, but I have a counter example of what you stated.

 

 

 

 

 

 

 

And the "Z" rule is nothing but the Corrisponding, Alt. Int. angle, and oppisite Int. Angles that are mentioned in my proofs.

 

 

 

 

 

 

 

(P.S. to all: saying that I was probably not the first person to figure this out doesn't dissapoint me. Chances are someone has done this before. I take pride in the fact that I figured this out with only the help of myself.)

 

 

 

 

 

 

 

And I say yet again, most other theorums are like this. For example, the vertical angle theorum is decided by both the angle addition postulate and the reflexive proporty. Therefore, stating that this is not a true therom would not be logical, as you used them in your diagram. Also, you must not know the geometry terms that I used. Not only that, but I have a counter example of what you stated.

 

 

 

 

 

 

 

And the "Z" rule is nothing but the Corrisponding, Alt. Int. angle, and oppisite Int. Angles that are mentioned in my proofs.

 

 

 

 

 

 

 

(P.S. to all: saying that I was probably not the first person to figure this out doesn't dissapoint me. Chances are someone has done this before. I take pride in the fact that I figured this out with only the help of myself.)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I didn't say that the theorem was not correct :? I said that it is not a new theorem because it is made up of two other basic theorems, ad then went on to add to your proof using a differing method to your own. I don't quite know what you are on about with the terminology as we don't use those terms in England.

 

 

 

 

 

 

 

Also, due to the phrasing of your past I can't quite figure out if you are being intentionally insulting, or whether it is just an unfortunate by-product of your determination to prove me wrong. If it is intentional, then trousers down to you, if it is not, then accept my apologies for the hostile nature of this post!

The thing that always gets me down regarding interesting ideas is that until you're way beyond university level in understanding of a subject like maths, someone has almost certainly had the same idea as you already. And it's "theorem", with an e -- not meaning to be pedantic, but really.

... I just typed out 4 whole paragraphs of flaming because there's so many idiots in the world...

 

 

 

Umm... that proof is obvious, but if you have to show work it can save you a min or so...

 

 

 

Challenge for you: Geometricaly tri-sect an angle. Do that and I'll send you a Mars bar by courier

 

 

 

 

 

 

 

You owe me a mars bar:

 

 

 

 

 

 

 

Given an angle CAB draw a circle with centre A so that AC and AB are radii of the circle. From C draw a line to cut the line (not segment) BA produced at E. Have this line cut the circle at F and have the property that EF is equal to the radius of the circle. This can be done in a mechanical way by marking a length equal to the radius of the circle on the ruler and moving it keeping one mark on BA produced and having the second mark on the circle. Move the ruler keeping one mark on the line and the other on the circle until the ruler passes through C. Then the line EC is constructed. Finally draw from A the radius AX of the circle with AX parallel to EC. Then AX trisects angle CAB.

 

 

 

 

 

 

 

Message me on here for my address...

... I just typed out 4 whole paragraphs of flaming because there's so many idiots in the world...

 

 

 

Umm... that proof is obvious, but if you have to show work it can save you a min or so...

 

 

 

Challenge for you: Geometricaly tri-sect an angle. Do that and I'll send you a Mars bar by courier

 

 

 

 

 

 

 

You owe me a mars bar:

 

 

 

 

 

 

 

Given an angle CAB draw a circle with centre A so that AC and AB are radii of the circle. From C draw a line to cut the line (not segment) BA produced at E. Have this line cut the circle at F and have the property that EF is equal to the radius of the circle. This can be done in a mechanical way by marking a length equal to the radius of the circle on the ruler and moving it keeping one mark on BA produced and having the second mark on the circle. Move the ruler keeping one mark on the line and the other on the circle until the ruler passes through C. Then the line EC is constructed. Finally draw from A the radius AX of the circle with AX parallel to EC. Then AX trisects angle CAB.

 

 

 

 

 

 

 

Message me on here for my address...

 

 

 

 

 

 

 

Wikipedia? Could you draw me that so that I can follow it exactly - it's early and my brain hurts!