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Hon1nbo

Wed Feb 10 2010, 10:54PM

Registered Member #902 Joined: Sun Jul 15 2007, 08:17PM
Location: North Texas
Posts: 1040

hey all, I took on this little project just to better myself at proofs, even though I know that supposedly SAS (or any other triangle congruency for that matter) can't be proven without assuming one of them is a postulate. Well, I tried anyways just to see what happened
I was shocked, i had it: then my teacher told me I can't use the Pythagorean Theorem because it requires SAS. I came up with alternatives but they just say "they don't work" without any explanation. The point of me doing this was to better understand WHY these proofs don't work. They say the proof of SAS will work if the Pythagorean Theorem issue can be fixed, or I find a workaround and not use it. When I tried to present this, they didn't know what to think.

the first method, which was strait up Polar Coordinates and the Distance Formula seemed to work, until it was pointed out to me that the Pythagorean Theorem requires SAS to prove. However, I tried to then prove the Pythagorean Theorem without triangle congruences, but Triangle similarities.
The proof I used is to take the triangle and form a square using triangles of that type with the same sides A, B, and angles Theta and Alpha (I do not specify that they have the same C as in the normal proof which would reason by SAS) and then use those to form a rectangle (which in the normal proof is a Square, but since I only know for a fact that one of the sides is C I cannot quite state it is a square, but it is a rectangle because the other hypotenuses are perpendicular to C)
The area for the whole this can be expressed as
A = 4*0.5*A*B + C*X (where X is the length of the sides perpendicular to C in the rectangle, but we don't know it is C) = (A+B)^2

= 2*A*B + C*X = A^2 + 2*A*B + B^2

= C*X = A^2 + B^2

However, I then realized that the angles of all the triangles are all congruent, and that since the at least one side is known for all triangles, then I can state that the ratio of C to X = ratio of A to A and therefore X = C
and using algebra C^2 = A^2 + B^2

from here the Pythagorean Theorem and sin/cos (which are Defined rather than proved/postulated)can be used to prove SAS for Triangles ABC and DEF with Angles A and D being congruent, and side AB = DE and side AC = DF

BC = Square Root((AB-AC*cos Theta)^2 + (AC*sin Theta)^2) = Square Root((DE-DF*cos Theta)^2 + (DF*sin Theta)^2) = EF

the angles are then found easily using trigonometry

I have been frustrated by the inability of people to explain why something is wrong with my proofs, and "you just can't do that" without an explanation is really annoying. However, my teacher wasn't able to make much out of this himself after cursory glances, he wants me to re-write everything and sit down with him sometime at the school, which I do not have much time to do

I thought I might try to put this out there to get an explanation on things

-Jimmy

------------------
Before I realized the usage of similar triangles in the Pythagorean Theorem, I would end up with what is below this line, which might be able to be used, but I got nowhere on it today as I had more trouble than I thought getting the X/Y out of there (p.s. if I did not make it clear enough, X is really BC and Y is really EF and unfortunately I cannot just state that even though that is what they are
-----------------
this modified formula for The Pythagorean Theorem is then used to find the length of the missing side of the SAS Triangles (this, in conjunction with the use of Sine and the Known Angle so that we form for triangles ABC and DEF): from here down two letters in a formula indicate a triangle side

(BC)*X = (AB - AC*cos Theta)^2 + (AC*sin Theta)^2 = (DE-DF*cos Theta)^2 + (DF*sin Theta)^2 = (EF)*Y

(BC)*X = (EF)*Y

Mattski

Thu Feb 11 2010, 07:42AM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

What are you trying to prove, that two triangles with the length of two sides and the angle between them defined as equal are equivalent triangles? It's a little hard to follow geometry proofs without pictures and a clear statement of what you are proving.

So without actually giving you feedback on your proof, which I wouldn't necessarily understand anyway, I would simply advise that you look at the Law of Cosines since I think that is about as general as you can get when working with triangles. Once you have that proven you have proven then if you have two triangles ABC, DEF, then with length AB=DE and BC=EF, with angles between those sides identical, then the third side and remaining two angles must be identical as determined by law of cosines.

Steve Conner

Thu Feb 11 2010, 08:53AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

What your teacher is trying to say is that ultimately all of the proofs depend on you assuming Euclidean geometry.

You can't prove that parallel lines never cross, you have to assume it, and then all of the high school proofs follow.

Alternative geometries are possible, though. If you assume that parallel lines DO cross somewhere, then you get a non-Euclidean geometry. It still works, but the angles in a triangle don't add up to 180 degrees any more. For more info google "non-Euclidean geometry" or stare at an Escher print for a while.

Hon1nbo

Thu Feb 11 2010, 02:31PM

Registered Member #902 Joined: Sun Jul 15 2007, 08:17PM
Location: North Texas
Posts: 1040

well, we do assume Euclidean geometry here in the USA - the SAS states that two triangles with two sides the same length and the angle between those lengths being the same, are the same triangle - it can be proven, if you assume one of the other triangle congruences are true (SSS, AAS, SSA, etc)

my proof works by revising a proof of the pythagorean theorem to make it so that you no longer need to assume another triangle congruence is true

I will try to get drawings up to explain, but I know that the proof of SAS holds if my proof of the pythagorean theorem using similar triangles holds, but no can seem to explain to me if it works - my teacher kind of looked stumped

Steve Conner

Thu Feb 11 2010, 03:21PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

This is my point though. The proofs of Pythagoras and SAS only work in Euclidean geometry, and that's why your teacher is stumped: there's no proof of Euclid's fifth postulate on parallel lines, and therefore there's no proof of Pythagoras or SAS either.

Euclid's parallel lines postulate is the hamster under the newly laid carpet of geometry. It's a lump that doesn't look quite right, you can try to stomp it down, but it'll just pop up somewhere else. If you don't assume either Pythagoras or SAS as a postulate, then you'll find yourself still having to assume the parallel lines postulate in some other form.

They still teach Euclidean geometry in high school, but that fifth postulate has never been proven. Since Einstein's general relativity, we now know that the fifth postulate is technically false, the geometry of real things is not quite Euclidean, and the angles in a triangle don't quite add up to 180 degrees, especially near a black hole or a really overweight Texan.

Hon1nbo

Thu Feb 11 2010, 04:42PM

Registered Member #902 Joined: Sun Jul 15 2007, 08:17PM
Location: North Texas
Posts: 1040

ok, maybe I am not getting this across: this is a high school class, and we have proofs of Pythagorean theorem assuming euclidean geometry

the reason for this different proof is because the normal proof of the pythagorean theorem requires SAS or another triangle congruence to prove, hence why SAS is assumed to be true. This has nothing to do with proving parallel lines as we are assuming euclidean geometry for this proof. All my proof does is to AVOID SAS completely in a proof of the pythagorean theorem. If this is done, then SAS can be proved without assuming and triangle congruences are a postulate. I understand completely that the Parallel Lines Theorem in Euclidian geometry is really a postulate you can't prove, but that is not what I am trying to prove
I will try to get pictures up today if you have trouble understanding my proof of the pythagoerean theorem

I am not trying to prove these for non-euclidean geometry. He suggested another geometry might have the answer, but that is not why my proofs are wrong (at least when he was able to make anything out) - I was going to give him a re-written copy today to see if things are better explained (he emailed me saying the main issue was that I needed to walk him through all of my steps) but we did not have school today due to snow (yes, we do have snow days here in Texas).

-Jimmy

Steve Conner

Thu Feb 11 2010, 04:57PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

As far as I understand it, there is no proof of Pythagoras, because it's logically equivalent to Euclid's fifth, and there is no proof of that.

Geometry without the fifth postulate is called "absolute geometry"

If you think you have a proof starting from absolute geometry and using high school algebra, it means that one of the logical equivalents of the (unproven) fifth postulate sneaked back in at some stage, so go back and examine your proof for these. Without Euclid's fifth, triangle similarity has no meaning, nor do the words "triangle", "rectangle", "right angle" and so on, so you can't use any of these concepts in a proof.

If you assume Euclidean geometry, then there's nothing to prove, because you already know that Pythagoras is just the fifth postulate in disguise.

Hon1nbo

Thu Feb 11 2010, 05:58PM

Registered Member #902 Joined: Sun Jul 15 2007, 08:17PM
Location: North Texas
Posts: 1040

do not see why the Pythagorean Theorem is essentially Euclid's fifth
the pythagorean theorem states that C^2 = A^2 + B^2 for a given right triangle ABC with C being the hypotenuse

in the US High School system the Pythagorean Theorem is commonly proved as early as freshman year, sometimes earlier

at a most basic of levels it may require the parallel postulate, but the whole point of Euclidean Geometry is that the parallel postulate is assumed, and therefore the Pythagorean Theorem can be proved, I am simply trying to prove it without assuming SAS (the parallel postulate is still assumed) so that SAS does not have to be assumed. If the Pythagorean theorem is proven with SAS, then SAS itself cannot be proven as it requires the Pythagorean Theorem to be proven

I really can't understand why this is a problem. Maybe we are just misunderstanding each other?

-Jimmy

Sulaiman

Thu Feb 11 2010, 08:08PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

If you use the Euclidian system any proof of SAS is trivial and of no value.
Your proof uses ASA to prove SAS !

Hon1nbo

Thu Feb 11 2010, 08:11PM

Registered Member #902 Joined: Sun Jul 15 2007, 08:17PM
Location: North Texas
Posts: 1040

Sulaiman wrote ...

If you use the Euclidian system any proof of SAS is trivial and of no value.
Your proof uses ASA to prove SAS !

where am I using ASA?

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