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jmolokach
Member since Jan1111

Feb2811, 09:34 AM (EST) 

""Proof of the Cosine Rule completely independent of the PT""

Alex, At your page: https://www.cuttheknot.org/pythagoras/cosine.shtml You mention "I'll be extremely curious to learn of any proof of the Cosine Rule completely independent of the Pythagorean Theorem." I have one. It is quite simple, and quite possibly someone has already done this. We have three equations resulting from drawing the three altitudes in a triangle ABC: 1) c = bcosA + acosB 2) b = acosC + ccosA 3) a = bcosC + ccosB From equation 2) we have cosA = (b  acosC)/c From equation 3) we have cosB = (a  bcosC)/c Substituting for cosA and cosB into equation 1) gives the Law of Cosines. molokach 

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jmolokach
Member since Jan1111

Mar0311, 05:23 PM (EST) 

2. "RE:"
In response to message #1

>Yeap, that's good. Many thanks, John. You're welcome. And thank you for posting this comment on your page: https://www.cuttheknot.org/pythagoras/CosLawMolokach.shtml I have a PWW that is a slight twist on my original post. Rather than solving for cosA and cosB, I took the equation: c = acosB + bcosA and multiplied both sides by c. Then I envisioned this as the square of side c being equal to the sum of two rectangles drawn off of sides a and b whose widths are c*cosB and c*cosA respectively. https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0Bwc0zPMWJ9wqNDBjZTNlNTYtMjZmZS00ZmE3LTlhMTMtYjRjNWQ1Nzg5YzY4&hl=en&authkey=CIPjm4oB This is similar to Don McConnell's PWW at: https://www.cuttheknot.org/pythagoras/DonMcConnell.shtml except it avoids the blue rectangular parts (which come out in the ensuing algebra after my diagram). In addition, the drawing of the arcs makes right angle symbols and marking segments congruent unnecessary. Also, I wonder if you would be willing to write a Java Applet to go along with this PWW. Consider points A and B fixed and C movable (perhaps only in the direction of side a). If one makes C a right angle, then the rectangles become squares (The Pythagorean Theorem). If one makes C an obtuse angle, then the rectangles become longer than the sides a and b due to the fact that the altitudes intersecting outside the triangle. When this happens also, the rectangles overlap, yet their sum has to be considered in the total for c^2. But I imagine the overlap could just change color or turn black or something. I have no knowledge of Java or programming in general, but I think it would make for a great applet and something you might take an interest in. Also, in doing this I knew I had seen something similar and this is when I found Don McConnell's PWW  I actually remember him posting it on the Exchange somewhere this past summer when I began this obsession of mine with proofs...
molokach 

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jmolokach
Member since Jan1111

Mar0611, 09:24 AM (EST) 

4. "RE:"
In response to message #3

>John, > >but I simply multiplied one equation by a, the other by b, >and the third by c. It is pretty simple. I suppose there are many simple ways to make the three equations into the Law of Cosines. I like what you did, just had a different view when I came up with the idea  especially when trying to think of how this might play out geometrically. >I've been playing with the idea of using > >https://www.jsxgraph.org > >Do have a look. It is extremely simple. Practically no >programming, only definitions of objects. I am kind of iffy on how this thing works. There is a download but no .exe file or canvas to create the objects in. It'seems as if you need to use GeoGebra or something to create the objects and then use this somehow to generate the embeddable code for what you created elsewhere. If I am wrong in this, please point me in the right direction. Also, I have attached what the obtuse case might look like. It is a crude drawing I wrote on my whiteboard at school, but illustrates what I wrote in an earlier post... https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0Bwc0zPMWJ9wqZDBmZmZhNjUtZjE2Mi00ZWQxLWIyN2YtOGI2MzhiNDgyYmM3&hl=en&authkey=CP_JvtYB
molokach 

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jmolokach
Member since Jan1111

Mar0711, 10:00 AM (EST) 

6. "RE:"
In response to message #5

"The drawing is right on an html page...." Yes, but how do I get one there to start with? OK still confused. There is no documentation on how to create objects from scratch. When I browse through the examples, some of them have the source code. So I could copy the code, but then I have no idea what to do with it (remember I am not a programmer). There still seems to be no way to create objects from scratch, i.e. an html page where I can create points, segments and such using a menu. Also none of the examples on the page come close to what I want, so I would be hard pressed to edit them anyway. I have read through the tutorials, and the best I can get is to copy html code for everything I want to create. So it appears that the only way to get an object is to create/copy the code for it. Still a lot of programming going on. What I was looking for was a drawing program, i.e. Geometer Sketchpad to create the objects and then something else to convert it to an applet. I was under the impression that jsxgraph was a drawing tool, but it appears that you have to piece you objects together from the drawings that someone else has already created. For instance, how does one create: https://www.cuttheknot.org/Curriculum/Geometry/HIAI019.shtml from scratch? molokach 

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alexb
Charter Member
2808 posts 
Mar0911, 10:05 AM (EST) 

7. "RE:"
In response to message #6

>Yes, but how do I get one there to start with? Just create two points and a line segment. > >There still seems to be >no way to create objects from scratch, i.e. an html page >where I can create points, segments and such using a menu. Right there is no menu. You have to type an html page. >What I was looking for was a drawing program, i.e. Geometer >Sketchpad to create the objects and then something else to >convert it to an applet. Try GeoGebra. >I was under the impression that jsxgraph was a drawing tool, No, it's a JavaScript library >For instance, how does one create: > >https://www.cuttheknot.org/Curriculum/Geometry/HIAI019.shtml This is Java.


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jmolokach
Member since Jan1111

Mar1111, 10:22 AM (EST) 

8. "RE:"
In response to message #3

>John, > >but I simply multiplied one equation by a, the other by b, >and the third by c. It is pretty simple. >Consider a right triangle with altitude drawn from the right angle to the hypotenuse. This is a special case of my proof where: a = c*cosB b = c*cosA c = a*cosB + b*cosA If we do as you have done, we get c^2  a^2  b^2 = 0 which is exactly the Pythagorean Theorem. Does this qualify as yet another proof, or too close to proof 6? molokach 

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