You're welcome. And thank you for posting this comment on your page:

https://www.cut-the-knot.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.cut-the-knot.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...

but I simply multiplied one equation by a, the other by b, and the third by c. It is pretty simple.

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 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

No, not at all.

The drawing is right on an html page. If you look at the examples there, there is all that is needed. No exe files, just inclusion of JavaScript they provide. You ac download Javascript and include it from your own server, or link to their copy.

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.cut-the-knot.org/Curriculum/Geometry/HIAI019.shtml

from scratch?

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.cut-the-knot.org/Curriculum/Geometry/HIAI019.shtml

This is Java.

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?

Do not know what other could say. Far as I am concerned, for this proof you do not even need trigonomentry. IMHO, the use of sine and cosine is entirely accidental and redundant.

How true. One does not need to restate the obvious. I suppose this is only a highlight of the fact that the PT is a generalization of the Cosine Law.

The Monthly now has this PWW in its 'filler archive' awaiting proper page space to be published. I'll let you know if/when it does.