>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?