> Use a triangle with the sides 3, 4, and 5. You know that the angles > are 30, 60, and 90 degrees. Use heron's formula and use another area > of the triangle formula and there is your proof.
Sorry, but this is not a proof. If Heron's formula was not true and this example disproved it, it would be a perfectly valid proof by counterexample that it is not true. But Heron's formula is a generalized statement-- it is a relationship between area and sides of ALL triangles. Therefore a proof of it MUST prove its validity for all triangles.It would have to use the definition of a triangle (a planar polygon with three sides), perhaps some of the characteristics of all triangles which may be be accepted because they were proven elsewhere (such as the sum of internal angles being 180 degrees).
I cannot try to formulate a proof now because I don't remember Heron's formula. However, I can suggest the following strategy which I think might be useful for a proof by construction:
Start with another method of finding the area of a general triangle (don't use the pythagorean theorem or trigonometry-- they require right triangles). Use the following methods to convert angles into side lengths (you may need to solve a system of simultanious equations):
For a triangle A B C with opposite angles a, b and c respectively,
Law of cosines:
A^2 = B^2 + C^2 - 2 A B cos( a )
Law of sines:
sin( a ) sin( b ) -------- = -------- A B
If I got one of these two laws wrong, somebody correct me. It's been a while and I don't have a reference available.
=================================================== Jim McCann Physics sophomore, Carnegie Mellon University firstname.lastname@example.org (412) 862 3003