# Heronian triangles

How to prove that all Heronian triangles can be found using formulas described here?

I understand that the described substitution will give Heronian triangle, but how to prove that using the described substitution will give all Heronian triangles (with scaling solutions, for example $(12,10,10)$ is obtained when multiplying $(6,5,5)$ by $2$)?

-
Perhaps related: math.stackexchange.com/questions/114112/… –  joriki Mar 1 '12 at 20:02
This appears to be the Carmichael publication referred to in the Wikipedia article. –  Peter Phipps May 1 '12 at 10:57

Well, understanding that Heronian triangles are cyclic is a start. So the scaled up($\times n$, say) version of a Heronian would fit precisely into a circle of twice the diameter.