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 $2$)?
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.