In $\triangle ABC$, altitude $AD = 18$, median $BE = 9\sqrt5$ and median $CF = 15$. Find $BC$.
By herons formula and the normal area ( = 1/2 baseheight), $$18z=\sqrt(x+y+z)(x+y-z)(x+z-y)(z+y-x)$$
I solved this system using wolfram alpha and it is giving the right answer ($z=10$ so $BC=20$). But needless to say , it is a very tedious task to solve this system of equations. So a more elegant solution will be apreciated.