<s>[INST] Given length of two medians and one altitude , find the length of one side. In $\triangle ABC$, altitude $AD = 18$, median $BE = 9\sqrt5$ and median $CF = 15$. Find $BC$.  


(Note that I've drawn median AG)
By appolonius theorem ,
$$2(15)^2+ 2x^2=(2y)^2+(2z)^2$$
$$2(9\sqrt5)^2+2y^2=(2x)^2+(2z)^2$$
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.
 [/INST] First, note that your diagram is somewhat inaccurate. The (undrawn) $\overline{EF}$ should be parallel to $\overline{BC}$, and half as long. (This is the "Midpoint Theorem for Triangles".) With that in mind ...

Let $P$ and $Q$ be the feet of perpendiculars from $E$ and $F$ to $\overline{BC}$. Then 
$$|\overline{EP}| = |\overline{FQ}| = \frac{1}{2}|\overline{AD}| = 9$$
By the Pythagorean Theorem in $\triangle EPB$ and $\triangle FQC$,
$$\begin{align}
|\overline{EP}|^2 + |\overline{BP}|^2 = |\overline{BE}|^2 \quad &\to \quad |\overline{BP}| = 18 \\
|\overline{FQ}|^2 + |\overline{CQ}|^2 = |\overline{CF}|^2 \quad &\to \quad |\overline{CQ}| = 12
\end{align}$$
Then, because
$$|\overline{BP}| + |\overline{CQ}| = |\overline{BC}| + |\overline{PQ}| = |\overline{BC}| + |\overline{EF}| = |\overline{BC}| + \frac{1}{2}|\overline{BC}| = \frac{3}{2}|\overline{BC}|$$
we have
$$\frac{3}{2}|\overline{BC}| = 18 + 12 = 30 \qquad \to \qquad |\overline{BC}| = 20$$
</s>