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In $\triangle ABC$, altitude $AD = 18$, median $BE = 9\sqrt5$ and median $CF = 15$. Find $BC$.

figure (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.

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Let $G$ the centroid of $\triangle ABC$, $GH$ the height of $\triangle GBC$ relative to $BC$ and $FK$ the height of $\triangle FBC$ relative to $BC$. Recall that: $$CG=\frac{2}{3} CF =10$$ and $$BG = \frac{2}{3} BE = 6 \sqrt{5}.$$ Note that $\triangle BFK \sim \triangle BAD$ and that $\triangle CGH \sim \triangle CFK$, hence: $$FK= \frac{1}{2} AD= 9$$ and $$GH = \frac{2}{3} FK = 6.$$ Using Pytagoras we get: $$BH = 12$$ and $$HC= 8.$$ Therefore $$BC=20.$$

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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 $$\frac{3}{2}|\overline{BC}| = 18 + 12 = 30 \qquad \to \qquad |\overline{BC}| = 20$$

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