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Geometry Difficulty 4.7 AIME Prove it United States

Problem:
In $\triangle ABC$ , $D$ and $E$ are the midpoints of $BC$ and $CA$ , respectively. $AD$ and $BE$ intersect at $G$ . Given that $GEC D$ is cyclic, $AB = 41$ , and $AC = 31$ , compute $BC$ .

Solution

Solution:

By Power of a Point,
$\frac{2}{3} AD^{2} = AD \cdot AG = AE \cdot AC = \frac{1}{2} \cdot 31^{2}$
so $AD^{2} = \frac{3}{4} \cdot 31^{2}$ . The median length formula yields
$AD^{2} = \frac{1}{4}\left(2 AB^{2} + 2 AC^{2} - BC^{2}\right)$
whence
$BC = \sqrt{2 AB^{2} + 2 AC^{2} - 4 AD^{2}} = \sqrt{2 \cdot 41^{2} + 2 \cdot 31^{2} - 3 \cdot 31^{2}} = 49$

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