27. [13] In $\triangle A B C, D$ and $E$ are the midpoints of $B C$ and $C A$, respectively. $A D$ and $B E$ intersect at $G$. Given that $G E C D$ is cyclic, $A B=41$, and $A C=31$, compute $B C$.
Answer:
49
Solution:
By Power of a Point,
$$
\frac{2}{3} A D^{2}=A D \cdot A G=A E \cdot A C=\frac{1}{2} \cdot 31^{2}
$$
so $A D^{2}=\frac{3}{4} \cdot 31^{2}$. The median length formula yields
$$
A D^{2}=\frac{1}{4}\left(2 A B^{2}+2 A C^{2}-B C^{2}\right),
$$
whence
$$
B C=\sqrt{2 A B^{2}+2 A C^{2}-4 A D^{2}}=\sqrt{2 \cdot 41^{2}+2 \cdot 31^{2}-3 \cdot 31^{2}}=49 .
$$