2. Let $A B C$ be a triangle with $\angle B A C=90^{\circ}$. Let $D, E$, and $F$ be the feet of altitude, angle bisector, and median from $A$ to $B C$, respectively. If $D E=3$ and $E F=5$, compute the length of $B C$.
Answer:
20
Solution 1:
Since $F$ is the circumcenter of $\triangle A B C$, we have that $A E$ bisects $\angle D A F$. So by the angle bisector theorem, we can set $A D=3 x$ and $A F=5 x$. Applying Pythagorean theorem to $\triangle A D E$ then gives
$$
(3 x)^{2}+(5+3)^{2}=(5 x)^{2} \Longrightarrow x=2 \text {. }
$$

So $A F=5 x=10$ and $B C=2 A F=20$.

Solution 2: Let $B F=F C=x$. We know that $\triangle B A D \sim \triangle A C D$ so $\frac{B A}{A C}=\frac{B D}{D A}=\frac{D A}{D C}$ and thus $\frac{B A}{A C}=\sqrt{\frac{B D}{D C}}=\sqrt{\frac{x-8}{x+8}}$. By Angle Bisector Theorem, we also have $\frac{A B}{A C}=\frac{B E}{E C}=\frac{x-5}{x+5}$, which means that
$$
\sqrt{\frac{x-8}{x+8}}=\frac{x-5}{x+5} \Longrightarrow(x-8)(x+5)^{2}=(x+8)(x-5)^{2}
$$
which expands to
$$
x^{3}+2 x^{2}-55 x-200=x^{3}-2 x^{2}-55 x+200 \Longrightarrow 4 x^{2}=400 .
$$

This solves to $x=10$, and so $B C=2 x=20$.