8. Triangle $A B C$ has a right angle at $B$. Point $D$ lies on side $B C$ such that $3 \angle B A D=$ $\angle B A C$. Given $A C=2$ and $C D=1$, compute $B D$.
Answer: $\frac{3}{8}$
Solution: Let $B D=x$. We reflect $D$ over $A B$ to $D^{\prime}$. Then $D D^{\prime}=2 x$, but $A D$ bisects $C A D^{\prime}$, so $4 x=A D^{\prime}=A D$. Also, $A D=\sqrt{x^{2}+A B^{2}}=\sqrt{x^{2}+A C^{2}-B C^{2}}=$ $\sqrt{x^{2}+4-(x+1)^{2}}=\sqrt{3-2 x}$. We have the quadratic $16 x^{2}=3-2 x$ which gives $x=3 / 8$.