1. [10] Triangle $A B C$ has $A B=5, B C=3 \sqrt{2}$, and $A C=1$. If the altitude from $B$ to $A C$ and the angle bisector of angle $A$ intersect at at $D$, what is $B D$ ?
Answer: $\sqrt{\frac{5}{3}}$ Let $E$ be the foot of the perpendicular from $B$ to line $A C$. By the Law of Cosines, $\cos \angle B A C=\frac{4}{5}$, and it follows that $B E=3$ and $A E=4$. Now, by the Angle Bisector Theorem, $\frac{B D}{B E}=\frac{A B}{A B+A E}=\frac{5}{9}$, so $B D=\frac{5}{3}$.