9. [20] Cyclic quadrilateral $A B C D$ has side lengths $A B=1, B C=2, C D=3$, and $A D=4$. Determine $A C / B D$.
Answer: $5 / \mathbf{7}$. Let the diagonals intersect at $P$. Note that triangles $A B P$ and $D C P$ are similar, so that $3 A P=D P$ and $3 B P=C P$. Additionally, triangles $B C P$ and $A D P$ are similar, so that $2 B P=A P$. It follows that
$$
\frac{A C}{B D}=\frac{A P+P C}{B P+P D}=\frac{2 B P+3 B P}{B P+6 B P}=\frac{5}{7} .
$$