Let $A, B, C, D$ be four points on a circle in that order. Also, $A B=3, B C=5, C D=6$, and $D A=4$. Let diagonals $A C$ and $B D$ intersect at $P$. Compute $\frac{A P}{C P}$.
Note that $\triangle A P B \sim \triangle D P C$ so $\frac{A P}{A B}=\frac{D P}{C D}$. Similarly, $\triangle B P C \sim \triangle A P D$ so $\frac{C P}{B C}=\frac{D P}{D A}$. Dividing these two equations yields \[ \frac{A P}{C P}=\frac{A B \cdot D A}{B C \cdot C D}=\frac{2}{5} \] $\fbox{\frac{2}{5}}$.