Berkeley Math Circle: Monthly Contest 6

Solution:

Since $P A$ is tangent to the circle and $\angle A B D$ is the inscribed angle opposite $A D$, $\angle P A D = \angle A B P$. Since $\angle A P B$ is shared, it follows from AA similarity that $\triangle P A D \sim \triangle P B A$. Thus we get
$$\frac{A B}{A D} = \frac{P B}{P A}$$
Similarly, $\triangle P C D \sim \triangle P B C$, so
$$\frac{B C}{C D} = \frac{P B}{P C}$$
Since $P A$ and $P C$ are tangents from the same point to the same circle, they are the same length. Thus,
$$\frac{A B}{A D} = \frac{B C}{C D} \Longrightarrow A B \cdot C D = B C \cdot D A$$