10. [8] Points $A$ and $B$ lie on circle $\omega$. Point $P$ lies on the extension of segment $A B$ past $B$. Line $\ell$ passes through $P$ and is tangent to $\omega$. The tangents to $\omega$ at points $A$ and $B$ intersect $\ell$ at points $D$ and $C$ respectively. Given that $A B=7, B C=2$, and $A D=3$, compute $B P$.
Answer:
9
Solution: Say that $\ell$ be tangent to $\omega$ at point $T$. Observing equal tangents, write
$$
C D=C T+D T=B C+A D=5 .
$$

Let the tangents to $\omega$ at $A$ and $B$ intersect each other at $Q$. Working from Menelaus applied to triangle $C D Q$ and line $A B$ gives
$$
\begin{aligned}
-1 & =\frac{D A}{A Q} \cdot \frac{Q B}{B C} \cdot \frac{C P}{P D} \\
& =\frac{D A}{B C} \cdot \frac{C P}{P C+C D} \\
& =\frac{3}{2} \cdot \frac{C P}{P C+5},
\end{aligned}
$$
from which $P C=10$. By power of a point, $P T^{2}=A P \cdot B P$, or $12^{2}=B P \cdot(B P+7)$, from which $B P=9$.