2. $[\mathbf{3}] A, B, C$, and $D$ are points on a circle, and segments $\overline{A C}$ and $\overline{B D}$ intersect at $P$, such that $A P=8$, $P C=1$, and $B D=6$. Find $B P$, given that $B P<D P$.
Answer: 2. Writing $B P=x$ and $P D=6-x$, we have that $B P<3$. Power of a point at $P$ gives $A P \cdot P C=B P \cdot P D$ or $8=x(6-x)$. This can be solved for $x=2$ and $x=4$, and we discard the latter.