2022 AMC 10A Problem 23

Let $P'$ be the reflection of $P$ across the perpendicular bisector of $\overline{BC}.$ This forms two new isosceles trapezoids: $CBPP'$ and $DAPP'.$ Therefore, we get $$\begin{gathered} P'A = PD = 4 \\ P'D = PA = 1 \\ P'C = PB = 2 \\ P'B = PC = 3. \end{gathered}$$ Using Ptolemy’s theorem, we know that the product of the diagonals is equal to the sum of the products of the opposite sides. Therefore: $$\begin{gathered} PP' \cdot AD + 1 = 16 \\ PP' \cdot BC + 4 = 9. \end{gathered}$$ This gets us $PP' \cdot AD = 15$ and $PP' \cdot BC = 5.$ Dividing these two equations yields $\dfrac{BC}{AD} = \dfrac{1}{3}.$ Thus, B is the correct answer.