Points $A, B, C, D$ lie on a circle in that order such that $\frac{A B}{B C}=\frac{D A}{C D}$. If $A C=3$ and $B D=B C=4$, find $A D$.
By Ptolemy's theorem, we have $A B \cdot C D+B C \cdot D A=A C \cdot B D=3 \cdot 4=12$. Since the condition implies $A B \cdot C D=B C \cdot D A$, we have $D A=\frac{6}{B C}=\frac{3}{2}$. $\fbox{\frac{3}{2}}$.