2024 AMC 12A Problem 19

In $\triangle ACD,$ the law of cosines gives $AC^2=9+25-2(15)\cos120^\circ$ $=34+15=49,$ so $AC=7.$ Since $ABCD$ is cyclic, $\angle ABC=180^\circ-120^\circ=60^\circ.$ In $\triangle ABC$ with $BC=3$ and $AC=7,$ the law of cosines gives $49=AB^2+9-3AB,$ so $AB=8.$ By Ptolemy, $AC\cdot BD=AB\cdot CD+BC\cdot DA$ $=8\cdot3+3\cdot5=39,$ hence $BD=\tfrac{39}{7}.$ This is shorter than $AC=7.$ Thus, the correct answer is D.