Let $M$ be the midpoint of $CD$. Since $BM$ is the bisector of $B$, the reflection $E$ of $C$ in $BM$ lies on the ray $BA$. Because $\angle MEB = \angle MCB$, by reflection, and $\angle MECB = \angle DAB$ by hypothesis, we have $\angle MEB = \angle DAB$.

Hence $ME \parallel DA$; in particular $E$ is on the side $AB$.

Furthermore $ME = MC$ by reflection, and $MC = MD$, so $MC = MD = ME$. Therefore triangle $CDE$ is right with $\angle CED = 90^\circ$. Then $DE \perp CE$ and since $MB \perp CE$, we obtain $BM \parallel ED$.

Triangles $BEM$ and $EAD$ have parallel sides, hence they are similar in ratio $EM = CM = CD/AD = 3/2$. Then $BE = 3/2\ AE$ and $AB = 5/2\ AE$. Hence $AB/BC = AE/BE = 5/3$.