Let $I$ be the incenter. By construction, $\angle AMB=\angle AIB,$ and thus $360^\circ-2\angle C=90^\circ +\frac{\angle C}{2},$ or $\angle C=108^\circ.$ Therefore $\angle A+\angle B=72^\circ;$ but by construction, $\angle IAB=\angle BAM=\angle MBA=\angle ABI,$ and therefore $\angle A=\angle B=36^\circ,$ $\angle DAC=54^\circ,$ and $\angle AMB=144^\circ.$ Now $CA\cdot CD=2R\sin 36^\circ\cdot 2R\sin 54^\circ =4R^2\sin 36^\circ\cos 36^\circ=2R^2\sin 72^\circ=2R\sin 72^\circ\cdot R=AB\cdot AM.$

Kind of easier. Let $I$ be the incenter of $\triangle ABC$. Since $CI$ and $IM$ both go through the midpoint of arc $AB$(since $CI$ is the angle bisector as well and $MA=MB$) we get that $C,I$ and $M$ are collinear. This means that $CA=CB$. So now since $\angle MDC= \angle B$, $MC=MD$ and $CA=CB$ we get that $\triangle CAB\sim \triangle MCD$ and so, $\tfrac {CD}{AM}=\tfrac {CD}{DM}=\tfrac {AB}{CA}$ and we are done.

Because $I$ is the reflection of $M$ about $AB$ and $M$ is on the perpendicular bisector of $AB$, thus $AI=IB$. Define $\angle CAI = \alpha$. Thus, $\angle CAI = \alpha = \angle IAB = \angle IBA = \angle IBC$. It follows that $CA = CB$. Now $\frac {CA}{AB} = \frac{CB}{AB} = \frac {sin(2\alpha)} {sin(4\alpha)}$ by Law of sines on $\triangle ABC$ Similarly $\angle CMD = 2 \angle CAD = 180^\circ - 4\alpha$. Thus, Law of sines on $\triangle CMD$ gives $\frac{AM}{CD} = \frac{MD}{CD} = \frac {sin(2\alpha)}{sin(4\alpha)} =\frac{CB}{AB} $, and the conclusion follows. =)