So my claim here is that if we fix $AB$, then there is either one or no possible value of $M$ such that $\angle BAM = \alpha \ne 90^{\circ}$. We see this by letting $AB$ be the angle bisector of $\angle CAD$. Thus if we were to choose a point $M$, it must be on $C$, $D$ or there does not exist such a point. However, since quadrilateral $CADB$ is cyclic, $\angle C = 180^{\circ} - \angle D$. Noting that all triangles $\triangle ABM$ are similar, we see that there cannot exist two points for $M$ unless $\angle C = \angle D = 90^{\circ}$. Thus our claim is proven if $\angle BAM \ne 90^{\circ}$. Now to deal with the length condition...