Rectangle $ABCD$ has $AB = \sqrt{2}$ and $BC = 1$. Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$. What is the degree measure of $\angle AMD$?

1. The angle condition makes $MD$ the angle bisector of $\angle AMC$. Drop $DE\perp CM$. The right triangles $DAM$ and $DEM$ are congruent, so $AD=DE$.
2. The right triangles $CDE$ and $MCB$ are congruent, so $CM=CD=AB=\sqrt2$ and $BM=BC=1$.
3. In right triangle $MCB$, $BM=BC=1$, so $\angle BMC=45^\circ$. Since $A,M,B$ are collinear, $\angle AMC=180^\circ-45^\circ=135^\circ$.
4. As $MD$ bisects $\angle AMC$, $\angle AMD=\frac{135^\circ}{2}=67.5^\circ$.