Solution:

Since $Z$ lies on diagonal $AC$, we have $\angle DAZ = 45^{\circ}$ and $\angle ZAB = 45^{\circ}$. Therefore $B$ lies on diagonal $AY$ of square $AXYZ$ and $\angle BAX = 45^{\circ}$.

Since $AB = AD$ and $AX = AZ$ and $\angle BAX = 45^{\circ} = \angle DAZ$, we have congruent triangles
$$\triangle DAZ \equiv \triangle BAX \qquad (SAS).$$
Therefore let $x = \angle ZDA = \angle XBA$ and $y = \angle AZD = \angle AXB$. By the angle sum in triangle $DAZ$ we have $\angle DAZ + \angle AZD + \angle ZDA = 180^{\circ}$. Therefore $x + y = 135^{\circ}$. Now by the angle sum in quadrilateral $DAXM$ we get $\angle DAX + \angle AXM + \angle XMD + \angle MDA = 360^{\circ}$. Therefore
$$\angle BMD = 90^{\circ}.$$
Hence $ABMCD$ is cyclic (the circle with diameter $BD$). Therefore
$$\angle DMC = \angle DAC = 45^{\circ}.$$
Also $AXYMZ$ is cyclic (the circle with diameter $XZ$). Therefore
$$\angle YMX = \angle YAX = 45^{\circ}.$$
Hence $\angle YMC = \angle YMX + \angle YMD + \angle DMC = 45^{\circ} + 90^{\circ} + 45^{\circ} = 180^{\circ}$ as required.