Let $N,U,V$ be the midpoints of $BC,AC,BD.$ Since $NU=NV=\frac{_1}{^2}AB,$ then the parallelogram $MUNV$ is obviously a rhombus. But from $\angle BMC=90^{\circ},$ we obtain $NM=\frac{_1}{^2}BC=NV=NU$ $\Longrightarrow$ $\triangle MUN$ and $\triangle MVN$ are both equilateral $\Longrightarrow$ $\angle UNV=120^{\circ}.$ Thus, if $Q \equiv AB \cap DC,$ we have $\angle BQC=120^{\circ}.$ Let $P$ be the intersection of the diagonals $AC,BD.$ Then simple angle chase yields $\angle CPD=\angle CBD+\angle BCA=\frac{_1}{^2}(\angle BCQ+\angle CBQ)=30^{\circ}.$

another solution: let point $X$ on $CM$ such that $CM=MX$. $\Rightarrow AXDC$ is parallelogram $\Rightarrow AX=CD$, $CM=MX,BM\perp CX\Rightarrow BC=BX=BA=CD=AX$. so $B$ is circumcenter of $CAX$. $\widehat{ABX}=60\Rightarrow \widehat{ACX}=30$. similar we find $\widehat{MBD}=30$. let $AC$ meets $BD$ at $O$. $\widehat{CAM}=\widehat{CMD}-30,\widehat{BDM}=\widehat{BMA}-30,\widehat{BMA}+\widehat{CMD}=90\Rightarrow AOD=150$ so we are done.