Austrian Mathematical Olympiad

We reflect the points $B$ and $C$ in $M$ and obtain the points $E$ and $F$, respectively. We clearly have $EC=BF=AB+AF=AB+CD=BC=EF$, therefore, the quadrilateral $BCEF$ is a rhombus. Since the diagonals in a rhombus are orthogonal, we get $BE\perp CF$ and we obtain $\angle BMC=90^{\circ }$ as desired.

Figure 1: Problem 2