69th Belarusian Mathematical Olympiad

Answer: no. Since $(\angle ABC+\angle BCD)+(\angle BAD+\angle CDA)=360^{\circ }$, one of these summands is not less than $180^{\circ }$. Without loss of generality, assume that $\angle ABC+\angle BCD\ge 180^{\circ }$. Denote the reflection of the triangle $MCD$ about $M$ by $MB{D}_1$. The inequality $$\angle ABM+\angle MBE=\angle ABC+\angle BCD\ge 180^{\circ }$$ implies that the point $B$ lies either on the segment $AE$ or inside the triangle $AME$. In both cases $AB+BE<AM+ME$, which implies $AB+CD<AM+DM$.