55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

Let the length of the segments be $AB=CD=3a$, then $AO=CO=a$ and $OB=OD=2a$. Since $\angle AOD=\angle COB$ as vertical (fig/ 17), then $△AOD=△COB$. Hence $\angle ADO=\angle CBO$. As $△BOD$ is isosceles, then $\angle BDO=\angle DBO$, consequently $\angle MDB=\angle MBD$ as the sums of the equal angles. In other words $△MDB$ is isosceles, from where the equality we need.