In the figure, lines $AB$ and $CD$ pass through the center $O$ of a circle.
Given: $\;$ $O$ is the center of the circle and line $CD$ passes through $O$.
$\implies$ $CD$ is a diameter of the given circle.
Then, $\;$ $\angle CED = 90^\circ$ $\;\;\;$ [angle in a semicircle]
Given: $\;$ $\angle OCE = 40^\circ$
Points $D$, $O$ and $C$ are collinear points.
$\therefore \;$ $\angle DCE = \angle OCE = 40^\circ$
Now, in right triangle $CED$,
$\angle DCE + \angle CED + \angle CDE = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]
i.e. $\;$ $40^\circ + 90^\circ + \angle CDE = 180^\circ$
$\implies$ $\angle CDE = 50^\circ$
Given: $\;$ $\angle AOD = 75^\circ$
Now, $\;$ $\angle COM = \angle AOD = 75^\circ$ $\;\;\;$ [vertically opposite angles are equal]
In $\triangle OCM$,
$\angle OCM + \angle CMO + \angle COM = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]
i.e. $\;$ $40^\circ + \angle CMO + 75^\circ = 180^\circ$
i.e. $\;$ $\angle CMO = 65^\circ$
Now, $\;$ $\angle EMB = \angle CMO = 65^\circ$ $\;\;\;$ [vertically opposite angles are equal]
In $\triangle MBE$,
$\angle CED = \angle EMB + \angle MBE$ $\;\;\;$ [exterior angle is equal to the sum of the two interior opposite angles]
$\therefore \;$ $\angle MBE = \angle CED - \angle EMB = 90^\circ - 65^\circ = 25^\circ$
But $\;$ $\angle MBE = \angle OBE$ $\;\;\;$ [Points $O$, $M$ and $B$ are collinear points]
$\implies$ $\angle OBE = 25^\circ$