Ray $BD$ is the bisector of $\angle ABC$ and ray $BE$ is the bisector of $\angle DBC$. If $\angle EBC = 19^{\circ},$ then find $\angle DBC, \angle ABC$ and $\angle ABE$.

(N/A) Given that ray $BE$ is the bisector of $\angle DBC$,we have $\angle EBC = \angle EBD = 19^{\circ}$.
Therefore,$\angle DBC = \angle EBC + \angle EBD = 19^{\circ} + 19^{\circ} = 38^{\circ}$.
Since ray $BD$ is the bisector of $\angle ABC$,we have $\angle ABD = \angle DBC = 38^{\circ}$.
Thus,$\angle ABC = \angle ABD + \angle DBC = 38^{\circ} + 38^{\circ} = 76^{\circ}$.
Finally,$\angle ABE = \angle ABD + \angle DBE = 38^{\circ} + 19^{\circ} = 57^{\circ}$.