In the figure,$\angle ADC = 130^{\circ}$ and chord $BC = $ chord $BE$. Find $\angle CBE$.

$(100^{\circ})$ In the given figure,$ABCD$ is a cyclic quadrilateral. The sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$.
Therefore,$\angle ADC + \angle ABC = 180^{\circ}$.
Given $\angle ADC = 130^{\circ}$,so $130^{\circ} + \angle ABC = 180^{\circ}$.
$\angle ABC = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
Now,consider $\Delta OBC$ and $\Delta OBE$:
$BC = BE$ (Given)
$OC = OE$ (Radii of the same circle)
$OB = OB$ (Common side)
By $SSS$ congruence rule,$\Delta OBC \cong \Delta OBE$.
Therefore,$\angle OBC = \angle OBE$ (by $CPCT$).
Since $\angle ABC = 50^{\circ}$,and $\angle ABC$ is not directly $\angle OBC$ (as $O$ is the center,$AB$ is a diameter),we note that $\angle OBC$ is the angle subtended by chord $BC$ at the center. However,based on the geometry,$\angle OBC$ is part of the triangle. Since $\Delta OBC \cong \Delta OBE$,$\angle OBC = \angle OBE$.
Actually,$\angle ABC$ is the angle subtended by arc $AC$ at the circumference. The angle $\angle OBC$ is $50^{\circ}$.
Thus,$\angle CBE = \angle OBC + \angle OBE = 50^{\circ} + 50^{\circ} = 100^{\circ}$.