Solve this:

Q). $\angle$ADC = 130$°$, chord BC = chord BE, find $\angle$CBE.

Hello Pratham, <ADC + <ABC = 180 since sum of opposite angles in cyclic quad. is 180
==> 130 + <ABC = 180
===> <ABC = 50
If AB is taken as diameter or AB is perpendicular to CE, then in triangles BOC and BOE. 
chord BC = chord BE (equal sides given)
OB = OB 
<BOC = <BOE = 90
Thus triangles are congruent
==> <CBO = <EBO
<EBC = 50 + 50 = 100 deg