Find $∠ADC$ given $A,D$ on circle with diameter $BC$ and $∠BAO = 50°$ If $BC$ is a diameter of the circle and $∠BAO = 50°$. Then find the value of $∠ADC$.

This stumped me a little, I think there is a rule that mentions something about the center of the circle and it's relation to angles on the circumfrence, but I'm not sure, could someone help me point out the laws that would help to solve this problem? (Do not give me the solution, since I want to approach this myself)
Thanks in advance! :)

See $ \angle  OAB = \angle OBA  = 50^\circ$ (why?)
$\angle AOC = 100^\circ  $ (why?)
Now see angle subtended in the centre is double the angle subtended at any other point on the circle. Join AC. Now $ \angle AOC = 2\angle ADC$

 $\angle ADC= 50^\circ $

