The correct option is
A 70∘from the figure
AOC is a straight line
∠AOB+∠BOC=180⟹∠BOC=180−80=100
In triangle BCO
OB=OC⟹∠OCB=∠OBC( Angle opposite to equal side are equal)
∠OBC+∠OCB=180−100=80⟹∠OCB=∠OBC=40∘
As DCBE is cyclic quadrilateral
∠EDC+∠CBE=180⟹∠EDC=180−120=60
AS BE||CD
∠DEB+60=180⟹∠DEC+50=180−60⟹∠DEC=120−50=70∘