Let O be the center of a circle and AC be its diameter BD is a chord intersecting AC at E.  Point A is joined to B and D.  If ∠BDS = 60° and ∠AOD = 100°, then ∠BED = ?

100° 40° 80° 60°

100°

∠COD = 80° (linear angle) In ΔBOD → OB = OD (radius) ∠OBD = ∠ODB ∠OBD = 180° - (80° + 60°) = 40° ∠BEC = 60° + 40° = 100° (exterior angle of ∠BOE)