Geometry Practice Test 3

OA = OB = OC = radius
∠BOC = 2∠BAC = 2 × 50°
⇒ ∠BOC = 100°
O is the circumcentre of ΔABC.
OD ⊥ BC

In ΔOBD and ΔOCD,
OB = OC = radius
OD = OD common
∠ODB = ∠ODC = 90°
⇒ ΔOBD ≅ ΔOCD (by Rule RHS)
⇒ ∠BOD = ∠COD = 50°
Now in ΔBOD,
∠BOD + ∠ODB + ∠OBD = 180°
∠OBD = 180° - (90° + 50°)
∠OBD = 40°
∠OBC = 40°