∠CBE = 180 - ∠ABC∠CBO = 1/2 ∠CBE (BO is the bisector of ∠CBE)∠CBO = 1/2 ( 180 - ∠ABC) 1/2 x 180 = 90 ∠CBO = 90 - 1/2 ∠ABC .............(1) 1/2 x ∠ABC = 1/2∠ABC∠BCD = 180 - ∠ACD∠BCO = 1/2 ∠BCD ( CO is the bisector os ∠BCD)∠BCO = 1/2 (180 - ∠ACD)∠BCO = 90 - 1/2∠ACD .............(2)∠BOC = 180 - (∠CBO + ∠BCO)∠BOC = 180 - (90 - 1/2∠ABC + 90 - 1/2∠ACD)∠BOC = 180 - 180 + 1/2∠ABC + 1/2∠ACD∠BOC = 1/2 (∠ABC + ∠ACD)∠BOC = 1/2 ( 180 - ∠BAC) (180 -∠BAC = ∠ABC + ∠ACD)∠BOC = 90 - 1/2∠BACHence proved