Hello experts pls answer this question
Dear student
∠CBE = 180 - ∠ABC
∠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 - (∠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∠BAC..... (3)
1) ∠BAC = 180° - 2∠BOC
2) ∠BAC = 180° - 2× 55°
= ∠BAC = 70°
1) ∠BAC = 180° - 2∠BOC
2) ∠BAC = 180° - 2× 55°
= ∠BAC = 70°
Regards..