In triangle ABC the sides AB and AC of triangle ABC are produced to points E and D, respectively if bisector BO and CO of angle CBE and angle BCD respectively at point O, then prove that angle BOC=90 degree-1/2 angle A

In triangle ABC the sides AB and AC of triangle ABC are produced to po

1.	In triangle ABC the sides AB and AC of triangle ABC are produced to points E and D, respectively if bisector BO and CO of angle CBE and angle BCD respectively at point O, then prove that angle BOC=90 degree-1/2 angle A
Answer» Hope it helps!! Mark this answer as brainliest if u found it USEFUL and follow me for quick and accurate answers...Step-by-step EXPLANATION: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∠BAC Hence proved*