Question

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 1

Answer:

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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*