Question

if in a triangle ABC sides AB and BC of triangle is produced to point E andD respectively. If biswctor BO and CO of angle CBE and BCD respectively meet at point O then prove that angle BOC
$= 90 - \frac{1}{2} angle bac$
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Answer 1

Answer: ∠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

Hope it helps uh cheerio