Question

In the given figure , sides AB and AC of ∆ABC are produced to D and E r respectively. BO and CO , the bisectors of angle CBD and angle BCE respectively, meet at point O. Prove that angle BOC = 90°1/2 angle BAC.​

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

1/2 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

∠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

@EVYAAN.