Question

7. In triangle ABC, the sides AB and AC are produced
to D and E respectively, so that exterior angles
CBD and BCE are formed. If bisectors
of CBD and BCE intersect at point 0,
prove that BOC = 90° - 1/2


There is a clear pic ​

Answer 1

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

Answer 2

Step-by-step explanation:

Hope it is helpful to you