Question

in the adjacent figures AB and AC of angle ABC are produced to points E and D respectively meet at point 0 then prove that angle BOC=90-1/2angle 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) 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

Answer:

hope this pic will help uh buddy