Question

In the figure , sides AB and AC of triangle ABC are produced to point E and D respectively . if bisectors BO and CO of angle CBE and angle BCD respectively meet at point O, then prove that angle BOC =90°-1/2 angle BAC .

​

Answer 1

Step-by-step explanation:

∠CBE = 180 - ∠ABC

∠CBO = 1/2 ∠CBE (BO is the bisector of ∠CBE)

∠CBO = 1/2 ( 180 - ∠ABC)                                              

∠CBO = 90 - 1/2 ∠ABC    .............(1)                                  

Also,

∠BCD = 180 - ∠ACD

∠BCO = 1/2 ∠BCD     ( CO is the bisector os ∠BCD)

∠BCO = 1/2 (180 - ∠ACD)

∠BCO = 90 - 1/2∠ACD    .............(2)

Now,

∠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

Hope it helps!

Answer 2

Answer:

hiu

Step-by-step explanation:

gufurururf the clothes from for the clothes I can get a table as rr f