Alright, lets get started.
Please refer the diagram, I have attached.
∠ABC is given as y.
So, the ∠EBC will be 180 - ∠ABC.
∠EBC = 180 - y
As given in question, BO is the bisector of ∠EBC, it means ∠OBC will be half of the ∠EBC.
∠OBC = [tex] \frac{1}{2} [/tex] of ∠EBC
∠OBC = [tex] \frac{1}{2} [/tex] (180 - y)
∠OBC = [tex] 90 - \frac{y}{2} [/tex]
Similarly,
the ∠DCB will be 180 - ∠ACB
∠DCB = 180 - z
As given in question, CO is the bisector of ∠DCB, it means ∠OCB will be half of the ∠DCB.
∠OCB = [tex] \frac{1}{2} [/tex] of ∠DCB
∠OCB = [tex] \frac{1}{2} [/tex] (180 - z)
∠OCB = [tex] 90 - \frac{z}{2} [/tex]
In triangle OBC, the sum of the angles of the triangle will be 180.
∠BOC + [tex] 90 - \frac{y}{2} [/tex] + [tex] 90 - \frac{z}{2} [/tex] = 180
∠BOC + [tex] 180 - \frac{y}{2} - \frac{z}{2} = 180 [/tex]
∠BOC = [tex] \frac{y}{2} + \frac{z}{2} [/tex] = [tex] \frac{y+z}{2} [/tex]
As per upper triangle ABC, [tex] x + y + z = 180 [/tex]
or [tex] y + z = 180 - x [/tex]
Putting this value in value of ∠BOC
∠BOC = [tex] \frac{180-x}{2} [/tex]
∠BOC = [tex] 90 - \frac{x}{2} [/tex]
∠BOC = 90 - [tex] \frac{1}{2} BAC [/tex] : Hence proved
Hope it will help :)
