Boc= 90-1/2 angle bac how to prove this ?

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 = of ∠EBC

∠OBC = (180 - y)

∠OBC =

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 = of ∠DCB

∠OCB = (180 - z)

∠OCB =

In triangle OBC, the sum of the angles of the triangle will be 180.

∠BOC + + = 180

∠BOC +

∠BOC = =

As per upper triangle ABC,

or

Putting this value in value of ∠BOC

∠BOC =

∠BOC =

∠BOC = 90 - : Hence proved

Hope it will help :)

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