Question

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

Answer 1

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 = $\frac{1}{2}$ of ∠EBC

∠OBC = $\frac{1}{2}$ (180 - y)

∠OBC = $90 - \frac{y}{2}$

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 = $\frac{1}{2}$ of ∠DCB

∠OCB = $\frac{1}{2}$ (180 - z)

∠OCB = $90 - \frac{z}{2}$

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

∠BOC + $90 - \frac{y}{2}$ + $90 - \frac{z}{2}$ = 180

∠BOC + $180 - \frac{y}{2} - \frac{z}{2} = 180$

∠BOC = $\frac{y}{2} + \frac{z}{2}$ = $\frac{y+z}{2}$

As per upper triangle ABC, $x + y + z = 180$

or $y + z = 180 - x$

Putting this value in value of ∠BOC

∠BOC = $\frac{180-x}{2}$

∠BOC = $90 - \frac{x}{2}$

∠BOC = 90 - $\frac{1}{2} BAC$ : Hence proved

Hope it will help :)