In the figure,the sides $AB$ and $AC$ of $\Delta ABC$ are produced to points $E$ and $D$ respectively. If the bisectors $BO$ and $CO$ of $\angle CBE$ and $\angle BCD$ respectively meet at point $O$,then prove that $\angle BOC = 90^o - \frac{1}{2} \angle BAC$.

(N/A) Ray $BO$ is the bisector of $\angle CBE$.
Therefore,$\angle CBO = \frac{1}{2} \angle CBE = \frac{1}{2}(180^o - y) = 90^o - \frac{y}{2}$ ......... $(1)$
Similarly,ray $CO$ is the bisector of $\angle BCD$.
Therefore,$\angle BCO = \frac{1}{2} \angle BCD = \frac{1}{2}(180^o - z) = 90^o - \frac{z}{2}$ ......... $(2)$
In $\Delta BOC$,$\angle BOC + \angle BCO + \angle CBO = 180^o$ ......... $(3)$
Substituting $(1)$ and $(2)$ in $(3)$,we get:
$\angle BOC + (90^o - \frac{z}{2}) + (90^o - \frac{y}{2}) = 180^o$
$\angle BOC + 180^o - \frac{1}{2}(y + z) = 180^o$
$\angle BOC = \frac{1}{2}(y + z)$ ......... $(4)$
In $\Delta ABC$,$x + y + z = 180^o$ (Angle sum property of a triangle).
Therefore,$y + z = 180^o - x$.
Substituting this in $(4)$:
$\angle BOC = \frac{1}{2}(180^o - x) = 90^o - \frac{x}{2} = 90^o - \frac{1}{2} \angle BAC$.