In $\Delta ABC$,the sides $AB$ and $AC$ are produced to $D$ and $E$ respectively,so that exterior angles $\angle CBD$ and $\angle BCE$ are formed. If bisectors of $\angle CBD$ and $\angle BCE$ intersect at point $O$,prove that $\angle BOC = 90^{\circ} - \frac{1}{2} \angle A$.

(N/A) Let $\angle ABC = y$ and $\angle ACB = z$. The exterior angles are $\angle CBD = 180^{\circ} - y$ and $\angle BCE = 180^{\circ} - z$.
Since $BO$ and $CO$ are bisectors of $\angle CBD$ and $\angle BCE$ respectively,we have:
$\angle CBO = \frac{1}{2} \angle CBD = \frac{1}{2} (180^{\circ} - y) = 90^{\circ} - \frac{y}{2}$
$\angle BCO = \frac{1}{2} \angle BCE = \frac{1}{2} (180^{\circ} - z) = 90^{\circ} - \frac{z}{2}$
In $\Delta BOC$,the sum of angles is $180^{\circ}$:
$\angle BOC + \angle CBO + \angle BCO = 180^{\circ}$
$\angle BOC + (90^{\circ} - \frac{y}{2}) + (90^{\circ} - \frac{z}{2}) = 180^{\circ}$
$\angle BOC + 180^{\circ} - \frac{1}{2}(y + z) = 180^{\circ}$
$\angle BOC = \frac{1}{2}(y + z)$
In $\Delta ABC$,$y + z + \angle A = 180^{\circ}$,so $y + z = 180^{\circ} - \angle A$.
Substituting this into the equation for $\angle BOC$:
$\angle BOC = \frac{1}{2}(180^{\circ} - \angle A) = 90^{\circ} - \frac{1}{2} \angle A$.