Bisectors of angles $B$ and $C$ of a triangle $ABC$ intersect each other at the point $O$. Prove that $\angle BOC = 90^{\circ} + \frac{1}{2} \angle A$.

(N/A) Consider the triangle $ABC$. By the angle sum property of a triangle,we have:
$\angle A + \angle ABC + \angle ACB = 180^{\circ}$
Dividing the entire equation by $2$,we get:
$\frac{1}{2} \angle A + \frac{1}{2} \angle ABC + \frac{1}{2} \angle ACB = 90^{\circ}$
Since $BO$ and $CO$ are the angle bisectors of $\angle B$ and $\angle C$ respectively,we can write $\angle OBC = \frac{1}{2} \angle ABC$ and $\angle OCB = \frac{1}{2} \angle ACB$. Substituting these into the equation:
$\frac{1}{2} \angle A + \angle OBC + \angle OCB = 90^{\circ} .......(1)$
Now,consider the triangle $OBC$. By the angle sum property of a triangle:
$\angle BOC + \angle OBC + \angle OCB = 180^{\circ} .......(2)$
From equation $(1)$,we have $(\angle OBC + \angle OCB) = 90^{\circ} - \frac{1}{2} \angle A$. Substituting this into equation $(2)$:
$\angle BOC + (90^{\circ} - \frac{1}{2} \angle A) = 180^{\circ}$
Therefore,$\angle BOC = 180^{\circ} - 90^{\circ} + \frac{1}{2} \angle A$
$\angle BOC = 90^{\circ} + \frac{1}{2} \angle A$.