In $\Delta ABC$,the bisectors of $\angle B$ and $\angle C$ intersect each other at $I$. Prove that $\angle BIC = 90^{\circ} + \frac{1}{2} \angle A$.

(N/A) In $\Delta ABC$,the sum of angles is $180^{\circ}$,so $\angle A + \angle ABC + \angle ACB = 180^{\circ}$.
$\therefore \angle ABC + \angle ACB = 180^{\circ} - \angle A \quad \dots(1)$
The bisectors of $\angle B$ and $\angle C$ intersect at $I$.
$\therefore \angle IBC = \frac{1}{2} \angle ABC$ and $\angle ICB = \frac{1}{2} \angle ACB$.
In $\Delta IBC$,the sum of angles is $180^{\circ}$,so $\angle BIC + \angle IBC + \angle ICB = 180^{\circ}$.
$\therefore \angle BIC = 180^{\circ} - (\angle IBC + \angle ICB)$.
Substituting the values of $\angle IBC$ and $\angle ICB$:
$\therefore \angle BIC = 180^{\circ} - \left[ \frac{1}{2} \angle ABC + \frac{1}{2} \angle ACB \right]$.
$\therefore \angle BIC = 180^{\circ} - \frac{1}{2} (\angle ABC + \angle ACB)$.
Using equation $(1)$:
$\therefore \angle BIC = 180^{\circ} - \frac{1}{2} (180^{\circ} - \angle A)$.
$\therefore \angle BIC = 180^{\circ} - 90^{\circ} + \frac{1}{2} \angle A$.
$\therefore \angle BIC = 90^{\circ} + \frac{1}{2} \angle A$.