Bisectors of angles $A, B$ and $C$ of a triangle $ABC$ intersect its circumcircle at $D, E$ and $F$ respectively. Prove that the angles of the triangle $DEF$ are $90^{\circ} - \frac{1}{2}A, 90^{\circ} - \frac{1}{2}B$ and $90^{\circ} - \frac{1}{2}C$.

(N/A) Given a triangle $ABC$ inscribed in a circle,where the angle bisectors of $\angle A, \angle B$,and $\angle C$ intersect the circumcircle at $D, E$,and $F$ respectively.
Join $DE, EF$,and $FD$.
Since angles subtended by the same arc in the same segment are equal:
$\angle FDA = \angle FCA = \frac{1}{2} \angle C$ (as $CF$ is the bisector of $\angle C$)
$\angle EDA = \angle EBA = \frac{1}{2} \angle B$ (as $BE$ is the bisector of $\angle B$)
Adding these two equations:
$\angle FDE = \angle FDA + \angle EDA = \frac{1}{2} \angle C + \frac{1}{2} \angle B = \frac{1}{2}(\angle B + \angle C)$
Since $\angle A + \angle B + \angle C = 180^{\circ}$,we have $\angle B + \angle C = 180^{\circ} - \angle A$.
Therefore,$\angle FDE = \frac{1}{2}(180^{\circ} - \angle A) = 90^{\circ} - \frac{1}{2} \angle A$.
Similarly,$\angle FED = 90^{\circ} - \frac{1}{2} \angle B$ and $\angle EFD = 90^{\circ} - \frac{1}{2} \angle C$.
Thus,the angles of $\Delta DEF$ are $90^{\circ} - \frac{A}{2}, 90^{\circ} - \frac{B}{2}$,and $90^{\circ} - \frac{C}{2}$.