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) The bisectors of $\angle A, \angle B$ and $\angle C$ of $\Delta ABC$ intersect the circumcircle of $\Delta ABC$ at $D, E$ and $F$ respectively.
$\angle FDE = \angle FDA + \angle EDA$ (Adjacent angles)
$= \angle FCA + \angle EBA$ (Angles in the same segment of the circle are equal)
$= \frac{1}{2} \angle C + \frac{1}{2} \angle B$ (Since $AD, BE, CF$ are angle bisectors)
$= \frac{1}{2}(\angle B + \angle C)$
$= \frac{1}{2}(180^{\circ} - \angle A)$ (Since $\angle A + \angle B + \angle C = 180^{\circ}$)
$= 90^{\circ} - \frac{1}{2} \angle A$
Thus,$\angle FDE = 90^{\circ} - \frac{1}{2} \angle A$.
Similarly,$\angle DEF = 90^{\circ} - \frac{1}{2} \angle B$ and $\angle EFD = 90^{\circ} - \frac{1}{2} \angle C$.
$\angle FDE = \angle FDA + \angle EDA$ (Adjacent angles)
$= \angle FCA + \angle EBA$ (Angles in the same segment of the circle are equal)
$= \frac{1}{2} \angle C + \frac{1}{2} \angle B$ (Since $AD, BE, CF$ are angle bisectors)
$= \frac{1}{2}(\angle B + \angle C)$
$= \frac{1}{2}(180^{\circ} - \angle A)$ (Since $\angle A + \angle B + \angle C = 180^{\circ}$)
$= 90^{\circ} - \frac{1}{2} \angle A$
Thus,$\angle FDE = 90^{\circ} - \frac{1}{2} \angle A$.
Similarly,$\angle DEF = 90^{\circ} - \frac{1}{2} \angle B$ and $\angle EFD = 90^{\circ} - \frac{1}{2} \angle C$.
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