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 triangles $DEF$ are ${90}^o-\frac{1}{2}A$, ${90}^o-\frac{1}{2}B$ and ${90}^o-\frac{1}{2}C$.

In $△DEF$,

$\angle D=\angle EDF$

But $\angle EDF=\angle EDA+\angle FDA$      ....angle addition property

Now, $\angle EDA=\angle EBA$ and $\angle FDA=\angle FCA$      .....Angles inscribed in the same arc

$\therefore \angle EDF=\angle EBA+\angle FCA$

$=\frac{1}{2}\angle B+\frac{1}{2}\angle C$

[Since $BE$ is bisector of $\angle B$ and $CF$ is bisector $\angle C$]

$\therefore \angle D=\frac{\angle B+\angle C}{2}$        ....(1)

Similarly, $\angle E=\frac{\angle C+\angle A}{2}$ and $\angle F=\frac{\angle A+\angle B}{2}$              ...(2)

Now, $\angle A+\angle B+\angle C=180^o$        ....angle sum property of triangle

$\therefore \angle B+\angle C=180^{\circ }-\angle A$          ...(3)

Similarly, $\angle C+\angle A=180^o-\angle B$ and  $\angle A+\angle B=180^o-\angle C$                   ...(4)

Substituting eq (3) in eq (1), we get

$\therefore \angle D=\frac{180^o-\angle A}{2}$

$\angle D=90^o-\frac{1}{2}\angle A$

Similarly, $\angle E=90^o-\frac{1}{2}\angle B$ and $\angle F=90^o-\frac{1}{2}\angle C$