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  • 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° – 1/2 A, 90° – 1/2 B and 90° – 1/2 C.

Q 8. Bisectors of angles $\mathrm{A}, \mathrm{B}$ and C of a triangle ABC intersect its circumcircle at $\mathrm{D}, \mathrm{E}$ and F respectively.
Prove that the angles of the triangle DEF are $90^{\circ}-\frac{1}{2} C,90^{\circ}-\frac{1}{2} B$ and  $90^{\circ}-\frac{1}{2} A$

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Given:   Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.

To prove: the angles of the triangle DEF are $90^{\circ}-\frac{1}{2} C,90^{\circ}-\frac{1}{2} B$ and  $90^{\circ}-\frac{1}{2} A$

Proof: 

$\angle 1$ and $\angle 3$ are angles in the same segment. Therefore,
$\angle 1=\angle 3$  (angles in same segment are equal)-------(1) 
and $\angle 2=\angle 4$-------2
Adding 1 and 2 , we have
$\angle 1+\angle 2=\angle 3+\angle 4$
$\Rightarrow \angle D=\frac{1}{2} \angle B+\frac{1}{2} \angle C$
$\Rightarrow \angle D=\frac{1}{2}(\angle B+\angle C)$

$\Rightarrow \angle D=\frac{1}{2}\left(180^{\circ}+\angle C\right)$
and $\Rightarrow \angle D=\frac{1}{2}\left(180^{\circ}-\angle A\right)$
$\Rightarrow \angle D=90^{\circ}-\frac{1}{2} \angle A$
Similarly,
$\angle E=90^{\circ}-\frac{1}{2} \angle B$ and $\angle F=90^{\circ}-\frac{1}{2} \angle C$

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