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

Solution:

Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D,

It is given that $B E$ is the bisector of $\angle B$.

$\therefore \angle \mathrm{ABE}=\frac{\angle \mathrm{B}}{2}$

However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)

$\Rightarrow \angle \mathrm{ADE}=\frac{\angle \mathrm{B}}{2}$

Similarly, $\angle \mathrm{ACF}=\angle \mathrm{ADF}=\frac{\angle \mathrm{C}}{2}$ (Angle in the same segment for chord AF)

$\angle D=\angle A D E+\angle A D F$

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

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

$=\frac{1}{2}\left(180^{\circ}-\angle \mathrm{A}\right)$

$=90^{\circ}-\frac{1}{2} \angle \mathrm{A}$

Similarly, it can be proved that

$\angle \mathrm{E}=90^{\circ}-\frac{1}{2} \angle \mathrm{B}$

$\angle \mathrm{F}=90^{\circ}-\frac{1}{2} \angle \mathrm{C}$
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