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

Given:

Bisectors of angles $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ of a triangle $\mathrm{ABC}$ intersect its circumcircle at $\mathrm{D}, \mathrm{E}$ and $F$ respectively.

To do:

We have to prove that the angles of the triangle $\mathrm{DEF}$ are $90^{\circ}-\frac{1}{2} \mathrm{~A}, 90^{\circ}-\frac{1}{2} \mathrm{~B}$ and $90^{\circ}-\frac{1}{2} \mathrm{C}$.

Solution:

$\angle EDF = \angle EDA + \angle ADF$

$\angle EDA$ and $\angle EBA$ are the angles in the same segment of the circle.

This implies,

$\angle EDA = \angle EBA$

Similarly,

$\angle ADF$ and $\angle FCA$ are the angles in the same segment of the circle.

This implies,

$\angle A D F=\angle F C A$

$\angle E D F= \angle EDA + \angle ADF$

$=\angle EBA+\angle F C A$

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

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

$=\frac{180^{\circ}-\angle A}{2}$          (Since $\angle A+\angle B+\angle C=180^{\circ}$)

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

Similarly,

$\angle E=\frac{\angle C+\angle A}{2}$

$\angle E=\frac{180^{\circ}-\angle B}{2}$          (Since $\angle A+\angle B+\angle C=180^{\circ}$)

$=90^o-\frac{\angle B}{2}$

$\angle E=\frac{\angle A+\angle B}{2}$

$\angle F=\frac{180^{\circ}-\angle C}{2}$            (Since $\angle A+\angle B+\angle C=180^{\circ}$)

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

Hence proved.

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Updated on: 10-Oct-2022

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