Question

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 ${90}^{\circ }-\frac{1}{2}C$

Answer 1

It is given that BE is the bisector of $\angle B$.
$\therefore \angle ABE=\frac{\angle B}{2}$
However, $\angle ADE=\angle ABE$ (Angles in the same segment of chord AE)
$\therefore \angle ADE=\frac{\angle B}{2}$
Similarly,$\angle ACF=\angle ADF=\frac{\angle C}{2}$ ( Angle in the same segment of chord AF)
$\angle D=\angle ADE+\angle ADF$
$=\frac{\angle B}{2}+\frac{\angle C}{2}$
$=\frac{1}{2}(\angle B+\angle C)$
$=\frac{1}{2}({180}^{\circ }-\angle A)$
$={90}^{\circ }-\frac{1}{2}\angle A$
Similarly, it can be proved that
$\angle E={90}^{\circ }-\frac{1}{2}\angle B$
$\angle F={90}^{\circ }-\frac{1}{2}\angle C$