Question

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

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 for chord AE)

$\therefore \angle ADE=\frac{\angle B}{2}$

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

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

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